3, Measure algebras

MEASURE THEORY Volume 3

D.H.Fremlin

By the same author: Topological Riesz Spaces and Measure Theory, Cambridge University Press, 1974. Consequences of Martin’s Axiom, Cambridge University Press, 1982. Companions to the present volume: Measure Theory, vol. 1, Torres Fremlin, 2000. Measure Theory, vol. 2, Torres Fremlin, 2001.

First printing May 2002

MEASURE THEORY Volume 3 Measure Algebras

D.H.Fremlin Reader in Mathematics, University of Essex

Dedicated by the Author to the Publisher

This book may be ordered from the publisher at the address below. For price and means of payment see the author’s Web page http://www.essex.ac.uk/maths/staff/fremlin/mtsales.htm, or enquire from [email protected].

First published in 2002 by Torres Fremlin, 25 Ireton Road, Colchester CO3 3AT, England c D.H.Fremlin 2002 ° The right of D.H.Fremlin to be identified as author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. This work is issued under the terms of the Design Science License as published in http://dsl.org/copyleft/dsl.txt. For the source files see http://www.essex. ac.uk/maths/staff/fremlin/mt3.2002/readme.txt. Library of Congress classification QA312.F72 AMS 2000 classification 28A99 ISBN 0-9538129-3-6 Typeset by AMS-TEX Printed in England by Biddles Short Run Books, King’s Lynn

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Contents General Introduction Introduction to Volume 3

10 11

Chapter 31: Boolean algebras Introduction 311 Boolean algebras

13 13

Boolean rings and algebras; ideals and ring homomorphisms to Z2 ; Stone’s theorem; the operations ∪ , ∩ , 4 , \ and the relation ⊆ ; topology of the Stone space; Boolean algebras as complemented distributive lattices.

312 Homomorphisms

21

Subalgebras; ideals; Boolean homomorphisms; the ordering determines the ring structure; quotient algebras; extension of homomorphisms; homomorphisms and Stone spaces.

313 Order-continuity

30

General distributive laws; order-closed sets; order-closures; Monotone Class Theorem; order-preserving functions; order-continuity; order-dense sets; order-continuous Boolean homomorphisms; regularly embedded subalgebras.

314 Order-completeness

40

Dedekind completeness and σ-completeness; quotients, subalgebras, principal ideals; order-continuous homomorphisms; extension of homomorphisms; Loomis-Sikorski representation of a σ-complete algebra as a quotient of a σ-algebra of sets; regular open algebras; Stone spaces; Dedekind completion of a Boolean algebra.

315 Products and free products

51

Simple product of Boolean algebras; free product of Boolean algebras; algebras of sets and their quotients.

316 Further topics

59

The countable chain condition; weak (σ, ∞)-distributivity; Stone spaces; atomic and atomless Boolean algebras.

Chapter 32: Measure algebras Introduction 321 Measure algebras

68 68

Measure algebras; elementary properties; the measure algebra of a measure space; Stone spaces.

322 Taxonomy of measure algebras

72

Totally finite, σ-finite, semi-finite and localizable measure algebras; relation to corresponding types of measure space; completions and c.l.d. versions of measures; semi-finite measure algebras are weakly (σ, ∞)-distributive; subspace measures; simple products of measure algebras; Stone spaces of localizable measure algebras; localizations of semi-finite measure algebras.

323 The topology of a measure algebra

82

Defining a topology and uniformity on a measure algebra; continuity of algebraic operations; order-closed sets; Hausdorff and metrizable topologies, complete uniformities; closed subalgebras.

324 Homomorphisms

88

Homomorphisms induced by measurable functions; order-continuous and continuous homomorphisms; the topology of a semi-finite measure algebra is determined by the algebraic structure; measure-preserving homomorphisms.

325 Free products and product measures

95

The measure algebra of a product measure; the localizable measure algebra free product of two semi-finite measure algebras; the measure algebra of a product of probability measures; the probability algebra free product of probability algebras; factorizing through subproducts.

326 Additive functionals on Boolean algebras

104

Additive, countably additive and completely additive functionals; Jordan decomposition; Hahn decomposition.

327 Additive functionals on measure algebras Absolutely continuous and continuous additive functionals; Radon-Nikod´ ym theorem; the standard extension of a continuous additive functional on a closed subalgebra.

116

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Chapter 33: Maharam’s theorem Introduction 331 Classification of homogeneous measure algebras

123 123

Relatively atomless algebras; one-step extension of measure-preserving homomorphisms; Maharam type of a measure algebra; Maharam-type-homogeneous probability algebras of the same Maharam type are isomorphic; the measure algebra of {0, 1}κ is homogeneous.

332 Classification of localizable measure algebras

132

Any localizable measure algebra is isomorphic to a simple product of homogeneous totally finite algebras; exact description of isomorphism types; closed subalgebras.

333 Closed subalgebras

142

Relative Maharam types; extension of measure-preserving Boolean homomorphisms; complete classification of closed subalgebras of probability algebras as triples (A, µ ¯, C).

334 Products

158

Maharam types of product measures.

Chapter 34: Liftings Introduction 341 The lifting theorem

162 162

Liftings and lower densities; strictly localizable spaces have lower densities; construction of a lifting from a density; complete strictly localizable spaces have liftings; liftings and Stone spaces.

342 Compact measure spaces

174

Inner regular measures; compact classes; compact and locally compact measures; perfect measures.

343 Realization of homomorphisms

182

Representing homomorphisms between measure algebras by functions; possible when target measure space is locally compact; countably separated measures and uniqueness of representing functions; the split interval; perfect measures.

344 Realization of automorphisms

190

Simultaneously representing countable groups of automorphisms of measure algebras by functions – countably separated measure spaces, measures on {0, 1}I ; characterization of Lebesgue measure as a measure space; strong homogeneity of usual measure on {0, 1}I .

345 Translation-invariant liftings

200

Translation-invariant liftings on Rr and {0, 1}I ; there is no t.-i. Borel lifting on R.

346 Consistent liftings

208

Liftings of product measures which respect the product structure; translation-invariant liftings on {0, 1}I ; products of Maharam-type-homogeneous probability spaces; lower densities respecting product structures; consistent liftings; the Stone space of Lebesgue measure.

Chapter 35: Riesz spaces Introduction 351 Partially ordered linear spaces

219 219

Partially ordered linear spaces; positive cones; suprema and infima; positive linear operators; ordercontinuous linear operators; Riesz homomorphisms; quotient spaces; reduced powers; representation of p.o.l.ss as subspaces of reduced powers of R; Archimedean spaces.

352 Riesz spaces

226

Riesz spaces; identities; general distributive laws; Riesz homomorphisms; Riesz subspaces; order-dense subspaces and order-continuous operators; bands; the algebra of complemented bands; the algebra of projection bands; principal bands; f -algebras.

353 Archimedean and Dedekind complete Riesz spaces

237

Order-dense subspaces; bands; Dedekind (σ)-complete spaces; order-closed Riesz subspaces; order units; f -algebras.

354 Banach lattices

246

Riesz norms; Fatou norms; the Levi property; order-continuous norms; order-unit norms; M -spaces; are isomorphic to C(X) for compact Hausdorff X; L-spaces; uniform integrability in L-spaces.

355 Spaces of linear operators Order-bounded linear operators; the space L∼ (U ; V ); order-continuous operators; extension of ordercontinuous operators; the space L× (U ; V ); order-continuous norms.

255

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356 Dual spaces

264

The spaces U ∼ , U × , U ∗ ; biduals, embeddings U →V × where V ⊆U ∼ ; perfect Riesz spaces; L- and M -spaces; uniformly integrable sets in the dual of an M -space; relative weak compactness in L-spaces.

Chapter 36: Function spaces Introduction 361 S

275 276

Additive functions on Boolean rings; the space S(A); universal mapping theorems for linear operators on S; the map Tπ : S(A)→S(B) induced by a ring homomorphism π : A → B; projection bands in S(A); identifying S(A) when A is a quotient of an algebra of sets.

362 S ∼

287

Bounded additive functionals on A identified with order-bounded linear functionals on S(A); the Lspace S ∼ and its bands; countably additive, completely additive, absolutely continuous and continuous functionals; uniform integrability in S ∼ .

363 L∞

298

The space L∞ (A), as an M -space and f -algebra; universal mapping theorems for linear operators on L∞ ; Tπ : L∞ (A)→L∞ (B); representing L∞ when A is a quotient of an algebra of sets; integrals with respect to finitely additive functionals; projection bands in L∞ ; (L∞ )∼ and its bands; Dedekind completeness of A and L∞ ; representing σ-complete M -spaces; the generalized Hahn-Banach theorem; the Banach-Ulam problem.

364 L0

314

The space L0 (A); representing L0 when A is a quotient of a σ-algebra of sets; algebraic operations on L0 ; action of Borel measurable functions on L0 ; identifying L0 (A) with L0 (µ) when A is a measure algebra; embedding S and L∞ in L0 ; suprema and infima in L0 ; Dedekind completeness in A and L0 ; multiplication in L0 ; projection bands; Tπ : L0 (A)→L0 (B); simple products; *regular open algebras; *the space C ∞ (X).

365 L1

335

366 Lp

351

R R The space L1 (A, µ ¯); identification with L1 (µ); a u; the Radon-Nikod´ ym theorem again; w×Tπ u d¯ ν= R u d¯ µ; the duality between L1 and L∞ ; additive functions on A and linear operators on L1 ; Tπ : L1 (A, µ ¯)→L1 (B, ν¯) and Pπ : L1 (B, ν¯)→L1 (A, µ ¯); conditional expectations; bands in L1 ; varying µ ¯. The spaces Lp (A, µ ¯); identification with Lp (µ); Lq as the dual of Lp ; the spaces M 0 and M 1,0 ; Tπ : M 0 (A, µ ¯)→M 0 (B, ν¯) and Pπ : M 1,0 (B, ν¯)→M 1,0 (A, µ ¯); conditional expectations; the case p = 2.

367 Convergence in measure

360

Order*-convergence of sequences in lattices; in Riesz spaces; in Banach lattices; in quotients of spaces of measurable functions; in C(X); Lebesgue’s Dominated Convergence Theorem and Doob’s Martingale Theorem; convergence in measure in L0 (A); and pointwise convergence; defined by the Riesz space structure; positive linear operators on L0 ; convergence in measure and the canonical projection (L1 )∗∗ →L1 .

368 Embedding Riesz spaces in L0

375

Extension of order-continuous Riesz homomorphisms into L0 ; representation of Archimedean Riesz spaces as subspaces of L0 ; Dedekind completion of Riesz spaces; characterizing L0 spaces as Riesz spaces; weakly (σ, ∞)-distributive Riesz spaces.

369 Banach function spaces

386

Riesz spaces separated by their order-continuous duals; representing U × when U ⊆L0 ; Kakutani’s repre0 sentation of L-spaces as L1 spaces; extended Fatou norms; associate norms; Lτ ∼ = (Lτ )× ; Fatou norms ∞,1 1,∞ τ τ 1 2 and convergence in measure; M and M , k k∞,1 and k k1,∞ ; L + L .

Chapter 37: Linear operators between function spaces Introduction 371 The Chacon-Krengel theorem

402 402 (0)

L∼ (U ; V ) = L× (U ; V ) = B(U ; V ) for L-spaces U and V ; the class Tµ,¯ ¯ ν of k k1 -decreasing, k k∞ -decreasing linear operators from M 1,0 (A, µ ¯) to M 1,0 (B, ν¯).

372 The ergodic theorem

407 (0)

The Maximal Ergodic Theorem and the Ergodic Theorem for operators in Tµ, ¯ µ ¯ ; for inverse-measurepreserving functions φ : X→X; limit operators as conditional expectations; applications to continued fractions; mixing and ergodic transformations.

8

373 Decreasing rearrangements

R R The classes T , T × ; the space M 0,∞ ; decreasing rearrangements u∗ ; ku∗ kp = kukp ; |T u×v| ≤ u∗ ×v ∗ if T ∈ T ; the very weak operator topology and compactness of T ; v is expressible as T u, where T ∈ T , R R R R (0) iff 0t v ∗ ≤ 0t u∗ for every t; finding T such that T u×v = u∗ ×v ∗ ; the adjoint operator from Tµ,¯ ¯ ν to

422

(0)

Tν¯,µ¯ .

374 Rearrangement-invariant spaces

440

T -invariant subspaces of M 1,∞ , and T -invariant extended Fatou norms; relating T -invariant norms on different spaces; rearrangement-invariant sets and norms; when rearrangement-invariance implies T -invariance.

375 Kwapien’s theorem

451

Linear operators on L0 spaces; if B is measurable, a positive linear operator from L0 (A) to L0 (B) can be assembled from Riesz homomorphisms.

376 Integral operators

457

Kernel operators; free products of measure algebras and tensor products of L0 spaces; tensor products of L1 spaces; abstract integral operators (i) as a band in L× (U ; V ) (ii) represented by kernels belonging b to L0 (A⊗B) (iii) as operators converting weakly convergent sequences into order*-convergent sequences; operators into M -spaces or out of L-spaces; disintegrations.

Chapter 38: Automorphism groups Introduction 381 Automorphism groups of Boolean algebras

477 477

Assembling an automorphism; elements supporting an automorphism; cyclic automorphisms; exchanging involutions; the support of an automorphism; full subgroups of Aut A; expressing an automorphism as a product of involutions; subgroups of Aut A with many involutions; normal subgroups of full groups with many involutions; simple groups.

382 Automorphism groups of measure algebras

488

Measure-preserving automorphisms as products of involutions; normal subgroups of Aut A and Autµ¯ A.

383 Outer automorphisms

492

If G ≤ Aut A, H ≤ Aut B have many involutions, any isomorphism between G to H arises from an isomorphism between A and B; if A is nowhere rigid, Aut A has no outer automorphisms; applications to localizable measure algebras.

384 Entropy

502

Entropy of a partition of unity in a probability algebra; conditional entropy; entropy of a measurepreserving homomorphism; calculation of entropy (Kolmogorov-Sinaˇı theorem); Bernoulli shifts; isomorphic homomorphisms and conjugacy classes in Autµ¯ A; almost isomorphic inverse-measure-preserving functions.

385 More about entropy

515

Periodic and aperiodic parts of an endomorphism; the Halmos-Rokhlin-Kakutani lemma; the ShannonMcMillan-Breiman theorem; various lemmas.

386 Ornstein’s theorem

528

Bernoulli partitions; finding Bernoulli partitions with elements of given measure (Sinaˇı’s theorem); adjusting Bernoulli partitions; Ornstein’s theorem (Bernoulli shifts of the same finite entropy are isomorphic); Ornstein’s and Sinaˇı’s theorems in the case of infinite entropy.

387 Dye’s theorem

551

Full subgroups of Aut A; and orbits of inverse-measure-preserving functions; induced automorphisms of principal ideals; von Neumann transformations; von Neumann transformations generating a given full subgroup; classification of full subgroups generated by a single automorphism.

Chapter 39: Measurable algebras Introduction 391 Kelley’s theorem

567 567

Measurable algebras; strictly positive additive functionals and weak (σ, ∞)-distributivity; additive functionals subordinate to or dominating a given functional; intersection numbers; existence of strictly positive additive functionals; σ-linked Boolean algebras; Gaifman’s example.

392 Submeasures Submeasures; exhaustive, uniformly exhaustive and Maharam submeasures; the Kalton-Roberts theorem (a strictly positive uniformly exhaustive submeasure provides a strictly positive additive functional).

575

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393 The Control Measure Problem

579

Forms of the Control Measure Problem; strictly positive Maharam submeasures, exhaustive submeasures, exhaustive submeasures on the clopen algebra of {0, 1}N ; absolute continuity of submeasures; countable atomless Boolean algebras; topologies on Boolean algebras; topologies on L0 spaces; vector measures; examples of submeasures.

394 Kawada’s theorem

595

Full local semigroups; τ -equidecomposability; fully non-paradoxical subgroups of Aut A; bb : ac and db : ae; invariant additive functions from A to L∞ (C), where C is the fixed-point algebra of a group; invariant additive functionals and measures; ergodic fully non-paradoxical groups.

395 The Hajian-Ito theorem

610

Invariant measures on measurable algebras; weakly wandering elements.

Appendix to Volume 3 Introduction 3A1 Set theory

613 613

Calculation of cardinalities; cofinal sets, cofinalities; notes on the use of Zorn’s Lemma; the natural numbers as finite ordinals; lattice homomorphisms; the Marriage Lemma.

3A2 Rings

615

Rings; subrings, ideals, homomorphisms, quotient rings, the First Isomorphism Theorem; products.

3A3 General topology

618

Hausdorff, regular, completely regular, zero-dimensional, extremally disconnected, compact and locally compact spaces; continuous functions; dense subsets; meager sets; Baire’s theorem for locally compact spaces; products; Tychonoff’s theorem; the usual topologies on {0, 1}I , RI ; cluster points of filters; topology bases; uniform convergence of sequences of functions; one-point compactifications.

3A4 Uniformities

622

Uniform spaces; and pseudometrics; uniform continuity; subspaces; product uniformities; Cauchy filters and completeness; extending uniformly continuous functions; completions.

3A5 Normed spaces

625

The Hahn-Banach theorem in analytic and geometric forms; cones and convex sets; weak and weak* topologies; reflexive spaces; Uniform Boundedness Theorem; completions; normed algebras; compact linear operators; Hilbert spaces.

3A6 Groups

628

Involutions; inner and outer automorphisms; normal subgroups.

References for Volume 3

629

Index to Volumes 1-3 Principal topics and results General index

633 641

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General introduction In this treatise I aim to give a comprehensive description of modern abstract measure theory, with some indication of its principal applications. The first two volumes are set at an introductory level; they are intended for students with a solid grounding in the concepts of real analysis, but possibly with rather limited detailed knowledge. As the book proceeds, the level of sophistication and expertise demanded will increase; thus for the volume on topological measure spaces, familiarity with general topology will be assumed. The emphasis throughout is on the mathematical ideas involved, which in this subject are mostly to be found in the details of the proofs. My intention is that the book should be usable both as a first introduction to the subject and as a reference work. For the sake of the first aim, I try to limit the ideas of the early volumes to those which are really essential to the development of the basic theorems. For the sake of the second aim, I try to express these ideas in their full natural generality, and in particular I take care to avoid suggesting any unnecessary restrictions in their applicability. Of course these principles are to to some extent contradictory. Nevertheless, I find that most of the time they are very nearly reconcilable, provided that I indulge in a certain degree of repetition. For instance, right at the beginning, the puzzle arises: should one develop Lebesgue measure first on the real line, and then in spaces of higher dimension, or should one go straight to the multidimensional case? I believe that there is no single correct answer to this question. Most students will find the one-dimensional case easier, and it therefore seems more appropriate for a first introduction, since even in that case the technical problems can be daunting. But certainly every student of measure theory must at a fairly early stage come to terms with Lebesgue area and volume as well as length; and with the correct formulations, the multidimensional case differs from the one-dimensional case only in a definition and a (substantial) lemma. So what I have done is to write them both out (§§114-115). In the same spirit, I have been uninhibited, when setting out exercises, by the fact that many of the results I invite students to look for will appear in later chapters; I believe that throughout mathematics one has a better chance of understanding a theorem if one has previously attempted something similar alone. As I write this Introduction (December 2001), the plan of the work is as follows: Volume Volume Volume Volume Volume

1: 2: 3: 4: 5:

The Irreducible Minimum Broad Foundations Measure Algebras Topological Measure Spaces Set-theoretic Measure Theory.

Volume 1 is intended for those with no prior knowledge of measure theory, but competent in the elementary techniques of real analysis. I hope that it will be found useful by undergraduates meeting Lebesgue measure for the first time. Volume 2 aims to lay out some of the fundamental results of pure measure theory (the Radon-Nikod´ ym theorem, Fubini’s theorem), but also gives short introductions to some of the most important applications of measure theory (probability theory, Fourier analysis). While I should like to believe that most of it is written at a level accessible to anyone who has mastered the contents of Volume 1, I should not myself have the courage to try to cover it in an undergraduate course, though I would certainly attempt to include some parts of it. Volumes 3 and 4 are set at a rather higher level, suitable to postgraduate courses; while Volume 5 will assume a wide-ranging competence over large parts of analysis and set theory. There is a disclaimer which I ought to make in a place where you might see it in time to avoid paying for this book. I make no attempt to describe the history of the subject. This is not because I think the history uninteresting or unimportant; rather, it is because I have no confidence of saying anything which would not be seriously misleading. Indeed I have very little confidence in anything I have ever read concerning the history of ideas. So while I am happy to honour the names of Lebesgue and Kolmogorov and Maharam in more or less appropriate places, and I try to include in the bibliographies the works which I have myself consulted, I leave any consideration of the details to those bolder and better qualified than myself. The work as a whole is not yet complete; and when it is finished, it will undoubtedly be too long to be printed as a single volume in any reasonable format. I am therefore publishing it one part at a time. However, drafts of most of the rest are available on the Internet; see http://www.essex.ac.uk/ maths/staff/fremlin/mt.htm for detailed instructions. For the time being, at least, printing will be in short runs. I hope that readers will be energetic in commenting on errors and omissions, since it should be possible to correct these relatively promptly. An inevitable consequence of this is that paragraph references may go out of date rather quickly. I shall be most flattered if anyone chooses to rely on this book as a source

Introduction to Volume 3

11

for basic material; and I am willing to attempt to maintain a concordance to such references, indicating where migratory results have come to rest for the moment, if authors will supply me with copies of papers which use them. I mention some minor points concerning the layout of the material. Most sections conclude with lists of ‘basic exercises’ and ‘further exercises’, which I hope will be generally instructive and occasionally entertaining. How many of these you should attempt must be for you and your teacher, if any, to decide, as no two students will have quite the same needs. I mark with a > those which seem to me to be particularly important. But while you may not need to write out solutions to all the ‘basic exercises’, if you are in any doubt as to your capacity to do so you should take this as a warning to slow down a bit. The ‘further exercises’ are unbounded in difficulty, and are unified only by a presumption that each has at least one solution based on ideas already introduced. Occasionally I add a final ‘problem’, a question to which I do not know the answer and which seems to arise naturally in the course of the work. The impulse to write this book is in large part a desire to present a unified account of the subject. Cross-references are correspondingly abundant and wide-ranging. In order to be able to refer freely across the whole text, I have chosen a reference system which gives the same code name to a paragraph wherever it is being called from. Thus 132E is the fifth paragraph in the second section of the third chapter of Volume 1, and is referred to by that name throughout. Let me emphasize that cross-references are supposed to help the reader, not distract her. Do not take the interpolation ‘(121A)’ as an instruction, or even a recommendation, to lift Volume 1 off the shelf and hunt for §121. If you are happy with an argument as it stands, independently of the reference, then carry on. If, however, I seem to have made rather a large jump, or the notation has suddenly become opaque, local cross-references may help you to fill in the gaps. Each volume will have an appendix of ‘useful facts’, in which I set out material which is called on somewhere in that volume, and which I do not feel I can take for granted. Typically the arrangement of material in these appendices is directed very narrowly at the particular applications I have in mind, and is unlikely to be a satisfactory substitute for conventional treatments of the topics touched on. Moreover, the ideas may well be needed only on rare and isolated occasions. So as a rule I recommend you to ignore the appendices until you have some direct reason to suppose that a fragment may be useful to you. During the extended gestation of this project I have been helped by many people, and I hope that my friends and colleagues will be pleased when they recognise their ideas scattered through the pages below. But I am especially grateful to those who have taken the trouble to read through earlier drafts and comment on obscurities and errors.

Introduction to Volume 3 One of the first things one learns, as a student of measure theory, is that sets of measure zero are frequently ‘negligible’ in the straightforward sense that they can safely be ignored. This is not quite a universal principle, and one of my purposes in writing this treatise is to call attention to the exceptional cases in which ‘negligible’ sets are important. But very large parts of the theory, including some of the topics already treated in Volume 2, can be expressed in an appropriately abstract language in which negligible sets have been factored out. This is what the present volume is about. A ‘measure algebra’ is a quotient of an algebra of measurable sets by an ideal of negligible sets; that is, the elements of the measure algebra are equivalence classes of measurable sets. At the cost of an extra layer of abstraction, we obtain a language which can give concise and elegant expression to a substantial proportion of the ideas of measure theory, and which offers insights almost everywhere in the subject. It is here that I embark wholeheartedly on ‘pure’ measure theory. I think it is fair to say that the applications of measure theory to other branches of mathematics are more often through measure spaces rather than measure algebras. Certainly there will be in this volume many theorems of wide importance outside measure theory; but typically their usefulness will be in forms translated back into the language of the first two volumes. But it is also fair to say that the language of measure algebras is the only reasonable way to discuss large parts of a subject which, as pure mathematics, can bear comparison with any. In the structure of this volume I can distinguish seven ‘working’ and two ‘accessory’ chapters. The ‘accessory’ chapters are 31 and 35. In these I develop the theories of Boolean algebras and Riesz spaces (= vector lattices) which are needed later. As in Volume 2 you have a certain amount of choice in the order in

12

Introduction to Volume 3

which you take the material. Everything except Chapter 35 depends on Chapter 31, and everything except Chapters 31 and 35 depends on Chapter 32. Chapters 33, 34 and 36 can be taken in any order, but Chapter 36 relies on Chapter 35. (I do not mean that Chapter 33 is never referred to in Chapter 34, nor even that no results from Chapter 33 are relied on in the later chapters. What I mean is that their most important ideas are accessible without learning the material of Chapter 33 properly.) Chapter 37 depends on Chapters 35 and 36. Chapter 38 would be difficult to make sense of without some notion of what has been done in Chapter 33. Chapter 39 uses fragments of Chapters 35 and 36. The first half of the volume follows almost the only line permitted by the structure of the subject. If we are going to study measure algebras at all, we must know the relevant facts about Boolean algebras (Chapter 31) and how to translate what we know about measure spaces into the new language (Chapter 32). Then we must get a proper grip on the two most important theorems: Maharam’s theorem on the classification of measure algebras (Chapter 33) and the von Neumann-Maharam lifting theorem (Chapter 34). Since I am now writing for readers who are committed – I hope, happily committed – to learning as much as they can about the subject, I take the space to push these ideas as far as they can easily go, giving a full classification of closed subalgebras of probability algebras, for instance (§333), and investigating special types of lifting (§§345-346). I mention here three sections interpolated into Chapter 34 (§§342-344) which attack a subtle and important question: when can we expect homomorphisms between measure algebras to be realizable in terms of transformations between measure spaces, as discussed briefly in §235 and elsewhere. Chapters 36 and 37 are devoted to re-working the ideas of Chapter 24 on ‘function spaces’ in the more abstract context now available, and relating them to the general Riesz spaces of Chapter 35. I am concerned here not to develop new structures, nor even to prove striking new theorems, but rather to offer new ways of looking at the old ones. Only in the Ergodic Theorem (§372) do I come to a really important new result. Chapter 38 looks at two questions, both obvious ones to ask if you have been trained in twentieth-century pure mathematics: what does the automorphism group of a measure algebra look like, and inside such an automorphism group, what do the conjugacy classes look like? (The second question is a fancy way of asking how to decide, given two automorphisms of one of the structures considered in this volume, whether they are really different, or just copies of each other obtained by looking at the structure a different way up.) Finally, in Chapter 39, I discuss what is known about the question of which Boolean algebras can appear as measure algebras. Concerning the prerequisites for this volume, we certainly do not need everything in Volume 2. The important chapters there are 21, 23, 24, 25 and 27. If you are approaching this volume without having read the earlier parts of this treatise, you will need the Radon-Nikod´ ym theorem and product measures (of arbitrary families of probability spaces), for Maharam’s theorem; a simple version of the martingale theorem, for the lifting theorem; and an acquaintance with Lp spaces (particularly, with L0 spaces) for Chapter 36. But I would recommend the results-only versions of Volumes 1 and 2 in case some reference is totally obscure. Outside measure theory, I call on quite a lot of terms from general topology, but none of the ideas needed are difficult (Baire’s and Tychonoff’s theorems are the deepest); they are sketched in §§3A3 and 3A4. We do need some functional analysis for Chapters 36 and 39, but very little more than was already used in Volume 2, except that I now call on versions of the Hahn-Banach theorem (§3A5). In this volume I assume that readers have substantial experience in both real and abstract analysis, and I make few concessions which would not be appropriate when addressing active researchers, except that perhaps I am a little gentler when calling on ideas from set theory and general topology than I should be with my own colleagues, and I continue to include all the easiest exercises I can think of. I do maintain my practice of giving proofs in very full detail, not so much because I am trying to make them easier, but because one of my purposes here is to provide a complete account of the ideas of the subject. I hope that the result will be accessible to most doctoral students who are studying topics in, or depending on, measure theory.

311B

Boolean algebras

13

Chapter 31 Boolean algebras The theory of measure algebras naturally depends on certain parts of the general theory of Boolean algebras. In this chapter I collect those results which will be useful later. Since many students encounter the formal notion of Boolean algebra for the first time in this context, I start at the beginning; and indeed I include in the Appendix (§3A2) a brief account of the necessary part of the theory of rings, as not everyone will have had time for this bit of abstract algebra in an undergraduate course. But unless you find the algebraic theory of Boolean algebras so interesting that you wish to study it for its own sake – in which case you should perhaps turn to Sikorski 64 or Koppelberg 89 – I do not think it would be very sensible to read the whole of this chapter before proceeding to the main work of the volume in Chapter 32. Probably §311 is necessary to get an idea of what a Boolean algebra looks like, and a glance at the statements of the theorems in §312 would be useful, but the later sections can wait until you have need of them, on the understanding that apparently innocent formal manipulations may depend on concepts which take some time to master. I hope that the cross-references will be sufficiently well-targeted to make it possible to read this material in parallel with its applications.

311 Boolean algebras In this section I try to give a sufficient notion of the character of abstract Boolean algebras to make the calculations which will appear on almost every page of this volume seem both elementary and natural. The principal result is of course M.H.Stone’s theorem: every Boolean algebra can be expressed as an algebra of sets (311E). So the section divides naturally into the first part, proving Stone’s theorem, and the second, consisting of elementary consequences of the theorem and a little practice in using the insights it offers. 311A Definitions (a) A Boolean ring is a ring (A, +, .) in which a2 = a for every a ∈ A. (b) A Boolean algebra is a Boolean ring A with a multiplicative identity 1 = 1A ; I allow 1 = 0 in this context. Remark For notes on those parts of the elementary theory of rings which we shall need, see §3A2. I hope that the rather arbitrary use of the word ‘algebra’ here will give no difficulties; it gives me the freedom to insist that the ring {0} should be accepted as a Boolean algebra. 311B Examples (a) For any set X, (PX, 4, ∩) is a Boolean algebra; its zero is ∅ and its multiplicative identity is X. P P We have to check the following, which are all easily established, using Venn diagrams or otherwise: A4B ⊆ X for all A, B ⊆ X, (A4B)4C = A4(B4C) for all A, B, C ⊆ X, so that (PX, 4) is a semigroup; A4∅ = ∅4A = A for every A ⊆ X, so that ∅ is the identity in (PX, 4); A4A = ∅ for every A ⊆ X, so that every element of PX is its own inverse in (PX, 4), and (PX, 4) is a group; A4B = B4A for all A, B ⊆ X, so that (PX, 4) is an abelian group; A ∩ B ⊆ X for all A, B ⊆ X, (A ∩ B) ∩ C = A ∩ (B ∩ C) for all A, B, C ⊆ X, so that (PX, ∩) is a semigroup; A ∩ (B4C) = (A ∩ B)4(A ∩ C), (A4B) ∩ C = (A ∩ C)4(B ∩ C) for all A, B, C ⊆ X, so that (PX, 4, ∩) is a ring;

14

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311B

A ∩ A = A for every A ⊆ X, so that (PX, 4, ∩) is a Boolean ring; A ∩ X = X ∩ A = A for every A ⊆ X, so that (PX, 4, ∩) is a Boolean algebra and X is its identity. Q Q (b) Recall that an ‘algebra of subsets of X’ (136E) is a family Σ ⊆ PX such that ∅ ∈ Σ, X \ E ∈ Σ for every E ∈ Σ, and E ∪ F ∈ Σ for all E, F ∈ Σ. In this case (Σ, 4, ∩) is a Boolean algebra with zero ∅ and identity X. P P If E, F ∈ Σ, then E ∩ F = X \ ((X \ E) ∪ (X \ F )) ∈ Σ, E4F = (E ∩ (X \ F )) ∪ (F ∩ (X \ E)) ∈ Σ. Because ∅ and X = X \ ∅ both belong to Σ, we can work through the identities in (a) above to see that Σ, like PX, is a Boolean algebra. Q Q (c) Consider the ring Z2 = {0, 1}, with its ring operations +2 , · given by setting 0 +2 0 = 1 +2 1 = 0,

0 +2 1 = 1 +2 0 = 1,

0 · 0 = 0 · 1 = 1 · 0 = 0,

1 · 1 = 1.

I leave it to you to check, if you have not seen it before, that this is a ring. Because 0 · 0 = 0 and 1 · 1 = 1, it is a Boolean algebra. 311C Proposition Let A be a Boolean ring. (a) a + a = 0, that is, a = −a, for every a ∈ A. (b) ab = ba for all a, b ∈ A. proof (a) If a ∈ A, then a + a = (a + a)(a + a) = a2 + a2 + a2 + a2 = a + a + a + a, so we must have 0 = a + a. (b) Now for any a, b ∈ A, a + b = (a + b)(a + b) = a2 + ab + ba + b2 = a + ab + ba + b, so 0 = ab + ba = ab + ab and ab = ba. 311D Lemma Let A be a Boolean ring, I an ideal of A (3A2E), and a ∈ A \ I. Then there is a ring homomorphism φ : A → Z2 such that φa = 1 and φd = 0 for every d ∈ I. proof (a) Let I be the family of those ideals J of A which include I and do not contain a. Then I has a maximal element K say. P PSApply Zorn’s lemma. Since I ∈ I, I 6= ∅. If J is a non-empty totally ordered subset of I, then set J ∗ = J . If b, c ∈ J ∗ and d ∈ A, then there are J1 , J2 ∈ J such that b ∈ J1 and c ∈ J2 ; now J = J1 ∪ J2 is equal to one of J1 , J2 , so belongs to J , and 0, b + c, bd all belong to J, so all belong to J ∗ . Thus J ∗ C A; of course I ⊆ J ∗ and a ∈ / J ∗ , so J ∗ ∈ I and is an upper bound for J in I. As J is arbitrary, the hypotheses of Zorn’s lemma are satisfied and I has a maximal element. Q Q (b) For b ∈ A set Kb = {d : d ∈ A, bd ∈ K}. The following are easy to check: (i) K ⊆ Kb for every b ∈ A, because K is an ideal. (ii) Kb C A for every b ∈ A. P P 0 ∈ K ⊆ Kb . If d, d0 ∈ Kb and c ∈ A then b(d + d0 ) = bd + bd0 , b(dc) = (bd)c belong to K, so d + d0 , dc ∈ Kb . Q Q (iii) If b ∈ A and a ∈ / Kb , then Kb ∈ I so Kb = K.

311Fc

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15

(iv) Now a2 = a ∈ / K, so a ∈ / Ka and Ka = K. (v) If b ∈ A \ K then b ∈ / Ka , that is, ba = ab ∈ / K, and a ∈ / Kb ; consequently Kb = K. (vi) If b, c ∈ A \ K then c ∈ / Kb so bc ∈ / K. (vii) If b, c ∈ A \ K then bc(b + c) = b2 c + bc2 = bc + bc = 0 ∈ K, so b + c ∈ Kbc . By (vi) and (v), Kbc = K so b + c ∈ K. (c) Now define φ : A → Z2 by setting φd = 0 if d ∈ K, φd = 1 if d ∈ A \ K. Then φ is a ring homomorphism. P P (i) If b, c ∈ K then b + c, bc ∈ K so φ(b + c) = 0 = φb +2 φc,

φ(bc) = 0 = φb φc.

(ii) If b ∈ K, c ∈ A \ K then c = (b + b) + c = b + (b + c) ∈ /K so b + c ∈ / K, while bc ∈ K, so φ(b + c) = 1 = φb +2 φc,

φ(bc) = 0 = φb φc.

φ(b + c) = 1 = φb +2 φc,

φ(bc) = 0 = φb φc

(iii) Similarly, if b ∈ A \ K and c ∈ K. (iv) If b, c ∈ A \ K, then by (b-vi) and (b-vii) we have b + c ∈ K, bc ∈ / K so φ(b + c) = 0 = φb +2 φc,

φ(bc) = 1 = φb φc.

Thus φ is a ring homomorphism. Q Q (d) Finally, if d ∈ I then d ∈ K so φd = 0; and φa = 1 because a ∈ / K. 311E M.H.Stone’s Theorem: first form Let A be any Boolean ring, and let Z be the set of ring homomorphisms from A onto Z2 . Then we have an injective ring homomorphism a 7→ b a : A → PZ, setting b a = {z : z ∈ Z, z(a) = 1}. If A is a Boolean algebra, then b 1A = Z. proof (a) If a, b ∈ A, then d = {z : z(a+b) = 1} = {z : z(a) +2 z(b) = 1} = {z : {z(a), z(b)} = {0, 1}} = b a+b a4bb, b = {z : z(ab) = 1} = {z : z(a)z(b) = 1} = {z : z(a) = z(b) = 1} = b ab a ∩ bb. Thus a 7→ b a is a ring homomorphism. (b) If a ∈ A and a 6= 0, then by 311D, with I = {0}, there is a z ∈ Z such that z(a) = 1, that is, z ∈ b a; so that b a 6= ∅. This shows that the kernel of a 7→ b a is {0}, so that the homomorphism is injective (3A2Db). (c) If A is a Boolean algebra, and z ∈ Z, then there is some a ∈ A such that z(a) = 1, so that z(1A )z(a) = z(1A a) 6= 0 and z(1A ) 6= 0; thus b 1A = Z. 311F Remarks (a) For any Boolean ring A, I will say that the Stone space of A is the set Z of non-zero ring homomorphisms from A to Z2 , and the canonical map a 7→ b a : A → PZ is the Stone representation. (b) Because the map a 7→ b a : A → PZ is an injective ring homomorphism, A is isomorphic, as Boolean ring, to its image E = {b a : a ∈ A}, which is a subring of PZ. Thus the Boolean rings PX of 311Ba are leading examples in a very strong sense. (c) I have taken the set Z of the Stone representation to be actually the set of homomorphisms from A onto Z2 . Of course we could equally well take any set which is in a natural one-to-one correspondence with Z; a popular choice is the set of maximal ideals of A, since a subset of A is a maximal ideal iff it is the kernel of a member of Z, which is then uniquely defined.

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311G

311G The operations ∪ , \ , 4 on a Boolean ring Let A be a Boolean ring. (a) Using the Stone representation, we can see that the elementary operations ∪, ∩, \, 4 of set theory all correspond to operations on A. If we set a ∪ b = a + b + ab,

a ∩ b = ab,

a \ b = a + ab,

a4b = a+b

for a, b ∈ A, then we see that ad ∪b = b a4bb4(b a ∩ bb) = b a ∪ bb, ad ∩b = b a ∩ bb, ac \b = b a \ bb, ad 4b = b a4bb. Consequently all the familiar rules for manipulation of ∩, ∪, etc. will apply also to ∩ , ∪ , and we shall have, for instance, a ∩ (b ∪ c) = (a ∩ b) ∪ (a ∩ c),

a ∪ (b ∩ c) = (a ∪ b) ∩ (a ∪ c)

for any members a, b, c of any Boolean ring A. (b) Still importing terminology from elementary set theory, I will say that a set A ⊆ A is disjoint if a ∩ b = 0, that is, ab = 0, for all distinct a, b ∈ A; and that an indexed family hai ii∈I in A is disjoint if ai ∩ aj = 0 for all distinct i, j ∈ I. (Just as I allow ∅ to be a member of a disjoint family of sets, I allow 0 ∈ A or ai = 0 in the present context.) (c) A partition of unity in A will be either a disjoint set C ⊆ A such that there is no non-zero a ∈ A such that a ∩ c = 0 for every c ∈ C or a disjoint family hci ii∈I in A such that there is no non-zero a ∈ A such that a ∩ ci = 0 for every i ∈ I. (In the first case I allow 0 ∈ C, and in the second I allow ci = 0.) (d) If C and D are two partitions of unity, I say that C refines D if for every c ∈ C there is a d ∈ D such that cd = d. Note that if C refines D and D refines E then C refines E. P P If c ∈ C, there is a d ∈ D such that cd = c; now there is an e ∈ E such that de = d; in this case, ce = (cd)e = c(de) = cd = c; as c is arbitrary, C refines E. Q Q 311H The order structure of a Boolean ring Again treating a Boolean ring A as an algebra of sets, we have a natural ordering on it, setting a ⊆ b if ab = a, so that a ⊆ b iff b a ⊆ bb. This translation makes it obvious that ⊆ is a partial ordering on A, with least element 0, and with greatest element 1 iff A is a Boolean algebra. Moreover, A is a lattice (definition: 2A1Ad), with a ∪ b = sup{a, b}, a ∩ b = inf{a, b} for all a, b ∈ A. Generally, for a0 , . . . , an ∈ A, supi≤n ai = a0 ∪ . . . ∪ an ,

inf i≤n ai = a0 ∩ . . . ∩ an ;

suprema and infima of finite subsets A correspond to unions and intersections of the corresponding families in the Stone space. (But suprema and infima of infinite subsets of A are a very different matter; see §313 below.) It may be obvious, but it is nevertheless vital to recognise that when A is a ring of sets then ⊆ agrees with ⊆. 311I The topology of a Stone space: Theorem Let Z be the Stone space of a Boolean ring A, and let T be {G : G ⊆ Z and for every z ∈ G there is an a ∈ A such that z ∈ b a ⊆ G}. Then T is a topology on Z, under which Z is a locally compact zero-dimensional Hausdorff space, and E = {b a : a ∈ A} is precisely the set of compact open subsets of Z. A is a Boolean algebra iff Z is compact.

311J

Boolean algebras

17

S proof (a) Because E is closed under ∩, and E = Z (recall that Z is the set of surjective homomorphisms from A to Z2 , so that every z ∈ Z is somewhere non-zero and belongs to some b a), E is a topology base, and T is a topology. (b) T is Hausdorff. P P Take any distinct z, w ∈ Z. Then there is an a ∈ A such that z(a) 6= w(a); let us take it that z(a) = 1, w(a) = 0. There is also a b ∈ A such that w(b) = 1, so that w(b + ab) = w(b) +2 w(a)w(b) = 1 and w ∈ (b + ab)b; also a(b + ab) = ab + a2 b = ab + ab = 0, so b a ∩ (b + ab)b = (a(b + ab))b = b 0 = ∅, and b a, (b + ab)b are disjoint members of T containing z, w respectively. Q Q (c) If a ∈ A then b a is compact. P P Let F be an ultrafilter on Z containing b a. For each b ∈ A, z0 (b) = limz→F z(b) must be defined in Z2 , since one of the sets {z : z(b) = 0}, {z : z(b) = 1} must belong to F. If b, c ∈ A, then the set F = {z : z(b) = z0 (b), z(c) = z0 (c), z(b + c) = z0 (b + c), z(bc) = z0 (bc)} belongs to F, so is not empty; take any z1 ∈ F ; then z0 (b + c) = z1 (b + c) = z1 (b) +2 z1 (c) = z0 (b) +2 z0 (c), z0 (bc) = z1 (bc) = z1 (b)z1 (c) = z0 (b)z0 (c). As b, c are arbitrary, z0 : A → Z2 is a ring homomorphism. Also z0 (a) = 1, because b a ∈ F , so z0 ∈ b a. Now let G be any open subset of Z containing z0 ; then there is a b ∈ A such that z0 ⊆ bb ⊆ G; since limz→F z(b) = z0 (b) = 1, we must have bb = {z : z(b) = 1} ∈ F and G ∈ F. Thus F converges to z0 . As F is arbitrary, b a is compact (2A3R). Q Q (d) This shows that b a is a compact open set for every a ∈ A. Moreover, since every point of Z belongs to some b a, every point of Z has a compact neighbourhood, and Z is locally compact. Every b a is closed (because it is compact, or otherwise), so E is a base for T consisting of open-and-closed sets, and T is zero-dimensional. (e) Now suppose that E ⊆ Z is an open compact set. If E = ∅ then E = b 0. Otherwise, set G = {b a : a ∈ A, b a ⊆ E}. S Then G is a family of open subsets S of Z and G = E, because E is open. But E is also compact, so there is a finite G0 ⊆ G such that E = G0 . Express G0 as {b a0 , . . . , b an }. Then E=b a0 ∪ . . . ∪ b an = (a0 ∪ . . . ∪ an )b. This shows that every compact open subset of Z is of the form b a for some a ∈ A. (f ) Finally, if A is a Boolean algebra then Z = b 1 is compact, by (c); while if Z is compact then (e) tells us that Z = b a for some a ∈ A, and of course this a must be a multiplicative identity for A, so that A is a Boolean algebra. 311J

We have a kind of converse of Stone’s theorem.

Proposition Let X be a locally compact zero-dimensional Hausdorff space. Then the set A of open-andcompact subsets of X is a subring of PZ. If Z is the Stone space of A, there is a unique homeomorphism θ : Z → X such that b a = θ−1 [a] for every a ∈ A. proof (a) Because X is Hausdorff, all its compact sets are closed, so every member of A is closed. Consequently a ∪ b, a \ b, a ∩ b and a4b belong to A for all a, b ∈ A, and A is a subring of PX. It will be helpful to know that A is a base for the topology of X. P P If G ⊆ X is open and x ∈ G, then (because X is locally compact) there is a compact set K ⊆ X such that x ∈ int K; now (because X is zero-dimensional) there is an open-and-closed set a ⊆ X such that x ∈ a ⊆ G ∩ int K; because a is a closed subset of a compact subset of X, it is compact, and belongs to A, while x ∈ a ⊆ G. Q Q

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311J

(b) Let R ⊆ Z × X be the relation {(z, x): for every a ∈ A, x ∈ a ⇐⇒ z(a) = 1}. Then R is the graph of a bijective function θ : Z → X. P P (i) If z ∈ Z and x, x0 ∈ X are distinct, then, because X is Hausdorff, there is an open set G ⊆ X containing x and not containing x0 ; because A is a base for the topology of X, there is an a ∈ A such that x ∈ a ⊆ G, so that x0 ∈ / a. Now either z(a) = 1 and (z, x0 ) ∈ / R, or z(a) = 0 and (z, x) ∈ / R. Thus R is the graph of a function θ with domain included in Z and taking values in X. (ii) If z ∈ Z, there is an a0 ∈ A such that z(a0 ) = 1. Consider A = {a : z(a) = 1}.TThis is a family of closed subsets ofTX containing the compact set a0 , and a ∩ b ∈ A for all a, b ∈ A. So A is not empty (3A3Db); take x ∈ A. Then x ∈ a whenever z(a) = 1. On the other hand, if z(a) = 0, then z(a0 \ a) = z(a0 4(a ∩ a0 )) = z(a0 ) +2 z(a0 )z(a) = 1, so x ∈ a0 \ a and x ∈ / a. Thus (z, x) ∈ R and θ(z) = x is defined. As z is arbitrary, the domain of θ is the whole of Z. (iii) If x ∈ X, define z : A → Z2 by setting z(a) = 1 if x ∈ a, 0 otherwise. It is elementary to check that z is a ring homomorphism form A to Z2 . To see that it takes the value 1, note that because A is a base for the topology of X there is an a ∈ A such that x ∈ a, so that z(a) = 1. So z ∈ Z, and of course (z, x) ∈ R. As x is arbitrary, θ is surjective. (iv) If z, z 0 ∈ Z and θ(z) = θ(z 0 ), then, for any a ∈ A, z(a) = 1 ⇐⇒ θ(z) ∈ a ⇐⇒ θ(z 0 ) ∈ a ⇐⇒ z 0 (a) = 1, so z = z 0 . Thus θ is injective. Q Q (c) For any a ∈ A, θ−1 [a] = {z : θ(z) ∈ a} = {z : z(a) = 1} = b a. It follows that P (i) If G ⊆ X is open, then (because A is a base for the topology S θ is a homeomorphism. P of X) G = {a : a ∈ A, a ⊆ G} and S S θ−1 [G] = {θ−1 [a] : a ∈ A, a ⊆ G} = {b a : a ∈ A, a ⊆ G} is an open subset of Z. As G is S arbitrary, θ is continuous. (ii) On theSother hand, if G ⊆ X and θ−1 [G] is −1 a for some A ⊆ A, so that G = A is an open set in X. Accordingly open, then θ [G] is of the form a∈A b θ is a homeomorphism. Q Q (d) Finally, I must check the uniqueness of θ. But of course if θ˜ : Z → X is any function such that −1 ˜ θ [a] = b a for every a ∈ A, then the graph of θ˜ must be R, so θ˜ = θ. 311K Remark Thus we have a correspondence between Boolean rings and zero-dimensional locally compact Hausdorff spaces which is (up to isomorphism, on the one hand, and homeomorphism, on the other) one-to-one. Every property of Boolean rings which we study will necessarily correspond to some property of zero-dimensional locally compact Hausdorff spaces. 311L Complemented distributive lattices I have introduced Boolean algebras through the theory of rings; this seems to be the quickest route to them from an ordinary undergraduate course in abstract algebra. However there are alternative approaches, taking the order structure rather than the algebraic operations as fundamental, and for the sake of an application in Chapter 35 I give the details of one of these. Proposition Let A be a lattice such that (i) (a ∨ b) ∧ c = (a ∧ c) ∨ (b ∧ c) for all a, b, c ∈ A; (ii) there is a bijection a 7→ a0 : A → A which is order-reversing, that is, a ≤ b iff b0 ≤ a0 , and such that a00 = a for every a; (iii) A has a least element 0 and a ∧ a0 = 0 for every a ∈ A. Then A has a Boolean algebra structure for which a ⊆ b iff a ≤ b.

311X

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19

proof (a) Write 1 for 00 ; if a ∈ A, then a0 ≥ 0 so a = a00 ≤ 00 = 1, and 1 is the greatest element of A. If a, b ∈ A then, because 0 is an order-reversing bijection, a0 ∨ b0 = (a ∧ b)0 . P P For c ∈ A, a0 ∨ b0 ≤ c ⇐⇒ a0 ≤ c & b0 ≤ c ⇐⇒ c0 ≤ a & c0 ≤ b ⇐⇒ c0 ≤ a ∧ b ⇐⇒ (a ∧ b)0 ≤ c. Q Q Similarly, a0 ∧ b0 = (a ∨ b)0 . If a, b, c ∈ A then (a ∧ b) ∨ c = ((a0 ∨ b0 ) ∧ c0 )0 = ((a0 ∧ c0 ) ∨ (b0 ∧ c0 ))0 = (a ∨ c) ∧ (b ∨ c). (b) Define addition and multiplication on A by setting a + b = (a ∧ b0 ) ∨ (a0 ∧ b),

ab = a ∧ b

for a, b ∈ A. (c)(i) If a, b ∈ A then (a + b)0 = (a0 ∨ b) ∧ (a ∨ b0 ) = (a0 ∧ a) ∨ (a0 ∧ b0 ) ∨ (b ∧ a) ∨ (b ∧ b0 ) = 0 ∨ (a0 ∧ b0 ) ∨ (b ∧ a) = (a0 ∧ b0 ) ∨ (a ∧ b). So if a, b, c ∈ A then (a + b) + c = ((a + b) ∧ c0 ) ∨ ((a + b)0 ∧ c) = (((a ∧ b0 ) ∨ (a0 ∧ b)) ∧ c0 ) ∨ (((a0 ∧ b0 ) ∨ (a ∧ b)) ∧ c) = (a ∧ b0 ∧ c0 ) ∨ (a0 ∧ b ∧ c0 ) ∨ (a0 ∧ b0 ∧ c) ∨ (a ∧ b ∧ c); as this last formula is symmetric in a, b and c, it is also equal to a + (b + c). Thus addition is associative. (ii) For any a ∈ A, a + 0 = 0 + a = (a0 ∧ 0) ∨ (a ∧ 00 ) = 0 ∨ (a ∧ 1) = a, so 0 is the additive identity of A. Also a + a = (a ∧ a0 ) ∨ (a0 ∧ a) = 0 ∨ 0 = 0 so each element of A is its own additive inverse, and (A, +) is a group. It is abelian because ∨, ∧ are commutative. (d) Because ∧ is associative and commutative, (A, ·) is a commutative semigroup; also 1 is its identity, because a ∧ 1 = a for every a ∈ A. As for the distributive law in A, ab + ac = (a ∧ b ∧ (a ∧ c)0 ) ∨ ((a ∧ b)0 ∧ a ∧ c) = (a ∧ b ∧ (a0 ∨ c0 )) ∨ ((a0 ∨ b0 ) ∧ a ∧ c) = (a ∧ b ∧ a0 ) ∨ (a ∧ b ∧ c0 ) ∨ (a0 ∧ a ∧ c) ∨ (b0 ∧ a ∧ c) = (a ∧ b ∧ c0 ) ∨ (b0 ∧ a ∧ c) = a ∧ ((b ∧ c0 ) ∨ (b0 ∧ c)) = a(b + c) for all a, b, c ∈ A. Thus (A, +, ·) is a ring; because a ∧ a = a for every a, it is a Boolean ring. (e) For a, b ∈ A, a ⊆ b ⇐⇒ ab = a ⇐⇒ a ∧ b = a ⇐⇒ a ≤ b, so the order relations of A coincide. Remark It is the case that the Boolean algebra structure of A is uniquely determined by its order structure, but I delay the proof to the next section (312L). 311X Basic exercises (a) Let A0 , . . . , An be sets. Show that A0 4 . . . 4An = {x : #({i : i ≤ n, x ∈ Ai }) is odd}.

20

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311Xb

(b) Let X be a set, and Σ ⊆ PX. Show that the following are equiveridical: (i) Σ is an algebra of subsets of X; (ii) Σ is a subring of PX (that is, contains ∅ and is closed under 4 and ∩) and contains X; (iii) ∅ ∈ Σ, X \ E ∈ Σ for every E ∈ Σ, and E ∩ F ∈ Σ for all E, F ∈ Σ. (c) Let A be any Boolean ring. Let a 7→ a0 be any bijection between A and a set B disjoint from A. Set B = A ∪ B, and extend the addition and multiplication of A to form binary operations on B by using the formulae a + b0 = a0 + b = (a + b)0 , a0 b = b + ab,

a0 + b0 = a + b,

ab0 = a + ab, a0 b0 = (a + b + ab)0 .

Show that B is a Boolean algebra and that A is an ideal in B. > (d) Let A be a Boolean ring, and K a finite subset of A. Show that the subring of A generated by K #(K) has at most 22 members, being the set of sums of products of members of K. > (e) Show that any finite Boolean ring is isomorphic to PX for some finite set X (and, in particular, is a Boolean algebra). (f ) Let A be any Boolean ring. Show that a ∪ (b ∩ c) = (a ∩ b) ∪ (a ∩ c), a ∪ (b ∩ c) = (a ∩ b) ∪ (a ∩ c) for all a, b, c ∈ A directly from the definitions in 311G, without using Stone’s theorem. >(g) Let A be any Boolean ring. Show that if we regard the Stone space Z of A as a subset of {0, 1}A , then the topology of Z (311I) is just the subspace topology induced by the ordinary product topology of {0, 1}A . (h) Let I be any set, and set X = {0, 1}I with its usual topology (3A3K). Show that for a subset E of X the following are equiveridical: (i) E is open-and-compact; (ii) E is determined by coordinates in a finite subset of I (definition: 254M); (iii) E belongs to the algebra of subsets of X generated by {Ei : i ∈ I}, where Ei = {x : x(i) = 1} for each i. (i) Let (A, ≤) be a lattice such that (α) A has a least element 0 and a greatest element 1 (β) for every a, b, c ∈ A, a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) and a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) (γ) for every a ∈ A there is an a0 ∈ A such that a ∨ a0 = 1 and a ∧ a0 = 0. Show that there is a Boolean algebra structure on A for which ≤ agrees with ⊆ . 311Y Further exercises (a) Let A be a Boolean ring, and B the Boolean algebra constructed by the method of 311Xc. Show that the Stone space of B can be identified with the one-point compactification (3A3O) of the Stone space of A. (b) Let (A, ∨, ∧, 0, 1) be such that (i) (A, ∨) is a commutative semigroup with identity 0 (ii) (A, ∧) is a commutative semigroup with identity 1 (iii) a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) for all a, b, c ∈ A (iv) a ∨ a = a ∧ a = a for every a ∈ A (v) for every a ∈ A there is an a0 ∈ A such that a ∨ a0 = 1, a ∧ a0 = 0. Show that there is a Boolean algebra structure on A for which ∨ = ∪ , ∧ = ∩ . 311 Notes and comments My aim in this section has been to get as quickly as possible to Stone’s theorem, since this is surely the best route to a picture of general Boolean algebras; they are isomorphic to algebras of sets. This means that all their elementary algebraic properties – indeed, all their first-order properties – can be effectively studied in the context of elementary set theory. In 311G-311H I describe a few of the ways in which the Stone representation suggests algebraic properties of Boolean algebras. You should not, however, come too readily to the conclusion that Boolean algebras will never be able to surprise you. In this book, in particular, we shall need to work a good deal with suprema and infima of infinite sets in Boolean algebras, for the ordering of 311H; and even though this corresponds to the ordering S ⊆ of ordinary sets, we find that (sup A)b is sufficiently different from a∈A b a to need new kinds of intuition.

312B

Homomorphisms

21

S (The point is that a∈A b a is an open set in the Stone space, but need not be compact if A is infinite, so may well be smaller than (sup A)b, even when sup A is defined in A.) There is also the fact that Stone’s theorem depends crucially on a fairly strong form of the axiom of choice (employed through Zorn’s Lemma in the argument of 311D). Of course I shall be using the axiom of choice without scruple throughout this volume. But it should be clear that such results as 312B-312C in the next section cannot possibly need the axiom of choice for their proofs, and that to use Stone’s theorem in such a context is slightly misleading. Nevertheless, it is so useful to be able to regard a Boolean algebra as an algebra of sets – especially when dealing with only finitely many elements of the algebra at a time – that henceforth I will almost always use the symbols 4 , ∩ for the addition and multiplication of a Boolean ring, and will use ∪ , \ , ⊆ without further comment, just as if I were considering ∪, \ and ⊆ in the Stone space. (In 311Gb I have given a definition of ‘disjointness’ in a Boolean algebra based on the same idea.) Even without the axiom of choice this approach can be justified, once we have observed that finitely-generated Boolean algebras are finite (311Xd), since relatively elementary methods show that any finite Boolean algebra is isomorphic to PX for some finite set X. I have taken a Boolean algebra to be a particular kind of commutative ring with identity. Of course there are other approaches. If we wish to think of the order relation as primary, then 311L and 311Xi are reasonably natural. Other descriptions can be based on a list of the properties of the binary operations ∪ , ∩ and the complementation operation a 7→ a0 = 1 \ a, as in 311Yb. I give extra space to 311L only because this is well adapted to an application in 352Q below.

312 Homomorphisms I continue the theory of Boolean algebras with a section on subalgebras, ideals and homomorphisms. From now on, I will relegate Boolean rings which are not algebras to the exercises; I think there is no need to set out descriptions of the trifling modifications necessary to deal with the extra generality. The first part of the section (312A-312K) concerns the translation of the basic concepts of ring theory into the language which I propose to use for Boolean algebras. 312L shows that the order relation on a Boolean algebra defines the algebraic structure, and in 312M-312N I give a fundamental result on the extension of homomorphisms. I end the section with results relating the previous ideas to the Stone representation of a Boolean algebra (312O-312S). 312A Subalgebras Let A be a Boolean algebra. I will use the phrase subalgebra of A to mean a subring of A containing its multiplicative identity 1 = 1A . 312B Proposition Let A be a Boolean algebra, and B a subset of A. Then the following are equiveridical, that is, if one is true so are the others: (i) B is a subalgebra of A; (ii) 0 ∈ B, a ∪ b ∈ B for all a, b ∈ B, and 1 \ a ∈ B for all a ∈ B; (iii) B 6= ∅, a ∩ b ∈ B for all a, b ∈ B, and 1 \ a ∈ B for all a ∈ B. proof (a)(i)⇒(iii) If B is a subalgebra of A, and a, b ∈ B, then of course we shall have 0, 1 ∈ B, so B 6= ∅, a ∩ b ∈ B,

1 \ a = 1 4 a ∈ B.

(b)(iii)⇒(ii) If (iii) is true, then there is some b0 ∈ B; now 1 \ b0 ∈ B, so 0 = b0 ∩ (1 \ b0 ) ∈ B. If a, b ∈ B, then a ∪ b = 1 \ ((1 \ a) ∩ (1 \ b)) ∈ B. So (ii) is true.

22

Boolean algebras

312B

(c)(ii)⇒(i) If (ii) is true, then for any a, b ∈ B, a ∩ b = 1 \ ((1 \ a) ∪ (1 \ b)) ∈ B, a 4 b = (a ∩ (1 \ b)) ∪ (b ∩ (1 \ a)) ∈ B, so (because also 0 ∈ B) B is a subring of A, and 1 = 1 \ 0 ∈ B, so B is a subalgebra. Remark Thus an algebra of subsets of a set X, as defined in 136E or 311Bb, is just a subalgebra of the Boolean algebra PX. 312C Ideals in Boolean algebras: Proposition If A is a Boolean algebra, a set I ⊆ A is an ideal of A iff 0 ∈ I, a ∪ b ∈ I for all a, b ∈ I, and a ∈ I whenever b ∈ I and a ⊆ b. proof (a) Suppose that I is an ideal. Then of course 0 ∈ I. If a, b ∈ I then a ∩ b ∈ I so a ∪ b = (a 4 b) 4 (a ∩ b) ∈ I. If b ∈ I and a ⊆ b then a = a ∩ b ∈ I. (b) Now suppose that I satisfies the conditions proposed. If a, b ∈ I then a4b ⊆ a∪b ∈ I so a 4 b ∈ I, while of course −a = a ∈ I, and also 0 ∈ I, by hypothesis; thus I is a subgroup of (A, 4). Finally, if a ∈ I and b ∈ A then a ∩ b ⊆ a ∈ I, so b ∩ a = a ∩ b ∈ I; thus I is an ideal. Remark Thus what I have called an ‘ideal of subsets of X’ in 232Xc is just an ideal in the Boolean algebra PX. 312D Principal ideals Of course, while an ideal I in a Boolean algebra A is necessarily a subring, it is not as a rule a subalgebra, except in the special case I = A. But if we say that a principal ideal of A is the ideal Aa generated by a single element a of A, we have a special phenomenon. 312E Proposition Let A be a Boolean algebra, and a any element of A. Then the principal ideal Aa of A generated by a is just {b : b ∈ A, b ⊆ a}, and (with the inherited operations ∩ ¹ Aa × Aa , 4 ¹ Aa × Aa ) is a Boolean algebra in its own right, with multiplicative identity a. proof b ⊆ a iff b ∩ a = a, so that Aa = {b : b ⊆ a} = {b ∩ a : b ∈ A} is an ideal of A, and of course it is the smallest ideal of A containing a. Being an ideal, it is a subring; the idempotent relation b ∩ b = b is inherited from A, so it is a Boolean ring; and a is plainly its multiplicative identity. 312F Boolean homomorphisms Now suppose that A and B are two Boolean algebras. I will use the phrase Boolean homomorphism to mean a function π : A → B which is a ring homomorphism (that is, π(a 4 b) = πa 4 πb, π(a ∩ b) = πa ∩ πb for all a, b ∈ A) which is uniferent, that is, π(1A ) = 1B . 312G Proposition Let A, B and C be Boolean algebras. (a) If π : A → B is a Boolean homomorphism, then π[A] is a subalgebra of B. (b) If π : A → B and θ : B → C are Boolean homomorphisms, then θπ : A → C is a Boolean homomorphism. (c) If π : A → B is a bijective Boolean homomorphism, then π −1 : B → A is a Boolean homomorphism. proof These are all immediate consequences of the corresponding results for ring homomorphisms (3A2D).

312K

Homomorphisms

23

312H Proposition Let A and B be Boolean algebras, and π : A → B a function. Then the following are equiveridical: (i) π is a Boolean homomorphism; (ii) π(a ∩ b) = πa ∩ πb and π(1A \ a) = 1B \ πa for all a, b ∈ A; (iii) π(a ∪ b) = πa ∪ πb and π(1A \ a) = 1B \ πa for all a, b ∈ A; (iv) π(a ∪ b) = πa ∪ πb and πa ∩ πb = 0B whenever a, b ∈ A and a ∩ b = 0A , and π(1A ) = 1B . proof (i)⇒(iv) If π is a Boolean homomorphism then of course π(1A ) = 1B ; also, given that a ∩ b = 0 in A, πa ∩ πb = π(a ∩ b) = π(0A ) = 0B , π(a ∪ b) = π(a 4 b) = πa 4 πb = πa ∪ πb. (iv)⇒(iii) Assume (iv), and take a, b ∈ A. Then πa = π(a ∩ b) ∪ π(a \ b),

πb = π(a ∩ b) ∪ π(b \ a),

so π(a ∪ b) = πa ∪ π(b \ a) = π(a ∩ b) ∪ π(a \ b) ∪ π(b \ a) = πa ∪ πb. Taking b = 1 \ a, we must have 1B = π(1A ) = πa ∪ π(1A \ a),

0B = πa ∩ π(1A \ a),

so π(1A \ a) = 1B \ πa. Thus (iii) is true. (iii)⇒(ii) If (iii) is true and a, b ∈ A, then π(a ∪ b) = π(1A \ ((1A \ a) ∩ (1A \ b))) = 1B \ ((1B \ πa) ∩ (1B \ πb))) = πa ∪ πb. So (ii) is true. (ii)⇒(i) If (ii) is true, then π(a 4 b) = π((1A \ ((1A \ a) ∩ (1A \ b)) ∩ (1A \ (a ∩ b))) = (1B \ ((1B \ πa) ∩ (1B \ πb)) ∩ (1B \ (πa ∩ πb))) = πa 4 πb for all a, b ∈ A, so π is a ring homomorphism; and now π(1A ) = π(1A \ 0A ) = 1B \ π(0A ) = 1B \ 0B = 1B , so that π is a Boolean homomorphism. 312I Proposition If A, B are Boolean algebras and π : A → B is a Boolean homomorphism, then πa ⊆ πb whenever a ⊆ b in A. proof a ⊆ b =⇒ a ∩ b = a =⇒ πa ∩ πb = πa =⇒ πa ⊆ πb. 312J Proposition Let A be a Boolean algebra, and a any member of A. Then the map b 7→ a ∩ b is a surjective Boolean homomorphism from A onto the principal ideal Aa generated by a. proof This is an elementary verification. 312K Quotient algebras: Proposition Let A be a Boolean algebra and I an ideal of A. Then the quotient ring A/I (3A2F) is a Boolean algebra, and the canonical map a 7→ a• : A → A/I is a Boolean homomorphism, so that (a 4 b)• = a• 4 b• , for all a, b ∈ A.

(a ∪ b)• = a• ∪ b• ,

(a ∩ b)• = a• ∩ b• ,

(a \ b)• = a• \ b•

24

Boolean algebras

312K

(b) The order relation on A/I is defined by the formula a• ⊆ b• ⇐⇒ a \ b ∈ I. For any a ∈ A, {u : u ⊆ a• } = {b• : b ⊆ a}. proof (a) Of course the map a 7→ a• = {a 4 b : b ∈ I} is a ring homomorphism (3A2F). Because (a• )2 = (a2 )• = a• for every a ∈ A, A/I is a Boolean ring; because 1• is a multiplicative identity, it is a Boolean algebra, and a 7→ a• is a Boolean homomorphism. The formulae given are now elementary. (b) We have a• ⊆ b• ⇐⇒ a• \ b• = 0 ⇐⇒ a \ b ∈ I. Now {u : u ⊆ a• } = {u ∩ a• : u ∈ A/I} = {(b ∩ a)• : b ∈ A} = {b• : b ⊆ a}. 312L The above results are both repetitive and nearly trivial. Now I come to something with a little more meat to it. Proposition If A and B are Boolean algebras and π : A → B is a bijection such that πa ⊆ πb whenever a ⊆ b, then π is a Boolean algebra isomorphism. proof (a) Because π is surjective, there must be c0 , c1 ∈ A such that πc0 = 0B , πc1 = 1B ; now π(0A ) ⊆ πc0 , πc1 ⊆ π(1A ), so we must have π(0A ) = 0B , π(1A ) = 1B . (b) If a ∈ A, then πa ∪ π(1A \ a) = 1B . P P There is a c ∈ A such that πc = 1B \ (πa ∪ π(1A \ a)). Now π(c ∩ a) ⊆ πc ∩ πa = 0B ,

π(c \ a) ⊆ πc ∩ π(1A \ a) = 0B ;

as π is injective, c ∩ a = c \ a = 0A and c = 0A , πc = 0B , πa ∪ π(1A \ a) = 1B . Q Q (c) If a ∈ A, then πa ∩ π(1A \ a) = 0B . P P It may be clear to you that this is just a dual form of (b). If not, I repeat the argument in the form now appropriate. There is a c ∈ A such that πc = 1B \ (πa ∩ π(1A \ a)). Now π(c ∪ a) ⊇ πc ∪ πa = 1B ,

π(c ∪ (1A \ a)) ⊇ πc ∪ π(1A \ a) = 1B ;

as π is injective, c ∪ a = c ∪ (1A \ a) = 1A and c = 1A , πc = 1B , πa ∩ π(1A \ a) = 0B . Q Q (d) Putting (b) and (c) together, we have π(1A \ a) = 1B \ πa for every a ∈ A. Now π(a ∪ b) = πa ∪ πb for every a, b ∈ A. P P Surely πa ∪ πb ⊆ π(a ∪ b). Let c ∈ A be such that πc = π(a ∪ b) \ (πa ∪ πb). Then π(c ∩ a) ⊆ πc ∩ πa = 0B ,

π(c ∩ b) ⊆ πc ∩ πb = 0B ,

so c ∩ a = c ∩ b = 0 and c ⊆ 1A \ (a ∪ b); accordingly πc ⊆ π(1A \ (a ∪ b)) = 1B \ π(a ∪ b); as also πc ⊆ π(a ∪ b), πc = 0B and π(a ∪ b) = πa ∪ πb. Q Q (e) So the conditions of 312H(iii) are satisfied and π is a Boolean homomorphism; being bijective, it is an isomorphism. 312M I turn next to a fundamental lemma on the construction of homomorphisms. We need to start with a proper description of a certain type of subalgebra. Lemma Let A be a Boolean algebra, and A0 a subalgebra of A; let c be any member of A. Then A1 = {(a ∩ c) ∪ (b \ c) : a, b ∈ A0 } is a subalgebra of A; it is the subalgebra of A generated by A0 ∪ {c}.

312N

Homomorphisms

25

proof We have to check the following: a = (a ∩ c) ∪ (a \ c) ∈ A1 for every a ∈ A0 , so A0 ⊆ A1 ; in particular, 0 ∈ A1 . 1 \ ((a ∩ c) ∪ (b \ c)) = ((1 \ a) ∩ c) ∪ ((1 \ b) \ c) ∈ A1 for all a, b ∈ A0 , so 1 \ d ∈ A1 for every d ∈ A1 . (a ∩ c) ∪ (b \ c) ∪ (a0 ∩ c) ∪ (b0 \ c) = ((a ∪ a0 ) ∩ c) ∪ ((b ∪ b0 ) \ c) ∈ A1 for all a, b, a0 , b0 ∈ A0 , so d ∪ d0 ∈ A1 for all d, d0 ∈ A1 . Thus A1 is a subalgebra of A (using 312B). c = (1 ∩ c) ∪ (0 \ c) ∈ A1 , so A1 includes A0 ∪ {c}; and finally it is clear that any subalgebra of A including A0 ∪ {c}, being closed under ∩ , ∪ and complementation, must include A1 , so that A1 is the subalgebra of A generated by A0 ∪ {c}. 312N Lemma Let A and B be Boolean algebras, A0 a subalgebra of A, π : A0 → B a Boolean homomorphism, and c ∈ A. If v ∈ B is such that πa ⊆ v ⊆ πb whenever a, b ∈ A0 and a ⊆ c ⊆ b, then there is a unique Boolean homomorphism π1 from the subalgebra A1 of A generated by A0 ∪ {c} such that π1 extends π and π1 c = v. proof (a) The basic fact we need to know is that if a, a0 , b, b0 ∈ A0 and (a ∩ c) ∪ (b \ c) = d = (a0 ∩ c) ∪ (b0 \ c), then (πa ∩ v) ∪ (πb \ v) = (πa0 ∩ v) ∪ (πb0 \ v). P P We have a ∩ c = d ∩ c = a0 ∩ c. Accordingly (a 4 a0 ) ∩ c = 0 and c ⊆ 1 \ (a 4 a0 ). Consequently (since a 4 a0 surely belongs to A0 ) v ⊆ π(1 \ (a 4 a0 )) = 1 \ (πa 4 πa0 ), and πa ∩ v = πa0 ∩ v. Similarly, b \ c = d \ c = b0 \ c, so (b 4 b0 ) \ c = 0,

b 4 b0 ⊆ c,

π(b 4 b0 ) ⊆ v

and πb \ v = πb0 \ v. Putting these together, we have the result. Q Q (b) Consequently, we have a function π1 defined by writing π1 ((a ∩ c) ∪ (b \ c)) = (πa ∩ v) ∪ (πb \ c) for all a, b ∈ A0 ; and 312M tells us that the domain of π1 is just A1 . Now π1 is a Boolean homomorphism. P P This amounts to running through the proof of 312M again. (i) If a, b ∈ A0 , then π1 (1 \ ((a ∩ c) ∪ (b \ c))) = π1 (((1 \ a) ∩ c) ∪ ((1 \ b) \ c)) = (π(1 \ a) ∩ v) ∪ (π(1 \ b) \ v) = ((1 \ πa) ∩ v) ∪ ((1 \ πb) \ v) = 1 \ ((πa ∩ v) ∪ (πb \ v)) = 1 \ π1 ((a ∩ c) ∪ (b \ c)).

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Boolean algebras

312N

So π1 (1 \ d) = 1 \ π1 d for every d ∈ A1 . (ii) If a, b, a0 , b0 ∈ A0 , then π1 ((a ∩ c) ∪ (b \ c) ∪ (a0 ∩ c) ∪ (b0 \ c)) = π1 (((a ∪ a0 ) ∩ c) ∪ ((b ∪ b0 ) \ c)) = (π(a ∪ a0 ) ∩ v) ∪ (π(b ∪ b0 ) \ v) = ((πa ∪ πa0 ) ∩ v) ∪ ((πb ∪ πb0 ) \ v) = (πa ∩ v) ∪ (πb \ v) ∪ (πa0 ∩ v) ∪ (πb0 \ v) = π1 ((a ∩ c) ∪ (b \ c)) ∪ π1 ((a0 ∩ v) ∪ (b0 \ v)). So π1 (d ∪ d0 ) = π1 d ∪ π1 d0 for all d, d0 ∈ A1 . By 312H(iii), π1 is a Boolean homomorphism. Q Q (c) If a ∈ A0 , then π1 a = π1 ((a ∩ c) ∪ (a \ c)) = (πa ∩ v) ∪ (πa \ v) = πa, so π1 extends π. As for the action of π1 on c, π1 c = π1 ((1 ∩ c) ∪ (0 \ c)) = (π1 ∩ v) ∪ (π0 \ v) = (1 ∩ v) ∪ (0 \ v) = v, as required. (d) Finally, the formula of (b) is the only possible definition for any Boolean homomorphism from A1 to B which will extend π and take c to v. So π1 is unique. 312O Homomorphisms and Stone spaces Because the Stone space Z of a Boolean algebra A (311E) can be constructed explicitly from the algebraic structure of A, it must in principle be possible to describe any feature of the Boolean structure of A in terms of Z. In the next few paragraphs I work through the most important identifications. Proposition Let A be a Boolean algebra, and Z its Stone space; write b a ⊆ Z for the open-and-closed set corresponding to a ∈ A. Then there is a one-to-one correspondence between ideals I of A and open sets G ⊆ Z, given by the formulae S a, I = {a : b a ⊆ G}. G = a∈I b S a; then H(I) is a union of open subsets of Z, so is open. proof (a) For any ideal I C A, set H(I) = a∈I b For any open set G ⊆ Z, set J(G) = {a : a ∈ A, b a ⊆ G}; then J(G) satisfies the conditions of 312C, so is an ideal of A. (b) If I C A, then J(H(I)) = I. P P (i) If a ∈ I, then b a ⊆ H(I) so a ∈ J(H(I)). (ii) If a ∈ J(H(I)), then S b b a is compact and all the b are open, there must be finitely many b0 , . . . , bn ∈ I b a ⊆ H(I) = b∈I b. Because b such that b a ⊆ bb0 ∪ . . . ∪ bbn . But now a ⊆ b0 ∪ . . . ∪ bn ∈ I, so a ∈ I. Q Q (c) If G ⊆ Z is open, then H(J(G)) = G. P P (i) If z ∈ G, then (because {b a : a ∈ A} is a base for the topology of Z) there is an a ∈ A such that z ∈ b a ⊆ G; now a ∈ J(G) and z ∈ H(J(G)). (ii) If z ∈ H(J(G)), there is an a ∈ J(G) such that z ∈ b a; now b a ⊆ G, so z ∈ G. Q Q This shows that the maps G 7→ J(G), I 7→ H(I) are two halves of a one-to-one correspondence, as required. 312P Theorem Let A, B be Boolean algebras, with Stone spaces Z, W ; write b a ⊆ Z, bb ⊆ W for the open-and-closed sets corresponding to a ∈ A, b ∈ B. Then we have a one-to-one correspondence between Boolean homomorphisms π : A → B and continuous functions φ : W → Z, given by the formula πa = b ⇐⇒ φ−1 [b a] = bb. proof (a) Recall that I have constructed Z, W as the sets of Boolean homomorphisms from A, B to Z2 (311E). So if π : A → B is any Boolean homomorphism, and w ∈ W , ψπ (w) = wπ is a Boolean homomorphism from A to Z2 (312Gb), and belongs to Z. Now ψπ−1 [b a] = π ca for every a ∈ A. P P

312R

Homomorphisms

27

ψπ−1 [b a] = {w : ψπ (w) ∈ b a} = {w : wπ ∈ b a} = {w : wπ(a) = 1} = {w : w ∈ π ca}. Q Q S Consequently ψπ is continuous. P P Let G be any open subset of Z. Then G = {b a:b a ⊆ G}, so S S ψπ−1 [G] = {ψπ−1 [b a] : b a ⊆ G} = {c πa : b a ⊆ G} is open. As G is arbitrary, ψπ is continuous. Q Q (b) If φ : W → Z is continuous, then for any a ∈ A the set φ−1 [b a] must be an open-and-closed set in W ; consequently there is a unique member of B, call it θφ a, such that φ−1 [b a] = θd φ a. Observe that, for any w ∈ W and a ∈ A, w(θφ a) = 1 ⇐⇒ w ∈ θd a ⇐⇒ (φ(w))(a) = 1, φ a ⇐⇒ φ(w) ∈ b so φ(w) = wθφ . Now θφ is a Boolean homomorphism. P P (i) If a, b ∈ A then c θφ (a ∪ b)b = φ−1 [(a ∪ b)b] = φ−1 [b a ∪ bb] = φ−1 [b a] ∪ φ−1 [bb] = θd φ a ∪ θφ b = (θφ a ∪ θφ b)b, so θφ (a ∪ b) = θφ a ∪ θφ b. (ii) If a ∈ A, then θφ (1 \ a)b = φ−1 [(1 \ a)b] = φ−1 [Z \ b a] = W \ φ−1 [b a] = W \ θd φ a = (1 \ θφ a)b, so θφ (1 \ a) = 1 \ θφ a. (iii) By 312H, θφ is a Boolean homomorphism. Q Q (c) For any Boolean homomorphism π : A → B, π = θψπ . P P For a ∈ A, (θψπ a)b = ψπ−1 [b a] = π ca, so θψπ a = a. Q Q (d) For any continuous function φ : W → Z, φ = ψθφ . P P For any w ∈ W , ψθφ (w) = wθφ = φ(w). Q Q (e) Thus π 7→ ψπ , φ 7→ θφ are the two halves of a one-to-one correspondence, as required. 312Q Theorem Let A, B, C be Boolean algebras, with Stone spaces Z, W and V . Let π : A → B and θ : B → C be Boolean homomorphisms, with corresponding continuous functions φ : W → Z and ψ : V → W . Then the Boolean homomorphism θπ : A → C corresponds to the continuous function φψ : V → Z. proof For any a ∈ A, d = (θ(πa))b = ψ −1 [c θπa π a] = ψ −1 [φ−1 [b a]] = (φψ)−1 [b a].

312R Proposition Let A and B be Boolean algebras, with Stone spaces Z and W , and π : A → B a Boolean homomorphism, with associated continuous function φ : W → Z. Then (a) π is injective iff φ is surjective; (b) π is surjective iff φ is injective. proof (a) If a ∈ A, then b a ∩ φ[W ] = ∅ ⇐⇒ φ(w) ∈ /b a for every w ∈ W ⇐⇒ (φ(w))(a) = 0 for every w ∈ W ⇐⇒ w(πa) = 0 for every w ∈ W ⇐⇒ πa = 0. Now W is compact, so φ[W ] is also compact, therefore closed, and

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Boolean algebras

312R

φ is not surjective ⇐⇒ Z \ φ[W ] 6= ∅ ⇐⇒ there is a non-zero a ∈ A such that b a ⊆ Z \ φ[W ] ⇐⇒ there is a non-zero a ∈ A such that πa = 0 ⇐⇒ π is not injective (3A2Db). (b)(i) If π is surjective and w, w0 are distinct members of W , then there is a b ∈ B such that w ∈ bb and w0 ∈ / bb. Now b = πa for some a ∈ A, so φ(w) ∈ b a and φ(w0 ) ∈ /b a, and φ(w) 6= φ(w0 ). As w and w0 are arbitrary, φ is injective. (ii) If φ is injective and b ∈ B,Sthen K = φ[bb], L = φ[W \ bb] are disjoint compact subsets of Z. Consider I = {a : a ∈ A, L ∩ b a = ∅}. Then a∈I b a = Z \ L ⊇ K. Because K is compact and every b a is open, there is a finite family a0 , . . . , an ∈ I such that K ⊆ b a0 ∪ . . . ∪ b an . Set a = a0 ∪ . . . ∪ an . Then b a=b a0 ∪ . . . ∪ b an includes K and is disjoint from L. So π ca = φ−1 [b a] includes bb and is disjoint from W \ bb; that is, π ca = bb and πa = b. As b is arbitrary, π is surjective. 312S Principal ideals If A is a Boolean algebra and a ∈ A, we have a natural surjective Boolean homomorphism b 7→ b ∩ a : A → Aa , the principal ideal generated by a (312J). Writing Z for the Stone space of A and Za for the Stone space of Aa , this homomorphism must correspond to an injective continuous function φ : Za → Z (312Rb). Because Za is compact and Z is Hausdorff, φ must be a homeomorphism between Za and its image φ[Za ] ⊆ Z (3A3Dd). To identify φ[Za ], note that it is compact, therefore closed, and that Z \ φ[Za ] = =

[ [

{bb : b ∈ A, bb ∩ φ[Za ] = ∅} [ {bb : φ−1 [bb] = ∅} = {bb : b ∩ a = 0} = Z \ b a,

so that φ[Za ] = b a. It is therefore natural to identify Za with the open-and-closed set b a ⊆ Z. 312X Basic exercises (a) Let A be a Boolean ring, and B a subset of A. Show that B is a subring of A iff 0 ∈ B and a ∪ b, a \ b ∈ B for all a, b ∈ B. (b) Let A be a Boolean algebra and B a subset of A. Show that B is a subalgebra of A iff 1 ∈ B and a \ b ∈ B for all a, b ∈ B. (c) Let A be a Boolean algebra. Suppose that I ⊆ A ⊆ A are such that 1 ∈ A, a ∩ b ∈ I for all a, b ∈ I and a \ b ∈ A whenever a, b ∈ A and b ⊆ a. Show that A includes the subalgebra of A generated by I. (Hint: 136Xf.) (d) Show that if A is a Boolean ring, a set I ⊆ A is an ideal of A iff 0 ∈ I, a ∪ b ∈ I for all a, b ∈ I, and a ∈ I whenever b ∈ I and a ⊆ b. (e) Let A and B be Boolean algebras, and φ : A → B a function such that (i) φ(a) ⊆ φ(b) whenever a ⊆ b (ii) φ(a) ∩ φ(b) = 0B whenever a ∩ b = 0A (iii) φ(a) ∪ φ(b) ∪ φ(c) = 1B whenever a ∪ b ∪ c = 1A . Show that φ is a Boolean homomorphism. (f ) Let A be a Boolean ring, and a any member of A. Show that the map b 7→ a∩b is a ring homomorphism from A onto the principal ideal Aa generated by a. (g) Let A1 and A2 be Boolean rings, and let B1 , B2 be the Boolean algebras constructed from them by the method of 311Xc. Show that any ring homomorphism from A1 to A2 has a unique extension to a Boolean homomorphism from B1 to B2 .

312 Notes

Homomorphisms

29

(h) Let A and B be Boolean rings, A0 a subalgebra of A, π : A0 → B a ring homomorphism, and c ∈ A. Show that if v ∈ B is such that πa \ v = πb ∩ v = 0 whenever a, b ∈ A0 and a \ c = b ∩ c = 0, then there is a unique ring homomorphism π1 from the subring A1 of A generated by A0 ∪ {c} such that π1 extends π0 and π1 c = v. (i) Let A be a Boolean ring, and Z its Stone space. Show that there S is a one-to-one correspondence between ideals I of A and open sets G ⊆ Z, given by the formulae G = a∈I b a, I = {a : b a ⊆ G}. (j) Let A be a Boolean algebra, and suppose that A is the subalgebra of itself generated by A0 ∪ {c}, where A0 is a subalgebra of A and c ∈ A. Let Z be the Stone space of A and Z0 the Stone space of A0 . Let ψ : Z → Z0 be the continuous surjection corresponding to the embedding of A0 in A. Show that ψ¹ b c and ψ¹Z \ b c are injective. Now let B be another Boolean algebra, with Stone space W , and π : A0 → B a Boolean homomorphism, with corresponding function φ : W → Z0 . Show that there is a continuous function φ1 : W → Z such that ψφ1 = φ iff there is an open-and-closed set V ⊆ W such that φ[V ] ⊆ ψ[b c] and φ[W \ V ] ⊆ ψ[Z \ b c]. (k) Let A be a Boolean algebra, with Stone space Z, and I an ideal of A, corresponding to an open set G ⊆ Z. Show that the Stone space of the quotient algebra A/I may be identified with Z \ G. 312Y Further exercises (a) Find a function φ : P{0, 1, 2} → Z2 such that φ(1 \ a) = 1 \ φa for every a ∈ P{0, 1, 2} and φ(a) ⊆ φ(b) whenever a ⊆ b, but φ is not a Boolean homomorphism. (b) Let A be the Boolean ring of finite subsets of N. Show that there is a bijection π : A → A such that πa ⊆ πb whenever a ⊆ b but π is not a ring homomorphism. (c) Let A, B be Boolean rings, with Stone spaces Z, W . Show that we have a one-to-one correspondence between ring homomorphisms π : A → B and continuous functions φ : H → Z, where H ⊆ W is an open set, such that φ−1 [K] is compact for every compact set K ⊆ Z, given by the formula πa = b ⇐⇒ φ−1 [b a] = bb. (d) Let A, B, C be Boolean rings, with Stone spaces Z, W and V . Let π : A → B and θ : B → C be ring homomorphisms, with corresponding continuous functions φ : H → Z and ψ : G → W . Show that the ring homomorphism θπ : A → C corresponds to the continuous function φψ : ψ −1 [H] → Z. (e) Let A and B be Boolean rings, with Stone spaces Z and W , and π : A → B a ring homomorphism, with associated continuous function φ : H → Z. Show that π is injective iff φ[H] is dense in Z, and that π is surjective iff φ is injective and H = W . (f ) Let A be a Boolean ring and a ∈ A. Show that the Stone space of the principal ideal Aa of A generated by a can be identified with the compact open set b a in the Stone space of A. Show that the identity map is a ring homomorphism from Aa to A, and corresponds to the identity function on b a. 312 Notes and comments The definitions of ‘subalgebra’ and ‘Boolean homomorphism’ (312A, 312F), like that of ‘Boolean algebra’, are a trifle arbitrary, but will be a convenient way of mandating appropriate treatment of multiplicative identities. I run through the work of 312A-312J essentially for completeness; once you are familiar with Boolean algebras, they should all seem obvious. 312L has a little bit more to it. It shows that the order structure of a Boolean algebra defines the ring structure, in a fairly strong sense. I call 312N a ‘lemma’, but actually it is the most important result in this section; it is the basic tool we have for extending a homomorphism from a subalgebra to a slightly larger one, and with Zorn’s Lemma (another ‘lemma’ which deserves a capital L) will provide us with general methods of constructing homomorphisms. In 312O-312S I describe the basic relationships between the Boolean homomorphisms and continuous functions on Stone spaces. 312P-312Q show that, in the language of category theory, the Stone representation provides a ‘contravariant functor’ from the category of Boolean algebras with Boolean homomorphisms to the category of topological spaces with continuous functions. Using 311I-311J, we know exactly which topological spaces appear, the zero-dimensional compact Hausdorff spaces; and we know also that the functor is faithful, that is, that we can recover Boolean algebras and homomorphisms from the corresponding topological spaces

30

Boolean algebras

312 Notes

and continuous functions. There is an agreeable duality in 312R. All of this can be done for Boolean rings, but there are some extra complications (312Yc-312Yf). To my mind, the very essence of the theory of Boolean algebras is the fact that they are abstract rings, but at the same time can be thought of ‘locally’ as algebras of sets. Consequently we can bring two quite separate kinds of intuition to bear. 312N gives an example of a ring-theoretic problem, concerning the extension of homomorphisms, which has a resolution in terms of the order relation, a concept most naturally described in terms of algebras-of-sets. It is very much a matter of taste and habit, but I myself find that a Boolean homomorphism is easiest to think of in terms of its action on finite subalgebras, which are directly representable as PX for some finite X (311Xe); the corresponding continuous map between Stone spaces is less helpful. I offer 312Xj, the Stone-space version of 312N, for you to test your own intuitions on.

313 Order-continuous homomorphisms Because a Boolean algebra has a natural partial order (311H), we have corresponding notions of upper bounds, lower bounds, suprema and infima. These are particularly important in the Boolean algebras arising in measure theory, and the infinitary operations ‘sup’ and ‘inf’ require rather more care than the basic binary operations ‘ ∪ ’, ‘ ∩ ’, because intuitions from elementary set theory are sometimes misleading. I therefore take a section to work through the most important properties of these operations, together with the homomorphisms which preserve them. 313A Relative complementation: Proposition Let A be a Boolean algebra, e a member of A, and A a non-empty subset of A. (a) If sup A is defined in A, then inf{e \ a : a ∈ A} is defined and equal to e \ sup A. (b) If inf A is defined in A, then sup{e \ a : a ∈ A} is defined and equal to e \ inf A. proof (a) Writing a0 for sup A, we have e \ a0 ⊆ e \ a for every a ∈ A, so e \ a0 is a lower bound for C = {e \ a : a ∈ A}. Now suppose that c is any lower bound for C. Then (because A is not empty) c ⊆ e, and a = (a \ e) ∪ (e \ (e \ a)) ⊆ (a0 \ e) ∪ (e \ c) for every a ∈ A. Consequently a0 ⊆ (a0 \ e) ∪ (e \ c) is disjoint from c and c = c ∩ e ⊆ e \ a0 . Accordingly e \ a0 is the greatest lower bound of C, as claimed. (b) This time set a0 = inf A, C = {e \ a : a ∈ A}. As before, e \ a0 is surely an upper bound for C. If c is any upper bound for C, then e \ c ⊆ e \ (e \ a) = e ∩ a ⊆ a for every a ∈ A, so e \ c ⊆ a0 and e \ a0 ⊆ c. As c is arbitrary, e \ a0 is indeed the least upper bound of C. Remark In the arguments above I repeatedly encourage you to treat ∩ , ∪ , \ , ⊆ as if they were the corresponding operations and relation of basic set theory. This is perfectly safe so long as we take care that every manipulation so justified has only finitely many elements of the Boolean algebra in hand at once. 313B General distributive laws: Proposition Let A be a Boolean algebra. (a) If e ∈ A and A ⊆ A is a non-empty set such that sup A is defined in A, then sup{e ∩ a : a ∈ A} is defined and equal to e ∩ sup A. (b) If e ∈ A and A ⊆ A is a non-empty set such that inf A is defined in A, then inf{e ∪ a : a ∈ A} is defined and equal to e ∪ inf A. (c) Suppose that A, B ⊆ A are non-empty and sup A, sup B are defined in A. Then sup{a ∩ b : a ∈ A, b ∈ B} is defined and is equal to sup A ∩ sup B. (d) Suppose that A, B ⊆ A are non-empty and inf A, inf B are defined in A. Then inf{a ∪ b : a ∈ A, b ∈ B} is defined and is equal to inf A ∪ inf B.

313C

Order-continuous homomorphisms

31

proof (a) Set B = {e \ a : a ∈ A}, C = {e \ b : b ∈ B} = {e ∩ a : a ∈ A}. Using 313A, we have inf B = e \ sup A,

sup C = e \ inf B = e ∩ sup A,

as required. (b) Set a0 = inf A, B = {e ∪ a : a ∈ A}. Then e ∪ a0 ⊆ e ∪ a for every a ∈ A, so e ∪ a0 is a lower bound for B. If c is any lower bound for B, then c \ e ⊆ a for every a ∈ A, so c \ e ⊆ a0 and c ⊆ e ∪ a0 ; thus e ∪ a0 is the greatest lower bound for B, as claimed. (c) By (a), we have a ∩ sup B = supb∈B a ∩ b for every a ∈ A, so supa∈A,b∈B a ∩ b = supa∈A (a ∩ sup B) = sup A ∩ sup B, using (a) again. (d) Similarly, using (b) twice, inf a∈A,b∈B a ∪ b = inf a∈A (a ∪ inf B) = inf A ∪ inf B. 313C As always, it is worth developing a representation of the concepts of sup and inf in terms of Stone spaces. Proposition Let A be a Boolean algebra, and Z its Stone space; for a ∈ A write b a for the corresponding open-and-closed subset of Z. S (a) If A ⊆ A and a0 ∈ A then a0 = sup A in A iff b a0 = a∈A b a. T a. (b) If A ⊆ A is non-empty and a0 ∈ A then a0 = inf A in A iff b a0 = int a∈A b T a is nowhere dense in Z. (c) If A ⊆ A is non-empty then inf A = 0 in A iff a∈A b proof (a) For any b ∈ A, b is an upper bound for A ⇐⇒ b a ⊆ bb for every a ∈ A [ [ ⇐⇒ b a ⊆ bb ⇐⇒ b a ⊆ bb a∈A

a∈A

S because bb is certainly closed in Z. It follows at once that if b a0 is actually equal to a∈A b a then a0 must be S S the least upper bound of A in A. On the other hand, if a0 = sup A, then a∈A b a⊆b a0 . ?? If b a0 6= a∈A b a, S then b a0 \ a∈A b a is a non-empty open set in Z, so includes bb for some non-zero b ∈ A; now b a⊆b a0 \ bb, so a ⊆ a0 \ b for every a ∈ A, and a0 \ b is an upper bound for A strictly less than a0 . X X Thus b a0 must be S exactly a∈A b a. (b) Take complements: setting a1 = 1 \ a0 , we have a0 = inf A ⇐⇒ a1 = sup 1 \ a a∈A

(by 313A) ⇐⇒ b a1 =

[ a∈A

⇐⇒ b a0 = Z \

Z \b a [ a∈A

(c) Since

T

a a∈A b

Z \b a = int

\

b a.

a∈A

is surely a closed set, it is nowhere dense iff it has empty interior, that is, iff 0 = inf A.

32

Boolean algebras

313D

313D I started the section with the results above because they are easily stated and of great importance. But I must now turn to some new definitions, and I think it may help to clarify the ideas involved if I give them in their own natural context, even though this is far more general than we have any immediate need for here. Definitions Let P be a partially ordered set and C a subset of P . (a) C is order-closed if sup A ∈ C whenever A is a non-empty upwards-directed subset of C such that sup A is defined in P , and inf A ∈ C whenever A is a non-empty downwards-directed subset of C such that inf A is defined in P . (b) C is sequentially order-closed if supn∈N pn ∈ C whenever hpn in∈N is a non-decreasing sequence in C such that supn∈N pn is defined in P , and inf n∈N pn ∈ C whenever hpn in∈N is a non-increasing sequence in C such that inf n∈N pn is defined in P . Remark I hope it is obvious that an order-closed set is sequentially order-closed. 313E Order-closed subalgebras and ideals Of course, in the very special cases of a subalgebra or ideal of a Boolean algebra, the concepts ‘order-closed’ and ‘sequentially order-closed’ have expressions simpler than those in 313D. I spell them out. (a) Let B be a subalgebra of a Boolean algebra A. (i) The following are equiveridical: (α) B is order-closed in A; (β) sup B ∈ B whenever B ⊆ B and sup B is defined in A; (β 0 ) inf B ∈ B whenever B ⊆ B and inf B is defined in A; (γ) sup B ∈ B whenever B ⊆ B is non-empty and upwards-directed and sup B is defined in A; (γ 0 ) inf B ∈ B whenever B ⊆ B is non-empty and downwards-directed and inf B is defined in A. P P Of course (β) ⇒ (γ). If (γ) is true and B ⊆ B is any set with a supremum in A, then B 0 = {0} ∪ {b0 ∪ . . . ∪ bn : b0 , . . . , bn ∈ B} is a non-empty upwards-directed set with the same upper bounds as B, so sup B = sup B 0 ∈ B. Thus (γ) ⇒ (β) and (β), (γ) are equiveridical. Next, if (β) is true and B ⊆ B is a set with an infimum in A, then B 0 = {1 \ b : b ∈ B} ⊆ B and sup B 0 = 1 \ inf B is defined, so sup B 0 and inf B belong to B . Thus (β) ⇒ (β 0 ). In the same way, (γ 0 ) ⇐⇒ (β 0 ) ⇒ (β) and (β), (β 0 ), (γ), (γ 0 ) are all equiveridical. But since we also have (α) ⇐⇒ (γ)&(γ 0 ), (α) is equiveridical with the others. Q Q Replacing the sets B above by sequences, the same arguments provide conditions for B to be sequentially order-closed, as follows.

A;

(ii) The following are equiveridical: (α) B is sequentially order-closed in A; (β) supn∈N bn ∈ B whenever hbn in∈N is a sequence in B and supn∈N bn is defined in A; (β 0 ) inf n∈N bn ∈ B whenever hbn in∈N is a sequence in B and inf n∈N bn is defined in A; (γ) supn∈N bn ∈ B whenever hbn in∈N is a non-decreasing sequence in B and supn∈N bn is defined in (γ 0 ) inf n∈N bn ∈ B whenever hbn in∈N is a non-increasing sequence in B and inf n∈N bn is defined in

A. (b) Now suppose that I is an ideal of A. Then if A ⊆ I is non-empty all lower bounds of A necessarily belong to I; so that I is order-closed iff sup A ∈ I whenever A ⊆ I is non-empty, upwards-directed and has a supremum in A; I is sequentially order-closed iff supn∈N an ∈ I whenever han in∈N is a non-decreasing sequence in I with a supremum in A. Moreover, because I is closed under ∪ , I is order-closed iff sup A ∈ I whenever A ⊆ I has a supremum in A; I is sequentially order-closed iff supn∈N an ∈ I whenever han in∈N is a sequence in I with a supremum in A.

313G

Order-continuous homomorphisms

33

(c) If A = PX is a power set, then a sequentially order-closed subalgebra of A is just a σ-algebra of sets, while a sequentially order-closed ideal of A is a what I have called a σ-ideal of sets (112Db). If A is itself a σ-algebra of sets, then a sequentially order-closed subalgebra of A is a ‘σ-subalgebra’ in the sense of 233A. Accordingly I will normally use the phrases σ-subalgebra, σ-ideal for sequentially order-closed subalgebras and ideals of Boolean algebras. 313F Order-closures and generated sets (a) It is an immediate consequence of the definitions that T (i) if S is any non-empty family of subalgebras of a Boolean algebra A, then S is a subalgebra T of A; (ii) if F is any non-empty family of order-closed subsets of a partially ordered set P , then F is an order-closed subset of P ; T (iii) if F is any non-empty family of sequentially order-closed subsets of a partially ordered set P , then F is a sequentially order-closed subset of P . (b) Consequently, given any Boolean algebra A and a subset B of A, we have a smallest subalgebra B of A including B, being the intersection of all the subalgebras of A which include B; a smallest σ-subalgebra Bσ of A including B, being the intersection of all the σ-subalgebras of A which include B; and a smallest order-closed subalgebra Bτ of A including B, being the intersection of all the order-closed subalgebras of A which include B. We call B, Bσ and Bτ the subalgebra, σ-subalgebra and order-closed subalgebra generated by B. (I shall return to this in 331E.) (c) If A is a Boolean algebra and B any subalgebra of A, then the smallest order-closed subset B of A which includes B is again a subalgebra of A (so is the order-closed subalgebra of A generated by B). P P (i) The set {b : 1 \ b ∈ B} is order-closed (use 313A) and includes B, so includes B; thus 1 \ b ∈ B for every b ∈ B. (ii) If c ∈ B, the set {b : b ∪ c ∈ B} is order-closed (use 313Bb) and includes B, so includes B; thus b ∪ c ∈ B whenever b ∈ B and c ∈ B. (iii) If c ∈ B, the set {b : b ∪ c ∈ B} is order-closed and includes B (by (ii)), so includes B; thus b ∪ c ∈ B whenever b, c ∈ B. (iv) By 312B, B is a subalgebra of A. Q Q 313G This is a convenient moment at which to spell out an abstract version of the Monotone Class Theorem (136B). Lemma Let A be a Boolean algebra. (a) Suppose that 1 ∈ I ⊆ A ⊆ A and that a ∩ b ∈ I for all a, b ∈ I, b \ a ∈ A whenever a, b ∈ A and a ⊆ b. Then A includes the subalgebra of A generated by I. (b) If moreover supn∈N an ∈ A for every non-decreasing sequence han in∈N in A with a supremum in A, then A includes the σ-subalgebra of A generated by I. (c) And if sup C ∈ A whenever C ⊆ A is an upwards-directed set with a supremum in A, then A includes the order-closed subalgebra of A generated by I. proof (a)(i) Let P be the family of all sets J such that IS⊆ J ⊆ A and a ∩ b ∈ J for all a, b ∈ J. Then I ∈ P and if Q ⊆ P is upwards-directed and not empty, Q ∈ P. By Zorn’s Lemma, P has a maximal element B. (ii) Now B = {c : c ∈ A, c ∩ b ∈ A for every b ∈ B}. P P If c ∈ B, then of course c ∩ b ∈ B ⊆ A for every b ∈ B, because B ∈ P. If c ∈ A \ B, consider J = B ∪ {c ∩ b : b ∈ B}. Then c = c ∩ 1 ∈ J so J properly includes B and cannot belong to P. On the other hand, if b1 , b2 ∈ B, b1 ∩ b2 ∈ B ⊆ J,

(c ∩ b1 ) ∩ b2 = b1 ∩ (c ∩ b2 ) = (c ∩ b1 ) ∩ (c ∩ b2 ) = c ∩ (b1 ∩ b2 ) ∈ J,

so c1 ∩ c2 ∈ J for all c1 , c2 ∈ J; and of course I ⊆ B ⊆ J. So J cannot be a subset of A, and there must be a b ∈ B such that c ∩ b ∈ / A. Q Q

34

Boolean algebras

313G

(iii) Consequently c \ b ∈ B whenever b, c ∈ B and b ⊆ c. P P If a ∈ B, then b ∩ a, c ∩ a ∈ B ⊆ A and b ∩ a ⊆ c ∩ a, so (c \ b) ∩ a = (c ∩ a) \ (b ∩ a) ∈ A by the hypothesis on A. By (ii), c \ b ∈ B. Q Q (iv) It follows that B is a subalgebra of A. P P If b ∈ B, then b ⊆ 1 ∈ I ⊆ B, so 1 \ b ∈ B. If a, b ∈ B, then a ∪ b = 1 \ ((1 \ a) ∩ (1 \ b)) ∈ B. 0 = 1 \ 1 ∈ B, so that the conditions of 312B(ii) are satisfied. Q Q Now the subalgebra of A generated by I is included in B and therefore in A, as required. (b) Now suppose that supn∈N an belongs to A whenever han in∈N is a non-decreasing sequence in A with a supremum in A. Then B, as defined in part (a) of the proof, is a σ-subalgebra of A. P P Let hbn in∈N be a non-decreasing sequence in B with a supremum c in A. Then for any b ∈ B, hbn ∩ bin∈N is a non-decreasing sequence in A with a supremum c ∩ b in A (313Ba). So c ∩ b ∈ A. As b is arbitrary, c ∈ B, by the criterion in (a-ii) above. As hbn in∈N is arbitrary, B is a σ-subalgebra, by 313Ea. Q Q Accordingly the σ-subalgebra of A generated by I is included in B and therefore in A. (c) Finally, if sup C ∈ A whenever C is a non-empty upwards-directed subset of A with a least upper bound in A, B is order-closed. P P Let C ⊆ B be a non-empty upwards-directed set with a supremum c in A. Then for any b ∈ B, {c ∩ b : c ∈ C} is a non-empty upwards-directed set in A with supremum c ∩ b in A. So c ∩ b ∈ A. As b is arbitrary, c ∈ B. As C is arbitrary, B is order-closed in A. Q Q Accordingly the order-closed subalgebra of A generated by I is included in B and therefore in A. 313H Definitions It is worth distinguishing various types of supremum- and infimum-preserving function. Once again, I do this in almost the widest possible context. Let P and Q be two partially ordered sets, and φ : P → Q an order-preserving function, that is, a function such that φ(p) ≤ φ(q) in Q whenever p ≤ q in P . (a) I say that φ is order-continuous if (i) φ(sup A) = supp∈A φ(p) whenever A is a non-empty upwardsdirected subset of P and sup A is defined in P (ii) φ(inf A) = inf p∈A φ(p) whenever A is a non-empty downwards-directed subset of P and inf A is defined in P . (b) I say that φ is sequentially order-continuous or σ-order-continuous if (i) φ(p) = supn∈N φ(pn ) whenever hpn in∈N is a non-decreasing sequence in P and p = supn∈N pn in P (ii) φ(p) = inf n∈N φ(pn ) whenever hpn in∈N is a non-increasing sequence in P and p = inf n∈N pn in P . Remark You may feel that one of the equivalent formulations in Proposition 313Lb gives a clearer idea of what is really being demanded of φ in the ordinary cases we shall be looking at. 313I Proposition Let P , Q and R be partially ordered sets, and φ : P → Q, ψ : Q → R order-preserving functions. (a) ψφ : P → R is order-preserving. (b) If φ and ψ are order-continuous, so is ψφ. (c) If φ and ψ are sequentially order-continuous, so is ψφ. (d) φ is order-continuous iff φ−1 [B] is order-closed for every order-closed B ⊆ Q. proof (a)-(c) I think the only point that needs remarking is that if A ⊆ P is upwards-directed, then φ[A] ⊆ Q is upwards-directed, because φ is order-preserving. So if sup A is defined in P and φ, ψ are order-continuous, we shall have ψ(φ(sup A)) = ψ(sup φ[A]) = sup ψ[φ[A]]. (d)(i) Suppose that φ is order-continuous and that B ⊆ Q is order-closed. Let A ⊆ φ−1 [B] be a nonempty upwards-directed set with supremum p ∈ P . Then φ[A] ⊆ B is non-empty and upwards-directed, because φ is order-preserving, and φ(p) = sup φ[A] because φ is order-continuous. Because B is order-closed,

313L

Order-continuous homomorphisms

35

φ(p) ∈ B and p ∈ φ−1 [B]. Similarly, if A ⊆ φ−1 [B] is non-empty and downwards-directed, and inf A is defined in P , then φ(inf A) = inf φ[A] ∈ B and inf A ∈ φ−1 [B]. Thus φ−1 [B] is order-closed; as B is arbitrary, φ satisfies the condition. (ii) Now suppose that φ−1 [B] is order-closed in P whenever B ⊆ Q is order-closed in Q. Let A ⊆ P be a non-empty upwards-directed subset of P with a supremum p ∈ P . Then φ(p) is an upper bound of φ[A]. Let q be any upper bound of φ[A] in Q. Consider B = {r : r ≤ q}; then B ⊆ Q is upwards-directed and order-closed, so φ−1 [B] is order-closed. Also A ⊆ φ−1 [B] is non-empty and upwards-directed and has supremum p, so p ∈ φ−1 [B] and φ(p) ∈ B, that is, φ(p) ≤ q. As q is arbitrary, φ(p) = sup φ[A]. Similarly, φ(inf A) = inf φ[A] whenever A ⊆ P is non-empty, downwards-directed and has an infimum in P ; so φ is order-continuous. 313J

It is useful to introduce here the following notion.

Definition Let A be a Boolean algebra. A set D ⊆ A is order-dense if for every non-zero a ∈ A there is a non-zero d ∈ D such that d ⊆ a. Remark Many authors use the simple word ‘dense’ where I have insisted on the phrase ‘order-dense’. In the work of this treatise it will be important to distinguish clearly between this concept of ‘dense’ set and the topological concept (2A3U); typically, in those contexts in which both appear, an order-dense set can be in some sense much smaller than a topologically dense set. 313K Lemma If A is a Boolean algebra and D ⊆ A is order-dense, then for any a ∈ A there is a disjoint C ⊆ D such that sup C = a; in particular, a = sup{d : d ∈ D, d ⊆ a} and there is a partition of unity C ⊆ D. proof Set Da = {d : d ∈ D, d ⊆ a}. Applying Zorn’s lemma to the family C of disjoint sets C ⊆ Da , we have a maximal C ∈ C. Now if b ∈ A and b 6⊇ a, there is a d ∈ D such that 0 6= d ⊆ a \ b. Because C is maximal, there must be a c ∈ C such that c ∩ d 6= 0, so that c 6⊆ b. Turning this round, any upper bound of C must include a, so that a = sup C. It follows at once that a = sup Da . Taking a = 1 we obtain a partition of unity included in D. 313L Proposition Let A and B be Boolean algebras and π : A → B a Boolean homomorphism. (a) π is order-preserving. (b) The following are equiveridical: (i) π is order-continuous; (ii) whenever A ⊆ A is non-empty and downwards-directed and inf A = 0 in A, then inf π[A] = 0 in B; (iii) whenever A ⊆ A is non-empty and upwards-directed and sup A = 1 in A, then sup π[A] = 1 in B; (iv) whenever A ⊆ A and sup A is defined in A, then π(sup A) = sup π[A] in B; (v) whenever A ⊆ A and inf A is defined in A, then π(inf A) = inf π[A] in B; (vi) whenever C ⊆ A is a partition of unity, then π[C] is a partition of unity in B. (c) The following are equiveridical: (i) π is sequentially order-continuous; (ii) whenever han in∈N is a non-increasing sequence in A and inf n∈N an = 0 in A, then inf n∈N πan = 0 in B; (iii) whenever A ⊆ A is countable and sup A is defined in A, then π(sup A) = sup π[A] in B; (iv) whenever A ⊆ A is countable and inf A is defined in A, then π(inf A) = inf π[A] in B; (v) whenever C ⊆ A is a countable partition of unity, then π[C] is a partition of unity in B. proof (a) This is 312I. (b)(i)⇒(ii) is trivial, as π0 = 0. (ii)⇒(iv) Assume (ii), and let A be any subset of A such that c = sup A is defined in A. If A = ∅, then c = 0 and sup π[A] = 0 = πc. Otherwise, set A0 = {a0 ∪ . . . ∪ an : a0 , . . . , an ∈ A},

C = {c \ a : a ∈ A0 }.

36

Boolean algebras

313L

Then A0 is upwards-directed and has the same upper bounds as A, so c = sup A0 and 0 = inf C, by 313Aa. Also C is downwards-directed, so inf π[C] = 0 in B. But now π[C] = {πc \ πa : a ∈ A0 } = {πc \ b : b ∈ π[A0 ]}, π[A0 ] = {πa0 ∪ . . . ∪ πan : a0 , . . . , an ∈ A} = {b0 ∪ . . . ∪ bn : b0 , . . . , bn ∈ π[A]}, because π is a Boolean homomorphism. Again using 313Aa and the fact that b ⊆ πc for every b ∈ π[A0 ], we get πc = sup π[A0 ] = sup π[A]. As A is arbitrary, (iv) is satisfied. (iv)⇒(v) If A ⊆ A and c = inf A is defined in A, then 1 \ c = supa∈A 1 \ a, so πc = 1 \ π(1 \ c) = 1 \ supa∈A π(1 \ a) = inf a∈A 1 \ π(1 \ a) = inf a∈A πa. (v)⇒(ii) is trivial, because π0 = 0. (iv)⇒(iii) is similarly trivial. (iii)⇒(vi) Assume (iii), and let C be a partition of unity in A. Then C 0 = {c0 ∪ . . . ∪ cn : c0 , . . . , cn ∈ C} is upwards-directed and has supremum 1, so sup π[C 0 ] = 1. But (because π is a Boolean homomorphism) π[C] and π[C 0 ] have the same upper bounds, so sup π[C] = 1, as required. (vi)⇒(ii) Assume (vi), and let A ⊆ A be a set with infimum 0. Set D = {d : d ∈ A, ∃ a ∈ A, d ∩ a = 0}. Then D is order-dense in A. P P If e ∈ A \ {0}, then there is an a ∈ A such that e 6⊆ a, so that e \ a is a non-zero member of D included in e. Q Q Consequently there is a partition of unity C ⊆ D, by 313K. But now if b is any lower bound for π[A] in B, we must have b ∩ πd = 0 for every d ∈ D, so πc ⊆ 1 \ b for every c ∈ C, and 1 \ b = 1, b = 0. Thus inf π[A] = 0. As A is arbitrary, (ii) is satisfied. (v)&(iv)⇒(i) is trivial. (c) We can use nearly identical arguments, remembering only to interpolate the word ‘countable’ from time to time. I spell out the new version of (ii)⇒(iii), even though it requires no more than an adaptation of the language. Assume (ii), and let A be a countable subset of A with a supremum c ∈ A. If A = ∅, then c = 0 so πc = 0 = sup π[A]. Otherwise, let han in∈N be a sequence running over A; set a0n = a0 ∪ . . . ∪ an and cn = c \ a0n for each n. Then ha0n in∈N is non-decreasing, with supremum c, and hcn in∈N is non-increasing, with infimum 0; so inf n∈N πcn = 0 and supn∈N πan = supn∈N πa0n = πc. For (v)⇒(ii), however, a different idea is involved. Assume (v), and suppose that han in∈N is a nonincreasing sequence in A with infimum 0. Set c0 = 1 \ a0 , cn = an−1 \ an for n ≥ 1; then C = {cn : n ∈ N} is a partition of unity in A (because if c ∩ cn = 0 for every n, then c ⊆ an for every n), so π[C] is a partition of unity in B. Now if b ≤ πan for every n, b ∩ πcn for every n, so b = 0; thus inf n∈N πan = 0. As han in∈N is arbitrary, (ii) is satisfied. 313M out.

The following result is perfectly elementary, but it will save a moment later on to have it spelt

Lemma Let A and B be Boolean algebras and π : A → B an order-continuous Boolean homomorphism. (a) If D is an order-closed subalgebra of B, then π −1 [D] is an order-closed subalgebra of A. (b) If C is the order-closed subalgebra of A generated by C ⊆ A, then the order-closed subalgebra D of B generated by π[C] includes π[C]. (c) Now suppose that π is surjective and that C ⊆ A is such that the order-closed subalgebra of A generated by C is A itself. Then the order-closed subalgebra of B generated by π[C] is B. proof (a) Setting C = π −1 [D]: if a, a0 ∈ C then π(a ∩ b) = πa ∩ πb, π(a 4 a0 ) = πa 4 πa0 ∈ D, so a ∩ a0 , a 4 a0 ∈ C; π1 = 1 ∈ D so 1 ∈ C; thus C is a subalgebra of A. By 313Id, C is order-closed.

313R

Order-continuous homomorphisms

37

(b) By (a), π −1 [D] is an order-closed subalgebra of A. It includes C so includes C, and π[C] ⊆ D. (c) In the language of (b), we have C = A, so D must be B. 313N Definition The phrase regular embedding is sometimes used to mean an injective ordercontinuous Boolean homomorphism; a subalgebra B of a Boolean algebra A is said to be regularly embedded in A if the identity map from B to A is order-continuous, that is, if whenever b ∈ B is the supremum (in B) of B ⊆ B, then b is also the supremum in A of B; and similarly for infima. One important case is when B is order-dense (313O); another is in 314G-314H below. 313O Proposition Let A be a Boolean algebra and B an order-dense subalgebra of A. Then B is regularly embedded in A. In particular, if B ⊆ B and c ∈ B then c = sup B in B iff c = sup B in A. proof I have to show that the identity homomorphism ι : B → A is order-continuous. ?? Suppose, if possible, otherwise. By 313L(b-ii), there is a non-empty set B ⊆ B such that inf B = 0 in B but B = ι[B] has a non-zero lower bound a ∈ A. In this case, however (because B is order-dense) there is a non-zero d ∈ B with d ⊆ a, in which case d is a non-zero lower bound for B in B. X X 313P The most important use of these ideas to us concerns quotient algebras (313Q); I approach by means of a superficially more general result. Theorem Let A and B be Boolean algebras and π : A → B a Boolean homomorphism with kernel I. (a)(i) If π is order-continuous then I is order-closed. (ii) If π[A] is regularly embedded in B and I is order-closed then π is order-continuous. (b)(i) If π is sequentially order-continuous then I is a σ-ideal. (ii) If π[A] is regularly embedded in B and I is a σ-ideal then π is sequentially order-continuous. proof (a)(i) If A ⊆ I is upwards-directed and has a supremum c ∈ A, then πc = sup π[A] = 0, so c ∈ I. As remarked in 313Eb, this shows that I is order-closed. (ii) We are supposing that the identity map from π[A] to B is order-continuous, so it will be enough to show that π is order-continuous when regarded as a map from A to π[A]. Suppose that A ⊆ A is non-empty and downwards-directed and that inf A = 0. ?? Suppose, if possible, that 0 is not the greatest lower bound of π[A] in π[A]. Then there is a c ∈ A such that 0 6= πc ⊆ πa for every a ∈ A. Now π(c \ a) = πc \ πa = 0 for every a ∈ A, so c \ a ∈ I for every a ∈ A. The set C = {c \ a : a ∈ A} is upwards-directed and has supremum c; because I is order-closed, c = sup C ∈ I, and πc = 0, contradicting the specification of c. X X Thus inf π[A] = 0 in either π[A] or B. As A is arbitrary, π is order-continuous, by the criterion (ii) of 313Lb. (b) Argue in the same way, replacing each set A by a sequence. 313Q Corollary Let A be a Boolean algebra and I an ideal of A; write π for the canonical map from A to A/I. (a) π is order-continuous iff I is order-closed. (b) π is sequentially order-continuous iff I is a σ-ideal. proof π[A] = A/I is surely regularly embedded in A/I. 313R For order-continuous homomorphisms, at least, there is an elegant characterization in terms of Stone spaces. Proposition Let A and B be Boolean algebras, and π : A → B a Boolean homomorphism. Let Z and W be their Stone spaces, and φ : W → Z the corresponding continuous function (312P). Then the following are equiveridical: (i) π is order-continuous; (ii) φ−1 [M ] is nowhere dense in W for every nowhere dense set M ⊆ Z; (iii) int φ[H] 6= ∅ for every non-empty open set H ⊆ W .

38

Boolean algebras

313R

proof (a)(i)⇒(iii) Suppose that π is order-continuous. ?? Suppose, if possible, that H ⊆ W is a non-empty open set and int φ[H] = ∅. Let b ∈ B \ {0} be such that bb ⊆ H. Then φ[bb] has empty interior; but also it S is a closed set, so its complement is dense. Set A = {a : a ∈ A, b a ∩ φ[bb] = ∅}. Then a∈A b a = Z \ φ[bb] is a dense open set, so sup A = 1 in A (313Ca). Because π is order-continuous, sup π[A] = 1 in B (313L(b-iii)), and there is an a ∈ A such that πa ∩ b 6= 0. But this means that bb ∩ φ−1 [b a] 6= ∅ and φ[bb] ∩ b a 6= ∅, contrary to the definition of A. X X Thus there is no such set H, and (iii) is true. (b)(iii)⇒(ii) Now assume (iii). If M ⊆ Z is nowhere dense, set N = φ−1 [M ], so that N ⊆ W is a closed set. If H = int N , then int φ[H] ⊆ int M = ∅, so (iii) tells us that H is empty; thus N and φ−1 [M ] are nowhere dense, as required by (ii). T (c)(ii)⇒(i) Assume (ii), and let A ⊆ A be a non-empty set such that inf A = 0 in A. Then M = a∈A b a −1 has empty interior in Z (313Cb), so (being closed) is nowhere dense, and φ [M ] is also nowhere dense. If b ∈ B \ {0}, then T −1 bb 6⊆ φ−1 [M ] = T [b a] = a∈A π ca, a∈A φ so b is not a lower bound for π[A]. This shows that inf π[A] = 0 in B. As A is arbitrary, π is order-continuous (313L(b-ii)). 313X Basic exercises (a) Use 313C to give alternative proofs of 313A and 313B. (b) Let P be a partially ordered set. Show that there is a topology on P for which the closed sets are just the order-closed sets. (c) Let P be a partially ordered set, Q ⊆ P an order-closed set, and R a subset of Q which is order-closed in Q when Q is given the partial ordering induced by that of P . Show that R is order-closed in P . > (d) Let A be a Boolean algebra. Suppose that 1 ∈ I ⊆ A and that a ∩ b ∈ I for all a, b ∈ I. (i) Let B be the intersection of all those subsets A of A such that I ⊆ A and b \ a ∈ A whenever a, b ∈ A and a ⊆ b. Show that B is a subalgebra of A. (ii) Let Bσ be the intersection of all those subsets A of A such that I ⊆ A, b \ a ∈ A whenever a, b ∈ A and a ⊆ b and supn∈N bn ∈ A whenever hbn in∈N is a non-decreasing sequence in A with a supremum in A. Show that Bσ is a σ-subalgebra of A. (iii) Let Bτ be the intersection of all those subsets A of A such that I ⊆ A, b \ a ∈ A whenever a, b ∈ A and a ⊆ b and sup B ∈ A whenever B is a non-empty upwards-directed subset of A with a supremum in A. Show that Bτ is an order-closed subalgebra of A. (iv) Hence give a proof of 313G not relying on Zorn’s Lemma or any other use of the axiom of choice. (e) Let A be a Boolean algebra, and B a subalgebra of A. Let Bσ be the smallest sequentially order-closed subset of A including B. Show that Bσ is a subalgebra of A. > (f ) Let X be a set, and A a subset of PX. Show that A is an order-closed subalgebra of PX iff it is of the form {f −1 [F ] : F ⊆ Y } for some set Y , function f : X → Y . (g) Let P and Q be partially ordered sets, and φ : P → Q an order-preserving function. Show that φ is sequentially order-continuous iff φ−1 [C] is sequentially order-closed in A for every sequentially order-closed C ⊆ B. (h) For partially ordered sets P and Q, let us call a function φ : P → Q monotonic if it is either order-preserving or order-reversing. State and prove definitions and results corresponding to 313H, 313I and 313Xg for general monotonic functions. >(i) Let A be a Boolean algebra. Show that the operations (a, b) 7→ a ∪ b and (a, b) 7→ a ∩ b are ordercontinuous operations from A×A to A, if we give A×A the product partial order, saying that (a, b) ≤ (a0 , b0 ) iff a ⊆ a0 and b ⊆ b0 .

313Yf

Order-continuous homomorphisms

39

(j) Let A be a Boolean algebra. Show that if a subalgebra of A is order-dense then it is dense in the topology of 313Xb. > (k) Let A be a Boolean algebra and A ⊆ A any disjoint set. Show that there is a partition of unity in A including A. > (l) Let A, B be Boolean algebras and π1 , π2 : A → B two order-continuous Boolean homomorphisms. Show that {a : π1 a = π2 a} is an order-closed subalgebra of A. (m) Let A and B be Boolean algebras and π1 , π2 : A → B two Boolean homomorphisms. Suppose that π1 and π2 agree on some order-dense subset of A, and that one of them is order-continuous. Show that they are equal. (Hint: if π1 is order-continuous, π2 a ⊇ π1 a for every a.) (n) Let A and B be Boolean algebras, A0 an order-dense subalgebra of A, and π : A → B a Boolean homomorphism. Show that π is order-continuous iff π¹ A0 : A0 → B is order-continuous. > (o) Let A be a Boolean algebra. For A ⊆ A set A⊥ = {b : a ∩ b = 0 ∀ a ∈ A}. (i) Show that A⊥ is an order-closed ideal of A. (ii) Show that a set A ⊆ A is an order-closed ideal of A iff A = A⊥⊥ . (iii) Show that if I ⊆ A is an order-closed ideal then {a• : a ∈ I ⊥ } is an order-dense ideal in the quotient algebra A/I. (p) Let A and B be Boolean algebras, with Stone spaces Z and W ; let π : A → B be a Boolean homomorphism, and φ : W → Z the corresponding continuous function. Show that the following are equiveridical: (i) π is order-continuous; (ii) int φ−1 [F ] = φ−1 [int F ] for every closed F ⊆ Z (iii) φ−1 [G] = φ−1 [G] for every open G ⊆ Z. 313Y Further exercises (a) Prove 313A-313C for general Boolean rings. (b) Let A and B be Boolean algebras, with Stone spaces Z and W , and π : A → B a Boolean homomorphism, with associated continuous function φ : W → Z. Show that π is sequentially order-continuous iff φ−1 [M ] is nowhere dense for every nowhere dense zero set M ⊆ Z. (c) Let P be any partially ordered set, and let T be the topology of 313Xb. (i) Show that a sequence hpn in∈N in P is T-convergent to p ∈ P iff every subsequence of hpn in∈N has a monotonic sub-subsequence with supremum or infimum equal to p. (ii) Show that a subset A of P is sequentially order-closed, in the sense of 313Db, iff the T-limit of any T-convergent sequence in A belongs to A. (iii) Suppose that A is an upwards-directed subset of P with supremum p0 ∈ P . For a ∈ A set Fa = {p : a ≤ p ∈ A}, and let F be the filter on P generated by {Fa : a ∈ A}. Show that F → p0 for T. (iv) Show that if Q is another partially ordered set, endowed with a topology S in the same way, then a monotonic function φ : P → Q is order-continuous iff it is continuous for the topologies T and S, and is sequentially order-continuous iff it is sequentially continuous for these topologies. (d) Let U be a Banach lattice (242G, 354Ab). Show that its norm is order-continuous in the sense of 242Yc (or 354Dc) iff its restriction to {u : u ≥ 0} is order-continuous in the sense of 313Ha. (e) Let A and B be Boolean algebras with Stone spaces Z and W respectively, π : A → B a Boolean homomorphism and φ : W → Z the corresponding continuous function. Show that π[A] is order-dense in B iff φ is irreducible, that is, φ[F ] 6= φ[W ] for any proper closed subset F of W . (f ) Let A and B be Boolean algebras with Stone spaces Z and W respectively, π : A → B a Boolean homomorphism and φ : W → Z the corresponding continuous function. Show that the following are equiveridical: (i) π is injective and order-continuous; (ii) for M ⊆ Z, M is nowhere dense iff φ−1 [M ] is nowhere dense.

40

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313 Notes

313 Notes and comments I give ‘elementary’ proofs of 313A-313B because I believe that they help to exhibit the relevant aspects of the structure of Boolean algebras; but various abbreviations are possible, notably if we allow ourselves to use the Stone representation (313Xa). 313A and 313Ba-b can be expressed by saying that the Boolean operations ∪ , ∩ and \ are (separately) order-continuous. Of course, \ is orderreversing, rather than order-preserving, in the second variable; but the natural symmetry in the concept of partial order means that the ideas behind 313H-313I can be applied equally well to order-reversing functions (313Xh). In fact, ∪ and ∩ can be regarded as order-continuous functions on the product space (313Bc-d, 313Xi). Clearly 313Bc-d can be extended into forms valid for any finite sequence A0 , . . . , An of subsets of A in place of A, B. But if we seek to go to infinitely many subsets of A we find ourselves saying something new; see 316G-316J below. Proposition 313C, and its companions 313R, 313Xp and 313Yb, are worth studying not only as a useful technique, but S also in order to understand the difference between sup A, where A is a set in a Boolean algebra, and A, where A is a family of sets. Somehow sup A can be larger, and inf A smaller, than one’s first intuition might suggest, corresponding to the fact that not every subset of the Stone space corresponds to an element of the Boolean algebra. I should like to use the words ‘order-closed’ and ‘sequentially order-closed’ to mean closed, or sequentially closed, for some more or less canonical topology. The difficulty is that while a great many topologies can be defined from a partial order (one is described in 313Xb and 313Yc, and another in 367Yc), none of them has such pre-eminence that it can be called ‘the’ order-topology. Accordingly there is a degree of arbitrariness in the language I use here. Nevertheless (sequentially) order-closed subalgebras and ideals are of such importance that they seem to deserve a concise denotation. The same remarks apply to (sequential) ordercontinuity. Concerning the term ‘order-dense’ in 313J, this has little to do with density in any topological sense, but the word ‘dense’, at least, is established in this context. With all these definitions, there is a good deal of scope for possible interrelations. The most important to us is 313Q, which will be used repeatedly (typically, with A an algebra of sets), but I think it is worth having the expanded version in 313P available. I take the opportunity to present an abstract form of an important lemma on σ-algebras generated by families closed under ∩ (136B, 313Gb). This time round I use the Zorn’s Lemma argument in the text and suggest the alternative, ‘elementary’ method in the exercises (313Xd). The two methods are opposing extremes in the sense that the Zorn’s Lemma argument looks for maximal subalgebras included in A (which are not unique, and have to be picked out using the axiom of choice) and the other approach seeks minimal subalgebras including I (which are uniquely defined, and can be described without the axiom of choice). Note that the concept of ‘order-closed’ algebra of sets is not particularly useful; there are too few orderclosed subalgebras of PX and they are of too simple a form (313Xf). It is in abstract Boolean algebras that the idea becomes important. In the most important partially ordered sets of measure theory, the sequentially order-closed sets are the same as the order-closed sets (see, for instance, 316Fb below), and most of the important order-closed subalgebras dealt with in this chapter can be thought of as σ-subalgebras which are order-closed because they happen to lie in the right kind of algebra.

314 Order-completeness The results of §313 are valid in all Boolean algebras, but of course are of most value when many suprema and infima exist. I now set out the most useful definitions which guarantee the existence of suprema and infima (314A) and work through their elementary relationships with the concepts introduced so far (314C314J). I then embark on the principal theorems concerning order-complete Boolean algebras: the extension theorem for homomorphisms to a Dedekind complete algebra (314K), the Loomis-Sikorski representation of a Dedekind σ-complete algebra as a quotient of a σ-algebra of sets (314M), the characterization of Dedekind complete algebras in terms of their Stone spaces (314S), and the idea of ‘Dedekind completion’ of a Boolean algebra (314T-314U). On the way I describe ‘regular open algebras’ (314O-314Q).

314Be

Order-completeness

41

314A Definitions Let P be a partially ordered set. (a) P is Dedekind complete, or order-complete, or conditionally complete if every non-empty subset of P with an upper bound has a least upper bound. (b) P is Dedekind σ-complete, or σ-order-complete, if (i) every countable non-empty subset of P with an upper bound has a least upper bound (ii) every countable non-empty subset of P with a lower bound has a greatest lower bound. 314B Remarks (a) I give these definitions in the widest possible generality because they are in fact of great interest for general partially ordered sets, even though for the moment we shall be concerned only with Boolean algebras. Indeed I have already presented the same idea in the context of Riesz spaces (241F). (b) You will observe that the definition in (a) of 314A is asymmetric, unlike that in (b). This is because the inverted form of the definition is equivalent to that given; that is, P is Dedekind complete (on the definition 314Aa) iff every non-empty subset of P with a lower bound has a greatest lower bound. P P (i) Suppose that P is Dedekind complete, and that B ⊆ P is non-empty and bounded below. Let A be the set of lower bounds for B. Then A has at least one upper bound (since any member of B is an upper bound for A) and is not empty; so a0 = sup A is defined. Now if b ∈ B, b is an upper bound for A, so a0 ≤ b; thus a0 ∈ A and must be the greatest member of A, that is, the greatest lower bound of B. (ii) Similarly, if every non-empty subset of P with a lower bound has a greatest lower bound, P is Dedekind complete. Q Q (c) In the special case of Boolean algebras, we do not need both halves of the definition 314Ab; in fact we have, for any Boolean algebra A, A is Dedekind σ-complete ⇐⇒ every non-empty countable subset of A has a least upper bound ⇐⇒ every non-empty countable subset of A has a greatest lower bound. P P Because A has a least element 0 and a greatest element 1, every subset of A has upper and lower bounds; so the two one-sided conditions together are equivalent to saying that A is Dedekind σ-complete. I therefore have to show that they are equiveridical. Now if A ⊆ A is a non-empty countable set, so is B = {1 \ a : a ∈ A}, and inf A = 1 \ sup B,

sup A = 1 \ inf B

whenever the right-hand-sides are defined (313A). So if the existence of a supremum (resp. infimum) of B is guaranteed, so is the existence of an infimum (resp. supremum) of A. Q Q The real point here is of course that (A, ⊆ ) is isomorphic to (A, ⊇ ). (d) Most specialists in Boolean algebra speak of ‘complete’, or ‘σ-complete’, Boolean algebras. I prefer the longer phrases ‘Dedekind complete’ and ‘Dedekind σ-complete’ because we shall be studying metrics on Boolean algebras and shall need the notion of metric completeness as well as that of order-completeness. (e) I have had to make some rather arbitrary choices in the definition here. The principal examples of partially ordered set to which we shall apply these definitions are Boolean algebras and Riesz spaces, which are all lattices. Consequently it is not possible to distinguish in these contexts between the property of Dedekind completeness, as defined above, and the weaker property, which we might call ‘monotone ordercompleteness’, (i) whenever A ⊆ P is non-empty, upwards-directed and bounded above then A has a least upper bound in P (ii) whenever A ⊆ P is non-empty, downwards-directed and bounded below then A has a greatest lower bound in P . (See 314Xa below. ‘Monotone order-completeness’ is the property involved in 314Ya, for instance.) Nevertheless I am prepared to say, on the basis of my own experience of working with other partially ordered sets, that ‘Dedekind completeness’, as I have defined it, is at least of sufficient importance to deserve a name. Note that it does not imply that P is a lattice, since it allows two elements of P to have no common upper bound.

42

Boolean algebras

314Bf

(f ) The phrase complete lattice is sometimes used to mean a Dedekind complete lattice with greatest and least elements; equivalently, a Dedekind complete partially ordered set with greatest and least elements. Thus a Dedekind complete Boolean algebra is a complete lattice in this sense, but R is not. (g) The most important Dedekind complete Boolean algebras (at least from the point of view of measure theory) are the ‘measure algebras’ of the next chapter. I shall not pause here to give other examples, but will proceed directly with the general theory. 314C Proposition Let A be a Dedekind σ-complete Boolean algebra and I a σ-ideal of A. Then the quotient Boolean algebra A/I is Dedekind σ-complete. proof I use the description in 314Bc. Let B ⊆ A/I be a non-empty countable set. For each u ∈ B, choose an au ∈ A such that u = a•u = au + I. Then c = supu∈B au is defined in A; consider v = c• in A/I. Because the map a 7→ a• is sequentially order-continuous (313Qb), v = sup B. As B is arbitrary, A/I is Dedekind σ-complete. 314D Corollary Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets of X. Then Σ ∩ I is a σ-ideal of the Boolean algebra Σ, and Σ/Σ ∩ I is Dedekind σ-complete. S proof Of course Σ is Dedekind σ-complete, because if hEn in∈N is any sequence in Σ then n∈N En is the least upper bound of {En : n ∈ N} in Σ. It is also easy to see that Σ ∩ I is a σ-ideal of Σ, since S F ∩ n∈N En ∈ I whenever F ∈ Σ and hEn in∈N is a sequence in Σ ∩ I. So 314C gives the result. 314E Proposition Let A be a Boolean algebra. (a) If A is Dedekind complete, then all its order-closed subalgebras and principal ideals are Dedekind complete. (b) If A is Dedekind σ-complete, then all its σ-subalgebras and principal ideals are Dedekind σ-complete. proof All we need to note is that if C is either an order-closed subalgebra or a principal ideal of A, and B ⊆ C is such that b = sup B is defined in A, then b ∈ C (see 313E(a-i-β)), so b is still the supremum of B in C; while the same is true if C is a σ-subalgebra and B ⊆ C is countable, using 313E(a-ii-β). 314F I spell out some further connexions between the concepts ‘order-closed set’, ‘order-continuous function’ and ‘Dedekind complete Boolean algebra’ which are elementary without being quite transparent. Proposition Let A and B be Boolean algebras and π : A → B a Boolean homomorphism. (a)(i) If A is Dedekind complete and π is order-continuous, then π[A] is order-closed in B. (ii) If B is Dedekind complete and π is injective and π[A] is order-closed then π is order-continuous. (b)(i) If A is Dedekind σ-complete and π is sequentially order-continuous, then π[A] is a σ-subalgebra of B. (ii) If B is Dedekind σ-complete and π is injective and π[A] is a σ-subalgebra of B then π is sequentially order-continuous. proof (a)(i) If B ⊆ π[A], then a0 = sup(π −1 [B]) is defined in A; now πa0 = sup(π[π −1 [B]]) = sup B in B (313L(b-iv)), and of course πa0 ∈ π[A]. By 313E(a-i-β) again, this is enough to show that π[A] is order-closed in B. (ii) Suppose that A ⊆ A and inf A = 0 in A. Then π[A] has an infimum b0 in B, which belongs to π[A] because π[A] is an order-closed subalgebra of B (313E(a-i-β 0 )). Now if a0 ∈ A is such that πa0 = b0 , we have π(a ∩ a0 ) = πa ∩ πa0 = πa for every a ∈ A, so (because π is injective) a ∩ a0 = a0 and a0 ⊆ a for every a ∈ A. But this means that a0 = 0 and b0 = π0 = 0. As A is arbitrary, π is order-continuous (313L(b-ii)). (b) Use the same arguments, but with sequences in place of the sets B, A above.

314J

Order-completeness

43

314G Corollary (a) If A is a Dedekind complete Boolean algebra and B is an order-closed subalgebra of A, then B is regularly embedded in A (definition: 313N). (b) Let A be a Dedekind complete Boolean algebra, B a Boolean algebra and π : A → B an ordercontinuous Boolean homomorphism. If C ⊆ A and C is the order-closed subalgebra of A generated by C, then π[C] is the order-closed subalgebra of B generated by π[C]. proof (a) Apply 314F(a-ii) to the identity map from B to A. (b) Let D be the order-closed subalgebra of B generated by π[C]. By 313Mb, π[C] ⊆ D. On the other hand, the identity homomorphism ι : C → A is order-continuous, by (a), so πι : C → B is order-continuous, and π[C] = πι[C] is order-closed in B, by 314F(a-i). But since π[C] is surely included in π[C], D is also included in π[C]. Accordingly π[C] = D, as claimed. 314H Corollary Let A be a Boolean algebra and B a subalgebra of A. (a) If A is Dedekind complete, then B is order-closed iff it is Dedekind complete in itself and is regularly embedded in A. (b) If A is Dedekind σ-complete, then B is a σ-subalgebra iff it is Dedekind σ-complete in itself and the identity map from B to A is sequentially order-continuous. proof Put 314E and 314F together. 314I Corollary (a) If A is a Boolean algebra and B is an order-dense subalgebra of A which is Dedekind complete in itself, then B = A. (b) If A is a Dedekind complete Boolean algebra, B is a Boolean algebra, π : A → B is an injective Boolean homomorphism and π[A] is order-dense in B, then π is an isomorphism. proof (a) Being order-dense, B is regularly embedded in A (313O), so this is a special case of 314Ha. (b) Because π[A] is order-dense, it is regularly embedded in B; also, the kernel of π is {0}, which is surely order-closed in A, so 313P(a-ii) tells us that π is order-continuous. By 314F(a-i), π[A] is order-closed in B; being order-dense, it must be the whole of B (313K). Thus π is surjective; being injective, it is an isomorphism. 314J When we come to applications of the extension procedure in 312N, the following will sometimes be needed. Lemma Let A be a Boolean algebra and A0 a subalgebra of A. Take any c ∈ A, and set A1 = {(a ∩ c) ∪ (b \ c) : a, b ∈ A0 }, the subalgebra of A generated by A0 ∪ {c} (312M). (a) Suppose that A is Dedekind complete. If A0 is order-closed in A, so is A1 . (b) Suppose that A is Dedekind σ-complete. If A0 is a σ-subalgebra of A, so is A1 . proof (a) Let D be any subset of A1 . Set E = {e : e ∈ A, there is some d ∈ D such that e ⊆ d}, A = {a : a ∈ A0 , a ∩ c ∈ E}, ∗

B = {b : b ∈ A0 , b \ c ∈ E}.

∗

Because A is Dedekind complete, a = sup A and b = sup B are defined in A; because A0 is order-closed, both belong to A0 , so d∗ = (a∗ ∩ c) ∪ (b∗ \ c) belongs to A1 . Now if d ∈ D, it is expressible as (a ∩ c) ∪ (b \ c) for some a, b ∈ A0 ; since a ∈ A and b ∈ B, we have a ⊆ a∗ and b ⊆ b∗ , so d ⊆ d∗ . Thus d∗ is an upper bound for D. On the other hand, if d0 is any other upper bound for D in A, it is also an upper bound for E, so we must have a∗ ∩ c = supa∈A a ∩ c ⊆ d0 ,

b∗ \ c = supb∈B b \ c ⊆ d0 ,

and d∗ ⊆ d0 . Thus d∗ = sup D. This shows that the supremum of any subset of A1 belongs to A1 , so that A1 is order-closed. (b) The argument is the same, except that we replace D by a sequence hdn in∈N , and A, B by sequences han in∈N , hbn in∈N in A0 such that dn = (an ∩ c) ∪ (bn \ c) for every n.

44

Boolean algebras

314K

314K Extension of homomorphisms The following is one of the most striking properties of Dedekind complete Boolean algebras. Theorem Let A be a Boolean algebra and B a Dedekind complete Boolean algebra. Let A0 be a Boolean subalgebra of A and π0 : A0 → B a Boolean homomorphism. Then there is a Boolean homomorphism π1 : A → B extending π0 . proof (a) Let P be the set of all Boolean homomorphisms π such that dom π is a Boolean subalgebra of A including A0 and π extends π0 . Identify each member of P with its graph, which is a subset of A × B, and order P by inclusion, so that π ⊆ θ means just that θ extends π. Then any non-empty totally ordered subset Q of P has an upper bound in P . P P Let π ∗ be the simple union of these graphs. (i) If (a, b) and 0 ∗ (a, b ) both belong to π , then there are π, π 0 ∈ Q such that πa = b, π 0 a = b0 ; now either π ⊆ π 0 or π 0 ⊆ π; in either case, θ = π ∪ π 0 ∈ Q, so that b = πa = θa = π 0 a = b0 . This shows that π ∗ is a function. (ii) Because Q 6= ∅, dom π0 ⊆ dom π ⊆ dom π ∗ for some π ∈ Q; thus π ∗ extends π0 (and, in particular, 0 ∈ dom π ∗ ). (iii) Now suppose that a, a0 ∈ dom(π ∗ ). Then there are π, π 0 ∈ Q such that a ∈ dom π, a0 ∈ dom π 0 ; once again, θ = π ∪π 0 ∈ Q, so that a, a0 ∈ dom θ, and a ∩ a0 ∈ dom θ ⊆ dom π ∗ ,

1 \ a ∈ dom θ ⊆ dom π ∗ ,

π ∗ (a ∩ a0 ) = θ(a ∩ a0 ) = θa ∩ θa0 = π ∗ a ∩ π ∗ a0 , π ∗ (1 \ a) = θ(1 \ a) = 1 \ θa = 1 \ π ∗ a. (iv) This shows that dom π ∗ is a subalgebra of A and that π ∗ is a Boolean homomorphism, that is, that π ∗ ∈ P ; and of course π ∗ is an upper bound for Q in P . Q Q (b) By Zorn’s Lemma, P has a maximal element π1 say. ?? Suppose, if possible, that A1 = dom π1 is not the whole of A; take c ∈ A \ A1 . Set A = {a : a ∈ A1 , a ⊆ c}. Because B is Dedekind complete, d = sup π1 [A] is defined in B. If a0 ∈ A and c ⊆ a0 , then of course a ⊆ a0 and π1 a ⊆ π1 a0 whenever a ∈ A, so that π1 a0 is an upper bound for π1 [A], and d ⊆ π1 a0 . But this means that there is an extension of π1 to a Boolean homomorphism π on the Boolean subalgebra of A generated by A1 ∪ {c} (312N). And this π must be a member of P properly extending π1 , which is supposed to be maximal. X X Thus dom π1 = A and π1 is an extension of π0 to A, as required. 314L The Loomis-Sikorski representation of a Dedekind σ-complete Boolean algebra The construction in 314D is not only the commonest way in which new Dedekind σ-complete Boolean algebras appear, but is adequate to describe them all. I start with an elementary general fact. Lemma Let X be any topological space, and write M for the family of meager subsets of X. Then M is a σ-ideal of subsets of X. proof The point is that if A ⊆ X is nowhere dense, so is every subset of A; this is obvious, since if B ⊆ A then B ⊆ A so int B ⊆ int A = ∅. So if B ⊆ A ∈ M, let hAn in∈N be a sequence of nowhere dense sets with union A; then hB ∩ An in∈N is a sequence of nowhere dense sets with union B, so B ∈ M. If hAn in∈N is a sequence in M with union A, then for each n we may choose a sequence hAnm im∈N of nowhere dense sets with union An ; then the countable family hAnm in,m∈N may be re-indexed as a sequence of nowhere dense sets with union A, so A ∈ M. Finally, ∅ is nowhere dense, so belongs to M. 314M Theorem Let A be a Dedekind σ-complete Boolean algebra, and Z its Stone space. Let E be the algebra of open-and-closed subsets of Z, and M the σ-ideal of meager subsets of Z. Then Σ = {E4A : E ∈ E, A ∈ M} is a σ-algebra of subsets of Z, M is a σ-ideal of Σ, and A is isomorphic, as Boolean algebra, to Σ/M.

314P

Order-completeness

45

proof (a) I start by showing that Σ is a σ-algebra. P P Of course ∅ = ∅4∅ ∈ Σ. If F ∈ Σ, express it as E4A where E ∈ E, A ∈ M; then Z \ F = (Z \ E)4A ∈ Σ. If hFn in∈N is a sequence in Σ, express each Fn as En 4An , where En ∈ E and An ∈ M. Now each En is expressible as b an , where an ∈ A. Because A is Dedekind σ-complete, a = supn∈N an is defined in A. Set S S E =b a ∈ E. By 313Ca, E = n∈N ESn , so the closed set E \ n∈N En has empty interior and is nowhere dense. Accordingly, setting A = E4 n∈N Fn , we have S S A ⊆ (E \ n∈N En ) ∪ n∈N An ∈ M, S so that n∈N Fn = E4A ∈ Σ. Thus Σ is closed under countable unions and is a σ-algebra. Q Q Evidently M ⊆ Σ, because ∅ ∈ E. (b) For each F ∈ Σ, there is exactly one E ∈ E such that F 4E ∈ M. P P There is surely some E ∈ E such that F is expressible as E4A where A ∈ M, so that F 4E = A ∈ M. If E 0 is any other member of E, then E 0 4E is a non-empty open set in X, while E 0 4E ⊆ A ∪ (F 4E 0 ); by Baire’s theorem for compact Hausdorff spaces (3A3G), A ∪ (F 4E 0 ) ∈ / M and F 4E 0 ∈ / M. Thus E is unique. Q Q (c) Consequently the maps E 7→ E • : E → Σ/M and a 7→ b a• : A → Σ/M are bijections. But since they are also Boolean homomorphisms, they are isomorphisms, and A ∼ = Σ/M, as claimed. 314N Corollary A Boolean algebra A is Dedekind σ-complete iff it is isomorphic to a quotient Σ/I where Σ is a σ-algebra of sets and I is a σ-ideal of Σ. proof Put 314D and 314M together. 314O Regular open algebras For Boolean algebras which are Dedekind complete in the full sense, there is another general method of representing them, which leads to further very interesting ideas. Definition Let X be a topological space. A regular open set in X is an open set G ⊆ X such that G = int G. Note that if F ⊆ X is any closed set, then G = int F is a regular open set, because G ⊆ G ⊆ F so G ⊆ int G ⊆ int F = G and G = int G. 314P Theorem Let X be any topological space, and write G for the set of regular open sets in X. Then G is a Dedekind complete Boolean algebra, with 1G = X and 0G = ∅, and with Boolean operations given by G 4G H = int G4H,

G ∩G H = G ∩ H,

G ∪G H = int G ∪ H,

with Boolean ordering given by G ⊆ G H ⇐⇒ G ⊆ H, and with suprema and infima given by sup H = int

S

H,

inf H = int

T

H = int

for all non-empty H ⊆ G. Remark I use the expressions ∩G

∪G

4G

\G

⊆G

in case the distinction between ∩

∪

4

\

⊆

∪

4

\

⊆

and ∩

is insufficiently marked.

T

H

G \G H = G \ H,

46

Boolean algebras

314P

proof I base the proof on the study of an auxiliary algebra of sets which involves some of the ideas already used in 314M. (a) Let I be the family of nowhere dense subsets of X. Then I is an ideal of subsets of X. P P Of course ∅ ∈ I. If A ⊆ B ∈ I then int A ⊆ int B = ∅. If A, B ∈ I and G is a non-empty open set, then G \ A is a non-empty open set and (G \ A) \ B is non-empty; accordingly G cannot be a subset of A ∪ B = A ∪ B. This shows that int A ∪ B = ∅, so that A ∪ B ∈ I. Q Q (b) For any set A ⊆ X, write ∂A for the boundary of A, that is, A \ int A. Set Σ = {E : E ⊆ X, ∂E ∈ I}. The Σ is an algebra of subsets of X. P P (i) ∂∅ = ∅ ∈ I so ∅ ∈ Σ. (ii) If A, B ⊆ X, then A ∪ B = A ∪ B, while int(A ∪ B) ⊇ int A ∪ int B; so ∂(A ∪ B) ⊆ ∂A ∪ ∂B. So if E, F ∈ Σ, ∂(E ∪ F ) ⊆ ∂E ∪ ∂F ∈ I and E ∪ F ∈ Σ. (iii) If A ⊆ X, then ∂(X \ A) = X \ A \ int(X \ A) = (X \ int A) \ (X \ A) = A \ int A = ∂A. So if E ∈ Σ, ∂(X \ E) = ∂E ∈ I and X \ E ∈ Σ. Q Q If A ∈ I, then of course ∂A = A ∈ I, so A ∈ Σ; accordingly I is an ideal in the Boolean algebra Σ, and we can form the quotient Σ/I. It will be helpful to note that every open set belongs to Σ, since if G is open then ∂G = G \ G cannot include any non-empty open set (since any open set meeting G must meet G). (c) For each E ∈ Σ, set VE = int E; then VE is the unique member of G such that E4VE ∈ I. P P (i) Being the interior of a closed set, VE ∈ G. Since int E ⊆ VE ⊆ E, E4VE ⊆ ∂E ∈ I. (ii) If G ∈ G is such that E4G ∈ I, then G \ VE ⊆ G \ VE ⊆ (G4E) ∪ (VE 4E) ∈ I, so G \ VE , being open, must be actually empty, and G ⊆ VE ; but this means that G ⊆ int VE = VE . Similarly, VE ⊆ G and VE = G. This shows that VE is unique. Q Q (d) It follows that the map G 7→ G• : G → Σ/I is a bijection, and we have a Boolean algebra structure on G defined by the Boolean algebra structure of Σ/I. What this means is that for each of the binary Boolean operations ∩G , 4G , ∪G , \G and for G, H ∈ G we must have G∗G H = int G ∗ H, writing ∗G for the operation on the algebra G and ∗ for the corresponding operation on Σ or PX. (e) Before working through the identifications, it will be helpful to observe that if H is any non-empty T T T subset of G, then int H = int H. P P Set G = int H. For every H ∈ H, G ⊆ H so G ⊆ int H = H; thus T T G ⊆ int H ⊆ int H = G, T T so G = int H. Q Q Consequently int H, being the interior of a closed set, belongs to G. (f )(i) If G, H ∈ G then their intersection in the algebra G is G ∩G H = int G ∩ H = int(G ∩ H) = G ∩ H, using (d) for the first equality and (e) for the second. (ii) Of course X ∈ G and X • = 1Σ/I , so X = 1G . (iii) If G ∈ G then its complement 1G \G G in G is int X \ G = int(X \ G) = X \ G. (iv) If G, H ∈ G, then the relative complement in G is G \G H = G ∩G (1G \G H) = G ∩ (X \ H) = G \ H = int(G \ H). (v) If G, H ∈ G, then G ∪G H = int G ∪ H and G 4G H = int G4H, by the remarks in (d). (g) We must note that for G, H ∈ G, G ⊆ G H ⇐⇒ G ∩G H = G ⇐⇒ G ∩ H = G ⇐⇒ G ⊆ H;

*314R

Order-completeness

47

that is, the ordering of the Boolean algebra G is just the partial ordering induced on G by the Boolean ordering ⊆ of PX or Σ. T S (h) If H is any non-empty subset of G, consider G0 = int H and G1 = int H. G0 = inf H in G. P P By (e), G0 ∈ G. Of course G0 ⊆ H for every H ∈ H, T so G0 is a lower bound for H. If G T is any lower bound for H in G, then G ⊆ H for every H ∈ H, so G ⊆ H; but also G is open, so G ⊆ int H = G0 . Thus G0 is the greatest lower bound for H. Q Q G1 = sup H in G. P P Being the interior of a closed set, G1 ∈ G, and of course S H = int H ⊆ int H = G1 for every H ∈ H, so G1 is an upper bound for H in G. If G is any upper bound for H in G, then S G = int G ⊇ int H = G1 ; thus G1 is the least upper bound for H in G. Q Q This shows that every non-empty H ⊆ G has a supremum and an infimum in G; consequently G is Dedekind complete, and the proof is finished. 314Q Remarks (a) G is called the regular open algebra of the topological space X. (b) Note that the map E 7→ VE : Σ → G of part (c) of the proof above is a Boolean homomorphism, if G is given its Boolean algebra structure. Its kernel is of course I; the induced map E • → 7 VE : Σ/I → G is just the inverse of the isomorphism G 7→ G• : G → Σ/I. *314R

I interpolate a lemma corresponding to 313R.

Lemma Let X and Y be topological spaces, and f : X → Y a continuous function such that f −1 [M ] is nowhere dense in X for every nowhere dense M ⊆ Y . Then we have an order-continuous Boolean homomorphism π from the regular open algebra RO(Y ) of Y to the regular open algebra RO(X) of X defined by setting πH = int f −1 [H] for every H ∈ RO(Y ). proof (a) By the remark in 314O, the formula for πH always defines a member of RO(X); and of course π is order-preserving. Observe that if H ∈ RO(Y ), then f −1 [H] is open, so f −1 [H] ⊆ πH. It will be convenient to note straight away that if V ⊆ Y is a dense open set then f −1 [V ] is dense in X. P P M = Y \ V is nowhere dense, so f −1 [M ] is nowhere dense and its complement f −1 [V ] is dense. Q Q (b) If H1 , H2 ∈ RO(Y ) then π(H1 ∩ H2 ) = πH1 ∩ πH2 . P P Because π is order-preserving, π(H1 ∩ H2 ) ⊆ πH1 ∩ πH2 . ?? Suppose, if possible, that they are not equal. Then (because π(H1 ∩ H2 ) is a regular open set) G = πH1 ∩ πH2 \ π(H1 ∩ H2 ) is non-empty. Set M = f [G]. Then f −1 [M ] ⊇ G is not nowhere dense, so H = int M must be non-empty. Now G ⊆ πH1 ⊆ f −1 [H1 ], so f [G] ⊆ f [f −1 [H1 ]] ⊆ f [f −1 [H1 ]] ⊆ H 1 , so M ⊆ H 1 and H ⊆ int H 1 = H1 . Similarly, H ⊆ H2 , and f −1 [H] ⊆ f −1 [H1 ∩ H2 ] ⊆ π(H1 ∩ H2 ). But also H ∩ f [G] is not empty, so ∅ 6= G ∩ f −1 [H] ⊆ G ∩ π(H1 ∩ H2 ), which is impossible. X XQ Q (c) If H ∈ RO(Y ) and H 0 = Y \ H is its complement in RO(Y ) then πH 0 = X \ πH is the complement of πH in RO(X). P P By (b), πH and πH 0 are disjoint. Now H ∪ H 0 is a dense open subset of Y , so πH ∪ πH 0 ⊇ f −1 [H] ∪ f −1 [H 0 ] = f −1 [H ∪ H 0 ] is dense in X, and the regular open set πH 0 must include the complement of πH in RO(X). Q Q Putting (b) and (c) together, we see that the conditions of 312H(ii) are satisfied, so that π is a Boolean homomorphism. (d)STo see that it is order-continuous, let H ⊆ RO(Y ) be a non-empty set with supremum Y . Then H0 = H is a dense open subset of Y (see the formula in 314P). So

48

Boolean algebras

S H∈H

πH ⊇

S H∈H

*314R

f −1 [H] = f −1 [H0 ]

is dense in X, and supH∈H πH = X in RO(X). By 313L(b-iii), π is order-continuous. 314S

It is now easy to characterize the Stone spaces of Dedekind complete Boolean algebras.

Theorem Let A be a Boolean algebra, and Z its Stone space; write E for the algebra of open-and-closed subsets of Z, and G for the regular open algebra of Z. Then the following are equiveridical: (i) A is Dedekind complete; (ii) Z is extremally disconnected (definition: 3A3Ae); (iii) E = G. proof (i)⇒(ii) If A is Dedekind complete, let G be any open set in Z. Set A = {a : a ∈ A, b a ⊆ G}, S a0 = sup A. Then G = {b a : a ∈ A}, because E is a base for the topology of Z, so b a0 = G, by 313Ca. Consequently G is open. As G is arbitrary, Z is extremally disconnected. (ii)⇒(iii) If E ∈ E, then of course E = E = int E, so E is a regular open set. Thus E ⊆ G. On the other hand, suppose that G ⊆ Z is a regular open set. Because Z is extremally disconnected, G is open; so G = int G = G is open-and-closed, and belongs to E. Thus E = G. (iii)⇒(i) Since G is Dedekind complete (314P), E and A are also Dedekind complete Boolean algebras. Remark Note that if the conditions above are satisfied, either 312L or the formulae in 314P show that the Boolean structures of E and G are identical. 314T I come now to a construction of great importance, both as a foundation for further constructions and as a source of insight into the nature of Dedekind completeness. Theorem Let A be a Boolean algebra, with Stone space Z; for a ∈ A let b a be the corresponding open-andb be the regular open algebra of Z (314P). closed subset of A. Let A (a) The map a 7→ b a is an injective order-continuous Boolean homomorphism from A onto an order-dense b subalgebra of A. (b) If B is any Dedekind complete Boolean algebra and π : A → B is an order-continuous Boolean b → B such that π1 b homomorphism, there is a unique order-continuous Boolean homomorphism π1 : A a = πa for every a ∈ A. proof (a)(i) Setting E = {b a : a ∈ A}, every member of E is open-and-closed, so is surely equal to the b for every a ∈ A. The formulae in 314P tell us interior of its closure, and is a regular open set; thus b a∈A b b that if a, b ∈ A, then b a ∩ b, taken in A, is just the set-theoretic intersection b a ∩ bb = (a ∩ b) b ; while 1 \ b a, b taken in A, is Z \b a = Z \b a = (1 \ a) b . b Thus the map a 7→ b b preserves ∩ and complementation, so And of course b 0 = ∅ is the zero of A. a:A→A is a Boolean homomorphism (312H). Of course it is injective. T (ii) If A ⊆ A is non-empty and inf A = 0, then a∈A b a is nowhere dense in Z (313Cc), so T inf{b a : a ∈ A} = int( a∈A b a) = ∅ b is order-continuous. (314P again). As A is arbitrary, the map a 7→ b a:A→A b is not empty, then there is a non-empty member of E included in it, by the definition of (iii) If G ∈ A b the topology of Z (311I). So E is an order-dense subalgebra of A. (b) Now suppose that B is a Dedekind complete Boolean algebra and π : A → B is an order-continuous b is an isomorphism between A and the Boolean homomorphism. Write ιa = b a for a ∈ A, so that ι : A → A −1 b Accordingly πι : E → B is an order-continuous Boolean homomorphism, order-dense subalgebra E of A. being the composition of the order-continuous Boolean homomorphisms π and ι−1 . By 314K, it has an b → B, and π1 ι = π, that is, π1 b extension to a Boolean homomorphism π1 : A a = πa for every a ∈ A. Now b b π1 is order-continuous. P P Suppose that H ⊆ A has supremum 1 in A. Set

314Xc

Order-completeness

49

H0 = {E : E ∈ E, E ⊆ H for some H ∈ H}. b Because E is order-dense in A, H = supE∈E,E⊆H E = supE∈H0 ,E⊆H E b It follows at once that sup H0 = 1 in E, so sup π1 [H0 ] = for every H ∈ H (313K), and sup H0 = 1 in A. −1 0 sup(πι )[H ] = 1. Since any upper bound for π1 [H] must also be an upper bound for π1 [H0 ], sup π1 [H] = 1 in B. As H is arbitrary, π1 is order-continuous (313L(b-iii)). Q Q 0 b → B is any other Boolean homomorphism such that π 0 b If π10 : A 1 a = πa for every a ∈ A, then π1 and π1 0 b G is the agree on E, and the argument just above shows that π1 is also order-continuous. But if G ∈ A, b supremum (in A) of F = {E : E ∈ E, E ⊆ G}, so π10 G = supE∈F π10 E = supE∈F π1 E = π1 G. As G is arbitrary, π10 = π1 . Thus π1 is unique. 314U The Dedekind completion of a Boolean algebra For any Boolean algebra A, I will say that b constructed in 314T is the Dedekind completion of A. the Boolean algebra A b and treat When using this concept I shall frequently suppress the distinction between a ∈ A and b a ∈ A, b A as itself an order-dense subalgebra of A. 314V Upper envelopes (a) Let A be a Boolean algebra, and C a subalgebra of A. For a ∈ A, write upr(a, C) = inf{c : c ∈ C, a ⊆ c} if the infimum is defined in C. (The most important cases are when A is Dedekind complete and C is orderclosed in A, so that C is Dedekind complete (314E) and upr(a, C) is defined for every a ∈ A; but others also arise.) (b) If A ⊆ A is such that upr(a, C) is defined for every a ∈ A, a0 = sup A is defined in A and c0 = supa∈A upr(a, C) is defined in C, then c0 = upr(a0 , C). P P If c ∈ C then c0 ⊆ c ⇐⇒ upr(a, C) ⊆ c for every a ∈ A ⇐⇒ a ⊆ c for every a ∈ A ⇐⇒ a0 ⊆ c. Q Q In particular, upr(a ∪ a0 , C) = upr(a, C) ∪ upr(a0 , C) whenever the right-hand side is defined. (c) If a ∈ A is such that upr(a, C) is defined, then upr(a ∩ c, C) = c ∩ upr(a, C) for every c ∈ C. P P For c0 ∈ C, a ∩ c ⊆ c0 ⇐⇒ a ⊆ c0 ∪ (1 \ c) ⇐⇒ upr(a, C) ⊆ c0 ∪ (1 \ c) ⇐⇒ c ∩ upr(a, C) ⊆ c0 . Q Q 314X Basic exercises (a) Let A be a Boolean algebra. (i) Show that the following are equiveridical: (α) A is Dedekind complete (β) every non-empty upwards-directed subset of A with an upper bound has a least upper bound (γ) every non-empty downwards-directed subset of A with a lower bound has a greatest lower bound. (ii) Show that the following are equiveridical: (α) A is Dedekind σ-complete (β) every nondecreasing sequence in A with an upper bound has a least upper bound (γ) every non-increasing sequence in A with a lower bound has a greatest lower bound. (b) Let A be a Boolean algebra. Show that any principal ideal of A is order-closed. Show that A is Dedekind complete iff every order-closed ideal is principal. >(c) Let A be a Dedekind σ-complete Boolean algebra, B a Boolean algebra and π : A → B a sequentially order-continuous Boolean homomorphism. If C ⊆ A and C is the σ-subalgebra of A generated by C, show that π[C] is the σ-subalgebra of B generated by π[C].

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(d) Let A be a Dedekind complete Boolean algebra, B an order-closed subalgebra of A, and a ∈ A; let Aa be the principal ideal of A generated by a. Show that {a ∩ b : b ∈ B} is an order-closed subalgebra of Aa . (e) Find a proof of 314Tb which does not appeal to 314K. > (f ) Let A be a Dedekind complete Boolean algebra and B an order-dense subalgebra of A. Show that A is isomorphic to the Dedekind completion of B. (g) Let A be a Dedekind complete Boolean algebra and C an order-closed subalgebra of A. Show that an element a of A belongs to C iff upr(1 \ a, C) = 1 \ upr(a, C) iff upr(1 \ a, C) ∩ upr(a, C) = 0. > (h) Let A be a Dedekind complete Boolean algebra, C an order-closed subalgebra of A, and a0 ∈ A, c0 ∈ C. Show that the following are equiveridical: (i) there is a Boolean homomorphism π : A → C such that πc = c for every c ∈ C and πa0 = c0 (ii) 1 \ upr(1 \ a0 , C) ⊆ c0 ⊆ upr(a0 , C). 314Y Further exercises (a) Let P be a Dedekind complete partially ordered set. Show that a set Q ⊆ P is order-closed iff sup R, inf R belong to Q whenever R ⊆ Q is a totally ordered subset of Q with upper and lower bounds in P . (Hint: show by induction on κ that if A ⊆ Q is upwards-directed and bounded above and #(A) ≤ κ then sup A ∈ Q.) (b) Let P be a lattice. Show that P is Dedekind complete iff every non-empty totally ordered subset of P with an upper bound in P has a least upper bound in P . (Hint: if A ⊆ P is non-empty and bounded below in P , let B be the set of lower bounds of A and use Zorn’s Lemma to find a maximal element of B.) (c) Give an example of a Boolean algebra A with an order-closed subalgebra A0 and an element c such that the subalgebra generated by A0 ∪ {c} is not order-closed. (d) Let X be any topological space. Let M be the σ-ideal of meager subsets of X, and set Bb = {G4A : G ⊆ X is open, A ∈ M}. b (i) Show that Bb is a σ-algebra of subsets of X, and that B/M is Dedekind complete. (Members of Bb are said b to be the subsets of X with the Baire property; B is the Baire property algebra of X.) (ii) Show that S b is dense, then A ∈ B. b (iii) Show that there is a largest open if A ⊆ X and {G : G ⊆ X is open, A ∩ G ∈ B} set V ∈ M. (iv) Let G be the regular open algebra of X. Show that the map G 7→ G• is an order-continuous b Boolean homomorphism from G onto B/M, so induces a Boolean isomorphism between the principal ideal b b of G generated by X \ V and B/M. (B/M is the category algebra of X; it is a Dedekind complete b Boolean algebra. X is called a Baire space if V = ∅; in this case G ∼ = B/M.) (e) Let A be a Dedekind σ-complete Boolean algebra, and han in∈N any sequence in A. For n ∈ N set En = {x : x ∈ {0, 1}N , x(n) = 1}, and let B be the σ-algebra of subsets of {0, 1}N generated by {En : n ∈ N}. (B is the ‘Borel σ-algebra’ of {0, 1}N ; see 4A2Wd.) Show that there is a unique sequentially order-continuous Boolean homomorphism θ : B → A such that θ(En ) = an for every n ∈ N. (Hint: define a suitable function φ from the Stone space Z of A to {0, 1}N , and consider {E : E ⊆ {0, 1}N , φ−1 [E] has the Baire property in Z}.) Show that θ[B] is the σ-subalgebra of A generated by {an : n ∈ N}. (f ) Let A be a Boolean algebra, and Z its Stone space. Show that A is Dedekind σ-complete iff G is open whenever G is a cozero set in Z. (Such spaces are called basically disconnected or quasi-Stonian.) (g) Let A, B be Dedekind complete Boolean algebras and D ⊆ A \ {0} an order-dense set. Suppose that φ : D → B is such that (i) φ[D] is order-dense in B (ii) for all d, d0 ∈ D, d ∩ d0 = 0 iff φd ∩ φd0 = 0. Show that φ has a unique extension to a Boolean isomorphism from A to B. (h) Let A be any Boolean algebra. Let J be the family of order-closed ideals in A. Show that (i) J is a Dedekind complete Boolean algebra with operations defined by the formulae I ∩ J = I ∩ J, 1 \ J = {a : a ∩ b = 0 for every b ∈ J} (ii) the map a 7→ Aa , the principal ideal generated by a, is an injective order-continuous Boolean homomorphism from A onto an order-dense subalgebra of J (iii) J is isomorphic to the Dedekind completion of A.

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314 Notes and comments At the risk of being tiresomely long-winded, I have taken the trouble to spell out a large proportion of the results in this section and the last in their ‘sequential’ as well as their ‘unrestricted’ forms. The point is that while (in my view) the underlying ideas are most clearly and dramatically expressed in terms of order-closed sets, order-continuous functions and Dedekind complete algebras, a large proportion of the applications in measure theory deal with sequentially order-closed sets, sequentially order-continuous functions and Dedekind σ-complete algebras. As a matter of simple technique, therefore, it is necessary to master both, and for the sake of later reference I generally give the statements of both versions in full. Perhaps the points to look at most keenly are just those where there is a difference in the ideas involved, as in 314Bb, or in which there is only one version given, as in 314M and 314T. If you have seen the Hahn-Banach theorem (3A5A), it may have been recalled to your mind by Theorem 314K; in both cases we use an order relation and a bit of algebra to make a single step towards an extension of a function, and Zorn’s lemma to turn this into the extension we seek. A good part of this section has turned out to be on the borderland between the theory of Boolean algebra and general topology; naturally enough, since (as always with the general theory of Boolean algebra) one of our first concerns is to establish connexions between algebras and their Stone spaces. I think 314T is the first substantial ‘universal mapping theorem’ in this volume; it is by no means the b is not just that we obtain a Dedekind complete Boolean algebra in last. The idea of the construction A which A is embedded as an order-dense subalgebra, but that we simultaneously obtain a theorem on the b of order-continuous Boolean homomorphisms defined on A. This characterization canonical extension to A b a 7→ b b becomes is enough to define the pair (A, a) up to isomorphism, so the exact method of construction of A of secondary importance. The one used in 314T is very natural (at least, if we believe in Stone spaces), but there are others (see 314Yh), with different virtues. 314K and 314T both describe circumstances in which we can find extensions of Boolean homomorphisms. Clearly such results are fundamental in the theory of Boolean algebras, but I shall not attempt any systematic presentation here. 314Ye can also be regarded as belonging to this family of ideas.

315 Products and free products I describe here two algebraic constructions of fundamental importance. They are very different in character, indeed may be regarded as opposites, despite the common use of the word ‘product’. The first part of the section (315A-315G) deals with the easier construction, the ‘simple product’; the second part (315H-315P) with the ‘free product’. Q 315A Products of Boolean algebras (a) Let hAi ii∈I be any family of Boolean algebras. Set A = i∈I Ai , with the natural ring structure a 4 b = ha(i) 4 b(i)ii∈I , a ∩ b = ha(i) ∩ b(i)ii∈I for a, b ∈ A. Then A is a ring (3A2H); it is a Boolean ring because a ∩ a = ha(i) ∩ a(i)ii∈I = a for every a ∈ A; and it is a Boolean algebra because if we set 1A = h1Ai ii∈I , then 1A ∩ a = a for every a ∈ A. I will call A the simple product of the family hAi ii∈I . I should perhaps remark that when I = ∅ then A becomes {∅}, to be interpreted as the singleton Boolean algebra. (b) The Boolean operations on A are now defined by the formulae a ∪ b = ha(i) ∪ b(i)ii∈I , for all a, b ∈ A.

a \ b = ha(i) \ b(i)ii∈I

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315B Theorem Let hAi ii∈I be a family of Boolean algebras, and A their simple product. (a) The maps a 7→ πi (a) = a(i) : A → Ai are all Boolean homomorphisms. (b) If B is any other Boolean algebra, then a map φ : B → A is a Boolean homomorphism iff πi φ : B → Ai is a Boolean homomorphism for every i ∈ I. proof Verification of these facts amounts just to applying the definitions with attention. 315C Products of partially ordered sets (a) It is perhaps worth spelling out the following elemenQ tary definition. If hPi ii∈I is any family of partially ordered sets, its product is the set P = i∈I Pi ordered by saying that p ≤ q iff p(i) ≤ q(i) for every i ∈ I; it is easy to check that P is now a partially ordered set. (b) The point is that if A is the simple product of a family hAi ii∈I of Boolean algebras, then the ordering of A is just the product partial order: a ⊆ b ⇐⇒ a ∩ b = a ⇐⇒ a(i) ∩ b(i) = a(i) ∀ i ∈ I ⇐⇒ a(i) ⊆ b(i) ∀ i ∈ I. Now we have the following elementary, but extremely useful, general facts about products of partially ordered sets. 315D Proposition Let hPi ii∈I be a family of non-empty partially ordered sets with product P . (a) For any non-empty set A ⊆ P and q ∈ P , (i) sup A = q in P iff supp∈A p(i) = q(i) in Pi for every i ∈ I, (ii) inf A = q in P iff inf p∈A p(i) = q(i) in Pi for every i ∈ I. (b) The coordinate maps p 7→ πi (p) = p(i) : P → Pi are all order-preserving and order-continuous. (c) For any partially ordered set Q and function φ : Q → P , φ is order-preserving iff πi φ is order-preserving for every i ∈ I. (d) For any partially ordered set Q and order-preserving function φ : Q → P , (i) φ is order-continuous iff πi φ is order-continuous for every i, (ii) φ is sequentially order-continuous iff πi φ is sequentially order-continuous for every i. (e)(i) P is Dedekind complete iff every Pi is Dedekind complete. (ii) P is Dedekind σ-complete iff every Pi is Dedekind σ-complete. proof All these are elementary verifications. Of course parts (b), (d) and (e) rely on (a). 315E Factor algebras as principal ideals Because Boolean algebras have least elements, we have a second type of canonical map associated with their products. If hAi ii∈I is a family of Boolean algebras with simple product A, define θi : Ai → A by setting (θi a)(i) = a, (θi a)(j) = 0Aj if i ∈ I, a ∈ Ai and j ∈ I \ {i}. Each θi is a ring homomorphism, and is a Boolean isomorphism between Ai and the principal ideal of A generated by θi (1Ai ). The family hθi (1Ai )ii∈I is a partition of unity in A. Associated with these embeddings is the following important result. 315F Proposition Let A be a Boolean algebra and hei ii∈I a partition of unity in A. Suppose either (i) that I is finite or (ii) that I is countable and A is Dedekind σ-complete or (iii) that A is Dedekind complete. Q Then the map a 7→ ha ∩ ei ii∈I is a Boolean isomorphism between A and i∈I Aei , writing Aei for the principal ideal of A generated by ei for each i. proof The given map is a Boolean homomorphism because each of the maps a 7→ a ∩ ei : A → Aei is (312J). It is injective because supi∈I ei = 1,Qso if a ∈ A \ {0} there is an i such that a ∩ ei 6= 0. It is surjective because hei ii∈I is disjoint and if c ∈ i∈I Aei then a = supi∈I c(i) is defined in A and a ∩ ej = supi∈I c(i) ∩ ej = c(j) for every j ∈ I (using 313Ba). The three alternative versions of the hypotheses of this proposition are designed to ensure that the supremum is always well-defined in A.

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315G Algebras of sets and their quotients The Boolean algebras of measure theory are mostly presented as algebras of sets or quotients of algebras of sets, so it is perhaps worth spelling out the ways in which the product construction applies to such algebras. Proposition Let hXi ii∈I be Q a family of sets, and Σi an algebra of subsets of Xi for each i. (a) The simple product i∈I Σi may be identified with the algebra Σ = {E : E ⊆ X, {x : (x, i) ∈ E} ∈ Σi for every i ∈ I} of subsets of X = {(x, i) : i ∈ I, x ∈ Xi }, with the canonical homomorphisms πi : Σ → Σi being given by πi E = {x : (x, i) ∈ E} for each E ∈ Σ. Q (b) Now suppose that Ji is an ideal of Σi for each i. Then i∈I Σi /Ji may be identified with Σ/J , where J = {E : E ∈ Σ, {x : (x, i) ∈ E} ∈ Ji for every i ∈ I}, and the canonical homomorphisms π ˜i : Σ/J → Σi /Ji are given by the formula π ˜i (E • ) = (πi E)• for every E ∈ Σ. Q proof (a) It is easy to check that Σ is a subalgebra of PX, and that the map E 7→ hπi Eii∈I : Σ → i∈I Σi is a Boolean isomorphism. (b) Again, it is easy to check that J is an ideal of Σ, that the proposed formula for π ˜i does Q indeed define a map from Σ/J to Σi /Ji , and that E • 7→ h˜ πi E • ii∈I is an isomorphism between Σ/J and i∈I Σi /Ji . 315H Free products I come now to the second construction of this section. (a) Definition Q Let hAi ii∈I be a family of Boolean algebras. For each i ∈ I, let Zi be the Stone space Then the free product of hAi ii∈I is the algebra A of of Ai . Set Z = i∈I Zi , with the product topology. N open-and-closed sets in Z; I will denote it by i∈I Ai . (b) For i ∈ I and a ∈ Ai , the set b a ⊆ Zi representing a is an open-and-closed subset of Zi ; because z 7→ z(i) : Z → Zi is continuous, εi (a) = {z : z(i) ∈ b a} is open-and-closed, so belongs to A. In this context I will call εi : Ai → A the canonical map. (c) The topological space Z may be identified with the Stone space of the Boolean algebra A. P P By Tychonoff’s theorem (3A3J), Z is compact. If z ∈ Z and G is an open subset of Z containing z, then there are J, hGj ii∈J such that J is a finite subset of I, Gj is an open subset of Zj for each j ∈ J, and z ∈ {w : w ∈ Z, w(j) ∈ Gj for every j ∈ J} ⊆ G. Because each Zj is zero-dimensional, we can find an open-and-closed set Ej ⊆ Zj such that z(j) ∈ Ej ⊆ Gj . Now T E = Z ∩ j∈J {w : w(j) ∈ Ej } is a finite intersection of open-and-closed subsets of Z, so is open-and-closed; and z ∈ E ⊆ G. As z and G are arbitrary, Z is zero-dimensional. Finally, Z, being the product of Hausdorff spaces, is Hausdorff. So the result follows from 311J. Q Q 315I Theorem Let hAi ii∈I be a family of Boolean algebras, with free product A. (a) The canonical map εi : Ai → A is a Boolean homomorphism for every i ∈ I. (b) For any Boolean algebra B and any family hφi ii∈I such that φi is a Boolean homomorphism from Ai to B for every i, there is a unique Boolean homomorphism φ : A → B such that φi = φεi for each i. proof These are both consequences of 312P-312Q. As in 315H, write Zi for the Stone space of A, and Z for Q Z i∈I i , identified with the Stone space of A, as observed in 315Hc. The maps εi : Ai → A are defined as the homomorphisms corresponding to the continuous maps z 7→ ε˜i (z) = z(i) : Z → Zi , so (a) is surely true. Now suppose that we are given a Boolean homomorphism φi : Ai → B for each i ∈ I. Let W be the Stone space of B, and let φ˜i : W → Zi be the continuous function corresponding to φi . By 3A3Ib, the map ˜ w 7→ φ(w) = hφ˜i (w)ii∈I : W → Z is continuous, and corresponds to a Boolean homomorphism φ : A → B;

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˜ φεi = φi for each i. Moreover, φ is the only Boolean homomorphism with this property, because φ˜i = ε˜i φ, because if ψ : A → B is a Boolean homomorphism such that ψεi = φi for every i, then ψ corresponds to a continuous map ψ˜ : W → Z, and we must have ε˜i ψ˜ = φ˜i for each i, so that ψ˜ = φ˜ and ψ = φ. This proves (b). 315J Of course 315I is the defining property of the free product (see 315Xg below). I list a few further basic facts. Proposition Let hAi ii∈I be a family of Boolean algebras, and A their free product; write εi : Ai → A for the canonical maps. S (a) A is the subalgebra of itself generated by i∈I εi [Ai ]. (b) Write C for the set of those members of A expressible in the form inf j∈J εj (aj ), where J ⊆ I is finite and aj ∈ Aj for every j. Then every member of A is expressible as the supremum of a disjoint finite subset of C. In particular, C is order-dense in A. (c) Every εi is order-continuous. (d) A = {0A } iff there is some i ∈ I such that Ai = {0Ai }. (e) Now suppose that Ai 6= {0Ai } for every i ∈ I. (i) εi is injective for every i ∈ I. (ii) If J ⊆ I is finite and aj is a non-zero member of Aj for each j ∈ J, then inf j∈J εj (aj ) 6= 0. (iii) If i, j are distinct members of I and a ∈ Ai , b ∈ Aj , then εi (a) = εj (b) iff either a = 0Ai and b = 0Aj or a = 1Ai and b = 1Aj . Q proof As usual, write Zi for the Stone space of Ai , and Z = i∈I Zi , identified with the Stone space of A (315Hc). S (a) Write A0 for the subalgebra of A generated by i∈I εi [Ai ]. Then εi : Ai → A0 is a Boolean homomorphism for each i, so by 315Ib there is a Boolean homomorphism φ : A → A0 such that φεi = εi for each i. Now, regarding φ as a Boolean homomorphism from A to itself, the uniqueness assertion of 315Ib (with B = A) shows that φ must be the identity, so that A0 = A. (b) Write D for the set of finite partitions of unity in A consisting of members of C, and A for the set of members of A expressible in the form sup D0 where D0 is a subset of a member of D. Then A is a subalgebra of A. P P (i) 1A ∈ C (set J = ∅ in the definition of members of C) so {1A } ∈ D and 0A , 1A ∈ A. (ii) Note that if c, d ∈ C then c ∩ d ∈ C. (iii) If a, b ∈ A, express them as sup D0 , sup E 0 where D0 ⊆ D ∈ D, E 0 ⊆ E ∈ D. Then F = {d ∩ e : d ∈ D, e ∈ E} ∈ D, so 1A \ a = sup D \ D0 ∈ A, a ∪ b = sup{f : f ∈ F, f ⊆ a ∪ b} ∈ A. Q Q Also, εi [Ai ] ⊆ A for each i ∈ I. P P If a ∈ Ai , then {εi (a), εi (1Ai \ a)} ∈ D, so εi (a) ∈ A. Q Q So (a) tells us that A = A, and every member of A is a finite disjoint union of members of C. (c) If i ∈ I and A ⊆ Ai and inf A = 0 in Ai , take any non-zero c ∈ A. By (b), we can find a finite J ⊆ I and a family haj ij∈J such that c0 = inf j∈J εj (aj ) ⊆ c and c0 6= 0. Regarding c0 as a subset of Z, we have a point z ∈ c0 . Adding i to J and setting ai = 1Ai if necessary, we may suppose that i ∈ J. Now c0 6= 0A so ai 6= 0Ai and there is an a ∈ A such that ai 6⊆ a, so there is a t ∈ b ai \ b a. In this case, setting w(i) = t, w(j) = z(j) for j 6= i, we have w ∈ c0 \ εi (a), and c0 , c are not included in εi (a). As c is arbitrary, this shows that inf εi [A] = 0. As A is arbitrary, εi is order-continuous. (d) The point is that A = {0A } iff Z = ∅, which is so iff some Zi is empty. (e)(i) Because no Zi is empty, all the coordinate maps from Z to Zi are surjective, so the corresponding homomorphisms εi are injective (312Ra). (ii) Because J is finite,

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inf j∈J εj (aj ) = {z : z ∈ Z, z(j) ∈ b aj for every j ∈ J} is not empty. (iii) If εi (a) = εj (b) = 0A then (using (i)) a = 0Ai and b = 0Aj ; if εi (a) = εj (b) = 1A then a = 1Ai and b = 1Aj . ?? If εi (a) = εj (b) ∈ A \ {0A , 1A }, then there are t ∈ b a and u ∈ Zj \ bb. Now there is a z ∈ Z such that z(i) = t and z(j) = u, so that z ∈ εi (a) \ εj (b). X X 315K Proposition Let hAi ii∈I be any family of Boolean algebras, and hJk ik∈K any partition of I. Then the free product A of hAi ii∈I is isomorphic to the free product B of hBk ik∈K , where each Bk is the free product of hAi ii∈Jk . proof Write εi : Ai → A, ε0i : A → Bk and δk : Bk → B for the canonical maps when k ∈ K, i ∈ Jk . Then the homomorphisms δk ε0i : Ai → B correspond to a homomorphism φ : A → B such that φεi = δk ε0i whenever i ∈ Jk . Next, for each k, the homomorphisms εi : Ai → A, for i ∈ Jk , correspond to a homomorphism ψk : Bk → A such that ψk ε0i = εi for i ∈ Jk ; and the family hψk ik∈K corresponds to a homomorphism ψ : B → A such that ψδk = ψk for k ∈ K. Consequently ψφεi = ψδk ε0i = ψk ²0i = ²i whenever k ∈ K, i ∈ Jk . Once again using the uniqueness assertion in 315Ib, ψφ is the identity homomorphism on A. On the other hand, if we look at φψ : B → B, then we see that φψδk ε0i = φψk ²0i = φ²i = δk ε0i whenever k ∈ K, i ∈ Jk . Now, for given k, {b : b ∈ Bk , φψδk b = δk b} is a subalgebra of Bk including S ε0 [A ], and must be the whole of Bk , by 315Ja. So {b : b ∈ B, φψb = b} is a subalgebra of B including Si∈Jk i i k∈K δk [Bk ], and is the whole of B. Thus φψ is the identity on B and φ, ψ are the two halves of an isomorphism between A and B. 315L Algebras of sets and their quotients Once again I devote a paragraph to spelling out the application of the construction to the algebras most important to us. Proposition Let hXi ii∈I Nbe a family of sets, and Σi an algebra of subsets of Xi for eachQi. (a) The free product i∈I Σi may be identified with the algebra Σ of subsets of X = i∈I Xi generated by the set {εi (E) : i ∈ I, E ∈ Σi }, where εi (E) = {x : x ∈ X,Nx(i) ∈ E}. (b) Now suppose that Ji is an ideal of Σi for each i. Then i∈I Σi /Ji may be identified with Σ/J , where J is the ideal of Σ generated by {εi (E) : i ∈ I, E ∈ Ji }; the corresponding canonical maps ε˜i : Σi /Ji → Σ/J being defined by the formula ε˜i (E • ) = (εi (E))• for i ∈ I, E ∈ Σi . proof I start by proving (b) in detail; the argument for (a) is then easy to extract. Write Ai = Σi /Ji , A = Σ/J . (i) Fix i ∈ I for the moment. By the definition of Σ, εi (E) ∈ Σ for E ∈ Σi , and it is easy to check that εi : Σi → Σ is a Boolean homomorphism. Again, because εi (E) ∈ J whenever E ∈ Ji , the kernel of the homomorphism E 7→ (εi (E))• : Σi → A includes Ji , so the formula for ε˜i defines a homomorphism from Ai to A. N Now let C = i∈I Ai be the free product, and write ε0i : Ai → C for the canonical homomorphisms. By 315I, there is a Boolean homomorphism φ : C → A such that φε0i = ε˜i for each i. The set {E : E ∈ Σ, E • ∈ φ[C]} is a subalgebra of Σ including εi [Σi ] for every i, so is Σ itself, and φ is surjective. (ii) We need a simple description of the ideal J , as follows: a set ES ∈ Σ belongs to J iff there are a finite K ⊆ I and a family hFk ik∈K such that Fk ∈ Jk for each k and E ⊆ k∈K εk (Fk ). For evidently such sets have to belong to J , since the εk (Fk ) will be in J , while the family of all these sets is an ideal containing εi (F ) whenever i ∈ I, F ∈ Ji . (iii) Now we can see that φ : C →QA is injective. P P Take any non-zero c ∈ C. By 315Jb, we can find a finite J ⊆ I and a family haj ij∈J in j∈J Aj such that 0 6= inf j∈J ε0j aj ⊆ c. Express each aj as Ej• , where T Ej ∈ Σj , and consider E = X ∩ j∈J εj (Ej ) ∈ Σ. Then

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E • = inf j∈J ε˜j aj = φ(inf j∈J ε0j aj ) ⊆ φ(c). Also, because ε0j aj 6= 0, Ej ∈ / Jj for each j. But it follows that E ∈ / J , because if K ⊆ I is finite and Fk ∈ Jk for each k ∈ K, set Ei = Xi for i ∈S I \ J, Fi = ∅ for i ∈ I \ K; then there is an x ∈ X such that x(i) ∈ Ei \ Fi for each i ∈ I, so that x ∈ E \ k∈K Fk . By the criterion of (ii), E ∈ / J . So 0 6= E • ⊆ φ(c). As c is arbitrary, the kernel of φ is {0}, and φ is injective. Q Q So φ : C → A is the required isomorphism. (iv) This proves (b). Reading through the arguments above, it is easy to see the simplifications which compose a proof of (a), reading Σi for Ai and {∅} for Ji . 315M Notation Free products are sufficiently surprising that I think it worth taking a moment to look at a pair of examples relevant to the kinds of application I wish to make of the concept in the next chapter. First let me introduce a somewhat more direct notation which seems appropriate for the free product of finitely many factors. If A and B are two Boolean algebras, I write A ⊗ B for their free product, and for a ∈ A, b ∈ B I write a ⊗ b for ε1 (a) ∩ ε2 (b), where ε1 : A → A ⊗ B, ε2 : B → A ⊗ B are the canonical maps. Observe that (a1 ⊗ b1 ) ∩ (a2 ⊗ b2 ) = (a1 ∩ a2 ) ⊗ (b1 ∩ b2 ), and that the maps a 7→ a ⊗ b0 , b 7→ a0 ⊗ b are always ring homomorphisms. Now 315J(e-ii) tells us that a ⊗ b = 0 only when one of a, b is 0. In the context of 315L, we can identify E ⊗ F with E × F for E ∈ Σ1 , F ∈ Σ2 , and E • ⊗ F • with (E × F )• . 315N Lemma Let A, B be Boolean algebras. (a) Any element of A ⊗ B is expressible as supi∈I ai ⊗ bi where hai ii∈I is a finite partition of unity in A. (b) If c ∈ A ⊗ B is non-zero there are non-zero a ∈ A, b ∈ B such that a ⊗ b ⊆ c. proof (a) Let C be the set of elements of A ⊗ B representable in this form. Then C is a subalgebra of A ⊗ B. P P (i) If hai ii∈I , ha0j ij∈J are finite partitions of unity in A, and bi , b0j members of B for i ∈ I and j ∈ J, the hai ∩ a0j ii∈I,j∈J is a partition of unity in A, and (sup ai ⊗ bi ) ∩ (sup a0j ⊗ b0j ) = sup (ai ⊗ bi ) ∩ (a0j ⊗ b0j ) i∈I

j∈J

i∈I,j∈J

= sup (ai ∩ a0j ) ⊗ (bi ∩ b0j ) ∈ C. i∈I,j∈J

0

0

So c ∩ c ∈ C for all c, c ∈ C. (ii) If hai ii∈I is a finite partition of unity in A and bi ∈ B for each i, then 1 \ supi∈I ai ⊗ bi = (supi∈I ai ⊗ 1) \ (supi∈I ai ⊗ bi ) = supi∈I ai ⊗ (1 \ bi ) ∈ C. Thus 1 \ c ∈ C for every c ∈ C. Q Q Since a ⊗ 1 = (a ⊗ 1) ∪ ((1 \ a) ⊗ 0) and 1 ⊗ b belong to C for every a ∈ A, b ∈ B, C must be the whole of A ⊗ B, by 315Ja. (b) Now this follows at once, just as in 315Jb. 315O Example A = PN ⊗ PN is not Dedekind σ-complete. P P Consider A = {{n} ⊗ {n} : n ∈ N} ⊆ A. ?? If A has a least upper bound c in A, then c is expressible as a supremum supj≤k aj ⊗bj , by 315Jb. Because k is finite, there must be distinct m, n such that {j : m ∈ aj } = {j : n ∈ aj }. Now {n} × {n} ⊆ c, so there is a j ≤ k such that (aj ∩ {n}) ⊗ (bj ∩ {n}) = ({n} ⊗ {n}) ∩ (aj ⊗ bj ) 6= 0, so that neither aj ∩ {n} nor bj ∩ {n} is empty, that is, n ∈ aj ∩ bj . But this means that m ∈ aj , so that (aj ⊗ bj ) ∩ ({m} ⊗ {n}) = (aj ∩ {m}) ⊗ (bj ∩ {n}) 6= 0, and c ∩ ({m} ⊗ {n}) 6= 0, even though a ∩ ({m} ⊗ {n}) = 0 for every a ∈ A. X X Thus we have found a countable subset of A with no supremum in A, and A is not Dedekind σ-complete. Q Q 315P Example Now let A be any non-trivial atomless Boolean algebra, and B the free product A ⊗ A. Then the identity homomorphism from A to itself induces a homomorphism φ : B → A given by setting

315Xj

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57

φ(a ⊗ b) = a ∩ b for every a, b ∈ A. The point I wish to make is that φ is not order-continuous. P P Let C be the set {a ⊗ b : a, b ∈ A, a ∩ b = 0}. Then φ(c) = 0A for every c ∈ C. If d ∈ B is non-zero, then by 315Nb there are non-zero a, b ∈ A such that a ⊗ b ⊆ d; now, because A is atomless, there is a non-zero a0 ⊆ a such that a \ a0 6= 0. At least one of b \ a0 , b \ (a \ a0 ) is non-zero; suppose the former. Then a0 ⊗ (b \ a0 ) is a non-zero member of C included in d. As d is arbitrary, this shows that sup C = 1B . So supc∈C φ(c) = 0A 6= 1A = φ(sup C), and φ is not order-continuous. Q Q Thus the free product (unlike the product, see 315Dd) does not respect order-continuity. 315X Basic exercises (a) Let hAi ii∈I be any family of Boolean algebras, with simple product A, and πi : A → Ai the coordinate homomorphisms. Suppose we have another Boolean algebra A0 , with homomorphisms πi0 : A0 → Ai , such that for every Boolean algebra B and every family hφi ii∈I of homomorphisms from B to the Ai there is a unique homomorphism φ : B → A0 such that φi = πi0 φ for every i. Show that there is a unique isomorphism ψ : A → A0 such that πi0 ψ = πi for every i ∈ I. (b) Let hPi ii∈I be a family of non-empty partially ordered sets, with product partially ordered set P . Show that P is a lattice iff every Pi is a lattice, and that in this case it is the product lattice in the sense that p ∨ q = hp(i) ∨ q(i)ii∈I , p ∧ q = hp(i) ∧ q(i)ii∈I for all p, q ∈ P . (c) Let hAi ii∈I be a family of Boolean algebras with simple product A. For each i ∈ I let Zi be the Stone space of Ai , and let Z be the Stone space of A. Show that the coordinate maps from A onto Ai induce homeomorphisms between the Zi and open-and-closed subsets Zi∗ of Z. Show that hZi∗ ii∈I is disjoint. Show S ∗ that i∈I Zi is dense in Z, and is equal to Z iff {i : Ai 6= {0}} is finite. (d) Let hAi ii∈I be a family of Boolean algebras, with simple product A. Suppose that for each i ∈ I we Q are given an ideal Ii ofQAi . Show that I = i∈I Ii is an ideal of A, and that A/I may be identified, as Boolean algebra, with i∈I Ai /Ii . (e) Let hXi ii∈I be any family of topological spaces. Let X be their disjoint union {(x, i) : i ∈ I, x ∈ Xi }, with the disjoint union topology; that is, a set G ⊆ X is open in X iff {x : (x, i) ∈ G} is open in Xi for every i ∈ I. (i) Show that the algebra of open-and-closed subsets of X can be identified, as Boolean algebra, with the simple product of the algebras of open-and-closed sets of the Xi . (ii) Show that the regular open algebra of X can be identified, as Boolean algebra, with the simple product of the regular open algebras of the Xi . (f ) Show that the topological product of any family of zero-dimensional spaces is zero-dimensional. (g) Let hAi ii∈I be any family of Boolean algebras, with free product A, and εi : Ai → A the canonical homomorphisms. Suppose we have another Boolean algebra A0 , with homomorphisms ε0i : Ai → A0 , such that for every Boolean algebra B and every family hφi ii∈I of homomorphisms from the Ai to B there is a unique homomorphism φ : A0 → B such that φi = φε0i for every i. Show that there is a unique isomorphism ψ : A → A0 such that ε0i = ψεi for every i ∈ I. (h) Let I be any set, and let A be the algebra of open-and-closed sets of {0, 1}I ; for each i ∈ I set ai = {x : x ∈ {0, 1}I , x(i) = 1} ∈ A. Show that for any Boolean algebra B, any family hbi ii∈I in B there is a unique Boolean homomorphism φ : A → B such that φ(ai ) = bi for every i ∈ I. (i) Let hAi ii∈I , hB Q j ij∈J beQtwo families Q of Boolean algebras. Show that there is a natural injective homomorphism φ : i∈I Ai ⊗ j∈J Bj → i∈I,j∈J Ai ⊗ Bj defined by saying that for a ∈

Q i∈I

Ai , b ∈

Q j∈J

φ(a ⊗ b) = ha(i) ⊗ b(j)ii∈I,j∈J Bj . Show that φ is surjective if I and J are finite.

Q (j) Let hJ(i)ii∈I be a family of sets, with product Q = i∈I N J(i). QLet hAij ii∈I,j∈J(i) Q be Na family of Boolean algebras. Describe a natural injective homomorphism φ : i∈I j∈J(i) Aij → q∈Q i∈I Ai,q(i) .

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(k) Let A and B be Boolean algebras with partitions of unity hai ii∈I , hbj ij∈J . Show that hai ⊗ bj ii∈I,j∈J is a partition of unity in A ⊗ B. (l) Let A and B be Boolean algebras and a ∈ A, b ∈ B. Write Aa , Bb for the corresponding principal ideals. Show that there is a canonical isomorphism between Aa ⊗ Bb and the principal ideal of A ⊗ B generated by a ⊗ b. N (m) Let hAi ii∈I be any family of Boolean algebras, with free product i∈I Ai , and εi : Ai → A the canonical maps. Show that εi [Ai ] is an order-closed subalgebra of A for every i. (n) Let A be a Boolean algebra. Let us say that a family hAi ii∈I of subalgebras of A is Booleanindependent if inf j∈J aj 6= 0 wheneverSJ ⊆ I is finite and aj ∈ Aj \ {0} for every Nj ∈ J. Show that in this case the subalgebra of A generated by i∈I Ai is isomorphic to the free product i∈I Ai . (o) Let hAi ii∈I and hBi ii∈I be two families of Boolean algebras, and suppose that for each i ∈ I we are given a Boolean homomorphism N Nφi : Ai → Bi with kernel Ki C Ai . Show that the φi induce a Boolean homomorphism φ : i∈I Ai → i∈I Bi with kernel generated by the images of the Ki . Show that if every φi is surjective, so is φ. (p) Let hAi ii∈I be any family of non-trivial Boolean algebras. Show that if J ⊆ N NI and Bj is a subalgebra of Aj for each j ∈ J, then j∈J Bj is canonically embedded as a subalgebra of i∈I Ai . (q) Let A and B be Boolean algebras, neither {0}. Show that any element of A⊗B is uniquely expressible as supi∈I ai ⊗ bi where hai ii∈I is a partition of unity in A, with no ai equal to 0, and bi 6= bj in B for i 6= j. 315Y Further exercises (a) Let hAi ii∈I and hBi ii∈I be two families of Boolean algebras, N and suppose N A → that we are given Boolean homomorphisms φi : Ai → Bi for each i; let φ : i i∈I Bi be the i∈I induced homomorphism. (i) Show that if every φi is order-continuous, so is φ. (ii) Show that if every φi is sequentially order-continuous, so is φ. (b) Let hZi ii∈I be any family of topological spaces with product Z. For i ∈ I, z ∈ Z set ε˜i (z) = z(i). Show that if M ⊆ Zi is nowhere dense in Zi then ε˜−1 i [M ] is nowhere dense in Z. Use this to prove 315Jc. (c) Let hAi ii∈IN be a family of Boolean N algebras, and suppose that we are given subalgebras Bi of Ai for each i; set A = i∈I Bi , and let φ : B → A be the homomorphism induced by the i∈I Ai and B = embeddings Bi ⊆ Ai . (i) Show that if every Bi is order-closed in Ai , then φ[B] is order-closed in A. (ii) Show that if every Bi is a σ-subalgebra of Ai , then φ[B] is a σ-subalgebra in A. (d) Let hXi ii∈I be a family of topological spaces, with product X. Let Gi , G be Nthe corresponding regular open algebras. Show that G can be identified with the Dedekind completion of i∈I Gi . (e) Use the ideas of 315Xh and 315L to give an alternative construction of ‘free product’, for which 315I and 315J(e-ii) are true, which does not depend on the concept of Stone space nor on any other use of the axiom of choice. (Hint: show that for any Boolean algebra A there is a canonical surjection from the algebra EA onto A, where EJ is the algebra of subsets of {0, 1}J generated by sets of the form {x : x(j) = 1}; show that for such N algebras EJ , at least, the method of 315H-315I can be used; now apply the method of 315L to describe i∈I Ai as a quotient of EJ where J = {(a, i) : i ∈ I, a ∈ Ai }. Finally check that if no Ai is trivial, then nor is the free product.) (f ) Let A and B be Boolean algebras. Show that A ⊗ B is Dedekind complete iff either A = {0} or B = {0} or A is finite and B is Dedekind complete or B is finite and A is Dedekind complete. (g) Let hPi ii∈I be any family of partially ordered spaces. (i) Give a construction of a partially ordered space P , together with a family of order-preserving maps εi : Pi → P , such that whenever Q is a partially ordered set and φi : Pi → Q is order-preserving for every i ∈ I, there is a unique order-preserving map φ : P → Q such that φi = φεi for every i. (ii) Show that φ will be order-continuous iff every φi is. (iii) Show that P will be Dedekind complete iff every Pi is, but (except in trivial cases) is not a lattice.

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315 Notes and comments In this section I find myself asking for slightly more sophisticated algebra than seems necessary elsewhere. The point is that simple products and free products are best regarded as defined by the properties described in 315B and 315I. That is, it is sometimes right to think of a simple product of a family hAi ii∈I of Boolean algebras as being a structure (A, hπi ii∈I ) where A is a Boolean algebra, πi : A → Ai is a homomorphism for every i ∈ I, and every family of homomorphisms from a Boolean algebra B to the Ai can be uniquely represented by a single homomorphism from B to A. Similarly, reversing the direction of the homomorphisms, we can speak of a free product (it would be natural to say ‘coproduct’) (A, hεi ii∈I ) of hAi ii∈I . On such definitions, it is elementary that any two simple products, or free products, are isomorphic in the obvious sense (315Xa, 315Xg), and very general arguments from abstract algebra, not restricted to Boolean algebras (see Bourbaki 68, IV.3.2), show that they exist. (But in order to prove such basic facts as that the πi are surjective, or that the εi are, except when the construction collapses altogether, injective, we do of course have to look at the special properties of Boolean algebras.) Now in the case of simple products, the Cartesian product construction is so direct and so familiar that there seems no need to trouble our imaginations with any other. But in the case of free products, things are more complicated. I have given primacy to the construction in terms of Stone spaces because I believe that this is the fastest route to effective mental pictures. But in some ways this approach seems to be inappropriate. If you take what in my view is a tenable position, and say that a Boolean algebra is best regarded as the limit of its finite subalgebras, then you might prefer a construction of a free product as a limit of free products of finitely many finite subalgebras. Or you might feel that it is wrong to rely on the axiom of choice to prove a result which certainly does not need it (see 315Ye). Because I believe that the universal mapping theorem 315I is the right basis for the study of free products, I am naturally led to use it as the starting point for proofs of theorems about free products, as in 315K. But 315J(e-ii) seems to lie deeper. (Note, for instance, that in 315L we do need the axiom of choice, in part Q (c) of the proof, since without it the product i∈I Xi could be empty.) Both ‘simple product’ and ‘free product’ are essentially algebraic constructions involving the category of Boolean algebras and Boolean homomorphisms, and any relationships with such concepts as order-continuity must be regarded as accidental. 315Cb and 315D show that simple products behave very straightforwardly when the homomorphisms involved are order-continuous. 315P, 315Xm and 315Ya-315Yc show that free products are much more complex and subtle. For finite products, we have a kind of distributivity; (A × B) ⊗ C can be identified with (A ⊗ C) × (B ⊗ C) (315Xi, 315Xj). There are contexts in which this makes it seem more natural to write A ⊕ B in place of A × B, and indeed I have already spoken of a ‘direct sum’ of measure spaces (214K) in terms which correspond closely to the simple product of algebras of sets described in 315Ga. Generally, the simple product corresponds to disjoint unions of Stone spaces (315Xc) and the free product to products of Stone spaces. But the simple product is indeed the product Boolean algebra, in the ordinary category sense; the universal mapping theorem 315B is exactly of the type we expect from products of topological spaces (3A3Ib) or partially ordered sets (315Dc), etc. It is the ‘free product’ which is special to Boolean algebras. The nearest analogy that I know of elsewhere is with the concept of ‘tensor product’ of linear spaces (cf. §253).

316 Further topics I introduce two special properties of Boolean algebras which will be of great importance in the rest of this volume: the countable chain condition (316A-316F) and weak (σ, ∞)-distributivity (316G-316K). I end the section with brief notes on atoms in Boolean algebras (316L-316M). 316A Definitions (a) A Boolean algebra A is ccc, or satisfies the countable chain condition, if every disjoint subset of A is countable. (b) A topological space X is ccc, or satisfies the countable chain condition, or has Souslin’s property, if every disjoint collection of open sets in X is countable.

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316B Theorem A Boolean algebra A is ccc iff its Stone space Z is ccc. proof (a) If A is ccc and G is a disjoint family of open sets in Z, then for each G ∈ G 0 = G \ {∅} we can find a non-zero aG ∈ A such that the corresponding open-and-closed set b aG is included in G. Now {aG : G ∈ G 0 } is a disjoint family in A, so is countable; since aG 6= aH for distinct G, H ∈ G 0 , G 0 and G must be countable. As G is arbitrary, Z is ccc. (b) If Z is ccc and A ⊆ A is disjoint, then {b a : a ∈ A} is a disjoint family of open subsets of Z, so must be countable, and A is countable. As A is arbitrary, A is ccc. 316C Proposition Let A be a Dedekind σ-complete Boolean algebra and I a σ-ideal of A. Then the quotient algebra B = A/I is ccc iff every disjoint family in A \ I is countable. proof (a) Suppose that B is ccc and that A is a disjoint family in A \ I. Then {a• : a ∈ A} is a disjoint family in B, therefore countable, and a• 6= b• when a, b are distinct members of A; so A is countable. (b) Now suppose that B is not ccc. Then there is an uncountable disjoint set B ⊆ B. Of course B \ {0} is still uncountable, so may be enumerated as hbξ iξ (d) Let X be a separable topological space. Show that X is ccc. > (e) Show that the regular open algebra of a topological space X is ccc iff X is ccc, so that, in particular, the regular open algebra of R is ccc. (f ) Show that if A is a Boolean algebra and B is an order-dense subalgebra of A, then A is ccc iff B is. (g) Let A be a Dedekind σ-complete Boolean algebra. Show that it is ccc iff there is no family haξ iξ (m) Show that if A is a weakly (σ, ∞)-distributive Boolean algebra and B is a subalgebra of A which is regularly embedded in A, then B is weakly (σ, ∞)-distributive. (n) Show that if A is a weakly (σ, ∞)-distributive Boolean algebra and I is an order-closed ideal of A, then A/I is weakly (σ, ∞)-distributive. > (o) (i) Show that if A is a Boolean algebra and B is an order-dense subalgebra of A, then A is weakly (σ, ∞)-distributive iff B is. (ii) Let X be a zero-dimensional compact Hausdorff space. Show that the regular open algebra of X is weakly (σ, ∞)-distributive iff the algebra of open-and-closed subsets of X is. (p) Let A be a Boolean algebra. Show that it is weakly (σ, ∞)-distributive iff whenever hCn in∈N is a sequence of partitions of unity in A, there is a partition C of unity such that for every c ∈ C, n ∈ N there is a finite set I ⊆ Cn such that c ⊆ sup I. > (q) Show that the algebra of open-and-closed subsets of {0, 1}N , with its usual topology, is not weakly (σ, ∞)-distributive. (r) Let A be a Boolean algebra and B an order-dense subalgebra of A. Show that A and B have the same atoms, so that A is atomless, or purely atomic, iff B is. (s) Let A be a Boolean algebra and B a regularly embedded subalgebra of A. Show that (i) every atom of A is included in an atom of B (ii) if A is purely atomic, so is B (iii) if B is atomless, so is A. >(t) Let A be a Dedekind complete purely atomic Boolean algebra. Show that it is isomorphic to PA, where A is the set of atoms of A. (u) Let A be a Boolean algebra and I an order-closed ideal of A. Show that (i) if A is atomless, so is A/I (ii) if A is purely atomic, so is A/I. (v) Let A be a Boolean algebra. Show that (i) if A is atomless, so is every principal ideal of A (ii) if A is purely atomic, so is every principal ideal of A. (w) Let hAi ii∈I be a family of Boolean algebras with simple product A. Show that (i) A is purely atomic iff every Ai is (ii) A is atomless iff every Ai is. > (x) Show that any purely atomic Boolean algebra is weakly (σ, ∞)-distributive. (y) Let A be a Boolean algebra. Show that there is a one-to-one correspondence between atoms a of A and order-continuous Boolean homomorphisms φ : A → Z2 , defined by saying that φ corresponds to a iff φ(a) = 1. 316Y Further exercises (a) Let I be any set. Show that {0, 1}I , with its usual topology, is ccc. (Hint: show that if E ⊆ {0, 1}I is a non-empty open-and-closed set, then µE > 0, where µ is the usual measure on {0, 1}I .) (b) Show that the Stone space of the regular open algebra of R is separable. More generally, show that if a topological space X is separable so is the Stone space of its regular open algebra. (c) Let A be a Boolean algebra and Z its Stone space. Show that A is ccc iff every nowhere dense subset of Z is included in a nowhere dense zero set.

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Further topics

65

(d) Let X be a zero-dimensional topological space. Show that X is ccc iff the regular open algebra of X is ccc iff the algebra of open-and-closed subsets of X is ccc. (e) Set X = {0, 1}ω1 , and for ξ < ω1 set Eξ = {x : x ∈ X, x(ξ) = 1}. Let Σ be the algebra of subsets of X generated by {Eξ : ξ < ω1 } ∪ {{x} : x ∈ X}, and I the σ-ideal of Σ generated by {Eξ ∩ Eη : ξ < η < ω1 } ∪ {{x} : x ∈ X}. Show that Σ/I is not ccc, but that there is no uncountable disjoint family in Σ \ I. (f ) Let X be a regular topological space and G its regular open algebra. Show that G is weakly (σ, ∞)distributive iff every meager set in X is nowhere dense. (g) Let A be a Boolean algebra. A is weakly σ-distributive if whenever hamn im,n∈N is a double sequence in A such that hamn in∈N is non-increasing and has infimum 0 for every m ∈ N, then inf{a : ∀ m ∈ N ∃ n ∈ N, amn ⊆ a} = 0. (Dedekind complete weakly σ-distributive algebras are also called ω ω -bounding.) A has the Egorov property if whenever hamn im,n∈N is a double sequence in A such that hamn in∈N is non-increasing and has infimum 0 for every m ∈ N, then there is a non-increasing sequence ham im∈N such that inf m∈N am = 0 and for every m ∈ N there is an n ∈ N such that am ⊇ amn . (i) Show that if A has the Egorov property it is weakly σ-distributive. (ii) Show that if A is weakly (σ, ∞)-distributive it is weakly σ-distributive. (iii) Show that if A is ccc then it is weakly (σ, ∞)-distributive iff it has the Egorov property iff it is weakly σ-distributive. (iv) Show that P(NN ) does not have the Egorov property, even though it is weakly (σ, ∞)distributive. (Hint: try amn = {f : f (m) ≥ n}.) (h) Let A be a Boolean algebra and Z its Stone space. (i) Show that A is weakly σ-distributive iff the union of any sequence of nowhere dense zero sets in Z is nowhere dense. (ii) Show that A has the Egorov property iff the union of any sequence of nowhere dense zero sets in Z is included in a nowhere dense zero set. (i) Let A be a Dedekind σ-complete weakly (σ, ∞)-distributive Boolean algebra, Z its Stone space, E the algebra of open-and-closed subsets of Z, M the σ-ideal of meager subsets of Z, and Σ the Baire property algebra {E4M : E ∈ E, M ∈ M}, as in 314M. (i) Suppose that f : Z → R is a Σ-measurable function. Show that there is a dense open set G ⊆ Z such that f ¹G is continuous. (ii) Now suppose that A is Dedekind complete. Show that if f : Z → R is a function such that f ¹G is continuous for some dense open set G ⊆ Z, then f is Σ-measurable; and that if f is also bounded, there is a continuous function g : Z → R such that {z : f (z) 6= g(z)} is meager. (Hint: the graph of g will be the closure of the graph of f ¹G; because Z is extremally disconnected, this is the graph of a function.) (j) (i) Let X be a non-empty separable Hausdorff space without isolated points. Show that its regular open algebra is not weakly (σ, ∞)-distributive. (ii) Let (X, ρ) be a non-empty metric space without isolated points. Show that its regular open algebra is not weakly (σ, ∞)-distributive. (iii) Let I be any infinite set. Show that the algebra of open-and-closed subsets of {0, 1}I is not weakly (σ, ∞)-distributive. Show that the regular open algebra of {0, 1}I is not weakly (σ, ∞)-distributive. (k) For any set X, write CX = {I : I ⊆ X is finite} ∪ {X \ I : I ⊆ X is finite}. (i) Show that CX is an algebra of subsets of X (the finite-cofinite algebra). (ii) Show that a Boolean algebra is purely atomic iff it has an order-dense subalgebra isomorphic to the finite-cofinite algebra of some set. (iii) Show that a Dedekind σ-complete Boolean algebra is purely atomic iff it has an order-dense subalgebra isomorphic to the countable-cocountable algebra of some set (211R). (l) Let hAi ii∈I be a family of Boolean algebras, none of them {0}, with free product A. (i) Show that A is purely atomic iff every Ai is purely atomic and {i : Ai 6= {0, 1}} is finite. (ii) Show that A is atomless iff either some Ai is atomless or {i : Ai 6= {0, 1}} is infinite. (m) Show that a Boolean algebra is isomorphic to the algebra of open-and-closed subsets of {0, 1}N iff it is countable, atomless and not {0}.

66

Boolean algebras

316Yn

(n) Show that a Boolean algebra is isomorphic to the regular open algebra of R iff it is atomless, Dedekind complete, has a countable order-dense subalgebra and is not {0}. (o) Let G be the regular open algebra of R. Show that there is an injective Boolean homomorphism π : G → PN. (Hint: the Stone space of G is separable.) Show that there is a Boolean homomorphism φ : PN → G such that φπ is the identity on PN. (Hint: 314K.) (p) Write [N] 0. This means that ν satisfies the conditions of 213O and must be strictly localizable. (d)(iv)⇒(ii) This is just 211Ld. b be the Dedekind completion 322O Proposition Let (A, µ ¯) be a semi-finite measure algebra, and let A b b µ of A (314U). Then there is a unique extension of µ ¯ to a functional µ ˜ on A such that (A, ˜) is a localizable b b µ measure algebra. The embedding A ⊆ A identifies the ideals {a : a ∈ A, µ ¯a < ∞} and {a : a ∈ A, ˜a < ∞}. b For c ∈ A, b set proof (I write the argument out as if A were actually a subalgebra of A.) µ ˜c = sup{¯ µa : a ∈ A, a ⊆ c}. b to [0, ∞] extending µ b µ Evidently µ ˜ is a function from A ¯, so µ ˜0 = 0. Because A is order-dense in A, ˜c > 0 b whenever c 6= 0, because any P such c includes a non-zero member of A. If hcn in∈N is a disjoint sequence in A ∞ ˜cn . P P Let A be the set of all members of A expressible as a = supn∈N an with supremum c, then µ ˜c = n=0 µ where an ∈ A, an ⊆ cn for every n ∈ N. Now ∞ X

sup µ ¯a = sup{ a∈A

µ ¯an : an ∈ A, an ⊆ cn for every n ∈ N}

n=0

=

∞ X n=0

sup{¯ µan : an ⊆ cn } =

∞ X

µ ˜cn .

n=0

b cn = sup{a : a ∈ A, a ⊆ cn } for each n, and sup A, taken in A, b must be Also, because A is order-dense in A, 0 0 0 0 b c. But this means that if a ∈ A, a ⊆ c then a = supa∈A a ∩ a in A and therefore also in A; so that µ ¯a0 = supa∈A µ ¯(a0 ∩ a) ≤ supa∈A µ ¯a. Accordingly

322Q

Taxonomy of measure algebras

µ ˜c = supa∈A µ ¯a =

P∞ n=0

79

µ ˜cn . Q Q

b µ This shows that (A, ˜) is a measure algebra. It is semi-finite because (A, µ ¯) is and every non-zero element b includes a non-zero element of A, which in turn includes a non-zero element of finite measure. Since A b of A b µ is Dedekind complete, (A, ¯) is localizable. If µ ¯a is finite, then surely µ ˜a = µ ¯a is finite. If µ ˜c is finite, then {A : a ∈ A, a ⊆ c} is upwards-directed b and supa∈A µ ¯A = µ ˜c is finite, so b = sup A is defined in A and µ ¯b = µ ˜c. Because A is order-dense in A, b = c (313K, 313O) and c ∈ A, with µ ¯c = µ ˜c. b µ 322P Definition Let (A, µ ¯) be any semi-finite measure algebra. I will call (A, ˜), as constructed above, the localization of (A, µ ¯). Of course it is unique just in so far as the Dedekind completion of A is. 322Q Further properties of Stone spaces: Proposition Let (A, µ ¯) be a semi-finite measure algebra and (Z, Σ, ν) its Stone space. (a) Meager sets in Z are nowhere dense; every E ∈ Σ is uniquely expressible as G4M where G ⊆ Z is open-and-closed and M is nowhere dense, and νE = sup{νH : H ⊆ E is open-and-closed}. (b) The c.l.d. version ν˜ of ν is strictly localizable, and has the same negligible sets as ν. (c) If (A, µ ¯) is totally finite then νE = inf{νH : H ⊇ E is open-and-closed} for every E ∈ Σ. proof (a) I have already remarked (in the proof of 322N) that A is weakly (σ, ∞)-distributive, so that meager sets in Z are nowhere dense. But we know that every member of Σ is expressible as G4M where G is open-and-closed and M is meager, therefore nowhere dense. Moreover, the expression is unique, because if G4M = G0 4M 0 then G4G0 ⊆ M ∪ M 0 is open and nowhere dense, therefore empty, so G = G0 and M = M 0. Now let a ∈ A be such that b a = G, and consider B = {b : b ∈ A, bb ⊆ E}. Then sup B = a in A. P P If b b ∈ B, then b \ b a ⊆ M is nowhere dense, therefore empty; so a is an upper bound for B. ?? If a is not the supremum of B, then there is a non-zero c ⊆ a such that b ⊆ a \ c for every b ∈ B. But now b c cannot be empty, so b c \ M is non-empty, and there is a non-zero d ∈ A such that db ⊆ b c \ M . In this case d ∈ B and d⊆ 6 a \ c. X X Thus a = sup B. Q Q It follows that νE = νG = µ ¯a = sup µ ¯b b∈B

= sup νbb ≤ sup{νH : H ⊆ E is open-and-closed} ≤ νE b∈B

and νE = sup{νH : H ⊆ E is open-and-closed}. (b) This is the same as part (c) of the proof of 322N. We have a disjoint family C of sets of finite measure for ν such that whenever E ∈ Σ, νE > 0 there is a C ∈ C such that µ(C ∩ E) > 0. Now if ν˜F is defined and not 0, there is an E ∈ Σ such that E ⊆ F and νE > 0 (213Fc), so that there is a C ∈ C such that ν(E ∩ C) > 0; since νC < ∞, we have ν˜(F ∩ C) ≥ ν˜(E ∩ C) = ν(E ∩ C) > 0. And of course ν˜C < ∞ for every C ∈ C. This means that C witnesses that ν˜ satisfies the conditions of 213O, so that ν˜ is strictly localizable. Any ν-negligible set is surely ν˜-negligible. If M is ν˜-negligible then it is nowhere dense. P P If G ⊆ Z is open and not empty then there is a non-empty open-and-closed set H1 ⊆ G, and now H1 ∈ Σ, so there is a non-empty open-and-closed set H ⊆ H1 such that νH is finite (because ν is semi-finite). In this case H ∩ M is ν-negligible, therefore nowhere dense, and H 6⊆ M . But this means that G 6⊆ M ; as G is arbitrary, M is nowhere dense. Q Q Accordingly M ∈ M and is ν-negligible. Thus ν and ν˜ have the same negligible sets. (c) Because νZ < ∞,

80

Measure algebras

322Q

νE = νZ − ν(Z \ E) = νZ − sup{νH : H ⊆ Z \ E is open-and-closed} = inf{ν(Z \ H) : H ⊆ Z \ E is open-and-closed} = inf{νH : H ⊇ E is open-and-closed}. 322X Basic exercises > (a) Let (A, µ ¯) be a measure algebra. Let I∞ be the set of those a ∈ A which are ‘purely infinite’, that is, µ ¯a = ∞ and µ ¯b = ∞ for every non-zero b ⊆ a. Show that I∞ is a σ-ideal of A. Show that there is a function µ ¯sf : A/I∞ → [0, ∞] defined by setting µ ¯sf a• = sup{¯ µb : b ⊆ a, µ ¯b < ∞} for every a ∈ A. Show that (A/I∞ , µ ¯sf ) is a semi-finite measure algebra. (b) Let (X, Σ, µ) be a measure space and let µsf be the ‘semi-finite version’ of µ, as defined in 213Xc. Let (A, µ ¯) be the measure algebra of (X, Σ, µ). Show that the measure algebra of (X, Σ, µsf ) is isomorphic to the measure algebra (A/I∞ , µ ¯sf ) of (a) above. ˜ µ (c) Let (X, Σ, µ) be a measure space and (X, Σ, ˜) its c.l.d. version. Let (A, µ ¯) and (A2 , µ ¯2 ) be the corresponding measure algebras, and π : A → A2 the canonical homomorphism, as in 322Db. Show that the kernel of π is the ideal I∞ , as described in 322Xa, so that A/I∞ is isomorphic, as Boolean algebra, to π[A] ⊆ A2 . Show that this isomorphism identifies µ ¯sf , as described in 322Xa, with µ ¯2 ¹π[A]. (d) Give a direct proof of 322G, not relying on 215B and 321J. > (e) Let (A, µ ¯) be any measure algebra, A a non-empty subset of A, and c ∈ A such that µ ¯c < ∞. Show that (i) c0 = sup{a ∩ c : a ∈ A} is defined in A (ii) there is a countable set B ⊆ A such that c0 = sup{a ∩ c : a ∈ B}. (f ) Let (X, Σ, µ) be a measure space and ν an indefinite-integral measure over µ (§234). Show that the measure algebra of ν can be identified, as Boolean algebra, with a principal ideal of the measure algebra of µ. (g) Let (X, Σ, µ) be a measure space and A any subset of X; let µA be the subspace measure on A and ΣA its domain. Write (A, µ ¯) for the measure algebra of (X, Σ, µ) and (AA , µ ¯A ) for the measure algebra of (A, ΣA , µA ). Show that the formula F • 7→ (F ∩ A)• defines a sequentially order-continuous Boolean homomorphism π : A → AA which has kernel I = {F • : F ∈ Σ, F ∩ A = ∅}. Show that for any a ∈ A, µ ¯A (πa) = min{¯ µb : b ∈ A, a \ b ∈ I}. (h) Let (A, µ ¯) be a measure algebra and B an order-closed subalgebra of A. Suppose that (B, µ ¯¹ B) is semi-finite. Show that (A, µ ¯) is semi-finite. (i) Let (A, µ ¯) be any measure algebra and (Z, Σ, ν) its Stone space. Show that the c.l.d. version of ν is strictly localizable. 322Y Further exercises (a) Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of Σ. Set ˆ = {E4N : E ∈ Σ, N ∈ N }. N = {N : ∃ F ∈ I, N ⊆ F }. Show that N is a σ-ideal of subsets of X. Set Σ ˆ is a σ-algebra of subsets of X and that Σ/N ˆ Show that Σ is isomorphic to Σ/I. (b) Let (A, µ ¯) be a semi-finite measure algebra, and (Z, Σ, ν) its Stone space. Let ν˜ be the c.l.d. version ˜ its domain. Show that Σ ˜ is precisely the Baire property algebra {G4A : G ⊆ Z is open, A ⊆ Z of ν, and Σ ˜ is meager}, so that Σ/M can be identified with the regular open algebra of Z (314Yd) and the measure algebra of ν˜ can be identified with the localization of A. (c) Give an example of a localizable measure algebra (A, µ ¯) with a σ-subalgebra B such that (B, µ ¯¹ B) is semi-finite and atomless, but A is not atomless. (d) Let (X, Σ, µ) be a measure space and A ⊆ X a subset; let µA be the subspace measure on A, A and AA the measure algebras of µ and µA , and π : A → AA the canonical homomorphism, as described in 322Xg. (i) Show that if µA is semi-finite, then π is order-continuous. (ii) Show that if µ is semi-finite but µA is not, then π is not order-continuous.

322 Notes

Taxonomy of measure algebras

81

(e) Show that if (A, µ ¯) is a measure algebra, with Stone space (Z, Σ, ν), then ν has locally determined negligible sets in the sense of 213I. (f ) Let (A, µ ¯) be a localizable measure algebra and (Z, Σ, ν) its Stone space. (i) Show that a function f : Z → R is Σ-measurable iff there is a conegligible set G ⊆ X such that f ¹G is continuous. (Hint: 316Yi.) (ii) Show that f : Z → [0, 1] is Σ-measurable iff there is a continuous function g : Z → [0, 1] such that f = g ν-a.e. 322 Notes and comments I have taken this leisurely tour through the concepts of Chapter 21 partly to recall them (or persuade you to look them up) and partly to give you practice in the elementary manipulations of measure algebras. The really vital result here is the correspondence between ‘localizability’ in measure spaces and measure algebras. Part of the object of this volume (particularly in Chapter 36) is to try to make sense of the properties of localizable measure spaces, as discussed in Chapter 24 and elsewhere, in terms of their measure algebras. I hope that 322Be has already persuaded you that the concept really belongs to measure algebras, and that the formulation in terms of ‘essential suprema’ is a dispensable expedient. I have given proofs of 322C and 322G depending on the realization of an arbitrary measure algebra as the measure algebra of a measure space, and the corresponding theorems for measure spaces, because this seems the natural approach from where we presently stand; but I am sympathetic to the view that such proofs must be inappropriate, and that it is in some sense better style to look for arguments which speak only of measure algebras (322Xd). For any measure algebra (A, µ ¯), the set Af of elements of finite measure is an ideal of A; consequently it is order-dense iff it includes a partition of unity (322E). In 322F we have something deeper: any semifinite measure algebra must be weakly (σ, ∞)-distributive when regarded as a Boolean algebra, and this has significant consequences in its Stone space, which are used in the proofs of 322N and 322Q. Of course a result of this kind must depend on the semi-finiteness of the measure algebra, since any Dedekind σ-complete Boolean algebra becomes a measure algebra if we give every non-zero element the measure ∞. It is natural to look for algebraic conditions on a Boolean algebra sufficient to make it ‘measurable’, in the sense that it should carry a semi-finite measure; this is an unresolved problem to which I will return in Chapter 39. Subspace measures, simple products, direct sums, principal ideals and order-closed subalgebras give no real surprises; I spell out the details in 322I-322M and 322Xg-322Xh. It is worth noting that completing a measure space has no effect on its measure algebra (322D, 322Ya). We see also that from the point of view of measure algebras there is no distinction to be made between ‘localizable’ and ‘strictly localizable’, since every localizable measure algebra is representable as the measure algebra of a strictly localizable measure space (322Kd). (But strict localizability does have implications for some processes starting in the measure algebra; see 322L.) It is nevertheless remarkable that the canonical measure on the Stone space of a semifinite measure algebra is localizable iff it is strictly localizable (322N). This canonical measure has many other interesting properties, which I skim over in 322Q, 322Xi, 322Yb and 322Yf. In Chapter 21 I discussed a number of methods of improving measure spaces, notably ‘completions’ (212C) and ‘c.l.d. versions’ (213E). Neither of these is applicable in any general way to measure algebras. But in fact we have a more effective construction, at least for semi-finite measure algebras, that of ‘localization’ (322O-322P); I say that it is more effective just because localizability is more important than completeness or local determinedness, being of vital importance in the behaviour of function spaces (241Gb, 243Gb, 245Ec, 363M, 364O, 365J, 367N, 369A, 369C). Note that the localization of a semi-finite measure algebra does in fact correspond to the c.l.d. b do not have the same Stone spaces, version of a certain measure (322Q, 322Yb). But of course A and A b can be effectively represented as the measure algebra of a measure on the Stone space of A. even when A b not just What is happening in 322Yb is that we are using all the open sets of Z to represent members of A, the open-and-closed sets, which correspond to members of A.

82

Measure algebras

§323 intro.

323 The topology of a measure algebra I take a short section to discuss one of the fundamental tools for studying totally finite measure algebras, the natural metric that each carries. The same ideas, suitably adapted, can be applied to an arbitrary measure algebra, where we have a topology corresponding closely to the topology of convergence in measure on the function space L0 . Most of the section consists of an analysis of the relations between this topology and the order structure of the measure algebra. 323A The pseudometrics ρa (a) Let (A, µ ¯) be a measure algebra. Write Af = {a : a ∈ A, µ ¯a < ∞}. f For a ∈ A and b, c ∈ A, write ρa (b, c) = µ ¯(a ∩ (b 4 c)). Then ρa is a pseudometric on A. P P (i) Because µ ¯a < ∞, ρa takes values in [0, ∞[. (ii) If b, c, d ∈ A then b 4 d ⊆ (b 4 c) ∪ (c 4 d), so ρa (b, d) = µ ¯(a ∩ (b 4 d)) ≤ µ ¯((a ∩ (b 4 c)) ∪ (a ∩ (c 4 d))) ≤µ ¯(a ∩ (b 4 c)) + µ ¯(a ∩ (c 4 d)) = ρa (b, c) + ρa (c, d). (iii) If b, c ∈ A then ρa (b, c) = µ ¯(a ∩ (b 4 c)) = µ ¯(a ∩ (c 4 b)) = ρa (c, b). Q Q (b) Now the measure-algebra topology of the measure algebra (A, µ ¯) is that generated by the family P = {ρa : a ∈ Af } of pseudometrics on A. Similarly the measure-algebra uniformity on A is that generated by P. (For a general discussion of topologies defined by pseudometrics, see 2A3F et seq. For the associated uniformities see §3A4.) (c) Note that P is upwards-directed, since ρa∪a0 ≥ max(ρa , ρa0 ) for all a, a0 ∈ Af . (d) When (A, µ ¯) is totally finite, it is more natural to work from the measure metric ρ = ρ1 , with ρ(a, b) = µ ¯(a 4 b), since this by itself defines the measure-algebra topology and uniformity. *(e) Even when (A, µ ¯) is not totally finite, we still have a metric ρ on Af defined by setting ρ(a, b) = f µ ¯(a 4 b) for all a, b ∈ A , which is sometimes useful (323Xg). Note however that the topology on Af defined from ρ is not in general the topology induced by the measure-algebra topology of A. 323B Proposition Let (A, µ ¯) be any measure algebra. Then the operations ∪ , ∩ , \ and 4 are all uniformly continuous. proof The point is that for any b, c, b0 , c0 ∈ A we have (b ∗ c) 4 (b0 ∗ c0 ) ⊆ (b 4 b0 ) ∪ (c 4 c0 ) for any of the operations ∗ = ∪ , ∩ etc.; so that if a ∈ Af then ρa (b ∗ c, b0 ∗ c0 ) ≤ ρa (b, b0 ) + ρa (c, c0 ). Consequently the operation ∗ must be uniformly continuous. 323C Proposition (a) Let (A, µ ¯) be a totally finite measure algebra. Then µ ¯ : A → [0, ∞[ is uniformly continuous. (b) Let (A, µ ¯) be a semi-finite measure algebra. Then µ ¯ : A → [0, ∞] is lower semi-continuous. (c) Let (A, µ ¯) be any measure algebra. If a ∈ A and µ ¯a < ∞, then b 7→ µ ¯(b ∩ a) : A → R is uniformly continuous. proof (a) For any a, b ∈ A, |¯ µa − µ ¯b| ≤ µ ¯(a 4 b) = ρ1 (a, b). (b) Suppose that b ∈ A and µ ¯b > α ∈ R. Then there is an a ⊆ b such that α < µ ¯a < ∞ (322Eb). If c ∈ A is such that ρa (b, c) < µ ¯a − α, then

323D

The topology of a measure algebra

83

µ ¯c ≥ µ ¯(a ∩ c) = µ ¯a − µ ¯(a ∩ (b \ c)) > α. Thus {b : µ ¯b > α} is open; as α is arbitrary, µ ¯ is lower semi-continuous. (c) |¯ µ(a 4 b) − µ ¯(a 4 c)| ≤ ρa (b, c) for all b, c ∈ A. 323D The following facts are basic to any understanding of the relationship between the order structure and topology of a measure algebra. Lemma Let (A, µ ¯) be a measure algebra. (a) Let B ⊆ A be a non-empty upwards-directed set. For b ∈ B set Fb = {c : b ⊆ c ∈ B}. (i) {Fb : b ∈ B} generates a Cauchy filter F(B ↑) on A. (ii) If sup B is defined in A, then it is a topological limit of F(B ↑); in particular, it belongs to the topological closure of B. (b) Let B ⊆ A be a non-empty downwards-directed set. For b ∈ B set Fb = {c : b ⊇ c ∈ B}. (i) {Fb : b ∈ B} generates a Cauchy filter F(B↓) on A. (ii) If inf B is defined in A, then it is a topological limit of F(B ↓); in particular, it belongs to the topological closure of B. (c)(i) Closed subsets of A are order-closed in the sense of 313D. (ii) An order-dense subalgebra of A must be dense in the topological sense. (d) Now suppose that (A, µ ¯) is semi-finite. (i) The sets {b : b ⊆ c}, {b : b ⊇ c} are closed for every c ∈ A. (ii) If B ⊆ A is non-empty and upwards-directed and e is a cluster point of F(B ↑), then e = sup B. (iii) If B ⊆ A is non-empty and downwards-directed and e is a cluster point of F(B↓), then e = inf B. proof I use the notations Af , ρa from 323A. (a)(i) (α) If b, c ∈ B then there is a d ∈ B such that b ∪ c ⊆ d, so that Fd ⊆ Fb ∩ Fc ; consequently F(B ↑) = {F : F ⊆ A, ∃ b ∈ B, Fb ⊆ F } is a filter on A. (β) Let a ∈ Af , ² > 0. Then there is a b ∈ B such that µ ¯(a ∩ c) ≤ µ ¯(a ∩ b) + 12 ² for every 0 0 0 c ∈ B, and Fb ∈ F(B ↑). If now c, c ∈ Fb , c 4 c ⊆ (c \ b) ∪ (c \ b), so ρa (c, c0 ) ≤ µ ¯(a ∩ c \ b) + µ ¯(a ∩ c0 \ b) = µ ¯(a ∩ c) + µ ¯(a ∩ c0 ) − 2¯ µ(a ∩ b) ≤ ². As a and ² are arbitrary, F(B ↑) is Cauchy. (ii) Suppose that e = sup B is defined in A. Let a ∈ Af , ² > 0. By 313Ba, a ∩ e = supb∈B a ∩ b; but {a ∩ b : b ∈ B} is upwards-directed, so µ ¯(a ∩ e) = supb∈B µ ¯(a ∩ b), by 321D. Let b ∈ B be such that µ ¯(a ∩ b0 ) ≥ µ ¯(a ∩ e) − ². Then for any c ∈ Fb , e 4 c ⊆ e \ b, so ρa (e, c) = µ ¯(a ∩ (e 4 c)) ≤ µ ¯(a ∩ (e \ b)) = µ ¯(a ∩ e) − µ ¯(a ∩ b) ≤ ². As a and ² are arbitrary, F(B ↑) → e. Because B ∈ F(B ↑), e surely belongs to the topological closure of B. (b) Either repeat the arguments above, with appropriate inversions, using 321F in place of 321D, or apply (a) to the set {1 \ b : b ∈ B}. (c)(i) This follows at once from (a) and (b) and the definition in 313D. (ii) If B ⊆ A is an order-dense subalgebra and a ∈ A, then B = {b : b ∈ B, b ⊆ a} is upwards-directed and has supremum a (313K); by (a-ii), a ∈ B ⊆ B. As a is arbitrary, B is topologically dense. (d)(i) Set F = {b : b ⊆ c}. If d ∈ A \ F , then (because (A, µ ¯) is semi-finite) there is an a ∈ Af such that δ=µ ¯(a ∩ d \ c) > 0; now if b ∈ F , ρa (d, b) ≥ µ ¯(a ∩ d \ b) ≥ δ, so that d cannot belong to the closure of F . As d is arbitrary, F is closed. Similarly, {b : b ⊇ c} is closed. (ii) (α) If b ∈ B, then e ∈ Fb , because Fb ∈ F(B ↑); but {c : b ⊆ c ∈ A} is a closed set including Fb , so contains e, and b ⊆ e. As b is arbitrary, e is an upper bound for B. (β) If d is an upper bound of B,

84

Measure algebras

323D

then {c : c ⊆ d} is a closed set belonging to F(B ↑), so contains e. As d is arbitrary, this shows that e is the supremum of B, as claimed. (iii) Use the same arguments as in (ii), but inverted. 323E Corollary Let (A, µ ¯) be a measure algebra. (a) If hbn in∈N is a non-decreasing sequence in A with supremum b, then hbn in∈N → b for the measurealgebra topology. (b) If hbn in∈N is a non-increasing sequence in A with infimum b, then hbn in∈N → b for the measure-algebra topology. proof I call this a ‘corollary’ because it is the special case of 323Da-323Db in which B is the set of terms of a monotonic sequence; but it is probably easier to work directly from the definition in 323A, and use 321Be or 321Bf to see that limn→∞ ρa (bn , b) = 0 whenever µ ¯a < ∞. 323F

The following is a useful calculation.

P∞ Lemma Let (A, µ ¯) be a measure algebra and hcn in∈N a sequence in A such that the sum n=0 µ ¯(cn 4 cn+1 ) is finite. Set d0 = supn∈N inf m≥n cm , d1 = inf n∈N supm≥n cm . Then d0 = d1 and, writing d for their common value, limn→∞ µ ¯(cn 4 d) = 0. P∞ proof Write αn = µ ¯(cn 4 cn+1 ), βn = k=n αk for n ∈ N; we are supposing that limn→∞ βn = 0. Set bn = supm≥n cm 4 cm+1 ; then P∞ µ ¯bn ≤ m=n µ ¯(cm 4 cm+1 ) = βn for each n. If m ≥ n, then cm 4 cn ⊆ supn≤k<m ck 4 ck+1 ⊆ bn , so cn \ bn ⊆ cm ⊆ cn ∪ bn . Consequently cn \ bn ⊆ inf k≥m ck ⊆ supk≥m ck ⊆ cn ∪ bn for every m ≥ n, and cn \ bn ⊆ d0 ⊆ d1 ⊆ cn ∪ bn , so that cn 4 d0 ⊆ bn ,

cn 4 d1 ⊆ bn ,

d1 \ d0 ⊆ bn .

As this is true for every n, limn→∞ µ ¯(cn 4 di ) ≤ limn→∞ µ ¯bn = 0 for both i, and µ ¯(d1 4 d0 ) ≤ inf n∈N µ ¯bn = 0, so that d1 = d0 . 323G The classification of measure algebras: Theorem Let (A, µ ¯) be a measure algebra, T its measure-algebra topology and U its measure-algebra uniformity. (a) (A, µ ¯) is semi-finite iff T is Hausdorff. (b) (A, µ ¯) is σ-finite iff T is metrizable, and in this case U is also metrizable. (c) (A, µ ¯) is localizable iff T is Hausdorff and A is complete under U. proof I use the notations Af , ρa from 323A. (a)(i) Suppose that (A, µ ¯) is semi-finite and that b, c are distinct members of A. Then there is an a ⊆ b 4 c such that 0 < µ ¯a < ∞, and now ρa (b, c) > 0. As b and c are arbitrary, T is Hausdorff (2A3L).

323G

The topology of a measure algebra

85

(ii) Suppose that T is Hausdorff and that b ∈ A has µ ¯b = ∞. Then b 6= 0 so there must be an a ∈ Af such that µ ¯(a ∩ b) = ρa (0, b) > 0; in which case a ∩ b ⊆ b and 0 < µ ¯(a ∩ b) < ∞. As b is arbitrary, µ ¯ is semi-finite. (b)(i) Suppose that µ ¯ is σ-finite. Let han in∈N be a non-decreasing sequence in Af with supremum 1. Set ρ(b, c) =

∞ X ρan (b, c) 1 + 2n µ ¯an n=0

for b, c ∈ A. Then ρ is a metric on A, because if ρ(b, c) = 0 then an ∩ (b 4 c) = 0 for every n, so b 4 c = 0 and b = c. If a ∈ Af and ² > 0, take n such that µ ¯(a \ an ) ≤ 12 ². If b, c ∈ A and ρ(b, c) ≤ ²/2(1 + 2n µ ¯an ), then 1 ρan (b, c) ≤ 2 ² so ρa (b, c) = ρa\an (b, c) + ρa∩an (b, c) ≤ µ ¯(a \ an ) + ρan (b, c) 1 2

≤ ² + (1 + 2n µ ¯an )ρ(b, c) ≤ ². In the other direction, given ² > 0, take n ∈ N such that 2−n ≤ 21 ²; then ρ(b, c) ≤ ² whenever ρan (b, c) ≤ ²/2(n + 1). This shows that U is the same as the metrizable uniformity defined by {ρ}; accordingly T is also defined by ρ. (ii) Now suppose that T is metrizable, and let ρ be a metric defining T. For each n ∈ N there must be an0 , . . . , ankn ∈ Af and δn > 0 such that ρani (b, 1) ≤ δn for every i ≤ kn =⇒ ρ(b, 1) ≤ 2−n . Set d = supn∈N,i≤kn ani . Then ρani (d, 1) = 0 for every n, i, so ρ(d, 1) ≤ 2−n for every n and d = 1. Thus 1 is the supremum of countably many elements of finite measure and (A, µ ¯) is σ-finite. (c)(i) Suppose that (A, µ ¯) is localizable. Then T is Hausdorff, by (a). Let F be a Cauchy filter on A. For each a ∈ Af , choose a sequence hFn (a)in∈N in F such that ρa (b, c) ≤ 2−n whenever b, c ∈ Fn (a) and n ∈ N. T Choose can ∈ k≤n Fk (a) for each n; then ρa (can , ca,n+1 ) ≤ 2−n for each n. Set da = supn∈N inf k≥n a ∩ cak . Then limn→∞ ρa (da , can ) = limn→∞ µ ¯(da 4 (a ∩ can )) = 0, by 323F. If a, b ∈ Af and a ⊆ b, then da = a ∩ db . P P For each n ∈ N, Fn (a) and Fn (b) both belong to F, so must have a point e in common; now ρa (da , db ) ≤ ρa (da , can ) + ρa (can , e) + ρa (e, cbn ) + ρa (cbn , db ) ≤ ρa (da , can ) + ρa (can , e) + ρb (e, cbn ) + ρb (cbn , db ) ≤ ρa (da , can ) + 2−n + 2−n + ρb (cbn , db ) → 0 as n → ∞. Consequently ρa (da , db ) = 0, that is, da = a ∩ da = a ∩ db . Q Q Set d = sup{db : b ∈ Af }; this is defined because A is Dedekind complete. Then F → d. P P If a ∈ Af and ² > 0, then a ∩ d = supb∈Af a ∩ db = supb∈Af a ∩ b ∩ da∪b = supb∈Af a ∩ b ∩ da = a ∩ da . So if we choose n ∈ N such that 2−n + ρa (can , da ) ≤ ², then for any e ∈ Fn (a) we shall have ρa (e, d) ≤ ρa (e, can ) + ρa (can , d) ≤ 2−n + ρa (can , da ) ≤ ². Thus

86

Measure algebras

323G

{e : ρa (d, e) ≤ ²} ⊇ Fn (a) ∈ F. As a, ² are arbitrary, F converges to d. Q Q As F is arbitrary, A is complete. (ii) Now suppose that T is Hausdorff and that A is complete under U. By (a), (A, µ ¯) is semi-finite. Let B be any non-empty subset of A, and set B 0 = {b0 ∪ . . . ∪ bn : b0 , . . . , bn ∈ B}, so that B 0 is upwardsdirected and has the same upper bounds as B. By 323Da, we have a Cauchy filter F(B 0 ↑); because A is complete, this is convergent; and because (A, µ ¯) is semi-finite, its limit must be sup B 0 = sup B, by 323Dd. As B is arbitrary, A is Dedekind complete, so (A, µ ¯) is localizable. 323H Closed subalgebras The ideas used in the proof of (c) above have many other applications, of which one of the most important is the following. You may find it helpful to read the next theorem first on the assumption that (A, µ ¯) is a probability algebra. Theorem Let (A, µ ¯) be a localizable measure algebra, and B a subalgebra of A. Then it is closed for the measure-algebra topology iff it is order-closed. proof (a) If B is closed, it must be order-closed, by 323Dc. (b) Now suppose that B is order-closed. I repeat the ideas of part (c-i) of the proof of 323G. Let e be any member of the closure of B in A. For each a ∈ Af , n ∈ N choose can ∈ B such that ρa (can , e) ≤ 2−n . Then ∞ X

µ ¯((a ∩ can ) 4 (a ∩ ca,n+1 )) =

n=0

≤

∞ X n=0 ∞ X

ρa (can , ca,n+1 ) ρa (can , e) + ρa (e, ca,n+1 ) < ∞.

n=0

So if we set ea = supn∈N inf k≥n cak , then ρa (ea , can ) = ρa (a ∩ ea , a ∩ can ) → 0 as n → ∞, by 323F, and ρa (e, ea ) = 0, that is, a ∩ ea = a ∩ e. Also, because B is order-closed, inf k≥n cak ∈ B for every n, and ea ∈ B. Because A is Dedekind complete, we can set e0a = inf{eb : b ∈ Af , a ⊆ b}; then e0a ∈ B and e0a ∩ a = inf b⊇a eb ∩ a = inf b⊇a eb ∩ b ∩ a = inf b⊇a e ∩ b ∩ a = e ∩ a. Now e0a ⊆ e0b whenever a ⊆ b, so B = {e0a : a ∈ Af } is upwards-directed, and sup B = sup{e0a ∩ a : a ∈ Af } = sup{e ∩ a : a ∈ Af } = e because (A, µ ¯) is semi-finite. Accordingly e ∈ B. As e is arbitrary, B is closed, as claimed. 323I Notation In the context of 323H, I will say simply that B is a closed subalgebra of A. 323J Proposition If (A, µ ¯) is a localizable measure algebra and B is a subalgebra of A, then the topological closure B of B in A is precisely the order-closed subalgebra of A generated by B. proof Write Bτ for the smallest order-closed subset of A including B. By 313Fc, Bτ is a subalgebra of A, and is the order-closed subalgebra of A generated by B. Being an order-closed subalgebra of A, it is topologically closed, by 323H, and must include B. On the other hand, B, being topologically closed, is order-closed (323D(c-i)), so includes Bτ . Thus B = Bτ is the order-closed subalgebra of A generated by B. 323K

I note some simple results for future reference.

Lemma If (A, µ ¯) is a localizable measure algebra and B is a closed subalgebra of A, then for any a ∈ A the subalgebra C of A generated by B ∪ {a} is closed. proof By 314Ja, C is order-closed.

323Y

The topology of a measure algebra

87

323L Proposition Let h(Ai , µ ¯i )ii∈I be aQfamily of measure algebras with simple product (A, µ ¯) (322K). Then the measure-algebra topology on A = i∈I Ai defined by µ ¯ is just the product of the measure-algebra topologies of the Ai . proof I use the notations Af , ρa from 323A. Write T for the measure-algebra topology of A and S for the product topology. For i ∈ I, d ∈ Afi define a pseudometric ρ˜di on A by setting ρ˜di (b, c) = ρd (b(i), c(i)) whenever b, c ∈ A; then S is defined by P = {˜ ρdi : i ∈ I, a ∈ Afi } (3A3Ig). Now each ρ˜di is one of the defining pseudometrics for T, since ρ˜di (b, c) = µ ¯(d˜ ∩ (b4c)) ˜ = d, d(j) ˜ = 0 for j 6= i. So S ⊆ T. where d(i) P Now suppose that a ∈ Af and ² > 0. Then i∈I µ ¯i a(i) = µ ¯a is finite, so there is a finite set J ⊆ I such P that i∈I\J µ ¯i a(i) ≤ 12 ². For each j ∈ J, τj = ρ˜a(j),j belongs to P, and ρa (b, c) =

X

µ ¯i (a(i) ∩ (b(i) 4 c(i))) ≤

i∈I

=

X j∈J

X j∈J

τj (b(j), c(j)) +

1 ² 2

1 2

µ ¯j (a(j) ∩ (b(j) 4 c(j))) + ²

≤²

whenever b, c are such that τj (b(j), c(j)) ≤ ²/(1 + 2#(J)) for every j ∈ J. By 2A3H, the identity map from (A, S) to (A, T) is continuous, that is, T ⊆ S. Putting these together, we see that S = T, as claimed. 323X Basic exercises (a) Let (A, µ ¯) be a semi-finite measure algebra. Show that the set {(a, b) : a ⊆ b} is a closed set in A × A. > (b) Let (X, Σ, µ) be a σ-finite measure space and (A, µ ¯) its measure algebra. (i) Show that if T is a σ-subalgebra of Σ, then {F • : F ∈ T} is a closed subalgebra of A. (ii) Show that if B is a closed subalgebra of A, then {F : F ∈ Σ, F • ∈ B} is a σ-subalgebra of Σ. (c) Let (A, µ ¯) be a localizable measure algebra, and C ⊆ A a set such that sup A, inf A belong to C for all non-empty subsets A of C. Show that C is closed for the measure-algebra topology. (d) (i) Show that if (A, µ ¯) is any measure algebra and B is a subalgebra of A, then its topological closure B is again a subalgebra. (ii) Use this fact instead of 313Fc to prove 323J. (e) Let (A, µ ¯) be a measure algebra, and e ∈ A; let Ae be the principal ideal of A generated by e, and µ ¯e its measure (322H). Show that the topology on Ae defined by µ ¯e is just the subspace topology induced by the measure-algebra topology of A. > (f ) Let (X, Σ, µ) be a measure space, and (A, µ ¯) its measure algebra. (i) Show that we have an injection χ : A → L0 (µ) (see §241) given by setting χ(E • ) = (χE)• for every E ∈ Σ. (ii) Show that χ is a homeomorphism between A and its image if A is given its measure-algebra topology and L0 (µ) is given its topology of convergence in measure (245A). (g) Let (A, µ ¯) be a measure algebra, and Af the ideal of elements of finite measure. For a, b ∈ Af set ρ(a, b) = µ ¯(a 4 b). Show that (Af , µ ¯) is a complete metric space and that the operations ∪ , ∩ , \ and 4 are uniformly continuous on Af , while µ ¯ : Af → R is also uniformly continuous. Show that the f embedding A ⊆ A is continuous for the measure-algebra topology on A. In the context of 323Xf, show that χ : Af → L0 (µ) is an isometry between Af and a subset of L1 (µ). 323Y Further exercises (a) Let (A, µ ¯) be a σ-finite measure algebra. Show that a set F ⊆ A is closed for the measure-algebra topology iff e ∈ F whenever there are non-empty sets B, C ⊆ A such that B is upwards-directed, C is downwards-directed, sup B = inf C = e and [b, c] ∩ F 6= ∅ for every b ∈ B, c ∈ C, writing [b, c] = {d : b ⊆ d ⊆ c}.

88

Measure algebras

323Yb

(b) Give an example to show that (a) is false for general localizable measure algebras. (c) Give an example of a semi-finite measure algebra (A, µ ¯) with an order-closed subalgebra which is not closed for the measure-algebra topology. (d) Let (A, µ ¯) be a probability algebra and write B for the family of closed subalgebras of A. For B, ¯(b 4 c) + supc∈C inf b∈B µ ¯(b 4 c). Show that (B, ρ) is a complete metric C ∈ B set ρ(B, C) = supb∈B inf c∈C µ space. (e) Let (A, µ ¯) be the measure algebra of Lebesgue measure on R. Show that it is separable in its measurealgebra topology. (Hint: 245Yj.) 323Z Problem Find a localizable measure algebra (A, µ ¯), a σ-subalgebra B of A and a sequence hbn in∈N in B which converges, for the measure-algebra topology, to a member of A \ B. 323 Notes and comments The message of this section is that the topology of a measure algebra is essentially defined by its order and algebraic structure; see also 324F-324H below. Of course the results are really about semi-finite measure algebras, and indeed this whole volume, like the rest of measure theory, has little of interest to say about others; they are included only because they arise occasionally and it is not absolutely essential to exclude them. We therefore expect to be able to describe such things as closed subalgebras and continuous homomorphisms in terms of the ordering, as in 323H and 324G. For σ-finite algebras, indeed, there is an easy description of the topology in terms of the order (323Ya). I think the result of this section which I shall most often wish to quote is 323I: in most contexts, there is no need to distinguish between ‘closed subalgebra’ and ‘order-closed subalgebra’. I conjecture, however, that a σ-subalgebra of a localizable measure algebra need not be topologically sequentially closed (323Z). It is also the case that the topology of a measure algebra corresponds very closely indeed to the topology of convergence in measure. A description of this correspondence is in 323Xf. Indeed all the results of this section have analogues in the theory of topological Riesz spaces. I will enlarge on the idea here in §367. For the moment, however, if you look back to Chapter 24, you will see that 323B and 323G are closely paralleled by 245D and 245E, while 323Ya is related to 245L. It is I think natural to ask whether there are any other topological Boolean algebras with the properties 323B-323D. In fact a question in this direction, the Control Measure Problem, is one of the most important questions outstanding in abstract measure theory. I will discuss it in §393; the particular form relevant to the present section is what I call ‘CM4 ’ (393J).

324 Homomorphisms In the course of Volume 2, I had occasion to remark that elementary measure theory was unusual among abstract topics in pure mathematics in not being dominated by any particular class of structure-preserving operators. We now come to what I think is one of the reasons for the gap: the most important operators of the theory are not between measure spaces at all, but between their measure algebras. In this section I run through the most elementary facts about Boolean homomorphisms between measure algebras. I start with results on the construction of such homomorphisms from functions between measure spaces (324A324E), then investigate continuity and order-continuity of homomorphisms (324F-324H) before turning to measure-preserving homomorphisms (324I-324O). 324A Theorem Let (X, Σ, µ) and (Y, T, ν) be measure spaces, and (A, µ ¯), (B, ν¯) their measure algebras. ˆ for the domain of the completion µ Write Σ ˆ of µ. Let D ⊆ X be a set of full outer measure (definition: ˆ D be the subspace σ-algebra on D induced by Σ. ˆ Let φ : D → Y be a function such that 132E), and let Σ −1 −1 ˆ D for every F ∈ T and φ [F ] is µ-negligible whenever νF = 0. Then there is a sequentially φ [F ] ∈ Σ order-continuous Boolean homomorphism π : B → A defined by the formula πF • = E • whenever F ∈ T, E ∈ Σ and (E ∩ D)4φ−1 [F ] is negligible.

324E

Homomorphisms

89

ˆ such that H ∩ D = φ−1 [F ]; now there is an E ∈ Σ such proof Let F ∈ T. Then there is an H ∈ Σ −1 that E4H is negligible, so that (E ∩ D)4φ [F ] is negligible. If E1 is another member of Σ such that (E1 ∩ D)4φ−1 [F ] is negligible, then (E4E1 ) ∩ D is negligible, so is included in a negligible member G of Σ. Since (E4E1 ) \ G belongs to Σ and is disjoint from D, it is negligible; accordingly E4E1 is negligible and E • = E1• in A. What this means is that the formula offered defines a map π : B → A. It is now easy to check that π is a Boolean homomorphism, because if (E ∩ D)4φ−1 [F ],

(E 0 ∩ D)4φ−1 [F 0 ]

are negligible, so are ((X \ E) ∩ D)4φ−1 [Y \ F ],

((E ∪ E 0 ) ∩ D)4φ−1 [F ∪ F 0 ].

To see that π is sequentially order-continuous, let hbn in∈N be a sequence in B. For each n we may • −1 choose S an Fn ∈ T such S that Fn = bn , and En ∈ Σ such that (En ∩ D)4φ [Fn ] is negligible; now, setting F = n∈N Fn , E = n∈N En , S (E ∩ D)4φ−1 [F ] ⊆ n∈N (En ∩ D)4φ−1 [Fn ] is negligible, so π(supn∈N bn ) = π(F • ) = E • = supn∈N En• = supn∈N πbn . (Recall that the maps E 7→ E • , F 7→ F • are sequentially order-continuous, by 321H.) So π is sequentially order-continuous (313L(c-iii)). 324B Corollary Let (X, Σ, µ) and (Y, T, ν) be measure spaces, and (A, µ ¯), (B, ν¯) their measure algebras. Let φ : X → Y be a function such that φ−1 [F ] ∈ Σ for every F ∈ T and µφ−1 [F ] = 0 whenever νF = 0. Then there is a sequentially order-continuous Boolean homomorphism π : B → A defined by the formula πF • = (φ−1 [F ])• for every F ∈ T. 324C Remarks (a) In §235 and elsewhere in Volume 2 I spent a good deal of time on functions between measure spaces which satisfy the conditions of 324A. Indeed, I take the trouble to spell 324A out in such generality just in order to catch these applications. Some of the results of the present chapter (322D, 322Jb) can also be regarded as special cases of 324A. (b) The question of which homomorphisms between the measure algebras of measure spaces (X, Σ, µ), (Y, T, ν) can be realized by functions between X and Y is important and deep; I will return to it in §§343-344. (c) In the simplified context of 324B, I have actually defined a contravariant functor; the relevant facts are the following. 324D Proposition Let (X, Σ, µ), (Y, T, ν) and (Z, Λ, λ) be measure spaces, with measure algebras ¯ Suppose that φ : X → Y and ψ : Y → Z satisfy the conditions of 324B, that is, (A, µ ¯), (B, ν¯), (C, λ). φ−1 [F ] ∈ Σ if F ∈ T,

µφ−1 [F ] = 0 if νF = 0,

ψ −1 [G] ∈ T if G ∈ Λ,

µψ −1 [G] = 0 if λG = 0.

Let πφ : B → A, πψ : C → B be the corresponding homomorphisms. Then ψφ : X → Z is another map of the same type, and πψφ = πφ πψ : C → A. proof The necessary checks are all elementary. 324E Stone spaces While in the context of general measure spaces the question of realizing homomorphisms is difficult, in the case of the Stone representation it is relatively straightforward. Proposition Let (A, µ ¯) and (B, ν¯) be measure algebras, with Stone spaces Z and W ; let µ, ν be the corresponding measures on Z and W , as described in 321J-321K, and Σ, T their domains. If π : B → A is

90

Measure algebras

324E

any order-continuous Boolean homomorphism, let φ : Z → W be the corresponding continuous function, as described in 312P. Then φ−1 [F ] ∈ Σ for every F ∈ T, µφ−1 [F ] = 0 whenever νF = 0, and (writing E ∗ for the member of A corresponding to E ∈ Σ) πF ∗ = (φ−1 [F ])∗ for every F ∈ T. proof Recall that E ∗ = a iff E4b a is meager, where b a is the open-and-closed subset of Z corresponding to a ∈ A. In particular, µE = 0 iff E is meager. Now the point is that φ−1 [F ] is nowhere dense in Z whenever F is a nowhere dense subset of W , by 313R. Consequently φ−1 [F ] is meager whenever F is meager in W , since F is then just a countable union of nowhere dense sets. Thus we see already that µφ−1 [F ] = 0 whenever νF = 0. If F is any member of T, there is an open-and-closed set F0 such that F 4F0 is meager; now φ−1 [F0 ] is open-and-closed, so φ−1 [F ] = φ−1 [F0 ]4φ−1 [F 4F0 ] belongs to Σ. Moreover, if b ∈ B is such that bb = F0 , and a = πb, then b a = φ−1 [F0 ], so πF ∗ = πb = a = (φ−1 [F0 ])∗ = (φ−1 [F ])∗ , as required. 324F

I turn now to the behaviour of order-continuous homomorphisms between measure algebras.

Theorem Let (A, µ ¯) and B, ν¯) be measure algebras and π : A → B a Boolean homomorphism. (a) π is continuous iff it is continuous at 0 iff it is uniformly continuous. (b) If (B, ν¯) is semi-finite and π is continuous, then it is order-continuous. (c) If (A, µ ¯) is semi-finite and π is order-continuous, then it is continuous. proof I use the notations Af , ρa from 323A. (a) Suppose that π is continuous at 0; I seek to show that it is uniformly continuous. Take b ∈ Bf and ² > 0. Then there are a0 , . . . , an ∈ Af and δ > 0 such that ν¯(b ∩ πc) = ρb (πc, 0) ≤ ² whenever maxi≤n ρai (c, 0) ≤ δ; setting a = supi≤n ai , ν¯(b ∩ πc) ≤ ² whenever µ ¯(a ∩ c) ≤ δ. 0

Now suppose that ρa (c, c ) ≤ δ. Then µ ¯(a ∩ (c 4 c0 )) ≤ δ, so ρb (πc, πc0 ) = ν¯(b ∩ (πc 4 πc0 )) = ν¯(b ∩ π(c 4 c0 )) ≤ ². As b, ² are arbitrary, π is uniformly continuous. The rest of the implications are elementary. (b) Let A be a non-empty downwards-directed set in A with infimum 0. Then 0 ∈ A (323D(b-ii)); because π is continuous, 0 ∈ π[A]. ?? If b is a non-zero lower bound for π[A] in B, then (because (B, ν¯) is semi-finite) there is a c ⊆ b with 0 < ν¯c < ∞; now ρc (πa, 0) = ν¯(c ∩ πa) = ν¯c > 0 for every a ∈ A, so 0 ∈ / π[A]. X X Thus inf π[A] = 0 in B; as A is arbitrary, π is order-continuous (313L(b-ii)). (c) By (a), it will be enough to show that π is continuous at 0. Let b ∈ Bf , ² > 0. ?? Suppose, if possible, that for every a ∈ Af , δ > 0 there is a c ∈ A such that µ ¯(a ∩ c) ≤ δ but ν¯(b ∩ πc) ≥ ². For each a ∈ Af , −n n ∈ N choose can such that µ ¯(a ∩ can ) ≤ 2 but ν¯(b ∩ πcan ) ≥ ². Set ca = inf n∈N supm≥n cam ; then P∞ µ ¯(a ∩ ca ) ≤ inf n∈N m=n µ ¯(a ∩ can ) = 0, so ca ∩ a = 0. On the other hand, because π is order-continuous, πca = inf n∈N supm≥n πcam , so that ν¯(b ∩ πca ) = limn→∞ ν¯(b ∩ supm≥n πcam ) ≥ ². This shows that ρb (1 \ a, 0) = ν¯(b ∩ π(1 \ a)) ≥ ν¯(b ∩ πca ) ≥ ². But now observe that A = {1 \ a : a ∈ Af } is a downwards-directed subset of A with infimum 0, because (A, µ ¯) is semi-finite. So π[A] is downwards-directed and has infimum 0, and 0 must be in the closure of π[A], by 323D(b-ii); while we have just seen that ρb (d, 0) ≥ ² for every d ∈ π[A]. X X

324K

Homomorphisms

91

Thus there must be a ∈ Af , δ > 0 such that ρb (πc, 0) = ν¯(b ∩ πc) ≤ ² whenever ρa (c, 0) = µ ¯(a ∩ c) ≤ δ. As b, ² are arbitrary, π is continuous at 0 and therefore continuous. 324G Corollary If (A, µ ¯) and (B, ν¯) are semi-finite measure algebras, a Boolean homomorphism π : A → B is continuous iff it is order-continuous. 324H Corollary If A is a Boolean algebra and µ ¯, ν¯ are two measures both rendering A a semi-finite measure algebra, then they endow A with the same uniformity (and, of course, the same topology). proof By 324G, the identity map from A to itself is continuous whichever of the topologies we place on A; and by 324F it is therefore uniformly continuous. 324I Definition Let (A, µ ¯) and (B, ν¯) be measure algebras. A Boolean homomorphism π : A → B is measure-preserving if ν¯(πa) = µ ¯a for every a ∈ A. ¯ be measure algebras, and π : A → B, θ : B → C 324J Proposition Let (A, µ ¯), (B, ν¯) and (C, λ) measure-preserving Boolean homomorphisms. Then θπ : A → C is a measure-preserving Boolean homomorphism. proof Elementary. 324K Proposition Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a measure-preserving Boolean homomorphism. (a) π is injective. (b) (A, µ ¯) is totally finite iff (B, ν¯) is, and in this case π is order-continuous, therefore continuous, and π[A] is a closed subalgebra of B. (c) If (A, µ ¯) is semi-finite and (B, ν¯) is σ-finite, then (A, µ ¯) is σ-finite. (d) If (A, µ ¯) is σ-finite and π is sequentially order-continuous, then (B, ν¯) is σ-finite. (e) If (A, µ ¯) is semi-finite and π is order-continuous, then (B, ν¯) is semi-finite. (f) If (A, µ ¯) is atomless and semi-finite, and π is order-continuous, then B is atomless. (g) If B is purely atomic and (A, µ ¯) is semi-finite, then A is purely atomic. proof (a) If a 6= 0 in A, then ν¯πa = µ ¯a > 0 so πa 6= 0. By 3A2Db, π is injective. (b) Because ν¯1B = ν¯π1A = µ ¯ 1A , (A, µ ¯) is totally finite iff (B, ν¯) is. Now suppose that A ⊆ A is downwards-directed and non-empty and that inf A = 0. Then inf a∈A ν¯πa = inf a∈A µ ¯a = 0 by 321F. So ν¯b = 0 for any lower bound b of π[A], and inf π[A] = 0. As A is arbitrary, π is order-continuous. By 324Fc, π is continuous. By 314Fa, π[A] is order-closed in B, that is, ‘closed’ in the sense of 323I. (c) I appeal to 322G. If C is a disjoint family in A \ {0}, then hπcic∈C is a disjoint family in B \ {0}, so is countable, and C must be countable, because π is injective. Thus A is ccc and (being semi-finite) (A, µ ¯) is σ-finite. (d) Let han in∈N be a sequence in A such that µ ¯an < ∞ for every n and supn∈N an = 1. Then ν¯πan < ∞ for every n and (because π is sequentially order-continuous) supn∈N πan = 1, so (B, ν¯) is σ-finite. (e) Setting Af = {a : µ ¯a < ∞}, sup Af = 1; because π is order-continuous, sup π[Af ] = 1 in B. So if ν¯b = ∞, there is an a ∈ Af such that πa ∩ b 6= 0, and now 0 < ν¯(b ∩ πa) < ∞.

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(f ) Take any non-zero b ∈ B. As in (e), there is an a ∈ A such that µ ¯a < ∞ and a ∩ b 6= 0. If a ∩ b 6= b, then surely b is not an atom. Otherwise, set C = {c : c ∈ A, c ⊆ a, b ⊆ πc}. Then C is downwards-directed and contains a, so c0 = inf C is defined in A (321F), and µ ¯c0 = inf c∈C µ ¯c ≥ ν¯b > 0, so c0 6= 0. Because A is atomless, there is a d ⊆ c0 such that neither d nor c0 \ d is zero, so that neither c0 \ d nor d can belong to C. But this means that b ∩ πd and b ∩ π(c0 \ d) are both non-zero, so that again b is not an atom. As b is arbitrary, B is atomless. (g) Take any non-zero a ∈ A. Then there is an a0 ⊆ a such that 0 < µ ¯a0 < ∞. Because B is purely 0 atomic, there is an atom b of B with b ⊆ πa . Set C = {c : c ∈ A, c ⊆ a0 , b ⊆ πc}. Then C is downwards-directed and contains a0 , so c0 = inf C is defined in A, and µ ¯c0 = inf c∈C µ ¯c ≥ ν¯b > 0, so c0 6= 0. If d ⊆ c0 , then b ∩ πd must be either b or 0. If b ∩ πd = b, then d ∈ C and d = c0 . If b ∩ πd = 0, then c0 \ d ∈ C and d = 0. Thus c0 is an atom in A. As a is arbitrary, A is purely atomic. 324L Corollary Let (A, µ ¯) be a totally finite measure algebra, (B, ν¯) a measure algebra, and π : A → B a measure-preserving homomorphism. If C ⊆ A and C is the closed subalgebra of A generated by C, then π[C] is the closed subalgebra of B generated by π[C]. proof This is a special case of 314Gb. 324M Proposition Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with measure algebras (A, µ ¯) and (B, ν¯). Let φ : X → Y be inverse-measure-preserving. Then we have a sequentially order-continuous measure-preserving Boolean homomorphism π : B → A defined by setting πF • = φ−1 [F ]• for every F ∈ T. proof This is immediate from 324B. 324N Proposition Let (A, µ ¯) and (B, ν¯) be measure algebras, with Stone spaces Z and W ; let µ, ν be the corresponding measures on Z and W . If π : B → A is an order-continuous measure-preserving Boolean homomorphism, and φ : Z → W the corresponding continuous function, then φ is inverse-measurepreserving. proof Use 324E. In the notation there, if F ∈ T, then νF = ν¯F ∗ = µ ¯πF ∗ = µ ¯φ−1 [F ]∗ = µφ−1 [F ]. 324O Proposition Let (A, µ ¯) and (B, ν¯) be totally finite measure algebras, A0 a topologically dense subalgebra of A, and π : A0 → B a Boolean homomorphism such that ν¯πa = µ ¯a for every a ∈ A0 . Then π has a unique extension to a measure-preserving homomorphism from A to B. proof Let ρ, σ be the standard metrics on A, B, as in 323Ad. Then for any a, a0 ∈ A0 σ(πa, πa0 ) = ν¯(πa4πa0 ) = ν¯π(a4a0 ) = µ ¯(a4a0 ) = ρ(a, a0 ); that is, π : A0 → B is an isometry. Because A0 is dense in the metric space (A, ρ), while B is complete under σ (323Gc), there is a unique continuous function π ˆ : A → B extending π (3A4G). Now the operations (a, a0 ) 7→ π ˆ (a ∪ a0 ),

(a, a0 ) 7→ π ˆa ∪ π ˆ a0 : A × A → B,

are continuous and agree on the dense subset A0 × A0 of A × A; because the topology of B is Hausdorff, they agree on A × A, that is, π ˆ (a ∪ a0 ) = π ˆa ∪ π ˆ a0 for all a, a0 ∈ A (2A3Uc). Similarly, the operations a 7→ π ˆ (1 \ a),

a 7→ 1 \ π ˆa : A → B

324X

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93

are continuous and agree on the dense subset A0 of A, so they agree on A, that is, π ˆ (1 \ a) = 1 \ a for every a ∈ A. Thus π ˆ is a Boolean homomorphism. To see that it is measure-preserving, observe that a 7→ µ ¯a = ρ(a, 0),

a 7→ ν¯(ˆ π a) = σ(ˆ π a, 0) : A → R

are continuous and agree on A0 , so agree on A. Finally, π ˆ is the only measure-preserving Boolean homomorphism extending π, because any such map must be continuous (324Kb), and π ˆ is the only continuous extension of π. *324P The following fact will be useful in §386, by which time it will seem perfectly elementary; for the moment, it may be a useful exercise. Proposition Let (A, µ ¯) and (B, ν¯) be totally finite measure algebras such that µ ¯1 = ν¯1. Suppose that A ⊆ A and φ : A → B are such that ν¯(inf i≤n φai ) = µ ¯(inf i≤n ai ) for all a0 , . . . , an ∈ A. Let C be the smallest closed subalgebra of A including A. Then φ has a unique extension to a measure-preserving Boolean homomorphism from C to B. proof (a) Let Ψ be the family of all functions ψ extending φ and having the same properties; that is, ψ is a function from a subset of A to B, and ν¯(inf i≤n ψai ) = µ ¯(inf i≤n ai ) for all a0 , . . . , an ∈ dom ψ. By Zorn’s Lemma, Ψ has a maximal member θ. Write D for the domain of θ. (b)(i) If c, d ∈ D then c ∩ d ∈ D. P P?? Otherwise, set D0 = D ∪ {c ∩ d} and extend θ to θ0 : D0 → B by 0 writing θ (c ∩ d) = θc ∩ θd. It is easy to check that θ0 ∈ Ψ, which is supposed to be impossible. X XQ Q Now ν¯(θc ∩ θd ∩ θ(c ∩ d)) = µ ¯(c ∩ d) = ν¯(θc ∩ θd) = ν¯θ(c ∩ d), so θ(c ∩ d) = θc ∩ θd. (ii) If d ∈ D then 1 \ d ∈ D. P P?? Otherwise, set D0 = D ∪ {1 \ d} and extend θ to D0 by writing θ (1 \ d) = 1 \ θd. Once again, it is easy to check that θ0 ∈ Ψ, which is impossible. X XQ Q Consequently (since D is certainly not empty, even if C is), D is a subalgebra of A (312B(iii)). 0

(iii) Since ν¯θ1 = µ ¯1 = ν¯1, θ1 = 1. If d ∈ D then ν¯θ(1 \ d) = µ ¯(1 \ d) = µ ¯1 − µ ¯d = ν¯1 − ν¯θd = ν¯(1 \ θd), while ν¯(θd ∩ θ(1 \ d)) = µ ¯(d ∩ (1 \ d)) = 0, so θd ∩ θ(1 \ d)) = 0, θ(1 \ d) ⊆ 1 \ θd and θ(1 \ d) must be equal to 1 \ θd. By 312H, θ : D → B is a Boolean homomorphism. (iv) Let D be the topological closure of D in A. Then it is an order-closed subalgebra of A (323J), so, with µ ¯¹ D, is a totally finite measure algebra in which D is a topologically dense subalgebra. By 324O, there is an extension of θ to a measure-preserving Boolean homomorphism from D to B; of course this extension belongs to Ψ, so in fact D = D is a closed subalgebra of A. (c) Since A ⊆ D, C ⊆ D and φ1 = θ¹ C is a suitable extension of φ. To see that φ1 is unique, let φ2 : C → B be any other measure-preserving Boolean homomorphism extending φ. Set C = {a : φ1 a = φ2 a}; then C is a topologically closed subalgebra of A including A, so is the whole of C, and φ2 = φ1 . 324X Basic exercises (a) Let A and B be Boolean algebras, of which A is Dedekind σ-complete, and φ : A → B a sequentially order-continuous Boolean homomorphism. Let I be an ideal of A included in the kernel of φ. Show that we have a sequentially order-continuous Boolean homomorphism π : A/I → B given by setting φ(a• ) = φa for every a ∈ A.

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(b) Let (A, µ ¯) be a measure algebra, and B a σ-subalgebra of A. Show that provided that (B, µ ¯¹ B) is semi-finite, then the topology of B induced by µ ¯¹ B is just the subspace topology induced by the topology of A. (Hint: apply 324Fc to the embedding B ⊆ A.) ˜ µ (c) Let (X, Σ, µ) be a measure space and (X, Σ, ˜) its c.l.d. version. Let A, A2 be the corresponding measure algebras and π : A → A2 the canonical homomorphism (see 322Db). Show that π is topologically continuous. (d) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a bijective measure-preserving Boolean homomorphism. Show that π −1 : B → A is a measure-preserving homomorphism. (e) Let µ ¯ be counting measure on PN. Show that (PN, µ ¯) is a σ-finite measure algebra. Find a measurepreserving Boolean homomorphism from PN to itself which is not sequentially order-continuous. 324Y Further exercises (a) Let A and B be Boolean algebras, of which A is Dedekind complete, and φ : A → B an order-continuous Boolean homomorphism. Let I be an ideal of A included in the kernel of φ. Show that we have an order-continuous Boolean homomorphism π : A/I → B given by setting φ(a• ) = φa for every a ∈ A. (b) Let A be a Dedekind σ-complete Boolean algebra, and Z its Stone space. Write E for the algebra of open-and-closed subsets of Z, and Z for the family of nowhere dense zero sets of Z; let Zσ be the σ-ideal of subsets of Z generated by Z. Show that Σ = {E4U : E ∈ E, U ∈ Zσ } is a σ-algebra of subsets of Z, and describe a canonical isomorphism between Σ/Zσ and A. (c) Let A and B be Dedekind σ-complete Boolean algebras, with Stone spaces Z and W . Construct Zσ ⊆ Σ ⊆ PZ as in 324Yb, and let Wσ ⊆ T ⊆ PW be the corresponding structure defined from B. Let π : B → A be a sequentially order-continuous Boolean homomorphism, and φ : Z → W the corresponding continuous map. Show that if E ∗ ∈ A corresponds to E ∈ Σ, then πF ∗ = φ−1 [F ]∗ for every F ∈ T. (Hint: 313Yb.) (d) Let A be a Boolean algebra, B a ccc Boolean algebra and π : A → B an injective Boolean homomorphism. Show that A is ccc. (e) Let A be a Dedekind complete Boolean algebra, B a Boolean algebra, and π : A → B an ordercontinuous Boolean homomorphism. Show that for every atom b ∈ B there is an atom a ∈ A such that πa ⊇ b. Hence show that if A is atomless so is B, and that if B is purely atomic and π is injective then A is purely atomic. (f ) Let (A, µ ¯) and (B, ν¯) be localizable measure algebras and A0 an order-dense subalgebra of A. Suppose that π : A0 → B is an order-continuous Boolean homomorphism such that ν¯πa = µ ¯a for every a ∈ A0 . Show that π has a unique extension to a measure-preserving Boolean homomorphism from A to B. (g) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1]. (i) Show that there is an injective order-preserving function f : A → PN. (Hint: take a countable topologically dense subset D of A, and define f : A → P(D × N) by setting f (a) = {(d, q) : µ ¯(a ∩ d) ≥ q}.) (ii) Show that there is an order-preserving function h : PN → A such that h(f (a)) = a for every a ∈ A. (Hint: set h(I) = sup{a : f (a) ⊆ I}.) Compare 316Yo. (h) Let (A, µ ¯) and (B, ν¯) be probability algebras, and f : A → B an isometry for the measure metrics. Show that a 7→ f (a) 4 f (1) is a measure-preserving Boolean homomorphism. 324 Notes and comments If you examine the arguments of this section carefully, you will see that rather little depends on the measures named. Really this material deals with structures (X, Σ, I) where X is a set, Σ is a σ-ideal of subsets of X, and I is a σ-ideal of Σ, corresponding to the family of measurable negligible sets. In this abstract form it is natural to think in terms of sequentially order-continuous homomorphisms, as in 324Yc. I have stated 324E in terms of order-continuous homomorphisms just for a slight gain in

325A

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simplicity. But in fact, when there is a difference, it is likely that order-continuity, rather than sequential order-continuity, will be the more significant condition. Note that when the domain algebra is σ-finite, the two concepts coincide, because it is ccc (316Fd, 322G). Of course I need to refer to measures when looking at such concepts as σ-finite measure algebra or measurepreserving homomorphism, but even here the real ideas involved are such notions as order-continuity and the countable chain condition, as you will see if you work through 324K. It is instructive to look at the translations of these facts into the context of inverse-measure-preserving functions; see 235Xe. 324H shows that we may speak of ‘the’ topology and uniformity of a Dedekind σ-complete Boolean algebra which carries any semi-finite measure; the topology of such an algebra is determined by its algebraic structure. Contrast this with the theory of normed spaces: two Banach spaces (e.g., `1 and `2 ) can be isomorphic as linear spaces, both being of algebraic dimension c, while they are not isomorphic as topological linear spaces. When we come to the theory of ordered linear topological spaces, however, we shall again find ourselves with operators whose algebraic properties guarantee continuity (355C, 367P).

325 Free products and product measures In this section I aim to describe the measure algebras of product measures as defined in Chapter 25. This will involve the concept of ‘free product’ set out in §315. It turns out that we cannot determine the measure algebra of a product measure from the measure algebras of the factors (325B), unless the product measure is localizable; but that there is nevertheless a general construction of ‘localizable measure algebra free product’, applicable to any pair of semi-finite measure algebras (325D), which represents the measure algebra of the product measure in the most important cases (325Eb). In the second part of the section (325I-325M) I deal with measure algebra free products of probability algebras, corresponding to the products of probability spaces treated in §254. 325A Theorem Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with measure algebras (A, µ ¯) and (B, ν¯). ¯ be the corresponding measure Let λ be the c.l.d. product measure on X × Y , and Λ its domain; let (C, λ) algebra. (a)(i) The map E 7→ E × Y : Σ → Λ induces an order-continuous Boolean homomorphism from A to C. (ii) The map F 7→ X × F : T → Λ induces an order-continuous Boolean homomorphism from B to C. (b) The map (E, F ) 7→ E × F : Σ × T → Λ induces a Boolean homomorphism ψ : A ⊗ B → C. (c) ψ[A ⊗ B] is topologically dense in C. (d) For every c ∈ C, ¯ = sup{λ(c ¯ ∩ ψ(a ⊗ b)) : a ∈ A, b ∈ B, µ λc ¯a < ∞, ν¯b < ∞}. ¯ (e) If µ and ν are semi-finite, ψ is injective and λψ(a ⊗ b) = µ ¯a · µ ¯b for every a ∈ A, b ∈ B. proof (a) By 251E, E × Y ∈ Λ for every E ∈ Σ, and λ0 (E × Y ) = 0 whenever µE = 0, where λ0 is the primitive product measure described in 251A-251C; consequently λ(E × Y ) = 0 whenever µE = 0 (251F). Thus E 7→ (E × Y )• : Σ → C is a Boolean homomorphism with kernel including {E : µE = 0}, so descends to a Boolean homomorphism ε1 : A → C. To see that ε1 is order-continuous, let A ⊆ A1 be a non-empty downwards-directed set with infimum 0. ?? If there is a non-zero lower bound c of ε1 [A], express c as W • where W ∈ Λ. We have λ(W ) > 0; by the definition of λ (251F), there are G ∈ Σ, H ∈ T such that µG < ∞, νH < ∞ and λ(W ∩ (G × H)) > 0. Of course inf a∈A a ∩ G• = 0 in A, so inf a∈A µ ¯(a ∩ G• ) = 0, by 321F; let a ∈ A be such that µ ¯(a ∩ G• ) · νH < • λ(W ∩ (G × H)). Express a as E , where E ∈ Σ. Then λ(W \ (E × Y )) = 0. But this means that λ(W ∩ (G × H)) ≤ λ((E ∩ G) × H) = µ(E ∩ G) · νH = µ ¯(a ∩ G• ) · νH, contradicting the choice of a. X X Thus inf ε1 [A] = 0 in C; as A is arbitrary, ε1 is order-continuous. Similarly ε2 : B → C, induced by F 7→ X × F : T → Λ, is order-continuous. (b) Now there must be a corresponding Boolean homomorphism ψ : A ⊗ B → C such that ψ(a ⊗ b) = ε1 a ∩ ε2 b for every a ∈ A, b ∈ B, that is,

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ψ(E • ⊗ F • ) = (E × Y )• ∩ (X × F )• = (E × F )• for every E ∈ Σ, F ∈ T (315I). ¯ < ∞ and ² > 0. Express c, e as U • , W • where U , W ∈ Λ. By 251Ie, there (c) Suppose that c, e ∈ C, λe S are E0 , . . . , En ∈ Σ, F0 , . . . , Fn ∈ T, all of finite measure, such that λ((U ∩ W )4 i≤n Ei × Fi ) ≤ ². Set S c1 = ( i≤n Ei × Fi )• ∈ ψ[A ⊗ B]; then

¯ ∩ (c 4 c1 )) = λ(W ∩ (U 4 S λ(e i≤n Ei × Fi )) ≤ ².

As c, e and ² are arbitrary, ψ[A ⊗ B] is topologically dense in C. (d) By the definition of λ, we have λW = sup{λ(W ∩ (E × F )) : E ∈ Σ, F ∈ T, µE < ∞, νF < ∞} for every W ∈ Λ; so all we have to do is express c as W • . (e) Now suppose that µ and ν are semi-finite. Then λ(E × F ) = µE · νF for any E ∈ Σ, F ∈ T (251J), ¯ so λψ(a ⊗ b) = µ ¯a · ν¯b for every a ∈ A, b ∈ B. To see that ψ is injective, take any non-zero c ∈ A ⊗ B; then there must be non-zero a ∈ A, b ∈ B such that a ⊗ b ⊆ c (315Jb), so that ¯ ≥ λψ(a ¯ λψc ⊗ b) = µ ¯a · ν¯b > 0 and ψc 6= 0. 325B Characterizing the measure algebra of a product space A very natural question to ask is, whether it is possible to define a ‘measure algebra free product’ of two abstract measure algebras in a way which will correspond to one of the constructions above. I give an example to show the difficulties involved. Example There are complete locally determined localizable measure spaces (X, µ), (X 0 , µ0 ), with isomorphic measure algebras, and a probability space (Y, ν) such that the measure algebras of the c.l.d. product measures on X × Y , X 0 × Y are not isomorphic. proof Let (X, Σ, µ) be the complete locally determined localizable not-strictly-localizable measure space described in 216E. Recall that, for E ∈ Σ, µE = #({γ : γ ∈ C, fγ ∈ E}) if this is finite, ∞ otherwise (216Eb), where C is a set of cardinal greater than c. The map E 7→ {γ : fγ ∈ E} : Σ → PC is surjective (216Ec), so descends to an isomorphism between A, the measure algebra of µ, and PC. Let (X 0 , Σ0 , µ0 ) be C with counting measure, so that its measure algebra (A0 , µ ¯0 ) is isomorphic to (A, µ ¯), while µ0 is of course strictly localizable. Let (Y, T, ν) be {0, 1}C with its usual measure. Let λ, λ0 be the c.l.d. product measures on X × Y , X 0 × Y ¯ (C0 , λ ¯ 0 ) the corresponding measure algebras. Then λ is not localizable (254U), so respectively, and (C, λ), ¯ (C, λ) is not localizable (322Be). On the other hand, λ0 , being the c.l.d. product of strictly localizable ¯ 0 ) is localizable, and is not isomorphic measures, is strictly localizable (251N), therefore localizable, so (C0 , λ ¯ to (C, λ). 325C Thus there can be no universally applicable method of identifying the measure algebra of a product measure from the measure algebras of the factors. However, you have no doubt observed that the example above involves non-σ-finite spaces, and conjectured that this is not an accident. In contexts in which we know that all the algebras involved are localizable, there are positive results available, such as the following. Theorem Let (X1 , Σ1 , µ1 ) and (X2 , Σ2 , µ2 ) be semi-finite measure spaces, with measure algebras (A1 , µ ¯1 ) ¯ the corresponding measure and (A2 , µ ¯2 ). Let λ be the c.l.d. product measure on X1 × X2 , and (C, λ) algebra. Let (B, ν¯) be a localizable measure algebra, and φ1 : A1 → B, φ2 : A2 → B order-continuous Boolean homomorphisms such that ν¯(φ1 (a1 ) ∩ φ2 (a2 )) = µ ¯1 a1 · µ ¯2 a2 for all a1 ∈ A1 , a2 ∈ A2 . Then there is a unique order-continuous measure-preserving Boolean homomorphism φ : C → B such that φ(ψ(a1 ⊗ a2 )) = φ1 (a1 ) ∩ φ2 (a2 ) for all a1 ∈ A1 , a2 ∈ A2 , writing ψ : A1 ⊗ A2 → C for the canonical map described in 325A.

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proof (a) Because ψ is injective, it is an isomorphism between A1 ⊗ A2 and its image in C. I trust it will cause no confusion if I abuse notation slightly and treat A1 ⊗ A2 as actually a subalgebra of C. Now the Boolean homomorphisms φ1 , φ2 correspond to a Boolean homomorphism θ : A1 ⊗ A2 → B. The point is ¯ for every c ∈ A ⊗ B. P that ν¯θc = λc P By 315Jb, every member of A1 ⊗ A2 is expressible as supi≤n ai ⊗ a0i , 0 0 where ai ∈ A1 , ai ∈ A2 and hai ⊗ ai ii≤n is disjoint. Now for each i we have ¯ i ⊗ a0 ), ν¯θ(ai ⊗ a0 ) = ν¯(φ1 (ai ) ∩ φ2 (a0 )) = µ ¯1 ai · µ ¯2 a0 = λ(a i

by 325Ad. So ν¯θ(c) =

i

Pn i=0

ν¯θ(ai ⊗ a0i ) =

i

Pn i=0

i

¯ Q ¯ i ⊗ a0 ) = λc. λ(a Q i

(b) The following fact will underlie many of the arguments below. If e ∈ B, ν¯e < ∞ and ² > 0, there are e1 ∈ Af1 , e2 ∈ Af2 such that ν¯(e \ θ(e1 ⊗ e2 )) ≤ ², writing Afi = {a : µ ¯i a < ∞}. P P Because (A1 , µ ¯1 ) is semi-finite, Af1 has supremum 1 in A1 ; because φ1 is order-continuous, sup{φ1 (a) : a ∈ Af1 } = 1 in B, and inf{e \ φ1 (a) : a ∈ Af1 } = 0 (313Aa). Because Af1 is upwards-directed, {e \ φ1 (a) : a ∈ Af1 } is downwards-directed, so inf{¯ ν (e \ φ(a) : a ∈ Af1 } = 0 (321F). Let e1 ∈ Af1 be such that ν¯(e \ φ1 (e1 )) ≤ 12 ². In the same way, there is an e2 ∈ Af2 such that ν¯(e \ φ2 (e2 )) ≤ 12 ². Consider e0 = e1 ⊗ e2 ∈ C. Then ν¯(e \ θe0 ) = ν¯(e \ (φ1 (e1 ) ∩ φ2 (e2 ))) ≤ ν¯(e \ φ1 (e1 )) + ν¯(e \ φ2 (e2 )) ≤ ². Q Q ¯ P (c) The next step is to check that θ is uniformly continuous for the uniformities defined by ν¯, λ. P Take ¯ 1 ⊗ e2 ) < ∞ and ν¯(e \ θ(e1 ⊗ e2 )) ≤ 1 ². Set any e ∈ Bf and ² > 0. By (b), there are e1 , e2 such that λ(e 2 ¯ 4 c0 ) ∩ e0 ) ≤ 1 ². Then e0 = e1 ⊗ e2 . Now suppose that c, c0 ∈ A1 ⊗ A2 and λ((c 2 ¯ 4 c0 ) ∩ e0 ) + 1 ² ≤ ². ν¯((θ(c) 4 θ(c0 )) ∩ e) ≤ ν¯θ((c 4 c0 ) ∩ e0 ) + ν¯(e \ θe0 ) ≤ λ((c 2

By 3A4Cc, θ is uniformly continuous for the subspace uniformity on A1 ⊗ A2 . Q Q (d) Recall that A1 ⊗ A2 is topologically dense in C (325Ab), while B is complete for its uniformity (323Gc). So there is a uniformly continuous function φ : C → B extending θ (3A4G). (e) Because θ is a Boolean homomorphism, so is φ. P P (i) The functions c 7→ φ(1 \ c), c 7→ 1 \ φ(c) are continuous and the topology of B is Hausdorff, so {c : φ(1 \ c) = 1 \ φ(c)} is closed; as it includes A1 ⊗ A2 , it must be the whole of C. (ii) The functions (c, c0 ) 7→ φ(c ∪ c0 ), (c, c0 ) 7→ φ(c) ∪ φ(c0 ) are continuous, so {(c, c0 ) : φ(c ∪ c0 ) = φ(c) ∪ φ(c0 )} is closed in C × C; as it includes (A1 ⊗ A2 ) × (A1 ⊗ A2 ), it must be the whole of C × C. Q Q (f ) Because θ is measure-preserving, so is φ. P P Take any e1 ∈ Af1 , e2 ∈ Af2 . Then the functions ¯ c 7→ λ(c ∩ (e1 ⊗ e2 )), c 7→ ν¯φ(c ∩ (e1 ⊗ e2 )) are continuous and equal on A1 ⊗ A2 , so are equal on C. The argument of (b) shows that for any b ∈ B, ν¯b = sup{¯ ν (b ∩ e) : e ∈ Bf } = sup{¯ ν (b ∩ φ(e1 ⊗ e2 )) : e1 ∈ Af1 , e2 ∈ Af2 }, so that ν¯φ(c) = sup{¯ ν φ(c ∩ (e1 ⊗ e2 )) : e1 ∈ Af1 , e2 ∈ Af2 } ¯ ∩ (e1 ⊗ e2 )) : e1 ∈ Af , e2 ∈ Af } = λc ¯ = sup{λ(c 1

2

for every c ∈ C. Q Q (g) To see that φ is order-continuous, take any non-empty downwards-directed set C ⊆ C with infimum 0. ?? If φ[C] has a non-zero lower bound b in B, let e ⊆ b be such that 0 < ν¯e < ∞. Let e0 ∈ C be such that ¯ 0 < ∞ and ν¯(e \ φ(e0 )) < ν¯e, as in (b) above, so that ν¯(e ∩ φ(e0 )) > 0. Now, because inf C = 0, there is a λe ¯ ∩ e0 ) < ν¯(e ∩ φ(e0 )). But this means that c ∈ C such that λ(c ¯ ∩ e0 ) < ν¯(e ∩ φ(e0 )) ≤ ν¯(b ∩ φ(e0 )), ν¯(b ∩ φ(e0 )) ≤ ν¯φ(c ∩ e0 ) = λ(c which is absurd. X X Thus inf φ[C] = 0 in B. As C is arbitrary, φ is order-continuous.

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325C

(h) Finally, to see that φ is unique, observe that any order-continuous Boolean homomorphism from C to B must be continuous (324Fc); so that if it agrees with φ on A1 ⊗ A2 it must agree with φ on C. 325D Theorem Let (A1 , µ ¯1 ) and (A2 , µ ¯2 ) be semi-finite measure algebras. ¯ together with order-continuous Boolean homomorphisms (a) There is a localizable measure algebra (C, λ), ψ1 : A1 → C, ψ2 : A2 → C such that whenever (B, ν¯) is a localizable measure algebra, and φ1 : A1 → B, φ2 : A2 → B are order-continuous Boolean homomorphisms and ν¯(φ1 (a1 ) ∩ φ2 (a2 )) = µ ¯1 a1 · µ ¯2 a2 for all a1 ∈ A1 , a2 ∈ A2 , then there is a unique order-continuous measure-preserving Boolean homomorphism φ : C → B such that φψj = φj for both j. ¯ ψ1 , ψ2 ) is determined up to isomorphism by this property. (b) The structure (C, λ, (c)(i) The Boolean homomorphism ψ : A1 ⊗ A2 → C defined from ψ1 and ψ2 is injective, and ψ[A1 ⊗ A2 ] is topologically dense in C. (ii) The order-closed subalgebra of C generated by ψ[A1 ⊗ A2 ] is the whole of C. ¯ (d) If j ∈ {1, 2} and (Aj , µ ¯j ) is localizable, then ψj [Aj ] is a closed subalgebra of (C, λ). proof (a)(i) We may regard (A1 , µ ¯1 ) as the measure algebra of (Z1 , Σ1 , µ1 ) where Z1 is the Stone space of A1 , Σ1 is the algebra of subsets of Z1 differing from an open-and-closed set by a meager set, and µ1 is an appropriate measure (321K). Note that in this representation, each a ∈ A1 becomes identified with b a• , where b a is the open-and-closed subset of Z1 corresponding to a. Similarly, we may think of (A2 , µ ¯2 ) as the measure algebra of (Z2 , Σ2 , µ2 ), where Z2 is the Stone space of A2 . (ii) Let λ be the c.l.d. product measure on Z1 × Z2 . The point is that λ is strictly localizable. P P By 322E, both A1 and A2 have partitions of unity consisting of elements of finite measure; let hci ii∈I , hdj ij∈J be such partitions. Then hb ci × dbj ii∈I,j∈J is a disjoint family of sets of finite measure in Z1 × Z2 . If W ⊆ Z1 × Z2 is such that λW > 0, there must be sets E1 , E2 of finite measure such that λ(W ∩ (E1 × E2 )) > 0. Because E1• = supi∈I E1• ∩ ci , we must have P P ci ). ¯1 (E1• ∩ ci ) = i∈I µ1 (E1 ∩ b µ1 E1 = µ ¯1 E1• = i∈I µ P Similarly, µ2 E2 = i∈J µ2 (E2 ∩ dbj ). But this means that there must be finite I 0 ⊆ I, J 0 ⊆ J such that P ci )µ2 (E2 ∩ dbj ) > µ1 E1 · µ2 E2 − λ(W ∩ (E1 × E2 )), i∈I 0 ,j∈J 0 µ1 (E1 ∩ b so that there have to be i ∈ I 0 , j ∈ J 0 such that λ(W ∩ (b ci × dbj )) > 0. Now this means that hb ci × dbj ii∈I,j∈J satisfies the conditions of 213O. Because λ is surely complete and locally determined, it is strictly localizable. Q Q ¯ to be just the measure algebra of λ. The maps ψ1 , ψ2 will be the (iii) We may therefore take (C, λ) canonical maps described in 325Aa, inducing the map ψ : A1 ⊗ B1 → C referred to in 325C; and 325C now gives the result. ¯ 0 , ψ 0 , ψ 0 ) with the same property. (b) This is nearly obvious. Suppose we had an alternative structure (C0 , λ 2 1 Then we must have an order-continuous measure-preserving Boolean homomorphism φ : C → C0 such that φψj = ψj0 for both j; and similarly we have an order-continuous measure-preserving Boolean homomorphism φ0 : C0 → C such that φ0 ψj0 = ψj for both j. Now φ0 φ : C → C is an order-continuous measure-preserving Boolean homomorphism such that φ0 ψj = ψj for both j. By the uniqueness assertion in (a), applied with B = C, φ0 φ must be the identity on C. In the same way, φφ0 is the identity on C0 . So φ and φ0 are the two halves of the required isomorphism. (c) In view of the construction for C offered in part (a) of the proof, (i) is just a consequence of 325Ac and 325Ae. Now (ii) follows by 323J. (d) If Aj is Dedekind complete then ψj [Aj ] is order-closed in C because ψj is order-continuous (314F(a-i)). ¯ together with embeddings ψ1 and ψ2 , 325E Remarks (a) We could say that a measure algebra (C, λ), as described in 325D, is a localizable measure algebra free product of (A1 , µ ¯1 ) and (A2 , µ ¯2 ); and its uniqueness up to isomorphism makes it safe, most of the time, to call it ‘the’ localizable measure algebra free product. Observe that it can equally well be regarded as the uniform space completion of the algebraic free product; see 325Yb.

*325H

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99

(b) As the example in 325B shows, the localizable measure algebra free product of the measure algebras of given measure spaces need not appear directly as the measure algebra of their product. But there is one context in which it must so appear: if the product measure is localizable, 325C tells us at once that it has the right measure algebra. For σ-finite measure algebras, of course, any corresponding measure spaces have to be strictly localizable, so again we can use the product measure directly. 325F I ought not to proceed to the next topic without giving another pair of examples to show the subtlety of the concept of ‘measure algebra free product’. ¯ the measure algebra Example Let (A, µ ¯) be the measure algebra of Lebesgue measure µ on [0, 1], and (C, λ) 2 ¯ of Lebesgue measure λ on [0, 1] . Then (C, λ) can be regarded as the localizable measure algebra free product of (A, µ ¯) with itself, by 251M and 325Eb. Let ψ : A ⊗ A → C be the canonical map, as described in 325A. Then ψ[A ⊗ A] is not order-dense in C, and ψ is not order-continuous. P∞ P∞ proof (a) Let h²n in∈N be a sequence in [0, 1] such that n=0 ²n = ∞, but n=0 ²2n < 1; for instance, we 1 could take ²n = n+2 . Let hEn in∈N be a stochastically independent sequence of measurable subsets of [0, 1] such that µEn = ²n for each n. In A set an = En• , and consider cn = supi≤n ai ⊗ ai ∈ A ⊗ A for each n. (b) We have supn∈N cn = 1 in A ⊗ A. P P?? Otherwise, there is a non-zero a ∈ A ⊗ A such that a ∩ an = 0 for every n, and now there are non-zero b, b0 ∈ A such that S b ⊗ b0 ⊆ a. Set I = {n : anP ∩ b = 0}, J = {n : 0 an ∩ b } = 0. Then hEn in∈I is an independent family and µ( E ) ≤ 1 − µ ¯ b < 1, so i n∈I n∈I µEn < ∞, by P P the Borel-Cantelli lemma (273K). Similarly n∈J µEn < ∞. Because n∈N µEn = ∞, there must be some n ∈ N \ (I ∪ J). Now an ∩ b and an ∩ b0 are both non-zero, so 0 6= (an ∩ b) ⊗ (an ∩ b0 ) = (an ⊗ an ) ∩ (b ⊗ b0 ) = 0, which is absurd. X XQ Q (c) On the other hand,

P∞ ¯ P∞ P∞ µan )2 = n=0 ²2n < ∞, n=0 λψ(cn ) ≤ n=0 (¯

by the choice of the ²n . So supn∈N ψ(cn ) cannot be 1 in C. Thus ψ is not order-continuous. (d) By 313P(a-ii) and 313O, ψ[A ⊗ A] cannot be order-dense in C; alternatively, (b) shows that there can be no non-zero member of ψ[A ⊗ A] included in 1 \ supn∈N ψ(cn ). (Both these arguments rely tacitly on the fact that ψ is injective, as noted in 325Ae.) 325G Since 325F shows that the free product and the localizable measure algebra free product are very different constructions, I had better repeat an idea from §315 in the new context. ¯ the measure Example Again, let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1], and (C, λ) 2 algebra of Lebesgue measure on [0, 1] . Then there is no order-continuous Boolean homomorphism φ : C → A such that φ(a ⊗ b) = a ∩ b for all a, b ∈ A. P P Let φ : C → A be a Boolean homomorphism such that φ(a⊗b) = a ∩ b for all a, b ∈ A. For i < 2n let ani be the equivalence class in A of the interval [2−n i, 2−n (i+1)], ¯ n = 2−n for each n, so inf n∈N cn = 0 in C; and set cn = supi 0. Express c as W • . Then by 254Fe i∈I Ai is topologically dense Sin C. P there are H0 , . . . , Hk ∈ C such that λ(W 4 j≤k Hj ) ≤ ². Now cj = Hj• ∈ C for each j, so S N c0 = supj≤k cj = ( j≤k Hj )• ∈ i∈I Ai , ¯ 4 c0 ) ≤ ². Q and λ(c Q Since B is complete for its uniformity (323Gc), there is a uniformly continuous function φ : C → B extending θ (3A4G). (e) Because θ is a Boolean homomorphism, so is φ. P P (i) The functions c 7→ φ(1 \ c), 1 \ φ(c) are continuous and the topology of B is Hausdorff, so {c : φ(1 \ c) = 1 \ φ(c)} is closed; as it includes A1 ⊗ A2 , it must be the whole of C. (ii) The functions (c, c0 ) 7→ φ(c ∪ c0 ), (c, c0 ) 7→ φ(c) ∪ φ(c0 ) are continuous, so {(c, c0 ) : φ(c ∪ c0 ) = φ(c) ∪ φ(c0 )} is closed in C × C; as it includes (A1 ⊗ A2 ) × (A1 ⊗ A2 ), it must be the whole of C × C. Q Q ¯ c 7→ ν¯φ(c) are continuous and (f ) Because θ is measure-preserving, so is φ. P P The functions c 7→ λc, equal on A1 ⊗ A2 , so are equal on C. Q Q (g) Finally, to see that φ is unique, observe that any measure-preserving Boolean homomorphism from N C to B must be continuous, so that if it agrees with φ on i∈I Ai it must agree with φ on C. 325J

Of course this leads at once to a result corresponding to 325D.

Theorem Let h(Ai , µ ¯i )ii∈I be a family of probability algebras. ¯ together with order-continuous Boolean homomorphisms ψi : (a) There is a probability algebra (C, λ), Ai → C such that whenever (B, Q ν¯) is a probability algebra, and φi : Ai → B are Boolean homomorphisms ¯i ai whenever J ⊆ I is finite and ai ∈ Ai for each i ∈ J, then there is a such that ν¯(inf i∈J φi (ai )) = i∈J µ unique measure-preserving Boolean homomorphism φ : C → B such that φψj = φj for every j. ¯ hψi ii∈I ) is determined up to isomorphism by this property. (b) The structure (C, λ, N N (c) The Boolean homomorphism ψ : i∈I Ai → C defined from the ψi is injective, and ψ[ i∈I Ai ] is topologically dense in C. proof For (a) and (c), all we have to do is represent each (Ai , µ ¯i ) as the measure algebra of a probability space, and apply 325I. The uniqueness of C and the ψi follows from the uniqueness of the homomorphisms φ, as in 325D. ¯ hψi ii∈I ) is a, or the, probability algebra free 325K Definition As in 325Ea, we can say that (C, λ, product of h(Ai , µ ¯i )ii∈I . 325L Independent subalgebras If (A, µ ¯) is a probability algebra, Q we say that a family hBi ii∈I of subalgebras of A is (stochastically) independent if µ ¯(inf i∈J bi ) = i∈J µ ¯bi whenever J ⊆ I is finite and bi ∈ Bi for each i. (Compare 272Ab.) In this case the embeddings Bi ⊆ A give rise to an embedding of the probability algebra free product of h(Bi , µ ¯¹ Bi )ii∈I into A. (Compare 272J, 315Xn.)

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325M

325M We can now make a general trawl through Chapters 25 and 27 seeking results which can be expressed in the language of this section. I give some in 325Xe-325Xh. Some ideas from §254 which are thrown into sharper relief by a reformulation are in the following theorem. ¯ hψi ii∈I ) their probability algebra Theorem Let h(Ai , µ ¯i )ii∈I be a family of probability algebras and (C, λ, S free product. For J ⊆ I let CJ be the closed subalgebra of C generated by i∈J ψi [Ai ]. ¯ CJ , hψi ii∈J ) is a probability algebra free product of h(Ai , µ (a) For any J ⊆ I, (CJ , λ¹ ¯i )ii∈J . (b) For any c ∈ C, there is a unique smallest J ⊆ I such that c ∈ C , and this J is countable. J T (c) For any non-empty family J ⊆ PI, J∈J CJ = CT J . proof (a) If (B, ν¯, hφi ii∈J ) is any probability algebra free product of h(Ai , µ ¯i )ii∈J , then we have a measurepreserving homomorphism ψ : B → C such that ψφ = ψ for every i ∈ J. Because the subalgebra i i S S B0 of B generated by i∈J φi [Ai ] is topologically dense in B (325Jc), and ψ is continuous (324Kb), i∈J ψi [Ai ] is topologically dense in S ψ[B]; also ψ[B] is closed in C (324Kb again). But this means that ψ[B] is just the topological closure of i∈I ψi [Ai ] and must be CJ . Thus ψ is an isomorphism, and ¯ CJ , hψi ii∈J ) = (ψ[B], ν¯ψ −1 , hψφi ii∈I ) (CJ , λ¹ is also a probability algebra free product of h(Ai , µ ¯i )ii∈J . (b) As in 325J, we may suppose that each (Ai , µ ¯i ) is the measure algebra of a probability space (Xi , Σi , µi ), and that C is the measure algebra of their product (X, Λ, λ). Let W ∈ Λ be such that c = W • . By 254Rd, there is a unique smallest K ⊆ I such that W 4U is negligible for some U ∈ ΛK , where ΛK is the set of members of Λ which are determined by coordinates in K; and K is countable. But if we look at any J ⊆ I, {x : x(i) ∈ E} ∈ ΛJ for every i ∈ J, E ∈ Σi ; so {U • : U ∈ ΛJ } is a closed subalgebra of C including ψi [Ai ] for every i ∈ J, and therefore including CJ . On the other hand, as observed in 254Ob, any member of ΛJ is approximated, in measure, by sets in the σ-algebra TJ generated by sets of the form {x : x(i) ∈ E} where i ∈ J, E S ∈ Σi . Of course TJ ⊆ ΛJ , so {W • : W ∈ ΛJ } = {W • : W ∈ TJ } is the closed subalgebra of C generated by i∈K ψi [Ai ], which is CJ . Thus K is also the unique smallest subset of I such that c ∈ CK . T T other hand, suppose (c) Of T course CK ⊆ CJ whenever K ⊆ J ⊆ I, so T J∈J CJ ⊇ C J . On the T . As c is arbitrary, C ; then by (b) there is some K ⊆ J such that c ∈ C ⊆ C that c ∈ K J J∈J J T T J∈J CJ = C J . *325N Notation In this context, I will say that an element c of C is determined by coordinates in J if c ∈ CJ . 325X Basic exercises (a) Let (A1 , µ ¯1 ), (A2 , µ ¯2 ) be two semi-finite measure algebras, and suppose that for each j we are given a closed subalgebra Bj of Aj such that (Bj , ν¯j ) is also semi-finite, where ν¯j = µ ¯j ¹ Bj . Show that the localizable measure algebra free product of (B1 , ν¯1 ) and (B2 , ν¯2 ) can be thought of as a closed subalgebra of the localizable measure algebra free product of (A1 , µ ¯1 ) and (A2 , µ ¯2 ). (b) Let (A1 , µ ¯1 ), (A2 , µ ¯2 ) be two semi-finite measure algebras, and suppose that for each j we are given a principal ideal Bj of Aj . Set ν¯j = µ ¯j ¹ Bj . Show that the localizable measure algebra free product of (B1 , ν¯1 ) and (B2 , ν¯2 ) can be thought of as a principal ideal of the localizable measure algebra free product of (A1 , µ ¯1 ) and (A2 , µ ¯2 ). > (c) Let h(Ai , µ ¯i )ii∈I and h(Bj , ν¯j )ij∈J be families of semi-finite measure algebras, with simple products b loc (B, ν¯) can be (A, µ ¯) and (B, ν¯) (322K). Show that the localizable measure algebra free product (A, µ ¯ )⊗ b identified with the simple product of the family h(Ai , µ ¯i )⊗loc (Bj , ν¯j )ii∈I,j∈J . ¯ hψi ii∈I ) their probability algebra >(d) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras, and (C, λ, free product. Suppose that for each i ∈ I we are given a measure-preserving Boolean homomorphism πi : Ai → Ai . Show that there is a unique measure-preserving Boolean homomorphism π : C → C such that πψi = ψi πi for every i ∈ I.

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> (e) Let (A, µ ¯) be a probability algebra. We say that a family hai ii∈I in A is (stochastically) indeQ pendent if µ ¯(inf i∈j aj ) = i∈J µ ¯ai for every non-empty finite J ⊆ I. Show that this is so iff hAi ii∈I is independent, where Ai = {0, ai , 1 \ ai , 1} for each i. (Compare 272F.) (f ) Let (A, µ ¯) be a probability algebra, and hAi ii∈I a stochastically independent family of closed subalgebras of A. Let hJ(k)ik∈K be S a disjoint family of subsets of I, and for each k ∈ K let Bk be the closed subalgebra of A generated by i∈J(k) Ai . Show that hBk ik∈K is independent. (Compare 272K.) (g) Let (A, µ ¯) be a probability algebra, and hAi ii∈I a stochastically independent family of closed subalS T gebras of A. For J ⊆ I let BJ be the closed subalgebra of A generated by i∈J Ai . Show that {BI\J : J is a finite subset of I} = {0, 1}. (Hint: For J ⊆ I, show that µ ¯(b ∩ c) = µ ¯b · µ ¯c for every b ∈ BI\J , c ∈ BJ . Compare 272O, 325M.) ¯ hψi ii∈I ). (h) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras with probability algebra free product (C, λ, For c ∈ C let J be the smallest subset of I such that c belongs to the closed subalgebra of C generated by c S ψ [A ]. Show that if c ⊆ d in C, then there is an e ∈ C such that c ⊆ e ⊆ d and J ⊆ J ∩ J . (Hint: e c d i∈Jc i i 254R.) 325Y Further exercises (a) Let µ be counting measure on X = {0}, µ0 the countable-cocountable measure on X 0 = ω1 , and ν counting measure on Y = ω1 . Show that the measure algebras of the primitive product measures on X × Y , X 0 × Y are not isomorphic. (b) Let A be a Boolean algebra, and µ : A → [0, ∞] a function such that µ0 = 0 and µ(a ∪ b) = µa + µb whenever a, b ∈ A and a ∩ b = 0; suppose that Af = {a : µa < ∞} is order-dense in A. For e ∈ Af , a, b ∈ A set ρe (a, b) = µ(e ∩ (a 4 b)). Give A the uniformity defined by {ρe : µe < ∞}. (i) Show that the completion b of A under this uniformity has a measure µ A ˆ, extending µ, under which it is a localizable measure algebra. b µ (ii) Show that if a ∈ A, ˆa < ∞ and ² > 0, there is a b ∈ A such that µ ˆ(a 4 b) ≤ ². (iii) Show that for every b there is a sequence han in∈N in A such that a ⊇ sup a∈A inf ˆa = µ ˆ(supn∈N inf m≥n am ). m≥n am and µ n∈N b b (iv) In particular, the set of infima in A of sequences in A is order-dense in A. (v) Explain the relevance of this construction to the embedding A1 ⊗ A2 ⊆ C in 325D. S (c) In 325F, set W = n∈N En × En . Show that if A, B are any non-negligible subsets of [0, 1], then W ∩ (A × B) is not negligible. (d) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1]. Show that A ⊗ A is ccc but not weakly (σ, ∞)-distributive. (Hint: (i) A ⊗ A is embeddable as a subalgebra of a probability algebra (ii) in the notation of 325F, look at cmn = supm≤i≤n ei ⊗ ei .) (e) Repeat 325F-325G and 325Yc-325Yd with an arbitrary atomless probability space in place of [0, 1]. (f ) Let (A, µ ¯) be a probability algebra and hai ii∈I a (stochastically) independent family in A. Show that for any a ∈ A, ² > 0 the set {i : i ∈ I, |¯ µ(a ∩ ai ) − µ ¯a · µ ¯ai | ≥ ²} is finite, so that {i : µ ¯(a ∩ ai ) 6= µ ¯a · µ ¯ai } is countable. (Hint: 272Yd.) 325 Notes and comments 325B shows that the measure algebra of a product measure may be irregular if we have factor measures which are not strictly localizable. But two facts lead the way to the ‘localizable measure algebra free product’ in 325D-325E. The first is that every semi-finite measure algebra is embeddable, in a canonical way, in a localizable measure algebra (322N); and the second is that the Stone representation of a localizable measure algebra is strictly localizable (322M). It is a happy coincidence that we can collapse these two facts together in the construction of 325D. Another way of looking at the localizable measure algebra free product of two localizable measure algebras is to express it as the simple product of measure algebra free products of totally finite measure algebras, using 325Xc and the fact that for σ-finite measure algebras there is only one reasonable measure algebra free product, being that provided by any representation of them as measure algebras of measure spaces (325Eb).

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Measure algebras

325 Notes

Yet a third way of approaching measure algebra free products is as the uniform space completions of algebraic free products, using 325Yb. This gives the same result as the construction of 325D because the algebraic free product appears as a topologically dense subalgebra of the localizable measure algebra free product (325Dc) which is complete as uniform space (325Dc). (I have to repeat such phrases as ‘topologically dense’ because the algebraic free product is emphatically not order-dense in the measure algebra free product (325F).) The results in 251I on approximating measurable sets for a c.l.d. product measure by combinations of measurable rectangles correspond to general facts about completions of finitely-additive measures (325Yb(ii), 325Yb(iii)). It is worth noting that the completion process can be regarded as made up of two steps; first take infima of sequences of sets of finite measure, and then take arbitrary suprema (325Yb(iv)). The idea of 325F appears S in many guises, and this is only the first time that I shall wish to call on it. The point of the set W = n∈N En × En is that it is a measurable subset of the square (indeed, by taking the En to be open sets we can arrange that W should be open), of measure strictly less than 1 (in fact, as small as we wish), such that its complement does not include any non-negligible ‘measurable rectangle’ G × H; indeed, W ∩ (A × B) is non-negligible for any non-negligible sets A, B ⊆ [0, 1] (325Yc). I believe ¨ s & Oxtoby 55; I learnt the method of 325F that the first published example of such a set was by Erdo from R.O.Davies. I include 325G as a kind of guard-rail. The relationship between preservation of measure and ordercontinuity is a subtle one, as I have already tried to show in 324K, and it is often worth considering the possibility that a result involving order-continuous measure-preserving homomorphisms has a form applying to all order-continuous homomorphisms. However, there is no simple expression of such an idea in the present context. In the context of infinite free products of probability algebras, there is a degree of simplification, since there is only one algebra which can plausibly be called the probability algebra free product, and this is produced by any realization of the algebras as measure algebras of probability spaces (325I-325K). The examples 325F-325G apply equally, of course, to this context. At this point I mention the concept of ‘(stochastically) independent’ family (325L, 325Xe) because we have the machinery to translate several results from §272 into the language of measure algebras (325Xe-325Xg). I feel that I have to use the phrase ‘stochastically independent’ here because there is the much weaker alternative concept of ‘Boolean independence’ (315Xn) also present. But I leave most of this as exercises, because the language of measure algebras offers few ideas to the probability theory already covered in Chapter 27. All it can do is formalise the ever-present principle that negligible sets often can and should be ignored.

326 Additive functionals on Boolean algebras I devote two sections to the general theory of additive functionals on measure algebras. As many readers will rightly be in a hurry to get on to the next two chapters, I remark that the only significant result needed for §§331-332 is the Hahn decomposition of a countably additive functional (326I), and that this is no more than a translation into the language of measure algebras of a theorem already given in Chapter 23. The concept of ‘standard extension’ of a countably additive functional from a subalgebra (327F-327G) will be used for a theorem in §333, and as preparation for Chapter 36. I begin with notes on the space of additive functionals on an arbitrary Boolean algebra (326A-326D), corresponding to 231A-231B, but adding a more general form of the Jordan decomposition of a bounded additive functional into positive and negative parts (326D). The next subsection (326E-326I) deals with countably additive functionals, corresponding to 231C-231F. In 326J-326P I develop a new idea, that of ‘completely additive’ functional, which does not match anything in the previous treatment. In 326Q I return to additive functionals, giving a fundamental result on the construction of additive functionals on free products. 326A Additive functionals: Definition Let A be a Boolean algebra. A functional ν : A → R is finitely additive, or just additive, if ν(a ∪ b) = νa + νb whenever a, b ∈ A and a ∩ b = 0. A non-negative additive functional is sometimes called a finitely additive measure or charge.

326D

Additive functionals on Boolean algebras

105

326B Elementary facts Let A be a Boolean algebra and ν : A → R a finitely additive functional. The following will I hope be obvious. (a) ν0 = 0 (because ν0 = ν0 + ν0). (b) If c ∈ A, then a 7→ ν(a ∩ c) is additive (because (a ∩ c) ∪ (b ∩ c) = (a ∪ b) ∩ c). (c) αν is an additive functional for any α ∈ R. If ν 0 is another finitely additive functional on A, then ν + ν 0 is additive. P (d) If hνi ii∈I is any family of finitely additive functionals such that ν 0 a = i∈I νi a is defined in R for for every a ∈ A, then ν 0 is additive. (e) If B is another Boolean algebra and π : B → A is a Boolean homomorphism, then νπ : B → R is additive. In particular, if B is a subalgebra of A, then ν¹ B : B → R is additive. (f ) ν is non-negative iff it is order-preserving – that is, νa ≥ 0 for every a ∈ A ⇐⇒ νb ≤ νc whenever b ⊆ c (because νc = νb + ν(c \ b) if b ⊆ c). 326C The space of additive functionals Let A be any Boolean algebra. From 326Bc we see that the set M of all finitely additive real-valued functionals on A is a linear space (a linear subspace of RA ). We give it the ordering induced by that of RA , so that ν ≤ ν 0 iff νa ≤ ν 0 a for every a ∈ A. This renders it a partially ordered linear space (because RA is). 326D The Jordan decomposition (I): Proposition Let A be a Boolean algebra, and ν a finitely additive real-valued functional on A. Then the following are equiveridical: (i) ν is bounded; (ii) supn∈N |νan | < ∞ for every disjoint sequence han in∈N in A; (iii) P limn→∞ |νan | = 0 for every disjoint sequence han in∈N in A; ∞ (iv) n=0 |νan | < ∞ for every disjoint sequence han in∈N in A; (v) ν is expressible as the difference of two non-negative additive functionals. proof (a)(i)⇒(v) Assume that ν is bounded. For each a ∈ A, set ν + a = sup{νb : b ⊆ a}. Because ν is bounded, ν + is real-valued. Now ν + is additive. P P If a, b ∈ A and a ∩ b = 0, then ν + (a ∪ b) = sup νc = c ⊆ a∪b

sup

ν(d ∪ e) =

d ⊆ a,e ⊆ b

sup

νd + νe

d ⊆ a,e ⊆ b

(because d ∩ e ⊆ a ∩ b = 0 whenever d ⊆ a, e ⊆ b) = sup νd + sup νe = ν + a + ν + b. Q Q d⊆a

e⊆b

Consequently ν − = ν + − ν is also additive (326Bc). Since 0 = ν0 ≤ ν + a,

νa ≤ ν + a

for every a ∈ A, ν + ≥ 0 and ν − ≥ 0. Thus ν = ν + − ν − is the difference of two non-negative additive functionals. (b)(v)⇒(iv) If ν is expressible as ν1 − ν2 , where ν1 and ν2 are non-negative additive functionals, and han in∈N is disjoint, then Pn i=0 νj ai = νj (supi≤n ai ) ≤ νj 1

106

Measure algebras

for every n, both j, so that P∞ i=0

|νai | ≤

P∞ i=0

ν1 ai +

P∞ i=0

326D

ν2 ai ≤ ν1 1 + ν2 1 < ∞.

(c)(iv)⇒(iii)⇒(ii) are trivial. (d) not-(i)⇒not-(ii) Suppose that ν is unbounded. Choose sequences han in∈N , hbn in∈N inductively, as follows. b0 = 1. Given that supa ⊆ bn |νa| = ∞, choose cn ⊆ bn such that |νcn | ≥ |νbn | + n; then |νcn | ≥ n and |ν(bn \ cn )| = |νbn − νcn | ≥ |νcn | − |νbn | ≥ n. We have ∞ = sup |νa| = sup |ν(a ∩ cn ) + ν(a \ cn )| a ⊆ bn

a ⊆ bn

≤ sup |ν(a ∩ cn )| + |ν(a \ cn )| ≤ a ⊆ bn

sup a ⊆ bn ∩cn

|νa| +

sup

|νa|,

a ⊆ bn \cn

so at least one of supa ⊆ bn ∩cn |νa|, supa ⊆ bn \cn |νa| must be infinite; take bn+1 to be one of cn , bn \ cn such that supa ⊆ bn+1 |νa| = ∞, and set an = bn \ bn+1 , so that |νan | ≥ n. Continue. On completing the induction, we have a disjoint sequence han in∈N such that |νan | ≥ n for every n, so that (ii) is false. Remark I hope that this reminds you of the decomposition of a function of bounded variation as the difference of monotonic functions (224D). 326E Countably additive functionals: Definition Let A be a Boolean algebra. A functional ν : P∞ A → R is countably additive or σ-additive if n=0 νan is defined and equal to ν(supn∈N an ) whenever han in∈N is a disjoint sequence in A and supn∈N an is defined in A. A warning is perhaps in order. It can happen that A is presented to us as a subalgebra of a larger algebra B; for instance, A might be an algebra of sets, a subalgebra of some σ-algebra Σ ⊆ PX. In this case, there may be sequences in A which have a supremum in A which is not a supremum in B (indeed, this will happen just when the embedding is not sequentially order-continuous). So we can have a countably additive functional ν : B → R such that ν¹ A is not countably additive in the sense used here. A similar phenomenon will arise when we come to the Daniell integral in Volume 4 (§434). 326F Elementary facts Let A be a Boolean algebra and ν : A → R a countably additive functional. (a) ν is finitely additive. (Setting an = 0 for every n, we see from the definition in 326E that ν0 = 0. Now, given a ∩ b = 0, set a0 = a, a1 = b, an = 0 for n ≥ 2 to see that ν(a ∪ b) = νa + νb.) (b) If han in∈N is a non-decreasing sequence in A with a supremum a ∈ A, then P∞ νa = νa0 + n=0 ν(an+1 \ an ) = limn→∞ νan . (c) If han in∈N is a non-increasing sequence in A with an infimum a ∈ A, then νa = νa0 − ν(a0 \ a) = νa0 − limn→∞ ν(a0 \ an ) = limn→∞ νan . (d) If c ∈ A, then a 7→ ν(a ∩ c) is countably additive. (For supn∈N an ∩ c = c ∩ supn∈N an whenever the right-hand-side is defined, by 313Ba.) (e) αν is a countably additive functional for any α ∈ R. If ν 0 is another countably additive functional on A, then ν + ν 0 is countably additive. (f ) If B is another Boolean algebra and π : B → A is a sequentially order-continuous Boolean homomorphism, then νπ is a countably additive functional on B. (For if hbn in∈N is a disjoint sequence in B with supremum b, then hπbn in∈N is a disjoint sequence with supremum πb.)

326I

Additive functionals on Boolean algebras

107

(g) If A is Dedekind σ-complete and B is a σ-subalgebra of A, then ν¹ B : B → R is countably additive. (For the identity map from B to A is sequentially order-continuous, by 314Hb.) 326G Corollary Let A be a Boolean algebra and ν a finitely additive real-valued functional on A. (a) ν is countably additive iff limn→∞ νan = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0 in A. (b) If ν 0 is an additive functional on A and |ν 0 a| ≤ νa for every a ∈ A, and ν is countably additive, then 0 ν is countably additive. (c) If ν is non-negative, then ν is countably additive iff it is sequentially order-continuous. proof (a)(i) If ν is countably additive and han in∈N is a non-increasing sequence in A with infimum 0, then limn→∞ νan = 0 by 326Fc. (ii) If ν satisfies the condition, and han in∈N is a disjoint sequence in A with supremum a, set bn = a \ supi≤n ai for each n ∈ N; then hbn in∈N is non-increasing and has infimum 0, so Pn νa − i=0 νai = νa − ν(supi≤n ai ) = νbn → 0 P∞ as n → ∞, and νa = n=0 νan ; thus ν is countably additive. (b) If han in∈N is a disjoint sequence in A with supremum a, set bn = supi≤n ai for each n; then νa = limn→∞ νbn , so limn→∞ |ν 0 a − ν 0 bn | = limn→∞ |ν 0 (a \ bn )| ≤ limn→∞ ν(a \ bn ) = 0, and

P∞ n=0

ν 0 an = limn→∞ ν 0 bn = ν 0 a.

(c) If ν is countably additive, then it is sequentially order-continuous by 326Fb-326Fc. If ν is sequentially order-continuous, then of course it satisfies the condition of (a), so is countably additive. 326H The Jordan decomposition (II): Proposition Let A be a Boolean algebra and ν a bounded countably additive real-valued functional on A. Then ν is expressible as the difference of two non-negative countably additive functionals. proof Consider the functional ν + a = supb ⊆ a νb defined in the proof of 326D. If han in∈N is a disjoint sequence in A with supremum a, and b ⊆ a, then P∞ P∞ νb = n=0 ν(b ∩ an ) ≤ n=0 ν + an . P∞ As b is arbitrary, ν + a ≤ n=0 ν + an . But of course Pn ν + a ≥ ν + (supi≤n ai ) = i=0 ν + ai P∞ for every n ∈ N, so ν + a = n=0 ν + an . As han in∈N is arbitrary, ν + is countably additive. Now ν − = ν + − ν is also countably additive, and ν = ν + − ν − is the difference of non-negative countably additive functionals. 326I The Hahn decomposition: Theorem Let A be a Dedekind σ-complete Boolean algebra and ν : A → R a countably additive functional. Then ν is bounded and there is a c ∈ A such that νa ≥ 0 whenever a ⊆ c, while νa ≤ 0 whenever a ∩ c = 0. first proof By 314M, there are a set X and a σ-algebra Σ of subsets of X and a sequentially order-continuous Boolean homomorphism π from Σ onto A. Set ν1 = νπ : Σ → R. Then ν1 is countably additive (326Ff). So ν1 is bounded and there is a set H ∈ Σ such that ν1 F ≥ 0 whenever F ∈ Σ and F ⊆ H and ν1 F ≤ 0 whenever F ∈ Σ and F ∩ H = ∅ (231E). Set c = πH ∈ A. If a ⊆ c, then there is an F ∈ Σ such that πF = a; now π(F ∩ H) = a ∩ c = a, so νa = ν1 (F ∩ H) ≥ 0. If a ∩ c = 0, then there is an F ∈ Σ such that πF = a; now π(F \ H) = a \ c = a, so νa = ν1 (F \ H) ≤ 0. P∞ second proof (a) Note first that ν is bounded. P P If han in∈N is a disjoint sequence in A, then n=0 νan must exist and be equal to ν(supn∈N an ); in particular, limn→∞ νan = 0. By 326D, ν is bounded. Q Q

108

Measure algebras

326I

(b)(i) We know that γ = sup{νa : a ∈ A} < ∞. Choose a sequence han in∈N in A such that νan ≥ γ −2−n for every n ∈ N. For m ≤ n ∈ N, set bmn = inf m≤i≤n ai . Then νbmn ≥ γ − 2 · 2−m + 2−n for every n ≥ m. P P Induce on n. For n = m, this is due to the choice of am = bmm . For the inductive step, we have bm,n+1 = bmn ∩ an+1 , while surely γ ≥ ν(an+1 ∪ bmn ), so γ + νbm,n+1 ≥ ν(an+1 ∪ bmn ) + ν(an+1 ∩ bmn ) = νan+1 + νbmn ≥ γ − 2−n−1 + γ − 2 · 2−m + 2−n (by the choice of an+1 and the inductive hypothesis) = 2γ − 2 · 2−m + 2−n−1 . Subtracting γ from both sides, νbm,n+1 ≥ γ − 2 · 2−m + 2−n−1 and the induction proceeds. Q Q (ii) Set bm = inf n≥m bmn = inf n≥m an . Then νbm = limn→∞ νbmn ≥ γ − 2 · 2−m , by 326Fc. Next, hbn in∈N is non-decreasing, so setting c = supn∈N bn we have νc = limn→∞ νbn ≥ γ; since νc is surely less than or equal to γ, νc = γ. If b ∈ A and b ⊆ c, then νc − νb = ν(c \ b) ≤ γ = νc, so νb ≥ 0. If b ∈ A and b ∩ c = 0 then νc + νb = ν(c ∪ b) ≤ γ = νc so νb ≤ 0. This completes the proof. 326J Completely additive functionals: Definition Let A be a Boolean algebra. A functional ν : A → R is completely additive or τ -additive if it is finitely additive and inf a∈A |νa| = 0 whenever A is a non-empty downwards-directed set in A with infimum 0. 326K Basic facts Let A be a Boolean algebra and ν a completely additive real-valued functional on A. (a) ν is countably additive. P P If han in∈N is a non-increasing sequence in A with infimum 0, then for any infinite I ⊆ N the set {ai : i ∈ I} is downwards-directed and has infimum 0, so inf i∈I |νai | = 0; which means that limn→∞ νan must be zero. By 326Ga, ν is countably additive. Q Q (b) Let A be a non-empty downwards-directed set in A with infimum 0. Then for every ² > 0 there is an a ∈ A such that |νb| ≤ ² whenever b ⊆ a. P P?? Suppose, if possible, otherwise. Set B = {b : |νb| ≥ ², ∃ a ∈ A, b ⊇ a}. If a ∈ A there is a b0 ⊆ a such that |νb0 | > ². Now {a0 \ b0 : a0 ∈ A, a0 ⊆ a} is downwards-directed and has infimum 0, so there is an a0 ∈ A such that a0 ⊆ a and |ν(a0 \ b0 )| ≤ |νb0 | − ². Set b = b0 ∪ a0 ; then a0 ⊆ b and |νb| = |νb0 + ν(a0 \ b0 )| ≥ |νb0 | − |ν(a0 \ b0 )| ≥ ², so b ∈ B. But also b ⊆ a. Thus every member of A includes some member of B. Since every member of B includes a member of A, B is downwards-directed and has infimum 0; but this is impossible, since inf b∈B |νb| ≥ ². X XQ Q (c) If ν is non-negative, it is order-continuous. P P (i) If A is a non-empty upwards-directed set with supremum a0 , then {a0 \ a : a ∈ A} is a non-empty downwards-directed set with infimum 0, so

326M

Additive functionals on Boolean algebras

109

supa∈A νa = νa0 − inf a∈A ν(a0 \ a) = νa0 . (ii) If A is a non-empty downwards-directed set with infimum a0 , then {a \ a0 : a ∈ A} is a non-empty downwards-directed set with infimum 0, so inf a∈A νa = νa0 + inf a∈A ν(a \ a0 ) = νa0 . Q Q (d) If c ∈ A, then a 7→ ν(a ∩ c) is completely additive. P P If A is a non-empty downwards-directed set with infimum 0, so is {a ∩ c : a ∈ A}, so inf a∈A |ν(a ∩ c)| = 0. Q Q (e) αν is a completely additive functional for any α ∈ R. If ν 0 is another completely additive functional on A, then ν + ν 0 is completely additive. P P We know from 326Bc that ν + ν 0 is additive. Let A be a non-empty downwards-directed set with infimum 0. For any ² > 0, (b) tells us that there are a, a0 ∈ A such that |νb| ≤ ² whenever b ⊆ a0 and |ν 0 b| ≤ ² whenever b ⊆ a0 . But now, because A is downwards-directed, there is a b ∈ A such that b ⊆ a ∩ a0 , which means that |νb + ν 0 b| ≤ |νb| + |ν 0 b| is at most 2². As ² is arbitrary, inf a∈A |(ν + ν 0 )(a)| = 0, and ν + ν 0 is completely additive. Q Q (f ) If B is another Boolean algebra and π : B → A is an order-continuous Boolean homomorphism, then νπ is a completely additive functional on B. P P By 326Be, νπ is additive. If B ⊆ B is a non-empty downwards-directed set with infimum 0 in B, then π[B] is a non-empty downwards-directed set with infimum 0 in A, because π is order-continuous, so inf b∈B |νπb| = 0. Q Q In particular, if B is a regularly embedded subalgebra of A, then ν¹ B is completely additive. (g) If ν 0 is another additive functional on A and |ν 0 a| ≤ νa for every a ∈ A, then ν 0 is completely additive. P P If A ⊆ A is non-empty and downwards-directed and inf A = 0, then inf a∈A |ν 0 a| ≤ inf a∈A νa = 0. Q Q 326L

I squeeze a useful fact in here.

Proposition If A is a ccc Boolean algebra, a functional ν : A → R is countably additive iff it is completely additive. proof If ν is completely additive it is countably additive, by 326Ka. If ν is countably additive and A is a non-empty downwards-directed set in A with infimum 0, then there is a (non-empty) countable subset B of A also with infimum 0 (316E). Let hbn in∈N be a sequence running over B, and choose han in∈N in A such that a0 = b0 , an+1 ⊆ an ∩ bn for every n ∈ N. Then han in∈N is a non-increasing sequence with infimum 0, so limn→∞ νan = 0 (326Fc) and inf a∈A |νa| = 0. As A is arbitrary, ν is completely additive. 326M The Jordan decomposition (III): Proposition Let A be a Boolean algebra and ν a completely additive real-valued functional on A. Then ν is bounded and expressible as the difference of two non-negative completely additive functionals. proof (a) I must first check that ν is bounded. P P Let han in∈N be a disjoint sequence in A. Set A = {a : a ∈ A, there is an n ∈ N such that ai ⊆ a for every i ≥ n}. Then A is closed under ∩ , and if b is any lower bound for A then b ⊆ 1 \ an ∈ A, so b ∩ an = 0, for every n ∈ N; but this means that 1 \ b ∈ A, so that b ⊆ 1 \ b and b = 0. Thus inf A = 0. By 326Kb, there is an a ∈ A such that |νb| ≤ 1 whenever b ⊆ a. By the definition of A, there must be an n ∈ N such that |νai | ≤ 1 for every i ≥ n. But this means that supn∈N |νan | is finite. As han in∈N is arbitrary, ν is bounded, by 326D(ii). Q Q (b) As in 326D and 326H, set ν + a = supb ⊆ a νb for every a ∈ A. Then ν + is completely additive. P P We know that ν + is additive. If A is a non-empty downwards-directed subset of A with infimum 0, then for every ² > 0 there is an a ∈ A such that |νb| ≤ ² whenever b ⊆ a; in particular, ν + a ≤ ². As ² is arbitrary, inf a∈A ν + a = 0; as A is arbitrary, ν + is completely additive. Q Q − + Consequently ν = ν −ν is completely additive (326Ke) and ν = ν + −ν − is the difference of non-negative completely additive functionals.

110

Measure algebras

326N

326N

I give an alternative definition of ‘completely additive’ which you may feel clarifies the concept.

Proposition Let A be a Boolean algebra, and ν : A → R a function. Then the following are equiveridical: (i) ν is completely additive; P (ii) ν1 = Pi∈I νai whenever hai ii∈I is a partition of unity in A; (iii) νa = i∈I νai whenever hai ii∈I is a disjoint family in A with supremum a. P proof (For notes on sums i∈I , see 226A.) (a)(i)⇒(ii) P If ν is completely additive and hai ii∈I is a partition of unity in A, then (inducing on #(J)) ν(supi∈J ai ) = i∈J νai for every finite J ⊆ I. Consider A = {1 \ supi∈J ai : J ⊆ I is finite}. Then A is non-empty and downwards-directed and has infimum 0, so for every ² > 0 there is an a ∈ A such that |νb| ≤ ² whenever b ⊆ a. Express a as 1 \ supi∈J ai where J ⊆ I is finite. If now K is another finite subset of I including J, P |ν1 − i∈K ai | = |ν(1 \ supi∈K ai )| ≤ ². P As remarked in 226Ad, this means that ν1 = i∈I νai , as claimed. (b)(ii)⇒(iii) Suppose that ν satisfies the condition (ii), and that hai ii∈I is a disjoint family with supremum a. Take any j ∈ / I, set J = I ∪ {j} and aj = 1 \ a; then hai ii∈J , (a, 1 \ a) are both partitions of unity, so P P ν(1 \ a) + νa = ν1 = i∈J νai = ν(1 \ a) + i∈I νai , P and νa = i∈I νai . (c)(iii)⇒(i) Suppose that ν satisfies (iii). Then ν is additive. α) ν is bounded. P (α P Let han in∈N be a disjoint sequence in A. Applying Zorn’s Lemma to the set C of all disjoint families C ⊆ A including {an : n ∈ N}, we find a partition of unity C ⊇ {an : n ∈ N}. Now P νc is defined in R, so supn∈N |νan | ≤ supc∈C |νc| is finite. By 326D, ν is bounded. Q Q c∈C β ) Define ν + from ν as in 326D. Then ν + satisfies the same condition as ν. P (β P Let hai ii∈I be a disjoint family in A with supremum a. Then for any b ⊆ a, we have b = supi∈I b ∩ ai , so P P νb = i∈I ν(b ∩ ai ) ≤ i∈I ν + ai . P Thus ν + a ≤ i∈I ν + ai . But of course X i∈I

X ν + ai = sup{ ν + ai : J ⊆ I is finite} i∈J +

= sup{ν (sup ai ) : J ⊆ I is finite} ≤ ν + a, i∈J

+

so ν a =

P i∈I

+

ν ai . Q Q

(γγ ) It follows that ν + is completely additive. P P If A is a non-empty downwards-directed set with infimum 0, then B = {b : ∃ a ∈ A, b ∩ a = 0} is order-dense in A, so there is a partition of unity hbi ii∈I lying in B (313K). Now if J ⊆ I is finite, there is an a ∈ A such that a ∩ supi∈J bi = 0 (because A is downwards-directed), and P ν + a + i∈J ν + bi ≤ ν + 1. P Since ν + 1 = supJ⊆I is finite i∈J ν + bi , inf a∈A ν + a = 0. As A is arbitrary, ν + is completely additive. Q Q (δδ ) Now consider ν − = ν + − ν. Of course P P P ν − a = ν + a − νa = i∈I ν + ai − i∈I νai = i∈I ν − ai whenever hai ii∈I is a disjoint family in A with supremum a. Because ν − is non-negative, the argument of (γ) shows that ν − = (ν − )+ is completely additive. So ν = ν + − ν − is completely additive, as required.

*326Q

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326O For completely additive functionals, we have a useful refinement of the Hahn decomposition. I give it in a form adapted to the applications I have in mind. Proposition Let A be a Dedekind σ-complete Boolean algebra and ν : A → R a completely additive functional. Then there is a unique element of A, which I will denote [[ν > 0]], ‘the region where ν > 0’, such that νa > 0 whenever 0 6= a ⊆ [[ν > 0]], while νa ≤ 0 whenever a ∩ [[ν > 0]] = 0. proof Set C1 = {c : c ∈ A \ {0}, νa > 0 whenever 0 6= a ⊆ c}, C2 = {c : c ∈ A, νa ≤ 0 whenever a ⊆ c}. Then C1 ∪ C2 is order-dense in A. P P There is a c0 ∈ A such that νa ≥ 0 for every a ⊆ c, νa ≤ 0 whenever a ∩ c = 0 (326I). Given b ∈ A \ {0}, then b \ c0 ∈ C2 , so if b \ c0 6= 0 we can stop. Otherwise, b ⊆ c0 . If b ∈ C1 we can stop. Otherwise, there is a non-zero c ⊆ b such that νc ≤ 0; but in this case νa ≥ 0, ν(c \ a) ≥ 0 so νa = 0 for every a ⊆ c, and c ∈ C2 . Q Q There is therefore a partition of unity D ⊆ C1 ∪ C2 . Now D ∩ C1 is countable. P P If d ∈ D ∩ C1 , νd > 0. Also #({d : d ∈ D, νd ≥ 2−n }) ≤ 2n supa∈A νa is finite for each n, so D ∩ C1 is the union of a sequence of finite sets, and is countable. Q Q Accordingly D ∩ C1 has a supremum e. If 0 6= a ⊆ e then P P νa = c∈D ν(a ∩ c) = c∈D∩C1 ν(a ∩ c) ≥ 0 by 326N. Also there must be some c ∈ D ∩ C1 such that a ∩ c 6= 0, in which case ν(a ∩ c) > 0, so that νa > 0. If a ∩ e = 0, then P P νa = c∈D ν(a ∩ c) = c∈D∩C2 ν(a ∩ c) ≤ 0. Thus e has the properties demanded of [[ν > 0]]. To see that e is unique, we need observe only that if e0 has the same properties then ν(e \ e0 ) ≤ 0 (because (e \ e0 ) ∩ e0 = 0), so e \ e0 = 0 (because e \ e0 ⊆ e). Similarly, e0 \ e = 0 and e = e0 . Thus we may properly denote e by the formula [[ν > 0]]. 326P Corollary Let A be a Dedekind σ-complete Boolean algebra and µ, ν two completely additive functionals on A. Then there is a unique element of A, which I will denote [[µ > ν]], ‘the region where µ > ν’, such that µa > νa whenever 0 6= a ⊆ [[µ > ν]], µa ≤ νa whenever a ∩ [[µ > ν]] = 0. proof Apply 326O to the functional µ − ν, and set [[µ > ν]] = [[µ − ν > 0]]. *326Q Additive functionals on free products In Volume 4, when we return to the construction of measures on product spaces, the following fundamental fact will be useful. Theorem Let hAi ii∈I be a non-empty family of Boolean algebras, with free product A; write εi : Ai → A for the canonical maps, and C = {inf j∈J εj (aj ) : J ⊆ I is finite, aj ∈ Aj for every j ∈ J}. Suppose that θ : C → R is such that θc = θ(c ∩ εi (a)) + θ(c ∩ εi (1 \ a)) whenever c ∈ C, i ∈ I and a ∈ Ai . Then there is a unique finitely additive functional ν : A → R extending θ. proof (a) It will help if I note at once that θ0 = 0. P P θ0 = θ(0 ∩ εi (0)) + θ(0 ∩ εi (1)) = 2θ0

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for any i ∈ I. Q Q (b) The key is of course the P following fact: Pnif hcr ir≤m and hds is≤n are two disjoint families in C with m the same supremum in A, then r=0 θcr = s=0 θds . P P Let J ⊆ I be a finite set and Bi ⊆ Ai a finite subalgebra, for each i ∈ J, such that every cr and every ds belongs to the subalgebra A0 of A generated by {εj (b) : j ∈ J, b ∈ Bj }. Next, if j ∈ J and b ∈ Bj , then Pm Pm Pm r=0 θcr = r=0 θ(cr ∩ εj (b)) + r=0 θ(cr \ εj (b)). We can therefore find a disjoint family hc0r ir≤m0 in C ∩ A0 such that Pm Pm0 0 supr≤m0 c0r = supr≤m cr , r=0 θcr , r=0 θcr = and whenever r ≤ m0 , j ∈ J and b ∈ Bj then either c0r ⊆ εj (b) or c0r ∩ εj (b)b = 0; that is, every c0r is either 0 or of the form inf j∈J εj (bj ) where bj is an atom of Bj for every j. Similarly, we can find hd0s is≤n0 such that Pn0 Pn 0 sups≤n0 d0s = sups≤n ds , s=0 θds = s=0 θds , and whenever s ≤ n0 , j ∈ J and b ∈ Bj then d0s is either 0 or of the form inf j∈J εj (bj ) where bj is an atom of Bj for every j. But we now have supr≤m0 c0r = sups≤n0 d0s while for any r ≤ m0 , s ≤ n0 either c0r = d0s or c0r ∩ d0s = 0. It follows that the non-zero terms in the finite sequence hc0r ir≤m0 are just a rearrangement of the non-zero terms in hd0s is≤n0 , so that Pn0 Pn Pm Pm0 0 0 s=0 θds , s=0 θds = r=0 θcr = r=0 θcr = as required. Q Q Pm (c) By 315Jb, this means that we have a functional ν : A → R such that ν(supr≤m cr ) = r=0 θcr whenever hcr ir≤m is a disjoint family in C. It is now elementary to check that ν is additive, and it is clearly the only additive functional on A extending θ. 326X Basic exercises (a) Let A be a Boolean algebra and ν : A → R a finitely additive functional. Show that (i) ν(a ∪ b) = νa + νb − ν(a ∩ b) (ii) ν(a ∪ b ∪ c) = νa + νb + νc − ν(a ∩ b) − ν(a ∩ c) − ν(b ∩ c) + ν(a ∩ b ∩ c) for all a, b, c ∈ A. Generalize these results to longer sequences in A. (b) Let A be a Boolean algebra and ν : A → R a finitely additive functional. Show that the following are equiveridical: (i) ν is countably additive; (ii) limn→∞ νan = νa whenever han in∈N is a non-decreasing sequence in A with supremum a. (c) Let A be a Dedekind σ-complete Boolean algebra and ν : A → R a finitely additive functional. Show that the following are equiveridical: (i) ν is countably additive; (ii) limn→∞ νan = 0 whenever han in∈N is a sequence in A and inf n∈N supm≥n am = 0; (iii) limn→∞ νan = νa whenever han in∈N is a sequence in A and a = inf n∈N supm≥n am = supn∈N inf m≥n am . (Hint: for (i)⇒(iii), consider non-negative ν first.) (d) Let X be any uncountable set, and J an infinite subset of X. Let A be the finite-cofinite algebra of X (316Yk), and for a ∈ A set νa = #(a ∩ J) if a is finite, −#(J \ a) if a is cofinite. Show that ν is countably additive and unbounded. > (e) Let A be the algebra of subsets of [0, 1] generated by the family of (closed) intervals. Show that there is a unique additive functional ν : A → R such that ν[α, β] = β − α whenever 0 ≤ α ≤ β ≤ 1. Show that ν is countably additive but not completely additive. (f ) (i) Let (X, Σ, µ) be any atomless probability space. Show that µ : Σ → R is a countably additive functional which is not completely additive. (ii) Let X be any uncountable set and µ the countablecocountable measure on X (211R). Show that µ is countably additive but not completely additive. P (g) Let A be a Boolean algebra and ν : A → R a function. (i) Show that ν is finitely additive P iff i∈I νai = ν1 for every finite partition of unity hai ii∈I . (ii) Show that ν is countably additive iff i∈I νai = ν1 for every countable partition of unity hai ii∈I .

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(h) Show that 326O can fail if ν is only countably additive, rather than completely additive. (Hint: 326Xf.) (i) Let A be a Boolean algebra and ν a finitely additive real-valued functional on A. Let us say that a ∈ A is a support of ν if (α) νb = 0 whenever b ∩ a = 0 (β) for every non-zero b ⊆ a there is a c ⊆ b such that νc 6= 0. (i) Check that ν can have at most one support. (ii) Show that if a is a support for ν and ν is bounded, then the principal ideal Aa generated by a is ccc. (iii) Show that if A is Dedekind σ-complete and ν is countably additive, then ν is completely additive iff it has a support, and that in the language of 326O this is [[ν > 0]] ∪ [[−ν > 0]]. (iv) Taking J = X in 326Xd, show that X is the support of the functional ν there. 326Y Further exercises (a) Let A be a Boolean algebra and ν a non-negative finitely additive functional on A. Show that the following are equiveridical: (i) for every ² > 0 there is a finite partition hai ii∈I of unity in A such that νai ≤ ² for every i ∈ I; (ii) whenever ν 0 is a non-zero finitely additive functional such that 0 ≤ ν 0 ≤ ν there is an a ∈ A such that ν 0 a and ν 0 (1 \ a) are both non-zero. (Such functionals are called atomless.) (b) Let A be a Boolean algebra and ν1 , ν2 atomless non-negative additive functionals on A. Show that ν1 + ν2 , αν1 are atomless for every α ≥ 0, and that ν is atomless whenever ν is additive and 0 ≤ ν ≤ ν1 . (c) Let A be a Dedekind σ-complete Boolean algebra and ν an atomless non-negative finitely additive functional on A. Show that there is a family hat it∈[0,1] in A such that as ⊆ at and νat = tν1 whenever s ≤ t ∈ [0, 1]. (d) Let A be a Dedekind σ-complete Boolean algebra and ν0 , . . . , νn atomless non-negative finitely additive functionals on A. Show that there is an a ∈ A such that νi a = 12 νi 1 for every i ≤ n. (Hint: it is enough to consider the case ν0 ≥ ν1 . . . ≥ νn . For the inductive step, use the inductive hypothesis to construct hat it∈[0,1] such that as ⊆ at , νi at = tνi 1 if i < n, 0 ≤ s ≤ t ≤ 1. Now show that t 7→ νn (at+ 12 \ at ) is continuous on [0, 21 ].) (e) Let A be a Dedekind σ-complete Boolean algebra. Let ν : A → Rr , where r ≥ 1, be additive (in the sense that ν(a ∪ b) = νa + νb whenever a ∩ b = 0) and atomless (in the sense that for every ² > 0 there is a finite partition of unity hai ii∈I such that kνak ≤ ² whenever i ∈ I and a ⊆ ai ). Show that {νa : a ∈ A} is a convex set in Rr . (This is a version of Liapounoff ’s theorem. I am grateful to K.P.S.Bhaskara Rao for showing it to me.) (f ) Let A be a Dedekind σ-complete Boolean algebra and ν : A → [0, ∞[ a countably additive functional. Show that ν is atomless iff whenever a ∈ A and νa 6= 0 there is a b ⊆ a such that 0 < νb < νa. (g) Show that there is a finitely additive functional ν : PN → R such that ν{n} = 1 for every n ∈ N, so that ν is not bounded. (Hint: Use Zorn’s Lemma to construct a maximal linearly independent subset of `∞ including {χ{n} : n ∈ N}, and hence to construct a linear map f : `∞ → R such that f (χ{n}) = 1 for every n.) (h) Let A be any infinite Boolean algebra. Show that there is an unbounded finitely additive functional ν : A → R. (Hint: let htn in∈N be a sequence of distinct points in the Stone space of A, and set νa = ν 0 {n : tn ∈ b a} for a suitable ν 0 .) (i) Let A be a Boolean algebra, and give RA its product topology. Show that the space of finitely additive functionals on A is a closed subset of RA , but that the space of bounded finitely additive functionals is closed only when A is finite. (j) Let A be a Boolean algebra, and M the linear space of all bounded finitely additive real-valued functionals on A. For ν, ν 0 ∈ M say that ν ≤ ν 0 if νa ≤ ν 0 a for every a ∈ A. Show that (i) ν + , as defined in the proof of 326D, is just sup{0, ν} in M ; (ii) M is a Dedekind complete Riesz space (241E-241F, 353G);

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(iii) for ν, ν 0 ∈ M , |ν| = ν ∨ (−ν), ν ∨ ν 0 , ν ∧ ν 0 are given by the formulae |ν|(a) = supb ⊆ a νb − ν(a \ b),

(ν ∨ ν 0 )(a) = supb ⊆ a νb + ν 0 (a \ b),

(ν ∧ ν 0 )(a) = inf b ⊆ a νb + ν 0 (a \ b); (iv) for any non-empty A ⊆ M , A is bounded above in M iff Pn sup{ i=0 νi ai : νi ∈ A for each i ≤ n, hai ii≤n is disjoint} is finite, and then sup A is defined by the formula Pn (sup A)(a) = sup{ i=0 νi ai : νi ∈ A for each i ≤ n, hai ii≤n is disjoint, supi≤n ai = a} for every a ∈ A; (v) setting kνk = |ν|(1), k k is an order-continuous norm on M under which M is a Banach lattice. (k) Let A be a Boolean algebra. A functional ν : A → C is finitely additive if its real and imaginary parts are. Show that P the space of bounded finitely additive functionals from A to C is a Banach space under n the norm kνk = sup{ i=0 |νai | : hai ii≤n is a partition of unity in A}. (l) Let A be a Boolean algebra, and give it the topology Tσ for which the closed sets are the sequentially order-closed sets. Show that a finitely additive functional ν : A → R is countably additive iff it is continuous for Tσ . (m) Let A be a Boolean algebra, and Mσ the set of all bounded countably additive real-valued functionals on A. Show that Mσ is a closed and order-closed linear subspace of the normed space M of all additive functionals on A (326Yj), and that |ν| ∈ Mσ whenever ν ∈ Mσ . (n) Let A be a Boolean algebra and ν a non-negative finitely additive functional on A. Set νσ a = inf{supn∈N νan : han in∈N is a non-decreasing sequence with supremum a} for every a ∈ A. Show that νσ is countably additive, and is sup{ν 0 : ν 0 ≤ ν is countably additive}. (o) Let A be a Dedekind σ-complete Boolean algebra and hνn in∈N a sequence of countably additive realvalued functionals on A such that νa = limn→∞ νn a is defined in R for every a ∈ A. Show that ν is countably additive. (Hint: use arguments from part (a) of the proof of 247C to see that limn→∞ supk∈N |νk an | = 0 for every disjoint sequence han in∈N in A, and therefore that limn→∞ supk∈N |νk an | = 0 whenever han in∈N is a non-increasing sequence with infimum 0.) (p) Let A be a Boolean algebra, and Mτ the set of all completely additive real-valued functionals on A. Show that Mτ is a closed and order-closed linear subspace of the normed space M of all additive functionals, and that |ν| ∈ Mτ whenever ν ∈ Mτ . (q) Let A be a Boolean algebra and ν a non-negative finitely additive functional on A. Set ντ b = inf{supa∈A νa : A is a non-empty upwards-directed set with supremum b} for every b ∈ A. Show that ντ is completely additive, and is sup{ν 0 : ν 0 ≤ ν is completely additive}. (r) Let A be a Boolean algebra, and give it the topology T for which the closed sets are the order-closed sets (313Xb). Show that a finitely additive functional ν : A → R is completely additive iff it is continuous for T. (s) Let X be a set, Σ any σ-algebra of subsets of X, and ν :P Σ → R a functional. ShowP that ν is completely ∞ ∞ additive iff there are sequences hxn in∈N , hαn in∈N such that n=0 |αn | < ∞ and νE = n=0 αn χE(xn ) for every E ∈ Σ. (t) Let A and B be Boolean algebras and µ, ν finitely additive functionals on A, B respectively. Show that there is a unique finitely additive functional λ : A ⊗ B → R such that λ(a ⊗ b) = µa · νb for all a ∈ A, b ∈ B.

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N (u) Let hAi ii∈I be a family of Boolean algebras, with free product ( i∈I Ai , hεi ii∈I ), and for each i ∈ I let νi be a finitely N additive functional on Ai such that νi 1Q= 1. Show that there is a unique finitely additive functional ν : i∈I Ai → R such that ν(inf i∈J εi (ai )) = i∈J νi ai whenever J ⊆ I is non-empty and finite and ai ∈ Ai for each i ∈ J. 326 Notes and comments I have not mentioned the phrase ‘measure algebra’ anywhere in this section, and in principle this material could have been part of Chapter 31; but countably additive functionals are kissing cousins of measures, and most of the ideas here surely belong to ‘measure theory’ rather than to ‘Boolean algebra’, in so far as such divisions are meaningful at all. I have given as much as possible of the theory in a general form because the simplifications which are possible when we look only at measure algebras are seriously confusing if they are allowed too much prominence. In particular, it is important to understand that the principal properties of completely additive functionals do not depend on Dedekind completeness of the algebra, provided we take care over the definitions. Similarly, the definition of ‘countably additive’ functional for algebras which are not Dedekind σ-complete needs a moment’s attention to the phrase ‘and supn∈N an is defined in A’. It can happen that a functional is countably additive mostly because there are too few such sequences (326Xd). The formulations I have chosen as principal definitions (326A, 326E, 326J) are those which I find closest to my own intuitions of the concepts, but you may feel that 326G(i), 326Xc(iii) and 326N, or 326Yl and 326Yr, provide useful alternative patterns. The point is that countable additivity corresponds to sequential order-continuity (326Fb, 326Fc, 326Ff), while complete additivity corresponds to order-continuity (326Kc, 326Kf); the difficulty is that we must consider functionals which are not order-preserving, so that the simple definitions in 313H cannot be applied directly. It is fair to say that all the additive functionals ν we need to understand are bounded, and therefore may be studied in terms of their positive and negative parts ν + , ν − , which are order-preserving (326Bf); but many of the most important applications of these ideas depend precisely on using facts about ν to deduce facts about ν + and ν − . It is in 326D that we seem to start getting more out of the theory than we have put in. The ideas here have vast ramifications. What it amounts to is that we can discover much more than we might expect by looking at disjoint sequences. To begin with, the conditions here lead directly to 326I and 326M: every completely additive functional is bounded, and every countably additive functional on a Dedekind σ-complete Boolean algebra is bounded. (But note 326Yg-326Yh.) Naturally enough, the theory of countably additive functionals on general Boolean algebras corresponds closely to the special case of countably additive functionals on σ-algebras of sets, already treated in §§231-232 for the sake of the Radon-Nikod´ ym theorem. This should make 326E-326I very straightforward. When we come to completely additive functionals, however, there is room for many surprises. The natural map from a σ-algebra of measurable sets to the corresponding measure algebra is sequentially order-continuous but rarely order-continuous, so that there can be completely additive functionals on the measure algebra which do not correspond to completely additive functionals on the σ-algebra. Indeed there are very few completely additive functionals on σ-algebras of sets (326Ys). Of course these surprises can arise only when there is a difference between completely additive and countably additive functionals, that is, when the algebra involved is not ccc (326L). But I think that neither 326M nor 326N is obvious. I find myself generally using the phrase ‘countably additive’ in preference to ‘completely additive’ in the context of ccc algebras, where there is no difference between them. This is an attempt at user-friendliness; the phrase ‘countably additive’ is the commoner one in ordinary use. But I must say that my personal inclination is to the other side. The reason why so many theorems apply to countably additive functionals in these contexts is just that they are completely additive. I have given two proofs of 326I. I certainly assume that if you have got this far you are acquainted with the Radon-Nikod´ ym theorem and the associated basic facts about countably additive functionals on σ-algebras of sets; so that the ‘first proof’ should be easy and natural. On the other hand, there are purist objections on two fronts. First, it relies on the Stone representation, which involves a much stronger form of the axiom of choice than is actually necessary. Second, the classical Hahn decomposition in 231E is evidently a special case of 326I, and if we need both (as we certainly do) then one expects the ideas to stand out more clearly if they are applied directly to the general case. In fact the two versions of the argument are so nearly identical that (as you will observe, if you have Volume 2 to hand) they can share nearly every word. You can take the ‘second proof’, therefore, as a worked example in the translation of ideas from the context of σ-algebras

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of sets to the context of Dedekind σ-complete Boolean algebras. What makes it possible is the fact that the only limit operations referred to involve countable families. Arguments not involving limit operations can generally, of course, be applied to all Boolean algebras; I have lifted some exercises (326Yj, 326Yn) from §231 to give you some practice in such generalizations. Almost any non-trivial measure provides an example of a countably additive functional on a Dedekind σ-complete algebra which is not completely additive (326Xf). The question of whether such a functional can exist on a Dedekind complete algebra is the ‘Banach-Ulam problem’, to which I will return in 363S. In this section I have looked only at questions which can be adequately treated in terms of the underlying algebras A, without using any auxiliary structure. To go much farther we shall need to study the ‘function spaces’ S(A) and L∞ (A) of Chapter 36. In particular, the ideas of 326Yg, 326Yj-326Yk and 326Ym-326Yq will make better sense when redeveloped in §362.

327 Additive functionals on measure algebras When we turn to measure algebras, we have a simplification, relative to the general context of §326, because the algebras are always Dedekind σ-complete; but there are also elaborations, because we can ask how the additive functionals we examine are related to the measure. In 327A-327C I work through the relationships between the concepts of ‘absolute continuity’, ‘(true) continuity’ and ‘countable additivity’, following 232A-232B, and adding ‘complete additivity’ from §326. These ideas provide a new interpretation of the Radon-Nikod´ ym theorem (327D). I then use this theorem to develop some machinery (the ‘standard extension’ of an additive functional from a closed subalgebra to the whole algebra, 327F-327G) which will be used in §333. 327A

I start with the following definition and theorem corresponding to 232A-232B.

Definition Let (A, µ ¯) be a measure algebra and ν : A → R a finitely additive functional. Then ν is absolutely continuous with respect to µ ¯ if for every ² > 0 there is a δ > 0 such that |νa| ≤ ² whenever µ ¯a ≤ δ. 327B Theorem Let (A, µ ¯) be a measure algebra, and ν : A → R a finitely additive functional. Give A its measure-algebra topology and uniformity (§323). (a) If ν is continuous, it is completely additive. (b) If ν is countably additive, it is absolutely continuous with respect to µ ¯. (c) The following are equiveridical: (i) ν is continuous at 0; (ii) ν is countably additive and whenever a ∈ A and νa 6= 0 there is a b ∈ A such that µ ¯b < ∞ and ν(a ∩ b) 6= 0; (iii) ν is continuous everywhere on A; (iv) ν is uniformly continuous. (d) If (A, µ ¯) is semi-finite, then ν is continuous iff it is completely additive. (e) If (A, µ ¯) is σ-finite, then ν is continuous iff it is countably additive iff it is completely additive. (f) If (A, µ ¯) is totally finite, then ν is continuous iff it is absolutely continuous with respect to µ ¯ iff it is countably additive iff it is completely additive. proof (a) If ν is continuous, and A ⊆ A is non-empty, downwards-directed and has infimum 0, then 0 ∈ A (323D(b-ii)), so inf a∈A |νa| = 0. (b) ?? Suppose, if possible, that ν is countably additive but not absolutely continuous. Then there is an ² > 0 such that for every δ > 0 there is an a ∈ A such that µ ¯a ≤ δ but |νa| ≥ ². For each n ∈ N we may choose a bn ∈ A such that µ ¯bn ≤ 2−n and |νbn | ≥ ². Consider b∗n = supk≥n bk , b = inf n∈N b∗n . Then we have P∞ µ ¯b ≤ inf n∈N µ ¯(supk≥n bk ) ≤ inf n∈N k=n 2−k = 0, so µ ¯b = 0 and b = 0. On the other hand, ν is expressible as a difference ν + − ν − of non-negative countably additive functionals (326H), each of which is sequentially order-continuous (326Gc), and

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0 = limn→∞ (ν + + ν − )b∗n ≥ inf n∈N (ν + + ν − )bn ≥ inf n∈N |νbn | ≥ ², which is absurd. X X (c)(i)⇒(ii) Suppose that ν is continuous. Then it is completely additive, by (a), therefore countably additive. If νa 6= 0, there must be an b of finite measure such that |νd| < |νa| whenever d ∩ b = ∅, so that |ν(a \ b)| < |νa| and ν(a ∩ b) 6= 0. Thus the conditions are satisfied. (ii)⇒(iv) Now suppose that ν satisfies the two conditions in (ii). Because A is Dedekind σ-complete, ν must be bounded (326I), therefore expressible as the difference ν + − ν − of countably additive functionals. Set ν1 = ν + + ν − . Set γ = sup{ν1 b : b ∈ A, µ ¯b < ∞}, and choose a sequence hbn in∈N of elements of A of finite measure such that limn→∞ ν1 bn = γ; set b∗ = supn∈N bn . If d ∈ A and d ∩ b∗ = ∅ then νd = 0. P P If b ∈ A and µ ¯b < ∞, then |ν(d ∩ b)| ≤ ν1 (d ∩ b) ≤ ν1 (b \ bn ) = ν1 (b ∪ bn ) − ν1 bn ≤ γ − ν1 bn for every n ∈ N, so ν(d ∩ b) = 0. As b is arbitrary, the second condition tells us that νd = 0. Q Q Setting b∗n = supk≤n bk for each n, we have limn→∞ ν1 (b∗ \ b∗n ) = 0. Take any ² > 0, and (using (b) above) let δ > 0 be such that |νa| ≤ ² whenever µ ¯a ≤ δ. Let n be such that ν1 (b∗ \ b∗n ) ≤ ². Then |νa| ≤ |ν(a ∩ b∗n )| + |ν(a ∩ (b∗ \ b∗n ))| + |ν(a \ b∗ )| ≤ |ν(a ∩ b∗n )| + ν1 (b∗ \ b∗n ) ≤ |ν(a ∩ b∗n )| + ² for any a ∈ A. Now if b, c ∈ A and µ ¯((b 4 c) ∩ b∗n ) ≤ δ then |νb − νc| ≤ |ν(b \ c)| + |ν(c \ b)| ≤ |ν((b \ c) ∩ b∗ )| + |ν((c \ b) ∩ b∗ )| + 2² ≤ ² + ² + 2² = 4² because µ ¯((b \ c) ∩ b∗n ), µ ¯((c \ b) ∩ b∗n ) are both less than or equal to δ. As ² is arbitrary, ν is uniformly continuous. (iv)⇒(iii)⇒(i) are trivial. (d) One implication is covered by (a). For the other, suppose that ν is completely additive. Then it is countably additive. On the other hand, if νa 6= 0, consider B = {b : b ⊆ a, µ ¯b < ∞}. Then B is upwardsdirected and sup B = a, because µ ¯ is semi-finite (322Eb), so {a \ b : b ∈ B} is downwards-directed and has infimum 0. Accordingly inf b∈B |ν(a \ b)| = 0, and there must be a b ∈ B such that νb 6= 0. But this means that condition (ii) of (c) is satisfied, so that ν is continuous. (e) Now suppose that (A, µ ¯) is σ-finite. In this case A is ccc (322G) so complete additivity and countable additivity are the same (326L) and we have a special case of (d). (f ) Finally, suppose that µ ¯1 < ∞ and that ν is absolutely continuous with respect to µ ¯. If A ⊆ A is non-empty and downwards-directed and has infimum 0, then inf a∈A µ ¯a = 0 (321F), so inf a∈A |νa| must be 0; thus ν is completely additive. With (b) and (e) this shows that all four conditions are equiveridical. 327C Proposition Let (X, Σ, µ) be a measure space and (A, µ ¯) its measure algebra. (a) There is a one-to-one correspondence between finitely additive functionals ν¯ on A and finitely additive functionals ν on Σ such that νE = 0 whenever µE = 0, given by the formula ν¯E • = νE for every E ∈ Σ. (b) In (a), ν¯ is absolutely continuous with respect to µ ¯ iff ν is absolutely continuous with respect to µ. (c) In (a), ν¯ is countably additive iff ν is countably additive; so that we have a one-to-one correspondence between the countably additive functionals on A and the absolutely continuous countably additive functionals on Σ. (d) In (a), ν¯ is continuous for the measure-algebra topology on A iff ν is truly continuous in the sense of 232Ab. (e) Suppose that µ is semi-finite. Then, in (a), ν¯ is completely additive iff ν is truly continuous.

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proof (a) This should be nearly obvious. If ν¯ : A → R is additive, then the formula defines a functional ν : Σ → R which is additive by 326Be. Also, of course, µE = 0 =⇒ E • = 0 =⇒ νE = 0. On the other hand, if ν is an additive functional on Σ which is zero on negligible sets, then, for E, F ∈ Σ, E • = F • =⇒ µ(E \ F ) = µ(F \ E) = 0 =⇒ ν(E \ F ) = ν(F \ E) = 0 =⇒ νF = νE − ν(E \ F ) + ν(F \ E) = νE, so we have a function ν¯ : A → R defined by the given formula. If E, F ∈ Σ and E • ∩ F • = 0, then ν¯(E • ∪ F • ) = ν¯(E ∪ F )• = ν(E ∪ F ) = ν(E \ F ) + νF = ν¯E • + ν¯F • because (E \ F )• = E • \ F • = E • . Thus ν¯ is additive, and the correspondence is complete. (b) This is immediate from the definitions. (c)(i) If ν is countably additive, and han in∈N is a disjoint sequence in A, we can express it as hEn in∈N S where hEn in∈N is a sequence in Σ. Setting Fn = En \ i α¯ µ¹ C]] for every α ∈ R. (b)(i) Rν extends ν for every ν ∈ Mσ (C). (ii) R is linear and order-preserving. (iii) R(¯ µ¹ C) = µ ¯. P∞ ¯c for every (iv) If hν i Pn∞n∈N is a sequence of non-negative functionals in Mσ (C) such that n=0 νn c = µ ¯a for every a ∈ A. c ∈ C, then n=0 (Rνn )(a) = µ Remarks When saying that C is ‘closed’, I mean, indifferently, ‘topologically closed’ or ‘order-closed’; see 323H-323I. For the notation ‘[[ν > α¯ µ]]’ see 326O-326P. proof (a)(i) By 321J-321K, we may represent (A, µ ¯) as the measure algebra of a measure space (X, Σ, µ); write π for the canonical map from Σ to A. Write T for {E : E ∈ Σ, πE ∈ C}. Because C is a σ-subalgebra of C and π is a sequentially order-continuous Boolean homomorphism, T is a σ-subalgebra of Σ. (ii) For each ν ∈ Mσ (C), νπ : T → R is countably additive and zero on {F : F ∈ T, µF = 0}, so we R can choose a T-measurable function fν : X → R such that F fν d(µ¹ T) = νπF for every F ∈ T. Of course we can now think of fν as a µ-integrable function (233B), so we get a corresponding countably additive R functional Rν : A → R defined by setting (Rν)(πE) = E fν for every E ∈ Σ (327D). (In this context, of course, countably additive functionals are completely additive, by 327Bf.) For α ∈ R, set Hα = {x : fν (x) > α} ∈ T. Then for any E ∈ Σ, E ⊆ Hα , µE > 0 =⇒ E ∩ Hα = ∅ =⇒

R

R E

E

fν > αµE,

fν ≤ αµE.

Translating into terms of elements of A, and setting cα = πHα ∈ C, we have 0 6= a ⊆ cα =⇒ (Rν)(a) > α¯ µa, a ∩ cα = 0 =⇒ (Rν)(a) ≤ α¯ µa.

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So [[Rν > α¯ µ]] = cα ∈ C. Of course we now have νc = (Rν)(c) > α¯ µc when c ∈ C, 0 6= c ⊆ cα , νc ≤ α¯ µc when c ∈ C, c ∩ cα = 0, so that cα is also equal to [[ν > µ ¯¹ C]]. Thus the functional Rν satisfies the declared formula. (iii) To see that Rν is uniquely defined, observe that if λ ∈ MσR(A) and [[λ > α¯ µ]] = [[Rν > α¯ µ]] for every α, then there is a Σ-measurable function g : X → R such that E g dµ = λπE for every E ∈ Σ; but in this case (just as in (ii)) [[λ > α¯ µ]] = πGα , where Gα = {x : g(x) > α}, for each α. So we must have πGα = πHα , that is, µ(Gα 4Hα ) = 0, for every α. Accordingly S {x : fν (x) 6= g(x)} = q∈Q Gq 4Hq R R is negligible; fν = g a.e., E fν dµ = E g dµ for every E ∈ Σ and λ = Rν. (b)(i) If ν ∈ Mσ (C), (Rν)(πF ) =

R F

fν dµ =

R F

fν d(µ¹ T) = νπF

for every F ∈ T, so Rν extends ν. (ii) If ν = ν1 + ν2 , we must have

R

F

fν = νπF = ν1 πF + ν2 πF =

R F

fν1 +

R F

fν2 =

R F

fν1 + fν2

for every F ∈ T, so fν = fν1 + fν2 a.e., and we can repeat the formulae (Rν)(πE) =

R

f = E ν

R

f + fν2 = E ν1

R

f + E ν1

R

E

fν2 = (Rν1 )(πE) + (Rν2 )(πE),

in a different order, for every E ∈ Σ, to see that Rν = Rν1 + Rν2 . Similarly, if ν ∈ Mσ (C) and γ ∈ R, fγν = γfν a.e. and R(γν) = γRν. If ν1 ≤ ν2 in Mσ (C), then

R

F

fν1 = ν1 πF ≤ ν2 πF =

R

F

fν2

for every F ∈ T, so fν1 ≤ fν2 a.e. (131Ha), and Rν1 ≤ Rν2 . Thus R is linear and order-preserving. (iii) If ν = µ ¯¹ C then

R F

fν = νπF = µF =

R F

1

for every F ∈ T, so fν = 1 a.e. and Rν = µ ¯. (iv) Now suppose that hνP n in∈N is a sequence in Mσ (C) such that, for every c ∈ C, νn c ≥ 0 for every n P ∞ n ¯c. Set gn = i=0 fνi for each n; then 0 ≤ gn ≤ gn+1 ≤ 1 a.e. for every n, and n=0 νn c = µ R Pn limn→∞ gn = limn→∞ i=0 νi 1 = µ ¯1. R R But this means that, setting g = limn→∞ gn , g ≤ 1 a.e. and g = 1, so that g = 1 a.e. and R P∞ n=0 (Rνi )(πE) = limn→∞ E gn = µE P∞ for every E ∈ Σ, so that n=0 (Rνi )(a) = µ ¯a for every a ∈ A. and

327G Definition In the context of 327F, I will call Rν the standard extension of ν to A. Remark The point of my insistence on the uniqueness of R, and on the formula in 327Fa, is that Rν really is defined by the abstract structure (A, µ ¯, C, ν), even though I have used a proof which runs through the representation of (A, µ ¯) as the measure algebra of a measure space (X, Σ, µ). 327X Basic exercises (a) Let (A, µ ¯) and (B, µ ¯0 ) be totally finite measure algebras, and π : A → B a measure-preserving Boolean homomorphism. Let C be a closed subalgebra of A, and ν a countably additive functional on the closed subalgebra π[C] of B. (i) Show that νπ is a countably additive functional on C. (ii) Show that if ν˜ is the standard extension of ν to B, then ν˜π is the standard extension of νπ to A.

327 Notes

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(b) Let (X, Σ, µ) be a probability space, and T a σ-subalgebra of Σ. Let (A, µ ¯) be the measure algebra of (X, Σ, µ). Show that C = {F • : F ∈ T} is a closed subalgebra of A. Identify the spaces Mσ (A), Mσ (C) of countably additive functionals with L1 (µ), L1 (µ¹ T), as in 327D. Show that the conditional expectation operator P : L1 (µ) → L1 (µ¹ T) (242Jd) corresponds to the map ν 7→ ν¹ C : Mσ (A) → Mσ (C). (c) Let (A, µ ¯) be a totally finite measure algebra and ν : A → R a countably additive functional. Show that, for any a ∈ A, νa =

R∞ 0

µ ¯(a ∩ [[ν > α¯ µ]])dα −

R0

−∞

µ ¯(a \ [[ν > α¯ µ]])dα,

the integrals being taken with respect to Lebesgue measure. (Hint: take (A, µ ¯) to be the measure algebra of (X, Σ, µ); represent ν by a µ-integrable function f ; apply Fubini’s theorem to the sets {(x, t) : x ∈ E, 0 ≤ t < f (x)}, {(x, t) : x ∈ E, f (x) ≤ t ≤ 0} in X × R, where a = E • .) (d) Let (A, µ ¯) be a totally finite measure algebra, C a closed subalgebra of A and ν : C → R a countably additive functional with standard extension ν˜ : A → R. Show that, for any a ∈ A, ν˜a =

R∞ 0

µ ¯(a ∩ [[ν > α¯ µ¹ C]])dα −

R0

−∞

µ ¯(a \ [[ν > α¯ µ¹ C]])dα.

(e) Let (A, µ ¯) be a probability algebra, and B, C stochastically independent closed subalgebras of A (definition: 325L). Let ν be a countably additive functional on C, and ν˜ its standard extension to A. Show that ν˜(b ∩ c) = µ ¯b · νc for every b ∈ B, c ∈ C. (f ) Let (X, Σ, µ) be a probability space, and T a σ-subalgebra of Σ. Let ν be a probability measure with domain T such that νE = 0 whenever E ∈ T and µE = 0. Show that there is a probability measure λ with domain Σ which extends ν. 327Y Further exercises (a) Let (A1 , µ ¯1 ) and (A2 , µ ¯2 ) be localizable measure algebras with localizable ¯ Show that if ν1 , ν2 are completely additive functionals on A1 , A2 measure algebra free product (C, λ). respectively, there is a unique completely additive functional ν : C → R such that ν(a1 ⊗ a2 ) = ν1 a1 · ν2 a2 for every a1 ∈ A1 , a2 ∈ A2 . (Hint: 253D.) (b) Let (A, µ ¯) be a totally finite measure algebra and C a closed subalgebra; let R : Mσ (C) → Mσ (A) be the standard extension operator (327G). Show (i) that R is order-continuous (ii) that R(ν + ) = (Rν)+ , kRνk = kνk for every ν ∈ Mσ (C), defining ν + and kνk as in 326Yj. (c) Let (A, µ ¯) be a totally finite measure algebra and C a closed subalgebra of A. For a countably additive functional ν on C write ν˜ for its standard extension to A. Show that if ν, hνn in∈N are countably additive functionals on C and limn→∞ νn c = νc for every c ∈ C, then limn→∞ ν˜n a = ν˜a for every a ∈ A. (Hint: use ideas from §§246-247, as well as from 327G and 326Yo.) 327 Notes and comments When we come to measure algebras, it is the completely additive functionals which fit most naturally into the topological theory (327Bd); they correspond to the ‘truly continuous’ functionals which I discussed in §232 (327Cc), and therefore to the Radon-Nikod´ ym theorem (327D). I will return to some of these questions in Chapter 36. I myself regard the form here as the best expression of the essence of the Radon-Nikod´ ym theorem, if not the one most commonly applied. The concept of ‘standard extension’ of a countably additive functional (or, as we could equally well say, of a completely additive functional, since in the context of 327F the two coincide) is in a sense dual to the concept of ‘conditional expectation’. If (X, Σ, µ) is a probability space and T is a σ-subalgebra of Σ, then we have a corresponding closed subalgebra C of the measure algebra (A, µ ¯) of µ, and identifications between the spaces Mσ (A), Mσ (C) of countably additive functionals and the spaces L1 (µ), L1 (µ¹ T). Now we have a natural embedding S of L1 (µ¹ T) as a subspace of L1 (µ) (242Jb), and a natural restriction map from Mσ (A) to Mσ (C). These give rise to corresponding operators between the opposite members of each pair; the standard extension operator R of 327G, and the conditional expectation operator P of 242Jd. (See 327Xb.) The fundamental fact

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327 Notes

P Sv = v for every v ∈ L1 (µ¹ T) (242Jg) is matched by the fact that Rν¹ C = ν for every ν ∈ Mσ (C). R The further identification of Rν in terms of integrals µ ¯(a ∩ [[ν > α¯ µ]])dα (327Xc) is relatively inessential, but is striking, and perhaps makes it easier to believe that R is truly ‘standard’ in the abstract contexts which will arise in §333 below. It is also useful in such calculations as 327Xe. The isomorphisms between Mτ spaces and L1 spaces described here mean that any of the concepts involving L1 spaces discussed in Chapter 24 can be applied to Mτ spaces, at least in the case of measure algebras. In fact, as I will show in Chapter 36, there is much more to be said here; the space of bounded additive functionals on a Boolean algebra is already an L1 space in an abstract sense, and ideas such as ‘uniform integrability’ are relevant and significant there, as well as in the spaces of countably additive and completely additive functionals. I hope that 326Yj, 326Ym-326Yn, 326Yp-326Yq and 327Yb will provide some hints to be going on with for the moment.

331B

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123

Chapter 33 Maharam’s theorem We are now ready for the astonishing central fact about measure algebras: there are very few of them. Any localizable measure algebra has a canonical expression as a simple product of measure algebras of easily described types. This complete classification necessarily dominates all further discussion of measure algebras; to the point that all the results of Chapter 32 have to be regarded as ‘elementary’, since however complex their formulation they have been proved by techniques not involving, nor providing, any particular insight into the special nature of measure algebras. The proof depends, of course, on developing methods of defining measure-preserving homomorphisms and isomorphisms; I give a number of results, progressively more elaborate, but all based on the same idea. These techniques are of great power, leading, for instance, to an effective classification of closed subalgebras and their embeddings. ‘Maharam’s theorem’ itself, the classification of localizable measure algebras, is in §332. I devote §331 to the definition and description of ‘homogeneous’ probability algebras. In §333 I turn to the problem of describing pairs (A, C) where A is a probability algebra and C is a closed subalgebra. Finally, in §334, I give some straightforward results on the classification of free products of probability algebras.

331 Classification of homogeneous measure algebras I embark directly on the principal theorem of this chapter (331I), split between 331B, 331D and 331I; 331B and 331D will be the basis of many of the results in later sections of this chapter. In 331E-331H I introduce the concepts of ‘Maharam type’ and ‘Maharam homogeneity’. I discuss the measure algebras of products {0, 1}κ , showing that these provide a complete set of examples of Maharam-type-homogeneous probability algebras (331J-331L). I end the section with a brief comment on ‘homogeneous’ Boolean algebras (331M-331N). 331A Definition The following idea is almost the key to the whole chapter. Let A be a Boolean algebra and B an order-closed subalgebra of A. A non-zero element a of A is a relative atom over B if every c ⊆ a is of the form a ∩ b for some b ∈ B; that is, {a ∩ b : b ∈ B} is the principal ideal generated by a. We say that A is relatively atomless over B if there are no relative atoms in A over B. (I’m afraid the phrases ‘relative atom’, ‘relatively atomless’ are bound to seem opaque at this stage. I hope that after the structure theory of §333 they will seem more natural. For the moment, note only that a is an atom in A iff it is a relative atom over the smallest subalgebra {0, 1}, and every element of A is a relative atom over the largest subalgebra A. In a way, a is a relative atom over B if its image is an atom in a kind of quotient A/B. But we are two volumes away from any prospect of making sense of this kind of quotient.) 331B

The first lemma is the heart of Maharam’s theorem.

Lemma Let (A, µ ¯) be a totally finite measure algebra and B a closed subalgebra of A such that A is relatively atomless over B. Let ν : B → R be an additive functional such that 0 ≤ νb ≤ µ ¯b for every b ∈ B. Then there is a c ∈ A such that νb = µ ¯(b ∩ c) for every b ∈ B. Remark Recall that by 323H we need not distinguish between ‘order-closed’ and ‘topologically closed’ subalgebras. proof (a) It is worth noting straight away that ν is necessarily countably additive. This is easy to check from first principles, but if you want to trace the underlying ideas they are in 313O (the identity map from B to A is order-continuous), 326Ff (so µ¹ B : B → R is countably additive) and 326Gb (therefore ν is countably additive). (b) For each a ∈ A set νa b = µ ¯(b ∩ a) for every b ∈ B; then νa is countably additive (326Fd). The key P Because idea is the following fact: for every non-zero a ∈ A there is a non-zero d ⊆ a such that νd ≤ 21 νa . P A is relatively atomless over B, there is an e ⊆ a such that e 6= a ∩ b for any b ∈ B. Consider the countably

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331B

additive functional λ = νa − 2νe : B → R. By 326I, there is a b0 ∈ B such that λb ≥ 0 whenever b ∈ B, b ⊆ b0 , while λb ≤ 0 whenever b ∈ B, b ∩ b0 = 0. If e ∩ b0 6= 0, try d = e ∩ b0 . Then 0 6= d ⊆ a, and for every b ∈ B 1 2

1 2

νd b = νe (b ∩ b0 ) = (νa (b ∩ b0 ) − λ(b ∩ b0 )) ≤ νa b (because λ(b ∩ b0 ) ≥ 0) so νd ≤ 21 νa . If e ∩ b0 = 0, then (by the choice of e) e 6= a ∩ (1 \ b0 ), so d = a \ (e ∪ b0 ) 6= 0, and of course d ⊆ a. In this case, for every b ∈ B, 1 2

1 2

νd b = νa (b \ b0 ) − νe (b \ b0 ) = (λ(b \ b0 ) + νa (b \ b0 )) ≤ νa b (because λ(b \ b0 ) ≤ 0), so once again νd ≤ 21 νa . Thus in either case we have a suitable d. Q Q (c) It follows at once, by induction on n, that if a is any non-zero element of A and n ∈ N then there is a non-zero d ⊆ a such that νd ≤ 2−n νa . (d) Now let C be the set {a : a ∈ A, νa ≤ ν}. Then 0 ∈ C, so C 6= ∅. If D ⊆ C is upwards-directed and not empty, then a = sup D is defined in A, and νsup D b = µ ¯(b ∩ sup D) = µ ¯(supd∈D b ∩ d) = supd∈D µ ¯(b ∩ d) = supd∈D νd b ≤ νb using 313Ba and 321C. So a ∈ C and is an upper bound for D in C. In particular, any non-empty totally ordered subset of C has an upper bound in C. By Zorn’s Lemma, C has a maximal element c say. (e) ?? Suppose, if possible, that νc 6= ν. Then there is some b∗ ∈ B such that νc b∗ 6= νb∗ ; since νc ≤ ν, νc b∗ < νb∗ . Let n ∈ N be such that νb∗ > νc b∗ + 2−n µ ¯b∗ , and set λb = νb − νc b − 2−n µ ¯b for every b ∈ B. By 326I (for the second time), there is a b0 ∈ B such that λb ≥ 0 for b ∈ B, b ⊆ b0 , while λb ≤ 0 when b ∈ B and b ∩ b0 = 0. We have µ ¯(b0 \ c) = µ ¯b0 − µ ¯(b0 ∩ c) ≥ νb0 − νc b0 ≥ λb0 = λb∗ + λ(b0 \ b∗ ) − λ(b∗ \ b0 ) ≥ λb∗ > 0, so b0 \ c 6= 0. (This is where I use the hypothesis that ν ≤ µ ¯¹ B.) By (c), there is a non-zero d ⊆ b0 \ c such that νd ≤ 2−n νb0 \c ≤ 2−n νb0 . Now d ∩ c = 0 so c ⊂ d ∪ c. Also, for any b ∈ B, νd∪c b = νd b + νc (b ∩ b0 ) + νc (b \ b0 ) ≤ 2−n νb0 \c b + ν(b ∩ b0 ) − 2−n µ ¯(b ∩ b0 ) − λ(b ∩ b0 ) + ν(b \ b0 ) ≤ 2−n µ ¯(b ∩ b0 \ c) + ν(b ∩ b0 ) − 2−n µ ¯(b ∩ b0 ) + ν(b \ b0 ) ≤ νb. But this means that d ∪ c ∈ C and c is not maximal in C. X X Thus c is the required element of A giving a representation of ν. 331C Corollary Let (A, µ ¯) be an atomless semi-finite measure algebra, and a ∈ A. Suppose that 0≤γ≤µ ¯a. Then there is a c ∈ A such that c ⊆ a and µ ¯c = γ. proof If γ = µ ¯a, take c = a. If γ < µ ¯a, there is a d ∈ A such that d ⊆ a and γ ≤ µ ¯d < ∞ (322Eb). Apply 331B to the principal ideal Ad generated by d, with B = {0, d} and νd = γ. (The point is that because A is atomless, no non-trivial principal ideal of Ad can be of the form {c ∩ b : b ∈ B} = {0, c}.) Remark Of course this is also an easy consequence of 215D.

331Fb

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125

331D Lemma Let (A, µ ¯), (B, ν¯) be totally finite measure algebras and C ⊆ A a closed subalgebra. Suppose that π : C → B is a measure-preserving Boolean homomorphism such that B is relatively atomless over π[C]. Take any a ∈ A, and let C1 be the subalgebra of A generated by C ∪ {a}. Then there is a measure-preserving homomorphism from C1 to B extending π. proof We know that π[C] is a closed subalgebra of B (324Kb), and that π is a Boolean isomorphism between C and π[C]. Consequently the countably additive functional c 7→ µ ¯(c ∩ a) : C → R is transferred to a countably additive functional λ : π[C] → R, writing λ(πc) = µ ¯(c ∩ a) for every c ∈ C. Of course λ(πc) ≤ µ ¯c = ν¯(πc) for every c ∈ C. So by 331B there is a b ∈ B such that λ(πc) = ν¯(b ∩ πc) for every c ∈ C. If c ∈ C, c ⊆ a then ν¯(b ∩ πc) = λ(πc) = µ ¯(a ∩ c) = µ ¯c = ν¯(πc), so πc ⊆ b. Similarly, if a ⊆ c ∈ C, then ν¯(b ∩ πc) = µ ¯(a ∩ c) = µ ¯(a ∩ 1) = ν¯(b ∩ π1) = ν¯b, so b ⊆ πc. It follows from 312N that there is a Boolean homomorphism π1 : C1 → B, extending π, such that π1 a = b. To see that π1 is measure-preserving, take any member of C1 . By 312M, this is expressible as e = (c1 ∩ a) ∪ (c2 \ a), where c1 , c2 ∈ C. Now ν¯(π1 e) = ν¯((πc1 ∩ b) ∪ (πc2 \ b)) = ν¯(πc1 ∩ b) + ν¯(πc2 ) − ν¯(πc2 ∩ b) =µ ¯(c1 ∩ a) + µ ¯ c2 − µ ¯(c2 ∩ a) = µ ¯e. As e is arbitrary, π1 is measure-preserving. 331E Generating sets For the sake of the next definition, we need a language a little more precise than I have felt the need to use so far. The point is that if A is a Boolean algebra and B is a subset of A, there is more than one subalgebra of A which can be said to be ‘generated’ by B, because we can look at any of the three algebras – B, the smallest subalgebra of A including B; – Bσ , the smallest σ-subalgebra of A including B; – Bτ , the smallest order-closed subalgebra of A including B. (See 313Fb.) Now I will say henceforth, in this context, that – B is the subalgebra of A generated by B, and B generates A if A = B; – Bσ is the σ-subalgebra of A generated by B, and B σ-generates A if A = Bσ ; – Bτ is the order-closed subalgebra of A generated by B, and B τ -generates or completely generates A if A = Bτ . There is a danger inherent in these phrases, because if we have B ⊆ A0 , where A0 is a subalgebra of A, it is possible that the smallest order-closed subalgebra of A0 including B might not be recoverable from the smallest order-closed subalgebra of A including B. (See 331Yb-331Yc.) This problem will not seriously interfere with the ideas below; but for definiteness let me say that the phrases ‘B σ-generates A’, ‘B τ generates A’ will always refer to suprema and infima taken in A itself, not in any larger algebra in which it may be embedded. 331F Maharam types (a) With the language of 331E established, I can now define the Maharam type or complete generation τ (A) of any Boolean algebra A; it is the smallest cardinal of any subset of A which τ -generates A. (I think that this is the first ‘cardinal function’ which I have mentioned in this treatise. All you need to know, to confirm that the definition is well-conceived, is that there is some set which τ -generates A; and obviously A τ -generates itself. For this means that the set A = {#(B) : B ⊆ A τ -generates A} is a non-empty class of cardinals, and therefore, assuming the axiom of choice, has a least member (2A1Lf). In 331Ye-331Yf I mention a further function, the ‘density’ of a topological space, which is closely related to Maharam type.) (b) A Boolean algebra A is Maharam-type-homogeneous if τ (Aa ) = τ (A) for every non-zero a ∈ A, writing Aa for the principal ideal of A generated by a.

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331Fc

(c) Let (X, Σ, µ) be a measure space, with measure algebra (A, µ ¯). Then the Maharam type of (X, Σ, µ), or of µ, is the Maharam type of A; and (X, Σ, µ), or µ, is Maharam-type-homogeneous if A is. Remark I should perhaps remark that the phrases ‘Maharam type’ and ‘Maharam-type-homogeneous’, while well established in the context of probability algebras, are not in common use for general Boolean algebras. But the cardinal τ (A) is important in the general context, and is such an obvious extension of Maharam’s idea (Maharam 42) that I am happy to propose this extension of terminology. 331G For the sake of those who have not mixed set theory and algebra before, I had better spell out some basic facts. Proposition Let A be a Boolean algebra, B a subset of A. Let B be the subalgebra of A generated by B, Bσ the σ-subalgebra of A generated by B, and Bτ the order-closed subalgebra of A generated by B. (a) B ⊆ Bσ ⊆ Bτ . (b) If B is finite, so is B, and in this case B = Bσ = Bτ . (c) For every a ∈ B, there is a finite B 0 ⊆ B such that a belongs to the subalgebra of A generated by B 0 . Consequently #(B) ≤ max(ω, #(B)). (d) For every a ∈ Bσ , there is a countable B 0 ⊆ B such that a belongs to the σ-subalgebra of A generated by B 0 . (e) If A is ccc, then Bσ = Bτ . proof (a) All we need to know is that Bσ is a subalgebra of A including B, and that Bτ is a σ-subalgebra of A including B. (b) Induce on #(B), using 312M for the inductive step, to see that B is finite. In this case it must be order-closed, so is equal to Bτ . S (c)(i) For I ⊆ B, let CI be the subalgebra of A generated by I. If I, J ⊆ B then CI ∪ CJ ⊆ CI∪J . So {CI : I ⊆ B is finite} is a subalgebra of A, and must be equal to B, as claimed. (ii) To estimate the size of B, recall that the set [B] δ µ ¯¹ C]]; then c0 ∈ C and 0 6= c0 ⊆ c. Given that an ∈ Ab , cn ∈ C and 0 6= cn ⊆ c, then there must be an+1 ∈ Ab , dn ∈ C such that dn ⊆ cn and νan+1 (dn ) > νan (dn ). Set cn+1 = dn ∩ [[νan+1 > νan ]], so that cn+1 ∈ C and 0 6= cn+1 ⊆ cn , and continue. There is some n ∈ N such that nδ ≥ 1. For any i < n, the construction ensures that 0 6= cn+1 ⊆ ci+1 ⊆ [[νai+1 > νai ]], so νai (cn+1 ) < νai+1 (cn+1 ); also cn+1 ⊆ c0 so Pn−1

µ ¯(ai ∩ cn+1 ) = νai (cn+1 ) ≥ νa0 (cn+1 ) > δ µ ¯cn+1 .

But this means that i=0 µ ¯(ai ∩ cn+1 ) > µ ¯cn+1 and there must be distinct j, k < n such that aj ∩ ak ∩ cn+1 is non-zero. Because aj , ak ∈ A there are d0 , d00 ∈ C such that aj ∩ ak = aj ∩ d0 = ak ∩ d00 ; set d = cn+1 ∩ d0 ∩ d00 , so that d ∈ C and aj ∩ d = aj ∩ ak ∩ cn+1 = ak ∩ d,

νaj (d) = µ ¯(aj ∩ ak ∩ cn+1 ) = νak (d).

But as 0 6= d ⊆ [[νai+1 > νai ]] for every i < n, νa0 (d) < νa1 (d) < . . . < νan (d), so this is impossible. X XQ Q (c) Now for a global, rather than local, version of the same idea. For every b ∈ A there is an a ∈ Ab such that and νa ≥ νe whenever e ∈ Ab . P P (i) By (b), the set C of those c ∈ C such that there is an a ∈ Ab such that νa ¹ Cc ≥ νe ¹ Cc for every e ∈ Ab is order-dense in C. Let hci ii∈I be a partition of unity in C consisting of members of C, and for each i ∈ I choose ai ∈ Ab such that νai ¹ Cci ≥ νe ¹ Cci for every e ∈ Ab . Consider a = supi∈I ai ∩ ci . (ii) If a0 ∈ A and a0 ⊆ a, then for each i ∈ I there is a di ∈ C such that ai ∩ a0 = ai ∩ di . Set d0 = supi∈I ci ∩ di ; then (because hci ii∈I is disjoint) a ∩ d0 = supi∈I ai ∩ ci ∩ di = supi∈I ai ∩ ci ∩ a0 = a ∩ a0 = a0 . As a0 is arbitrary, this shows that a ∈ A. (iii) Of course a ∩ b = 0, so a ∈ Ab . Now take any e ∈ Ab , d ∈ C. Then P P νa (d) = i∈I νai (ci ∩ d) ≥ i∈I νe (ci ∩ d) = νe (d).

333K

Closed subalgebras

149

So this a has the required property. Q Q (d) Choose han in∈N inductively in A so that, for each n, an ∩ supi 0]]. We may therefore choose c0 , . . . , cn+1 ∈ Cc \ {0} and k(0), . . . , k(n) ∈ N such that c0 = c and, for i ≤ n, ci ∩ a0i ∩ ak(i) 6= 0 (choosing k(i), recalling that 0 6= ci ⊆ c ⊆ [[µ0i > 0]]), ci+1 ∈ C, ci+1 ⊆ ci , ci+1 ∩ a0i = ci+1 ∩ ak(i) = ci ∩ a0i ∩ ak(i) (choosing ci+1 , using the fact that a0i and ak(i) both belong to A – see the penultimate sentence in part (b) of the proof.) On reaching cn+1 , we have 0 6= cn+1 ⊆ c so µn (cn+1 ) < µ0n (cn+1 ). On the other hand, for each i ≤ n, cn+1 ∩ a0i ∩ ak(i) = cn+1 ∩ ci+1 ∩ a0i ∩ ak(i) = cn+1 ∩ a0i = cn+1 ∩ ak(i) , so µn (cn+1 ) < µ0n (cn+1 ) ≤ µ0i (cn+1 ) = µ ¯(cn+1 ∩ a0i ) = µ ¯(cn+1 ∩ ak(i) ) = µk(i) (cn+1 ), and k(i) must be less than n. There are therefore distinct i, j ≤ n such that k(i) = k(j). But in this case cn+1 ∩ a0i = cn+1 ∩ ak(i) = cn+1 ∩ ak(j) = cn+1 ∩ a0j 6= 0 because 0 6= cn+1 ⊆ [[µ0j > 0]]. So a0i , a0j cannot be disjoint, breaking one of the rules of the construction. X X Thus µn = µ0n for every n ∈ N. This completes the proof. 333K Theorem Let (A, µ ¯) be a totally finite measure algebra and C a closed subalgebra of A. Then there are unique families hµn in∈N , hµκ iκ∈K such that K is a countable set of infinite cardinals, P for i ∈ N ∪ K, µi is a non-negative countably additive functional on C, and i∈N∪K µi c = µ ¯c for every c ∈ C, µn+1 ≤ µn for every n ∈ N, and µκ 6= 0 for κ ∈ K, setting ei = [[µi > 0]] ∈ C, and giving the principal ideal Cei generated by ei the measure µi ¹ Cei for each i ∈ N ∪ K, and writing (Bκ , ν¯κ ) for the measure algebra of the usual measure on {0, 1}κ for κ ∈ K, we have a measure algebra isomorphism Q Q b κ π : A → n∈N Cen × κ∈K Ceκ ⊗B such that πc = (hc ∩ en in∈N , h(c ∩ eκ ) ⊗ 1iκ∈K ) b κ of c ∈ Ceκ . for each c ∈ C, writing c ⊗ 1 for the canonical image in Ceκ ⊗B proof (a) I aim to use the construction of 333H, but taking much more care over the choice of hai ii∈I in part (a) of the proof there. We start by taking han in∈N as in 333J, and setting µn c = µ ¯(an ∩ c) for every n ∈ N, c ∈ C; then these an will deal with the part in sup A, as defined in the proof of 333J.

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333K

(b) The further idea required here concerns the treatment of infinite κ. Let hbi ii∈I be any partition of unity in A consisting of non-zero members of A which are relatively Maharam-type-homogeneous over C, and hκi ii∈I the corresponding cardinals, so that κi = 0 iff bi ∈ A. Set I1 = {i : i ∈ I, κi ≥ ω}. Set K = {κi : i ∈ I1 }, so that K is a countable set of infinite cardinals, and for κ ∈ K set Jκ = {i : κi = κ}, aκ = supi∈Jκ bi for κ ∈ K. Now every aκ is relatively Maharam-type-homogeneous over C. P P (Compare 332H.) Jκ must be countable, because A is ccc. If 0 6= a ⊆ aκ , there is some i ∈ Jκ such that a ∩ bi 6= 0; now τCa (Aa ) ≥ τCa∩bi (Aa∩bi ) = κi = κ. At the same time, S for each i ∈ Jκ , there is a set Di ⊆ Abi such that #(Di ) = κ and Cbi ∪ Di τ -generates Abi . Set D = i∈Jκ Di ∪ {bi : i ∈ Jκ }; then #(D) ≤ max(ω, #(Jκ ), supi∈K #(Di )) = κ. Let B be the closed subalgebra of Aaκ generated by Caκ ∪ D. Then Cbi ∪ Di ⊆ {b ∩ bi : b ∈ B} = B ∩ Abi so B ⊇ Abi for each i ∈ Jκ , and B = Aaκ . Thus Caκ ∪ D τ -generates Aaκ , and τCaκ (Aaκ ) ≤ κ ≤ min06=a ⊆ aκ τCa (Aa ). This shows that aκ is relatively Maharam-type-homogeneousover C, with τCaκ (Aaκ ) = κ. Q Q Since evidently hJκ iκ∈K and haκ iκ∈K are disjoint, and supκ∈K aκ = supi∈I1 bi , this process yields a partition hai ii∈N∪K of unity in A. Now the arguments of 333H show that we get an isomorphism π of the kind described. (c) To see that the families hµn in∈N , hµκ iκ∈K (and therefore the ei and the (Cei , µi ¹ Cei ), but not π) are uniquely defined, argue as follows. Take families h˜ µn in∈N , h˜ µκ iκ∈K˜ which correspond to an isomorphism Q Q b κ, π ˜ : A → D = n∈N Ce˜n × κ∈K˜ Ce˜κ ⊗B Q ˜ In the simple product Q b κ , we have a partition writing e˜i = [[˜ µi > 0]] for i ∈ N ∪ K. ˜ Ce˜κ ⊗B n∈N Ce˜n × κ∈K of unity he∗i ii∈N∪K˜ corresponding to the product structure. Now for d ⊆ e∗i , we have τπ˜ [C]d (Dd ) = 0 if i ∈ N, ˜ = κ if i = κ ∈ K. ˜ must be So K {κ : κ ≥ ω, ∃ a ∈ A, τCa (Aa ) = κ} = K, ˜ and for κ ∈ K, π ˜ −1 e∗κ = sup{a : a ∈ A, τCa (Aa ) = κ} = aκ , so that µ ˜κ = µκ . On the other hand, h˜ π −1 e∗n in∈N must be a disjoint sequence with supremum sup A, and the corresponding functionals µ ˜n are supposed to form a non-increasing sequence, so must be equal to the µn by 333J. 333L Remark Thus for the classification of structures (A, µ ¯, C), where (A, µ ¯) is a totally finite measure algebra and C is a closed subalgebra, it will be enough to classify objects (C, µ ¯, hµn in∈N , hµκ iκ∈K ), where (C, µ ¯) is a totally finite measure algebra, hµn in∈N is a non-increasing sequence of non-negative countably additive functionals on C, K is a countable set of infinite cardinals (possibly empty), hµκ iκ∈K is a family of non-zero non-negative countably additive functionals on C, P∞ P ¯. n=0 µn + κ∈K µκ = µ To do this we need the concept of ‘standard extension’ of a countably additive functional on a closed subalgebra of a measure algebra, treated in 327F-327G, together with the following idea.

333M

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333M Lemma Let (C, µ ¯) be a totally finite measure algebra and hµi ii∈I a family of countably additive functionals on C. For i ∈ I, α ∈ R set eiα = [[µi > α¯ µ]] (326P), and let C0 be the closed subalgebra of C generated by {eiα : i ∈ I, α ∈ R}. Write Σ for the σ-algebra of subsets of RI generated by sets of the form Eiα = {x : x(i) > α} as i runs through I, α runs over R. Then (a) there is a measure µ, with domain Σ, such that there is a measure-preserving isomorphism π : Σ/Nµ → • C0 for which πEiα = eiα for every i ∈ I, α ∈ R, writing Nµ for µ−1 [{0}]; (b) this formula determines both µ and π; (c) for every E ∈ Σ, i ∈ I, we have µi πE • =

R

E

x(i)µ(dx);

(d) for every i ∈ I, µi is the standard extension of µi ¹ C0 to C; (e) for every i ∈ I, µi ≥ 0 iff x(i) ≥ 0 for µ-almost every x; (f) for every i, j ∈ I, µi ≥ µj iff x(i) ≥ x(j) for µ-almost every x; (g) for every i ∈ I, µi = 0 iff x(i) = 0 for µ-almost every x. proof (a) Express (C, µ ¯) as the measure algebra of a measure space (Y, T, ν); write φ : T → C for the corresponding homomorphism. For each i ∈ I let fi : Y → R be a T-measurable, ν-integrable function such R that H fi = µi φH for every H ∈ T. Define ψ : Y → RI by setting ψ(y) = hfi (y)ii∈I ; then ψ −1 [Eiα ] ∈ Σ, and eiα = φ(ψ −1 [Eiα ]) for every i ∈ I, α ∈ R. (See part (a) of the proof of 327F.) So {E : E ⊆ RI , ψ −1 [E] ∈ T}, which is a σ-algebra of subsets of RI , contains every Eiα , and therefore includes Σ; that is, ψ −1 [E] ∈ T for every E ∈ Σ. Accordingly we may define µ by setting µE = νψ −1 [E] for every E ∈ Σ, and µ will be a measure on RI with domain Σ. The Boolean homomorphism E 7→ φψ −1 [E] : Σ → C has kernel Nµ , so descends to a homomorphism π : Σ/Nµ → C, which is measure-preserving. To see that π[Σ/Nµ ] = C0 , observe that because Σ is the σ-algebra generated by {Eiα : i ∈ I, α ∈ R}, π[Σ/Nµ ] must be the closed • subalgebra of C generated by {πEiα : i ∈ I, α ∈ R} = {eiα : i ∈ I, α ∈ R}, which is C0 . (b) Now suppose that µ0 , π 0 have the same properties. Consider A = {E : E ∈ Σ, πE • = π 0 E ◦ }, where I write E • for the equivalence class of E in Σ/Nµ , and E ◦ for the equivalence class of E in Σ/Nµ0 . Then A is a σ-subalgebra of Σ, because E 7→ πE • , E 7→ π 0 E ◦ are both sequentially order-continuous Boolean homomorphisms, and contains every Eiα , so must be the whole of Σ. Consequently µE = µ ¯πE • = µ ¯π 0 E ◦ = µ0 E for every E ∈ Σ, and µ0 = µ; it follows at once that π 0 = π. So µ and π are uniquely determined. (c) If E ∈ Σ and i ∈ I, Z

Z x(i)µ(dx) =

Z x(i)χE(x)µ(dx) =

ψ(y)(i)χE(ψ(y))ν(dy)

E

(applying 235I to the inverse-measure-preserving function ψ : Y → RI ) Z = fi (y)ν(dy) ψ −1 [E]

(by the definition of ψ) = µi φ(ψ −1 [E]) (by the choice of fi ) = µi πE • by the definition of π. (d) Because [[µi > α¯ µ]] ∈ C0 , it is equal to [[µi ¹ C0 > α¯ µ¹ C0 ]], for every α ∈ R. So µi must be the standard extension of µi ¹ C0 (327F). (e)-(g) The point is that, because the standard-extension operator is order-preserving (327F(b-ii)),

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µi ≥ 0 ⇐⇒ µi ¹ C0 ≥ 0 Z ⇐⇒ x(i)µ(dx) ≥ 0 for every E ∈ Σ E

⇐⇒ x(i) ≥ 0 µ-a.e., µi ≥ µj ⇐⇒ µi ¹ C0 ≥ µj ¹ C0 Z Z ⇐⇒ x(i)µ(dx) ≥ x(j)µ(dx) for every E ∈ Σ E

E

⇐⇒ x(i) ≥ x(j) µ-a.e., µi = 0 ⇐⇒ µi ¹ C0 = 0 Z ⇐⇒ x(i)µ(dx) = 0 for every E ∈ Σ E

⇐⇒ x(i) = 0 µ-a.e.. 333N A canonical form for closed subalgebras We now have all the elements required to describe a canonical form for structures (A, µ ¯, C), where (A, µ ¯) is a totally finite measure algebra and C is a closed subalgebra of A. The first step is the matching of such structures with structures (C, µ ¯, hµn in∈N , hµκ iκ∈K ), where (C, µ ¯) is a totally finite measure algebra, hµn in∈N is a non-increasing sequence of non-negative countably additive functionals on C, K is a countable set of cardinals, hµκ iκ∈K is a family of non-zero Pinfinite P ∞ ¯; this is the burden of 333K. non-negative countably additive functionals on C, and n=0 µn + κ∈K µκ = µ Next, given any structure of this second kind, we have a corresponding closed subalgebra C0 of C, a measure µ on RI , where I = N ∪ K, and an isomorphism π from the measure algebra C∗0 of µ to C0 , all uniquely defined from the family hµi ii∈I by the process of 333M. For any E belonging to the domain Σ of µ, and i ∈ I, we have µi πE • =

R

E

x(i)µ(dx)

(333Mc), so that µi ¹ C0 is fixed by π and µ. Moreover, the functionals µi can be recovered from their restrictions to C0 by the formulae of 327F (333Md). Thus from (C, µ ¯, hµi ii∈I ) we are led, by a canonical and reversible process, to the structure (C, µ ¯, C0 , I, µ, π). But the extension C of C0 = π[C∗0 ] can be described, up to isomorphism, by the same process as before; that is, it corresponds to a sequence h¯ νn in∈N and a family h¯ νκ iκ∈L of countably additive functionals on C0 satisfying the conditions of 333K. We can transfer these to C∗0 , where they correspond to families hνn in∈N , hνκ iκ∈L of absolutely continuous countably additive functionals defined on Σ, setting νj E = ν¯j πE • for E ∈ Σ, j ∈ N∪L. This process too is reversible; every absolutely continuous countably additive functional ν on Σ corresponds to countably additive functionals on C∗0 and C0 . Let me repeat that the results of 327F mean that the whole structure (C, µ ¯, hµi ii∈I ) can be recovered from (C0 , µ ¯¹ C0 , hµi ¹ C 0 ii∈I ) if we can get the P description of (C, µ ¯) right, and that the requirements µi ≥ 0, µn ≥ µn+1 , µκ 6= 0, i∈I µi = µ ¯ imposed in 333K will survive the process (327F(b-iv)). Putting all this together, a structure (A, µ ¯, C) leads, in a canonical and (up to isomorphism) reversible way, to a structure (K, µ, L, hνκ iκ∈N∪L ) such that

333P

Closed subalgebras

153

K and L are countable sets of infinite cardinals, µ is a totally finite measure on RI , where I = N ∪ K, and its domain Σ is precisely the σ-algebra of subsets of RI defined by the coordinate functionals, forPµ-almost every x ∈ RI we have x(i) ≥ 0 for every i ∈ I, x(n) ≥ x(n + 1) for every n ∈ N and i∈I x(i) = 1, for κ ∈ K, µ{x : x(κ) > 0} > 0, P (these two sections corresponding to the requirements µi ≥ 0, µn ≥ µn+1 , ¯, µκ 6= 0 – see i∈I µi = µ 333M(e)-(g)) for j ∈ J = N ∪ L, νj is a non-negative countably additive functional on Σ, P νn ≥ νn+1 for every n ∈ N, νκ 6= 0 for every κ ∈ L, j∈J νj = µ. 333O Remark I do not envisage quoting the result above very often. Indeed I do not claim that its final form adds anything to the constituent results 333K, 327F and 333M. I have taken the trouble to spell it out, however, because it does not seem to me obvious that the trail is going to end quite as quickly as it does. We need to use 333K twice, but only twice. The most important use of the ideas expressed here, I suppose, is in constructing examples to strengthen our intuition for the structures (A, µ ¯, C) under consideration, and I hope that you will experiment in this direction. 333P At the risk of trespassing on the province of Chapter 38, I turn now to a special type of closed subalgebra, in which there is a particularly elegant alternative form for a canonical description. The first step is an important result concerning automorphisms of homogeneous probability algebras. Proposition Let (B, ν¯) be a homogeneous probability algebra. Then there is a measure-preserving automorphism φ : B → B such that limn→∞ ν¯(c ∩ φn (b)) = ν¯c · ν¯b for all b, c ∈ B. proof (a) The case B = {0, 1} is trivial (φ is, and must be, the identity map) so we may take it that B is the measure algebra of {0, 1}κ with its usual measure νκ , where κ is an infinite cardinal. Because #(κ × Z) = max(ω, κ) = κ, there must be a bijection θ : κ → κ such that every orbit of θ in κ is infinite (take θ to correspond to the bijection (ξ, n) 7→ (ξ, n + 1) : κ × Z → κ × Z). This induces a ˆ bijection θˆ : {0, 1}κ → {0, 1}κ through the formula θ(x) = xθ for every x ∈ {0, 1}κ , and of course θˆ is an κ automorphism of the measure space ({0, 1} , νκ ). It therefore induces a corresponding automorphism φ of B, setting φE • = (θˆ−1 [E])• for every E in the domain Σ of νκ . (b) Let Σ0 be the family of subsets E of {0, 1}κ which are determined by coordinates in finite sets, ˜ for some finite set J ⊆ κ and some E ˜ ⊆ {0, 1}J ; that is, are expressible in the form E = {x : x¹J ∈ E} equivalently, expressible as a finite union of basic cylinder sets {x : x¹J = y}. Then Σ0 is a subalgebra of Σ, so C = {E • : E ∈ Σ0 } is a subalgebra of B. (c) Now if b, c ∈ C, there is an n ∈ N such that ν¯(c ∩ φm (b)) = ν¯c · ν¯b for every m ≥ n. P P Express b, c • • ˜ ˜ as E , F where E = {x : x¹J ∈ E}, F = {x : x¹K ∈ F } and J, K are finite subsets of κ. For ξ ∈ K, all the θn (ξ) are distinct, so only finitely many of them can belong to J; as K is also finite, there is an n such that θm [J] ∩ K = ∅ for every m ≥ n. Fix m ≥ n. Then φm (b) = H • where ˜ = {x : x¹L ∈ H}, ˜ H = {x : xθm ∈ E} = {x : xθm ¹J ∈ E} ˜ = {zθ−m : z ∈ E}. ˜ So ν¯(c ∩ φm (b)) = ν(F ∩ H). But L and K are disjoint, because where L = θm [J] and H m ≥ n, so F and H must be independent (cf. 272K), and ν¯(c ∩ φm (b)) = νF · νH = νF · νE = ν¯c · ν¯b, as claimed. Q Q (d) Now recall that for every E ∈ Σ, ² > 0 there is an E 0 ∈ Σ0 such that ν(E4E 0 ) ≤ ² (254Fe). So, given b, c ∈ B and ² > 0, we can find b0 , c0 ∈ C such that ν¯(b 4 b0 ) ≤ ² and ν¯(c 4 c0 ) ≤ ², and in this case

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Maharam’s theorem

333P

lim sup|¯ ν (c ∩ φn (b)) − ν¯c · ν¯b| n→∞

≤ lim sup |¯ ν (c ∩ φn (b)) − ν¯(c0 ∩ φn (b0 ))| n→∞

+ |¯ ν (c0 ∩ φn (b0 )) − ν¯c0 · ν¯b0 | + |¯ ν c · ν¯b − ν¯c0 · ν¯b0 | = lim sup |¯ ν (c ∩ φn (b)) − ν¯(c0 ∩ φn (b0 ))| + |¯ ν c · ν¯b − ν¯c0 · ν¯b0 | n→∞

≤ lim sup ν¯(c 4 c0 ) + ν¯(φn (b) 4 φn (b0 )) n→∞

+ ν¯c|¯ ν b − ν¯b0 | + |¯ ν c − ν¯c0 |¯ ν b0 ≤ ν¯(c 4 c0 ) + ν¯(b 4 b0 ) + ν¯c · ν¯(b 4 b0 ) + ν¯(c 4 c0 )¯ ν b0 ≤ 4². As ² is arbitrary, limn→∞ ν¯(c ∩ φn (b)) = ν¯c · ν¯b, as required. Remark Automorphisms of this type are called mixing (see 372P below). 333Q Corollary Let (C, µ ¯0 ) be a totally finite measure algebra and (B, ν¯) a probability algebra which is either homogeneous or purely atomic with finitely many atoms all of the same measure. Let (A, µ ¯) be the localizable measure algebra free product of (C, µ ¯0 ) and (B, ν¯). Then there is a measure-preserving automorphism π : A → A such that {a : a ∈ A, πa = a} = {c ⊗ 1 : c ∈ C}. Remark I am following 315M in using the notation c ⊗ b for the intersection in A of the canonical images of c ∈ C and b ∈ B. By 325Dc I need not distinguish between the free product C ⊗ B and its image in A. proof (a) Let me deal with the case of atomic B first. In this case, if B has n atoms b0 , . . . , bn−1 , let φ : B → B be the measure-preserving homomorphism cyclically permuting these atoms, so that φb0 = b1 , . . . , φbn−1 = b0 . Because φ is an automorphism of (B, ν¯), it induces an automorphism π of (A, µ ¯); any member of A is uniquely expressible as a = supi µn ]] = 0. P P?? Otherwise, set c = [[µ0 > µn ]] ∩ e0n . Then µ0 c > 0 so c ∩ a0 6= 0. By the last remark in (γ), there is a π ∈ G∗ such that c ∩ a0 ∩ πan 6= 0. Now there is a c0 ∈ C such that c ∩ a0 ∩ πan = c0 ∩ a0 , and of course we may suppose that c0 ⊆ c. But this means that π(c0 ∩ an ) = c0 ∩ πan ⊇ c0 ∩ a0 ∩ πan = c0 ∩ a0 , so that µn c0 = µ ¯(c0 ∩ an ) = µ ¯π(c0 ∩ an ) ≥ µ ¯(c0 ∩ a0 ) = µ0 c0 , which is impossible, because 0 6= c0 ⊆ [[µ0 > µn ]]. X XQ Q So µ0 c ≤ µn c whenever c ∈ C and c ⊆ e0n . Because the µk have been chosen to be a non-increasing sequence, we must have µ0 c = µ1 c = . . . = µn c for every c ⊆ e0n . P 1 (²²) Recalling now that i∈I 0 µi = µ ¯¹ C, we see that µ0 c ≤ n+1 µ ¯c for every c ⊆ e0n . It follows that if ∗ 0 ∗ ∗ e = inf n∈N en , µ0 e = 0; but this must mean that e = 0. Consequently, setting I = I 0 \ {0}, en = e0n−1 \ e0n for n ≥ 1, eκ = aκ for κ ∈ K, we find that hei ii∈I is a partition of unity in C. Moreover, for n ≥ 1 and c ⊆ en , we must have P P µ ¯c = i∈I 0 µi c = k 0. We have τ (C) ≤ max(ω, #(I), τ (A)) = max(#(I), τ (A)), by 334C. (b) Fix on b ∈ A \ {0, 1}. For each i ∈ I, let ψi : A → C be the canonical measure-preserving homomorphism corresponding to the inverse-measure-preserving function x 7→ x(i) : X I → X. For each n ∈ N, there is a set J ⊆ I of cardinal n, and now the finite subalgebra of C generated by {ψi b : i ∈ J} has atoms of measure at most δ n , where δ = max(¯ µb, 1 − µ ¯b) < 1. Consequently C can have no atom of measure greater than δ n , for any n, and is therefore atomless.

160

Maharam’s theorem

334E

(c) Because I is infinite, there is a bijection between I and I × N; that is, there is a partition hJi ii∈I of I into countably infinite sets. Now (X I , λ) can be identified with the product of the family h(X Ji , λi )ii∈I , where λi is the product measure on X Ji (254N). By (b), every λi is atomless, so there are sets Ei ⊆ X Ji of measure 21 . The sets Ei0 = {x : x¹Ji ∈ Ei } are now stochastically independent in X. Accordingly we have an inverse-measure-preserving function f : X → {0, 1}I , endowed with its usual measure νI , defined by setting f (x)(i) = 1 if x ∈ Ei0 , 0 otherwise, and therefore a measure-preserving Boolean homomorphism π : BI → C, writing BI for the measure algebra of νI . Now if c ∈ C \ {0} and Cc is the corresponding ideal, b 7→ c ∩ πb : BI → Cc is an order-continuous Boolean homomorphism. It follows that τ (Cc ) ≥ #(I) (331J). (d) Again take any non-zero c ∈ C. For each i ∈ I, set ai = inf{a : ψi a ⊇ c}. Writing Aai for the corresponding principal ideal of A, we have an order-continuous Boolean homomorphism ψi0 : Aai → Cc , given by the formula ψi0 a = ψi a ∩ c for every a ∈ Aai . Now ψi0 is injective, so is a Boolean isomorphism between Aai and its image ψi0 [Aai ], which by 314F(a-i) is a closed subalgebra of Cc . So τ (Aai ) = τ (ψi0 [Aai ]) ≤ τ (Cc ) by 332Tb. For any finite J ⊆ I, ¯ ≤ λ(inf ¯ 0 < λc i∈J ψi ai ) =

Q i∈J

¯ i ai ) = Q ¯ai . λ(ψ i∈J µ

So for any δ < 1, {i : µ ¯ai ≤ δ} must be finite, and supi∈I µ ¯ai = 1. In particular, supi∈I ai = 1 in A. But this means that if ζ is any cardinal such that the Maharam-type-ζ component eζ of A is non-zero, then eζ ∩ ai 6= 0 for some i ∈ I, so that ζ ≤ τ (Aeζ ∩ai ) ≤ τ (Aai ) ≤ τ (Cc ). As ζ is arbitrary, τ (A) ≤ max(ω, τ (Cc )) (332S). (e) Putting (a)-(d) together, we have max(τ (A), #(I)) ≤ max(ω, τ (Cc )) = τ (Cc ) ≤ τ (C) ≤ max(τ (A), #(I)) for every non-zero c ∈ C; so C is homogeneous. 334X Basic exercises (a) Let (X, Σ, µ) and (Y, T, ν) be complete locally determined measure spaces with c.l.d. product (X × Y, Λ, λ). Let A, C be the measure algebras of µ, λ respectively. Show that if νY > 0 then τ (A) ≤ τ (C). (b) Let X be a set and µ and ν two totally finite measures on X with the same domain Σ; then λ = µ + ν ¯ for the three measure algebras. Show is also a totally finite measure. Write (A, µ ¯), (B, ν¯) and (C, λ) that (i) there is a surjective order-continuous Boolean homomorphism from C onto A; (ii) C is isomorphic to a closed subalgebra of the localizable measure algebra free product of A and B; (iii) τ (A) ≤ τ (C) ≤ max(ω, τ (B), τ (C)). ¯ > (c) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras, with probability algebra free product (C, λ). Show that τ (Ai ) ≤ τ (C) for every i, and that #({i : i ∈ I, τ (Ai ) > 0}) ≤ τ (C). (d) Let X be a set and hµn in∈N a sequence of totally finite measures on X all with the same domain P∞ Σ. Let hα i be a sequence of non-negative real numbers such that αn µn X < ∞, and set λE = n n∈N n=0 P∞ ¯n ) for the measure algebra of µn and n=0 αn µn E for E ∈ Σ. Check that λ is a measure. Write (An , µ ¯ for the measure algebra of λ. Show that τ (C) ≤ max(ω, sup (C, λ) n∈N τ (An )). (e) Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces, and λ the product measure on X × Y . Show that λ is Maharam-type-homogeneous iff one of µ, ν is Maharam-type-homogeneous with Maharam type at least as great as the Maharam type of the other.

334 Notes

Products

161

(f ) Show that the product of any family of Maharam-type-homogeneous probability spaces is again Maharam-type-homogeneous. > (g) Let (X, Σ, µ) be a probability space of Maharam type κ, and I any set of cardinal at least max(ω, κ). Show that the product measure on X × {0, 1}I is Maharam-type-homogeneous, with Maharam type #(I). 334Y Further exercises (a) Let h(Xi , Σi , µi )ii∈I be an infinite family of probability spaces, with product (X, Λ, λ). Let κi be the Maharam type of µi for each i; set κ = max(#(I), supi∈I κi ). Show that either λ is Maharam-type-homogeneous, with Maharam type κ, or there are κ0 < κ, Xi0 ∈ Σi such that P 0 0 0 i∈I µi (Xi \ Xi ) < ∞ and the Maharam type of the subspace measure on Xi is at most κ for every i ∈ I 0 and either κ = 0 or #(I) < κ. 334 Notes and comments The results above are all very natural ones; I have spelt them out partly for completeness and partly for the sake of an application in §346 below. But note the second alternative in 334Ya; it is possible, even in an infinite product, for a kernel of relatively small Maharam type to be preserved.

162

Liftings

Chapter 34 The lifting theorem Whenever we have a surjective homomorphism φ : P → Q, where P and Q are mathematical structures, we can ask whether there is a right inverse of φ, a homomorphism ψ : Q → P such that φψ is the identity on Q. As a general rule, we expect a negative answer; those categories in which epimorphisms always have right inverses (e.g., the category of linear spaces) are rather special, and elsewhere the phenomenon is relatively rare and almost always important. So it is notable that we have a case of this at the very heart of the theory of measure algebras: for any complete probability space (X, Σ, µ) (in fact, for any complete strictly localizable space of non-zero measure) the canonical homomorphism from Σ to the measure algebra of µ has a right inverse (341K). This is the von Neumann-Maharam lifting theorem. Its proof, together with some essentially elementary remarks, takes up the whole of of §341. As a first application of the theorem (there will be others in Volume 4) I apply it to one of the central problems of measure theory: under what circumstances will a homomorphism between measure algebras be representable by a function between measure spaces? Variations on this question are addressed in §343. For a reasonably large proportion of the measure spaces arising naturally in analysis, homomorphisms are representable (343B). New difficulties arise if we ask for isomorphisms of measure algebras to be representable by isomorphisms of measure spaces, and here we have to work rather hard for rather narrowly applicable results; but in the case of Lebesgue measure and its closest relatives, a good deal can be done, as in 344H and 344I. Returning to liftings, there are many difficult questions concerning the extent to which liftings can be required to have special properties, reflecting the natural symmetries of the standard measure spaces. For instance, Lebesgue measure is translation-invariant; if liftings were in any sense canonical, they could be expected to be automatically translation-invariant in some sense. It seems sure that there is no canonical lifting for Lebesgue measure – all constructions of liftings involve radical use of the axiom of choice – but even so we do have many translation-invariant liftings (§345). We have less luck with product spaces; here the construction of liftings which respect the product structure is fraught with difficulties. I give the currently known results in §346.

341 The lifting theorem I embark directly on the principal theorem of this chapter (341K, ‘every non-trivial complete strictly localizable measure space has a lifting’), using the minimum of advance preparation. 341A-341B give the definition of ‘lifting’; the main argument is in 341F-341K, using the concept of ‘lower density’ (341C-341E) and a theorem on martingales from §275. In 341P I describe an alternative way of thinking about liftings in terms of the Stone space of the measure algebra. 341A Definition Let (X, Σ, µ) be a measure space, and A its measure algebra. By a lifting of A (or of (X, Σ, µ), or of µ) I shall mean either a Boolean homomorphism θ : A → Σ such that (θa)• = a for every a ∈ A or a Boolean homomorphism φ : Σ → Σ such that (i) φE = ∅ whenever µE = 0 (ii) µ(E4φE) = 0 for every E ∈ Σ. 341B Remarks (a) I trust that the ambiguities permitted by this terminology will not cause any confusion. The point is that there is a natural one-to-one correspondence between liftings θ : A → Σ and liftings φ : Σ → Σ given by the formula θE • = φE for every E ∈ Σ. P P (i) Given a lifting θ : A → Σ, the formula defines a Boolean homomorphism φ : Σ → Σ such that φ∅ = θ0 = ∅,

(E 4 φE)• = E • 4 (θE • )• = 0 ∀ E ∈ Σ,

341E

The lifting theorem

163

so that φ is a lifting. (ii) Given a lifting φ : Σ → Σ, the kernel of φ includes {E : µE = 0}, so there is a Boolean homomorphism θ : A → Σ such that θE • = φE for every E (3A2G), and now (θE • )• = (φE)• = E • for every E ∈ Σ, so θ is a lifting. Q Q I suppose that the word ‘lifting’ applies most naturally to functions from A to Σ; but for applications in measure theory the other type of lifting is used at least equally often. (b) Note that if φ : Σ → Σ is a lifting then φ2 = φ. P P For any E ∈ Σ, φ2 E 4 φE = φ(E 4 φE) = ∅. Q Q If φ is associated with θ : A → Σ, then φθa = θa for every a ∈ A. P P φθa = θ((θa)• ) = θa. Q Q 341C Definition Let (X, Σ, µ) be a measure space, and A its measure algebra. By a lower density of A (or of (X, Σ, µ), or of µ) I shall mean either a function θ : A → Σ such that (i) (θa)• = a for every a ∈ A (ii) θ0 = ∅ (iii) θ(a ∩ b) = θa ∩ θb for all a, b ∈ A or a function φ : Σ → Σ such that (i) φE = φF whenever E, F ∈ Σ and µ(E4F ) = 0 (ii) µ(E4φE) = 0 for every E ∈ Σ (iii) φ∅ = ∅ (iv) φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ. 341D Remarks (a) As in 341B, there is a natural one-to-one correspondence between lower densities θ : A → Σ and lower densities φ : Σ → Σ given by the formula θE • = φE for every E ∈ Σ. (For the requirement φE = φF whenever E • = F • in A means that every φ corresponds to a function θ, and the other clauses match each other directly.) (b) As before, if φ : Σ → Σ is a lower density then φ2 = φ. If φ is associated with θ : A → Σ, then φθ = θ. (c) It will be convenient, in the course of the proofs of 341F-341H below, to have the following concept available. If (X, Σ, µ) is a measure space with measure algebra A, a partial lower density of A is a function θ : B → Σ such that (i) the domain B of θ is a subalgebra of A (ii) (θb)• = b for every b ∈ B (iii) θ0 = ∅ (iv) θ(a ∩ b) = θa ∩ θb for all a, b ∈ B. Similarly, if T is a subalgebra of Σ, a function φ : T → Σ is a partial lower density if (i) φE = φF whenever E, F ∈ T and µ(E4F ) = 0 (ii) µ(E4φE) = 0 for every E ∈ T (iii) φ∅ = ∅ (iv) φ(E∩F ) = φE∩φF for all E, F ∈ T. (d) Note that lower densities and partial lower densities are order-preserving; if a ⊆ b in A, and θ is a lower density of A, then θa = θ(a ∩ b) = θa ∩ θb ⊆ θb. (e) Of course a Boolean homomorphism from A to Σ, or from Σ to itself, is a lifting iff it is a lower density. 341E Example Let µ be Lebesgue measure on Rr , where r ≥ 1, and Σ its domain. For E ∈ Σ set φE = {x : x ∈ Rr , limδ↓0

µ(E∩B(x,δ)) µB(x,δ)

= 1}.

(Here B(x, δ) is the closed ball with centre x and radius δ.) Then φ is a lower density for µ; we may call it lower Lebesgue density. P P (You may prefer at first to suppose that r = 1, so that B(x, δ) = [x − δ, x + δ] and µB(x, δ) = 2δ.) By 261Db (or 223B, for the one-dimensional case) φE4E is negligible for every E; in particular, φE ∈ Σ for every E ∈ Σ. If E4F is negligible, then µ(E ∩ B(x, δ)) = µ(F ∩ B(x, δ)) for every x and δ, so φE = φF . If E ⊆ F , then µ(E ∩ B(x, δ)) ≤ µ(F ∩ B(x, δ)) for every x, δ, so φE ⊆ φF ; consequently φ(E ∩ F ) ⊆ φE ∩ φF for all E, F ∈ Σ. If E, F ∈ Σ and x ∈ φE ∩ φF , then

164

Liftings

341E

µ(E ∩ F ∩ B(x, δ)) = µ(E ∩ B(x, δ)) + µ(F ∩ B(x, δ)) − µ((E ∪ F ) ∩ B(x, δ)) ≥ µ(E ∩ B(x, δ)) + µ(F ∩ B(x, δ)) − µ(B(x, δ)) for every δ, so µ(E∩F ∩B(x,δ)) µB(x,δ)

≥

µ(E∩B(x,δ)) µB(x,δ)

+

µ(F ∩B(x,δ)) µB(x,δ)

−1→1

as δ ↓ 0, and x ∈ φ(E ∩ F ). Thus φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ, and φ is a lower density. Q Q 341F The hard work of this section is in the proof of 341H below. To make it a little more digestible, I extract two parts of the proof as separate lemmas. Lemma Let (X, Σ, µ) be a probability space and A its measure algebra. Let B be a closed subalgebra of A and θ : B → Σ a partial lower density. Then for any e ∈ A there is a partial lower density θ1 , extending θ, defined on the subalgebra B1 of A generated by B ∪ {e}. proof (a) Because B is order-closed, v = upr(e, B) = inf{a : a ∈ B, a ⊇ e},

w = upr(1 \ e, B)

are defined in B (314V). Let E ∈ Σ be such that E = e. •

(b) We have a function θ1 : B1 → Σ defined by writing ¡ ¢ ¡ ¢ θ1 ((a ∩ e) ∪ (b \ e)) = θ((a ∩ v) ∪ (b \ v)) ∩ E ∪ θ((a \ w) ∪ (b ∩ w)) \ E for a, b ∈ B. P P By 312M, every element of B1 is expressible as (a ∩ e) ∪ (b \ e) for some a, b ∈ B. If a, a0 , 0 b, b ∈ B are such that (a ∩ e) ∪ (b \ e) = (a0 ∩ e) ∪ (b0 \ e), then a ∩ e = a0 ∩ e and b \ e = b0 \ e, that is, a 4 a0 ⊆ 1 \ e ⊆ w,

b 4 b0 ⊆ e ⊆ v.

This means that e ⊆ 1 \ (a 4 a0 ) ∈ B and 1 \ e ⊆ 1 \ (b 4 b0 ) ∈ B. So we also have v ⊆ 1 \ (a 4 a0 ) and w ⊆ 1 \ (b 4 b0 ). Accordingly a ∩ v = a0 ∩ v,

b ∩ w = b0 ∩ w.

a \ w = a0 \ w,

b \ v = b0 \ v.

But this means that ¡ ¢ ¡ ¢ θ((a ∩ v) ∪ (b \ v)) ∩ E ∪ θ((a \ w) ∪ (b ∩ w)) \ E ¡ ¢ ¡ ¢ = θ((a0 ∩ v) ∪ (b0 \ v)) ∩ E ∪ θ((a0 \ w) ∪ (b0 ∩ w)) \ E . Thus the formula given defines θ1 uniquely. Q Q (c) Now θ1 is a lower density. P P(i) If a, b ∈ B, ¢ ¡ ¢¢• θ((a ∩ v) ∪ (b \ v)) ∩ E ∪ θ((a \ w) ∪ (b ∩ w)) \ E ¡ ¢ ¡ ¢ = ((a ∩ v) ∪ (b \ v)) ∩ e ∪ ((a \ w) ∪ (b ∩ w)) \ e

(θ1 ((a ∩ e) ∪ (b \ e)))• =

¡¡

= (a ∩ e) ∪ (b \ e). So (θ1 c)• = c for every c ∈ B1 . (ii)

¡ ¢ ¡ ¢ θ1 (0) = θ((0 ∩ v) ∪ (0 \ v)) ∩ E ∪ θ((0 \ w) ∪ (0 ∩ w)) \ E = ∅.

(iii) If a, a0 , b, b0 ∈ B, then

341G

The lifting theorem

165

θ1 (((a ∩ e) ∪ (b \ e)) ∩ ((a0 ∩ e) ∪ (b0 \ e))) = θ1 ((a ∩ a0 ∩ e) ∪ (b ∩ b0 \ e)) ¡ ¢ ¡ ¢ = θ((a ∩ a0 ∩ v) ∪ (b ∩ b0 \ v)) ∩ E ∪ θ((a ∩ a0 \ w) ∪ (b ∩ b0 ∩ w)) \ E ¡ ¢ = θ(((a ∩ v) ∪ (b \ v)) ∩ ((a0 ∩ v) ∪ (b0 \ v))) ∩ E ¡ ¢ ∪ θ(((a \ w) ∪ (b ∩ w)) ∩ ((a0 \ w) ∪ (b0 ∩ w))) \ E ¡ ¢ = θ((a ∩ v) ∪ (b \ v)) ∩ θ((a0 ∩ v) ∪ (b0 \ v)) ∩ E ¡ ¢ ∪ θ((a \ w) ∪ (b ∩ w)) ∩ θ((a0 \ w) ∪ (b0 ∩ w)) \ E ¢ ¡ ¢¢ ¡¡ = θ((a ∩ v) ∪ (b \ v)) ∩ E ∪ θ((a \ w) ∪ (b ∩ w)) \ E ¡¡ ¢ ¡ ¢¢ ∩ θ((a0 ∩ v) ∪ (b0 \ v)) ∩ E ∪ θ((a0 \ w) ∪ (b0 ∩ w)) \ E = θ1 ((a ∩ e) ∪ (b \ e)) ∩ θ1 ((a0 ∩ e) ∪ (b0 \ e)). Q So θ1 (c ∩ c0 ) = θ1 (c) ∩ θ1 (c0 ) for all c, c0 ∈ B1 . Q (d) If a ∈ B, then θ1 (a) = θ1 ((a ∩ e) ∪ (a \ e)) ¡ ¢ ¡ ¢ = θ((a ∩ v) ∪ (a \ v)) ∩ E ∪ θ((a \ w) ∪ (a ∩ w)) \ E ¡ ¢ ¡ ¢ = θ(a) ∩ E ∪ θ(a) \ E = θa. Thus θ1 extends θ, as required. 341G Lemma Let (X, Σ, µ) be a probability space and (A, µ ¯) its measure algebra. Suppose we have a sequence hθn in∈N of partial lower densities such that, for each n, (i) the domain Bn of θn is a closed subalgebra of A (ii) Bn ⊆ Bn+1 and θn+1 extends θn . Let B be the closed subalgebra of A generated by S B . Then there is a partial lower density θ, with domain B, extending every θn . n n∈N proof (a) For each n, set Σn = {E : E ∈ Σ, E • ∈ Bn }, and set Σ∞ = {E : E ∈ Σ, E • ∈ B}. Then (because all the Bn , B are σ-subalgebras of A, and E 7→ E • is sequentially order-continuous) all the Σn , Σ∞ are is just the σ-algebra Σ∗∞ of subsets of X S σ-subalgebras of Σ. We need to know that Σ∞ S generated by n∈N Σn . P P Because Σ∞ is a σ-algebra including n∈N Σn , Σ∗∞ ⊆ Σ∞ . On the other hand, ∗ • ∗ B = {E : E ∈ Σ∞ } is a σ-subalgebra of A including Bn for every n ∈ N. Because A is ccc, B∗ is (order-)closed (316Fb), so includes B. This means that if E ∈ Σ∞ there must be an F ∈ Σ∗∞ such that E • = F • . But now (E4F )• = 0 ∈ B0 , so E4F ∈ Σ0 ⊆ Σ∗∞ , and E also belongs to Σ∗∞ . This shows that Σ∞ ⊆ Σ∗∞ and the two algebras are equal. Q Q (b) For each n ∈ N, we have the partial lower density θn : Bn → Σ. Since (θn a)• = a ∈ Bn for every a ∈ Bn , θn takes all its values in Σn . For n ∈ N, let φn : Σn → Σn be the lower density corresponding to θn (341Ba), that is, φn E = θn E • for every E ∈ Σn . (c) For a ∈ A, n ∈ N choose Ga ∈ Σ, gan such that G•a = a and gan is a conditional expectation of χGa on Σn ; that is,

R

E

gan =

R

E

χGa = µ(E ∩ Ga ) = µ ¯(E • ∩ a)

for every E ∈ Σn . As remarked in 233Db, such a function gan can always be found, and moreover we may take it to be Σn -measurable and defined everywhere on X, by 232He. Now if a ∈ B, limn→∞ gan (x) exists and is equal to χGa (x) for almost every x. P P By L´evy’s martingale theorem (275I), limn→∞ S gan is defined almost everywhere and is a conditional expectation of χGa on the σ-algebra generated by n∈N Σn . As

166

Liftings

341G

observed in (a), this is just Σ∞ ; and as χGa is itself Σ∞ -measurable, it is also a conditional expectation of itself on Σ∞ , and must be equal almost everywhere to limn→∞ gan . Q Q (d) For a ∈ B, k ≥ 1, n ∈ N set Hkn (a) = {x : x ∈ X, gan (x) ≥ 1 − 2−k } ∈ Σn , θa =

T

S k≥1

T n∈N

m≥n

˜ kn (a) = φn (Hkn (a)), H

˜ km (a). H

The rest of the proof is devoted to showing that θ : B → Σ has the required properties. ˜ kn (0) (e) G0 is negligible, so every g0n is zero almost everywhere, every Hkn (0) is negligible and every H is empty; so θ0 = ∅. P Ga \ Gb is negligible, gan ≤ gbn almost everywhere for every n, every (f ) If a ⊆ b in B, then θa ⊆ θb. P ˜ kn (a) ⊆ H ˜ kn (b) for every n and k, and θa ⊆ θb. Q Hkn (a) \ Hkn (b) is negligible, H Q (g) If a, b ∈ B then θ(a ∩ b) = θa ∩ θb. P P χGa∩b ≥ χGa + χGb − 1 a.e. so ga∩b,n ≥ gan + gbn − 1 a.e. for every n. Accordingly Hk+1,n (a) ∩ Hk+1,n (b) \ Hkn (a ∩ b) is negligible, and (because φn is a lower density) ˜ kn (a ∩ b) ⊇ φn (Hk+1,n (a) ∩ Hk+1,n (b)) = H ˜ k+1,n (a) ∩ H ˜ k+1,n (b) H for all k ≥ 1, n ∈ N. Now, if x ∈ θa ∩ θb, then, for any k ≥ 1, there are n1 , n2 ∈ N such that T ˜ ˜ k+1,m (a), x ∈ T x ∈ m≥n1 H m≥n2 Hk+1,m (b). But this means that x∈

T m≥max(n1 ,n2 )

˜ km (a ∩ b). H

As k is arbitrary, x ∈ θ(a ∩ b); as x is arbitrary, θa ∩ θb ⊆ θ(a ∩ b). We know already from (f) that θ(a ∩ b) ⊆ θa ∩ θb, so θ(a ∩ b) = θa ∩ θb. Q Q P hgan in∈N → χGa a.e., so setting (h) If a ∈ B, then θa• = a. P T S T Va = k≥1 n∈N m≥n Hkm (a) = {x : lim inf n→∞ gan (x) ≥ 1}, Va 4Ga is negligible, and Va• = a; but θa4Va ⊆

S k≥1,n∈N

˜ kn (a) Hkn (a)4H

is also negligible, so θa• is also equal to a. Q Q Thus θ is a partial lower density with domain B. P If a ∈ Bn , then Ga ∈ Σm for every m ≥ n, so gam = χGa (i) Finally, θ extends θn for every n ∈ N. P a.e. for every m ≥ n; Hkm (a)4Ga is negligible for k ≥ 1, m ≥ n; ˜ km = φm Ga = θm a = θn a H for k ≥ 1, m ≥ n (this is where I use the hypothesis that θm+1 extends θm for every m); and θa =

\ [ \

˜ km (a) H

k≥1 r∈N m≥r

=

\ [ \

k≥1 r≥n m≥r

˜ km (a) = H

\ [

θn a = θn a. Q Q

k≥1 r≥n

The proof is complete. 341H

Now for the first main theorem.

Theorem Let (X, Σ, µ) be any strictly localizable measure space. Then it has a lower density φ : Σ → Σ. If µX > 0 we can take φX = X.

341I

The lifting theorem

167

proof : Part A I deal first with the case of probability spaces. Let (X, Σ, µ) be a probability space, and (A, µ ¯) its measure algebra. (a) Set κ = #(A) and enumerate A as haξ iξ 0 for every i ∈ I. For each i ∈ I, let φi : Σi → Σi be a lower density of µi , where Σi = Σ ∩ PXi and µi = µ¹Σi , such that φi Xi = Xi . Then it is easy to check that we have a lower density φ : Σ → Σ given by setting S φE = i∈I φi (E ∩ Xi ) for every E ∈ Σ, and that φX = X. 341I The next step is to give a method of moving from lower densities to liftings. I start with an elementary remark on lower densities on complete measure spaces. Lemma Let (X, Σ, µ) be a complete measure space with measure algebra A. (a) Suppose that θ : A → Σ is a lower density and θ1 : A → PX is a function such that θ1 0 = ∅, θ1 (a ∩ b) = θ1 a ∩ θ1 b for all a, b ∈ A and θ1 a ⊇ θa for all a ∈ A. Then θ1 is a lower density. If θ1 is a Boolean homomorphism, it is a lifting. (b) Suppose that φ : Σ → Σ is a lower density and φ1 : Σ → PX is a function such that φ1 E = φ1 F whenever E4F is negligible, φ1 ∅ = ∅, φ1 (E ∩ F ) = φ1 E ∩ φ1 F for all E, F ∈ Σ and φ1 E ⊇ φE for all E ∈ Σ. Then φ1 is a lower density. If φ1 is a Boolean homomorphism, it is a lifting. proof (a) All I have to check is that θ1 a ∈ Σ and (θ1 a)• = a for every a ∈ A. But θa ⊆ θ1 a, So

θ(1 \ a) ⊆ θ1 (1 \ a),

θ1 a ∩ θ1 (1 \ a) = θ1 0 = ∅.

168

Liftings

341I

θa ⊆ θ1 a ⊆ X \ θ(1 \ a). Since (θa)• = a = (X \ θ(1 \ a))• , and µ is complete, θ1 is a lower density. If it is a Boolean homomorphism, then it is also a lifting (341De). (b) This follows by the same argument, or by looking at the functions from A to Σ defined by φ and φ1 and using (a). 341J Proposition Let (X, Σ, µ) be a complete measure space such that µX > 0, and A its measure algebra. (a) If θ : A → Σ is any lower density, there is a lifting θ : A → Σ such that θa ⊇ θa for every a ∈ A. (b) If φ : Σ → Σ is any lower density, there is a lifting φ : Σ → Σ such that φE ⊇ φE for every E ∈ Σ. proof (a) For each x ∈ θ1, set Ix = {a : a ∈ A, x ∈ θ(1 \ a)}. Then Ix is a proper ideal of A. P P We have 0 ∈ Ix , because x ∈ θ1, if b ⊆ a ∈ Ix then b ∈ Ix , because x ∈ θ(1 \ a) ⊆ θ(1 \ b), if a, b ∈ Ix then a ∪ b ∈ Ix , because x ∈ θ(1 \ a) ∩ θ(1 \ b) = θ(1 \ (a ∪ b)), 1∈ / Ix because x ∈ / ∅ = θ0. Q Q For x ∈ X \ θ1, set Ix = {0}; this is also a proper ideal of A, because A 6= {0}. By 311D, there is a surjective Boolean homomorphism πx : A → {0, 1} such that πx d = 0 for every d ∈ Ix . Define θ : A → PX by setting θa = {x : x ∈ X, πx (a) = 1} for every a ∈ A. It is easy to check that, because every πx is a surjective Boolean homomorphism, θ is a Boolean homomorphism. Now for any a ∈ A, x ∈ X, x ∈ θa =⇒ 1 \ a ∈ Ix =⇒ πx (1 \ a) = 0 =⇒ πx a = 1 =⇒ x ∈ θa. Thus θa ⊇ θa for every a ∈ A. By 341I, θ is a lifting, as required. (b) Repeat the argument above, or apply it, defining θ by setting θ(E • ) = φE for every E ∈ Σ, and φ by setting φE = θ(E • ) for every E. 341K The Lifting Theorem Every complete strictly localizable measure space of non-zero measure has a lifting. proof By 341H, it has a lower density, so by 341J it has a lifting. 341L Remarks If we count 341F-341K as a single argument, it may be the longest proof, after Carleson’s theorem (§286), which I have yet presented in this treatise, and perhaps it will be helpful if I suggest ways of looking at its components. (a) The first point is that the theorem should be thought of as one about probability spaces. The shift to general strictly localizable spaces (Part B of the proof of 341H) is purely a matter of technique. I would not have presented it if I did not think that it’s worth doing, for a variety of reasons, but there is no significant idea needed, and if – for instance – the result were valid only for σ-finite spaces, it would still be one of the great theorems of mathematics. So the rest of these remarks will be directed to the ideas needed in probability spaces. (b) All the proofs I know of the theorem depend in one way or another on an inductive construction. We do not, of course, need a transfinite induction written out in the way I have presented it in 341H above. Essentially the same proof can be presented as an application of Zorn’s Lemma; if we take P to be the set of partial lower densities, then the arguments of 341G and part (A-d) of the proof of 341H can be adapted

341Lf

The lifting theorem

169

to prove that any totally ordered subset of P has an upper bound in P , while the argument of 341F shows that any maximal element of P must have domain A. I think it is purely a matter of taste which form one prefers. I suppose I have used the ordinal-indexed form largely because that seemed appropriate for Maharam’s theorem in the last chapter. (c) There are then three types of inductive step to examine, corresponding to 341F, 341G and (A-d) in 341H. The first and last are easier than the second. Seeking the one-step extension of θ : B → Σ to θ1 : B1 → Σ, the natural model to use is the one-step extension of a Boolean homomorphism presented in 312N. The situation here is rather more complicated, as θ1 is not fully specified by the value of θ1 e, and we do in fact have more freedom at this point than is entirely welcome. The formula used in the proof of 341F ¨ cker 76. is derived from Graf & Weizsa (d) At this point I must call attention to the way in which the whole proof is dominated by the choice of closed subalgebras as the domains of our partial liftings. This is what makes the inductive step to a S limit ordinal ξ of countable cofinality difficult, because Aξ will ordinarily be larger than η 0, then b = H • 6= 0 and bb is a non-empty open set in Z. Because M is nowhere dense, there is a non-zero a ∈ A such that b a ⊆ bb \ M . Now µ(f −1 [bb]4H) = 0, so f −1 [b a] \ H is negligible, and f −1 [b a] ∩ H is a non-negligible measurable −1 set disjoint from E ∩ f [M ] and included in H; which is impossible. X X Thus H and E ∩ f −1 [M ] are negligible. This is true for every measurable set E of finite measure. Because µ is complete and locally determined, f −1 [M ] ∈ Σ and µf −1 [M ] = 0. So f −1 [F ] = f −1 [b a]4f −1 [M ] is measurable, and µf −1 [F ] = µf −1 [b a] = µθa = µ ¯a = νb a = νF . As F is arbitrary, f is inverse-measure-preserving. Q Q It follows at once that for any F ∈ T, f −1 [F ]• = a = πF • where a is that element of A such that M = F 4a is meager, because in this case f −1 [b a]• = a, by (a), while −1 f [M ] is negligible. So π is the homomorphism induced by f . (ii) Now suppose that f : X → Z is an inverse-measure-preserving function such that f −1 [F ]• = πF • for every F ∈ T. Then, in particular, f −1 [b a]• = πb a• = a for every a ∈ A, so that f satisfies the conditions of (a). 341Q Corollary Let (X, Σ, µ) be a strictly localizable measure space, (A, µ ¯) its measure algebra, and Z the Stone space of A; suppose that µX > 0. For E ∈ Σ write E ∗ for the open-and-closed subset of Z corresponding to E • ∈ A. Then there is a function f : X → Z such that E4f −1 [E ∗ ] is negligible for every E ∈ Σ. If µ is complete, then f is inverse-measure-preserving. ˆ its domain. Then we can identify (A, µ proof Let µ ˆ be the completion of µ, and Σ ¯) with the measure ˆ algebra of µ ˆ (322Da). Let θ : A → Σ be a lifting, and f : X → Z the corresponding function. If E ∈ Σ then ˆ = Σ, then f E∗ = b a where a = E • , so E4f −1 [E ∗ ] = E4θE • is negligible. If µ is itself complete, so that Σ is inverse-measure-preserving, by 341Pb. 341X Basic exercises (a) Let (X, Σ, µ) be a measure space and φ : Σ → Σ a function. Show that φ is a lifting iff it is a lower density and φE ∪ φ(X \ E) = X for every E ∈ Σ. > (b) Let µ be the usual measure on X = {0, 1}N , and Σ its domain. For x ∈ X and n ∈ N set Un (x) = {y : y ∈ X, y¹n = x¹n}. For E ∈ Σ set φE = {x : limn→∞ 2n µ(E ∩ Un (x)) = 1}. Show that φ is a lower density of (X, Σ, µ). > (c) Let P be the set of all lower densities of a complete measure space (X, Σ, µ), with measure algebra A, ordered by saying that θ ≤ θ0 if θa ⊆ θ0 a for every a ∈ A. Show that any non-empty totally ordered subset of P has an upper bound in P . Show that if θ ∈ P and a ∈ A and x ∈ X \ (θa ∪ θ(1 \ a)), then θ0 : A → Σ is a lower density, where θ0 b = θb ∪ {x} if either a ⊆ b or there is a c ∈ A such that x ∈ θc and a ∩ c ⊆ b, and θ0 b = θb otherwise. Hence prove 341J. (d) Let (X, Σ, µ) and (Y, T, ν) be measure spaces and suppose that there is an inverse-measure-preserving function f : X → Y such that the associated homomorphism from the measure algebra of ν to that of µ is an isomorphism. Show that for every lifting φ of (Y, T, ν) we have a corresponding lifting ψ of (X, Σ, µ) defined uniquely by the formula ψ(f −1 [F ]) = f −1 [φF ] for every F ∈ T.

341 Notes

The lifting theorem

173

(e) Let (X, Σ, µ) be a measure space, and write L∞ (Σ) for the linear space of all bounded Σ-measurable functions from X to R. Show that for any lifting φ : Σ → Σ of µ there is a unique linear operator T : L∞ (µ) → L∞ (Σ) such that T (χE)• = χ(φE) for every E ∈ Σ and T u ≥ 0 in L∞ (Σ) whenever u ≥ 0 in L∞ (µ). Show that (i) (T u)• = u and supx∈X |(T u)(x)| = kuk∞ for every u ∈ L∞ (µ) (ii) T (u × v) = T u × T v for all u, v ∈ L∞ (µ). 341Y Further exercises (a) Let X be a set, Σ an algebra of subsets of X and I an ideal of Σ; let A be the quotient Boolean algebra Σ/I. We say that a function θ : A → Σ is a lifting if it is a Boolean homomorphism and (θa)• = a for every a ∈ A, and that θ : A → Σ is a lower density if θ0 = ∅, θ(a ∩ b) = θa ∩ θb for all a, b ∈ A, and (θa)• = a for every a ∈ A. Show that if (X, Σ, I) is ‘complete’ in the sense that F ∈ Σ whenever F ⊆ E ∈ I, and if X ∈ / I, and θ : A → Σ is a lower density, then there is a lifting θ : A → Σ such that θa ⊆ θa for every a ∈ A. (b) Let X be a non-empty Baire space, Bb the σ-algebra of subsets of X with the Baire property (314Yd) b and M the ideal of meager subsets of X. Show that there is a lifting θ from B/M to Bb such that θG• ⊇ G • for every open G ⊆ X. (Hint: in 341Ya, set θ(G ) = G for every regular open set G.) (c) Let (X, Σ, µ) be a Maharam-type-homogeneous probability space with Maharam type κ ≥ ω. Let Σ be the Baire σ-algebra of Y = {0, 1}κ , that is, the σ-algebra of subsets of Y generated by the family {{x : x(ξ) = 1} : ξ < κ}, and let ν be the restriction to Σ of the usual measure on {0, 1}κ . Show that there is an inverse-measure-preserving function f : X → Y which induces an isomorphism between their measure algebras. (d) Let (X, Σ, µ) be a complete Maharam-type-homogeneous probability space with Maharam type κ ≥ ω, and let ν be the usual measure on {0, 1}κ . Show that there is an inverse-measure-preserving function f : X → Y which induces an isomorphism between their measure algebras. (e) Let (X, Σ, µ) be a semi-finite measure space which is not purely atomic. Write L1strict for the linear space of integrable functions f : X → R. Show that there is no operator T : L1 (µ) → L1strict such that (i) (T u)• = u for every u ∈ L1 (µ) (ii) T u ≥ T v whenever u ≥ v in L1 (µ). (Hint: Suppose first that µ is the usual measure on {0, 1}N . Let F be the countable set of continuous functions f : {0, 1}N → N. Show that if T satisfies (i) then there is an x ∈ {0, 1}N such that T (f • )(x) = f (x) for every f ∈ F ; find a sequence hfn in∈N in F such that {fn• : n ∈ N} is bounded above in L1 (µ) but supn∈N fn (x) = ∞. Now transfer this argument to some atomless fragment of X.) 341Z Problems (a) Can we construct, using the ordinary axioms of mathematics (including the axiom of choice, but not the continuum hypothesis), a probability space (X, Σ, µ) with no lifting? (b) Set κ = ω3 . (There is a reason for taking ω3 here; see Volume 5, when it appears, or Fremlin 89.) Let Σ be the Baire σ-algebra of X = {0, 1}κ (as in 341Yc), and let µ be the restriction to Σ of the usual measure on {0, 1}κ . Does (X, Σ, µ) have a lifting? 341 Notes and comments Innumerable variations of the proof of 341K have been devised, as each author has struggled with the technical complications. I have discussed the reasons for my own choices in 341L. The theorem has a curious history. It was originally announced by von Neumann, but he seems never to have written his proof down, and the first published proof is that of Maharam 58. That argument is based on Maharam’s theorem, 341Xd and 341Yd, which show that it is enough to find liftings for every {0, 1}κ ; this requires most of the ideas presented above, but feels more concrete, and some of the details are slightly simpler. The argument as I have written it owes a great deal to Ionescu Tulcea & Ionescu Tulcea 69. The lifting theorem and Maharam’s theorem are the twin pillars of modern abstract measure theory. But there remains a degree of mystery about the lifting theorem which is absent from the other. The first point is that there is nothing canonical about the liftings we can construct, except in the quite exceptional case of Stone spaces (341O). Even when there is a more or less canonical lower density present (341E, 341Xb), the conversion of this into a lifting requires arbitrary choices, as in 341J. While we can distinguish some

174

The lifting theorem

341 Notes

liftings as being somewhat more regular than others, I know of no criterion which marks out any particular lifting of Lebesgue measure, for instance, among the rest. Perhaps associated with this arbitrariness is the extreme difficulty of deciding whether liftings of any given type exist. Neither positive nor negative results are easily come by (I will present a few in the later sections of this chapter), and the nature of the obstacles remains quite unclear.

342 Compact measure spaces The next three sections amount to an extended parenthesis, showing how the Lifting Theorem can be used to attack one of the fundamental problems of measure theory: the representation of Boolean homomorphisms between measure algebras by functions between appropriate measure spaces. This section prepares for the main idea by introducing the class of ‘locally compact’ measures (342Ad), with the associated concepts of ‘compact’ and ‘perfect’ measures (342Ac, 342K). These depend on the notions of ‘inner regularity’ (342Aa, 342B) and ‘compact class’ (342Ab, 342D). I list the basic permanence properties for compact and locally compact measures (342G-342I) and mention some of the compact measures which we have already seen (342J). Concerning perfect measures, I content myself with the proof that a locally compact measure is perfect (342L). I end the section with two examples (342M, 342N). 342A Definitions (a) Let (X, Σ, µ) be a measure space. If K ⊆ PX, I will say that µ is inner regular with respect to K if µE = sup{µK : K ∈ K ∩ Σ, K ⊆ E} for every E ∈ Σ. Of course µ is inner regular with respect to K iff it is inner regular with respect to K ∩ Σ. T (b) A family K of sets is a compact class if K0 6= ∅ whenever K0 ⊆ K has the finite intersection property. Note that any subset of a compact class is again a compact class. (In particular, it is convenient to allow the empty set as a compact class.) (c) A measure space (X, Σ, µ), or a measure µ, is compact if µ is inner regular with respect to some compact class of subsets of X. Allowing ∅ as a compact class, and interpreting sup ∅ as 0 in (a) above, µ is a compact measure whenever µX = 0. (d) A measure space (X, Σ, µ), or a measure µ, is locally compact if the subspace measure µE is compact whenever E ∈ Σ and µE < ∞. Remark I ought to point out that the original definitions of ‘compact class’ and ‘compact measure’ (Marczewski 53) correspond to what I call ‘countably compact class’ and ‘countably compact measure’ in Volume 4. For another variation on the concept of ‘compact class’ see condition (β) in 343B(ii)-(iii). For examples of compact measure spaces see 342J. 342B

I prepare the ground with some straightforward lemmas.

Lemma Let (X, Σ, µ) be a measure space, and K ⊆ Σ a set such that whenever E ∈ Σ and µE > 0 there is a K ∈ K such that K ⊆ E and µK > 0. Let E ∈ Σ. S (a) There is a countable disjoint set K1 ⊆ K such that K ⊆ E for every K ∈ K1 and µ( K1 ) = µE. S (b) If µE < ∞ then µ(E \ K1 ) = 0. (c) S In any case, there is for any γ < µE a finite disjoint K0 ⊆ K such that K ⊆ E for every K ∈ K0 and µ( K0 ) ≥ γ.

342F

Compact measure spaces

175

proof Set K0 = {K : K ∈ K, K ⊆ E, µK > 0}. Let K∗ be a maximal disjoint subfamily of K0 . If K∗ is uncountable, then there is someSn ∈ N such that {K : K ∈ K∗ , µK ≥ 2−n } is infinite, so that there is a countable K1 ⊆ K∗ such that µ( K1 ) = ∞ = µE. S If K∗ is countable, set K1 = K∗ . Then F = K1 is measurable, and F ⊆ E. Moreover, there is no member of K0 disjoint from F ; but this means that E \ F must be negligible. So µF = µE, and (a) is true. Now (b) and (c) follow at once, because S S µ( K1 ) = sup{µ( K0 ) : K0 ⊆ K1 is finite}. Remark This lemma can be thought of as two more versions of the principle of exhaustion; compare 215A. 342C Corollary Let (X, Σ, µ) be a measure space and K ⊆ PX a family of sets such that (α) K ∪K 0 ∈ K whenever K, K 0 ∈ K and K ∩ K 0 = ∅ (β) whenever E ∈ Σ and µE > 0, there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. Then µ is inner regular with respect to K. proof Apply 342Bc to K ∩ Σ. 342D Lemma Let X be a set and K a family of subsets of X. (a) The following are equiveridical: (i) K is a compact class; (ii) there is a topology T on X such that X is compact and every member of K is a closed set for T. T (b) If K is a compact class, so are the families K1 = {K0 ∪ . . . ∪ Kn : K0 , . . . , Kn ∈ K} and K2 = { K0 : ∅ 6= K0 ⊆ K}. proof (a)(i)⇒(ii) Let T be the topology generated by {X \ K : K ∈ K}. Then of course every member of K is closed for T. Let F be an ultrafilter on X. Then K ∩ F has T the finite intersection property; because K is a compact class, it has non-empty intersection; take x ∈ X (K ∩ F). The family {G : G ⊆ X, either G ∈ F or x ∈ / G} is easily seen to be a topology on X, and contains X \ K for every K ∈ K (because if X \ K ∈ / F then K ∈ F and x ∈ K), so includes T; but this just means that every T-open set containing x belongs to F, that is, that F → x. As F is arbitrary, X is compact for T (2A3R). (ii)⇒(i) Use 3A3Da. (b) Let T be a topology on X such that X is compact and every member of K is closed for T; then the same is true of every member of K1 or K2 . 342E Corollary Suppose that (X, Σ, µ) is a measure space and that K is a compact class such that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. Then µ is compact. proof Set K1 = {K0 ∪ . . . ∪ Kn : K0 , . . . , Kn ∈ K}. By 342Db, K1 is a compact class, and by 342C µ is inner regular with respect to K1 . 342F Corollary A measure space (X, Σ, µ) is compact iff there is a topology on X such that X is compact and µ is inner regular with respect to the closed sets. proof (a) If µ is inner regular with respect to a compact class K, then there is a compact topology on X such that every member of K is closed; now the family F of closed sets includes K, so µ is also inner regular with respect to F. (b) If there is a compact topology on X such that µ is inner regular with respect to the family K of closed sets, then this is a compact class, so µ is a compact measure.

176

The lifting theorem

342G

342G Now I look at the standard questions concerning preservation of the properties of ‘compactness’ or ‘local compactness’ under the usual manipulations. Proposition (a) Any measurable subspace of a compact measure space is compact. (b) The completion and c.l.d. version of a compact measure space are compact. (c) A semi-finite measure space is compact iff its completion is compact iff its c.l.d. version is compact. (d) The direct sum of a family of compact measure spaces is compact. (e) The c.l.d. product of two compact measure spaces is compact. (f) The product of any family of compact probability spaces is compact. proof (a) Let (X, Σ, µ) be a compact measure space, and E ∈ Σ. If K is a compact class such that µ is inner regular with respect to K, then KE = K ∩ PE is a compact class (just because it is a subset of K) and the subspace measure µE is inner regular with respect to KE . ˇ µ (b) Let (X, Σ, µ) be a compact measure space. Write (X, Σ, ˇ) for either the completion or the c.l.d. version of (X, Σ, µ). Let K ⊆ PX be a compact class such that µ is inner regular with respect to K. Then ˇ and γ < µ µ ˇ is also inner regular with respect to K. P P If E ∈ Σ ˇE there is an E 0 ∈ Σ such that E 0 ⊆ E 0 0 and µE > γ; if µ ˇ is the c.l.d. version of µ, we may take µE to be finite. There is a K ∈ K ∩ Σ such that ˇ Q K ⊆ E 0 and µK ≥ γ. Now µ ˇK = µK ≥ γ and K ⊆ E and K ∈ K ∩ Σ. Q ˇ µ (c) Now suppose that (X, Σ, µ) is semi-finite; again write (X, Σ, ˇ) for either its completion or its c.l.d. version. We already know that if µ is compact, so is µ ˇ. IfTµ ˇ is compact, let K ⊆ PX be a compact class such that µ ˇ is inner regular with respect to K. Set K∗ = { K0 : ∅ 6= K0 ⊆ K}; then K∗ is a compact class (342Db). Now µ is inner regular with respect to K∗ . P P Take E ∈ Σ and γ < µE. Choose hEn in∈N , hKn in∈N as follows. Because µ is semi-finite, there is an E0 ⊆ E such that E0 ∈ Σ and γ < µE0 < ∞. Given En ∈ Σ ˇ such that Kn ⊆ En and µ such that µEn > γ, there is a Kn ∈ K ∩ Σ ˇKn > γ. Now there ∈ Σ such Tis an En+1 T that En+1 ⊆ Kn and µEn+1 > γ. Continue. On completing the induction, set K = n∈N Kn = n∈N En , so that K ∈ K∗ ∩ Σ and K ⊆ E and µK = limn→∞ µEn ≥ γ. As E, γ are arbitrary, µ is inner regular with respect to K∗ . Q Q As K∗ is a compact class, µ is a compact measure. (d) Let h(Xi , Σi , µi )ii∈I be a family of compact measure spaces, with direct sum (X, Σ, µ). We may suppose that each Xi is actually a subset of X, with µi the subspace measure. For S each i ∈ I let Ki ⊆ PXi be a compact class such that µi is inner regular with respect to Ki . Then K = i∈I Ki is a compact class, for if K0 ⊆ K has the finite intersection property, then K0 ⊆ Ki for some i, so has non-empty intersection. Now if E ∈ Σ, µE > 0 there is some i ∈ I such that µi (E ∩ Xi ) > 0, and we can find a K ∈ Ki ∩ Σi ⊆ K ∩ Σ such that K ⊆ E ∩ Xi and µi K > 0, in which case µK > 0. By 342E, µ is compact. (e) Let (X, Σ, µ) and (Y, T, ν) be two compact measure spaces, with c.l.d. product measure (X × Y, Λ, λ). Let T, S be topologies on X, Y respectively such that X and Y are compact spaces and µ, ν are inner regular with respect to the closed sets. Then the product topology on X × Y is compact (3A3J). The point is that λ is inner regular with respect to the family K of closed subsets of X × Y . P P Suppose that W ∈ Λ and λW > γ. Then there are E ∈ Σ, F ∈ T such that µE < ∞, νF < ∞ and λ(W ∩(E×F )) > γ (251F). Now there are sequences hEn in∈N , hFn in∈N in Σ, T respectively such that S (E × F ) \ W ⊆ n∈N En × Fn , P∞ n=0

µEn · νFn < λ((E × F ) \ W ) + λ((E × F ) ∩ W ) − γ = λ(E × F ) − γ

(251C). Set W 0 = (E × F ) \

S n∈N

E n × Fn =

T

n∈N ((E

× (F \ Fn )) ∪ ((E \ En ) × F )).

Then W 0 ⊆ W , and

S P∞ λ((E × F ) \ W 0 ) ≤ λ( n∈N En × Fn ) ≤ n=0 µEn · νFn < λ(E × F ) − γ,

so λW 0 > γ. Set ² = 41 (λW 0 − γ)/(1 + µE + µF ). For each n, we can find closed measurable sets Kn , Kn0 ⊆ X and Ln , L0n ⊆ Y such that Kn ⊆ E,

µ(E \ Kn ) ≤ 2−n ²,

342H

Compact measure spaces

L0n ⊆ F \ Fn ,

ν((F \ Fn ) \ L0n ) ≤ 2−n ²,

Kn0 ⊆ E \ En ,

µ((E \ En ) \ Kn0 ) ≤ 2−n ²,

Ln ⊆ F , Set V =

T

n∈N (Kn

177

ν(F \ Ln ) ≤ 2−n ².

× L0n ) ∪ (Kn0 × Ln ) ⊆ W 0 ⊆ W .

Now W0 \ V ⊆

[

((E \ Kn ) × F ) ∪ (E × ((F \ Fn ) \ L0n ))

n∈N

∪ (((E \ En ) \ Kn0 ) × F ) ∪ (E × (F \ Ln )), so λ(W 0 \ V ) ≤

∞ X

µ(E \ Kn ) · νF + µE · ν((F \ Fn ) \ L0n )

n=0

+ µ((E \ En ) \ Kn0 ) · νF + µE · ν(F \ Ln ) ≤

∞ X

2−n ²(2µE + 2µF ) ≤ λW 0 − γ,

n=0

and λV ≥ γ. But V is a countable intersection of finite unions of products of closed measurable sets, so is itself a closed measurable set, and belongs to K ∩ Λ. Q Q Accordingly the product topology on X × Y witnesses that λ is a compact measure. (f ) The same method works. In detail: let h(Xi , Σi , µi )ii∈I be a family of compact probability spaces, with product (X, Λ, λ). For each i, let Ti be a topology on Xi such that Xi is compact and µi is inner regular with respect to the closed sets. Give X the product topology; this is compact. If W ∈ Λ and S ² > 0, let hCn in∈N be a sequence of measurable cylinder sets (in the sense of 254A) such that X \ W ⊆ n∈N Cn and Q P∞ i∈I Eni where Eni ∈ Σi for each i and Jn = {i : Eni 6= Xi } n=0 λCn ≤ λ(X \ W ) + ². Express each Cn as −n is finite. For n ∈ N, i ∈ Jn set ²ni = 2 ²/(1 + #(Jn )). Choose closed measurable sets Kni ⊆ Xi \ Eni such that µi ((Xi \ Eni ) \ Kni ) ≤ ²ni for n ∈ N, i ∈ Jn . For each n ∈ N, set S Vn = i∈Jn {x : x ∈ X, x(i) ∈ Kni }, so that Vn is a closed measurable subset of X. Observe that X \ Vn = {x : x(i) ∈ X \ Kni for i ∈ Jn } includes Cn , and that

P P λ(X \ (Vn ∪ Cn )) ≤ i∈Jn λ{x : x(i) ∈ Xi \ (Kni ∪ Eni )} ≤ i∈Jn ²ni ≤ 2−n ². T Now set V = n∈N Vn ; then V is again a closed measurable set, and S X \ V ⊆ n∈N Cn ∪ (X \ (Cn ∪ Vn ))

has measure at most

P∞ n=0

λCn + 2−n ² ≤ 1 − λW + ² + 2²,

so λV ≥ λW − 3². As W and ² are arbitrary, λ is inner regular with respect to the closed sets, and is a compact measure. 342H Proposition (a) A compact measure space is locally compact. (b) A strictly localizable locally compact measure space is compact. (c) Let (X, Σ, µ) be a measure space. Suppose that whenever E ∈ Σ and µE > 0 there is an F ∈ Σ such that F ⊆ E, µF > 0 and the subspace measure on F is compact. Then µ is locally compact. proof (a) This is immediate from 342Ga and the definition of ‘locally compact’ measure space.

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342H

(b) Suppose that (X, Σ, µ) is a strictly localizable locally compact measure space. Let hXi ii∈I be a decomposition of X, and for each i ∈ I let µi be the subspace measure on Xi . Then µi is compact. Now µ can be identified with the direct sum of the µi , so itself is compact, by 342Gd. (c) Write F for the set of measurable sets F ⊆ X such that the subspace measures µF are compact. Take E ∈ Σ with µE < ∞. S By 342Bb, there is a countable disjoint 0family hFi ii∈I in F such that Fi ⊆ E 0 for each i, and F = E \ i∈I Fi is negligible; now this means that F ∈ F (342Ac), so we may take it that S E = i∈I Fi . In this case µE is isomorphic to the direct sum of the measures µFi and is compact. As E is arbitrary, µ is locally compact. 342I Proposition (a) Any measurable subspace of a locally compact measure space is locally compact. (b) A measure space is locally compact iff its completion is locally compact iff its c.l.d. version is locally compact. (c) The direct sum of a family of locally compact measure spaces is locally compact. (d) The c.l.d. product of two locally compact measure spaces is locally compact. proof (a) Trivial: if (X, Σ, µ) is locally compact, and E ∈ Σ, and F ⊆ E is a measurable set of finite measure for the subspace measure on E, then F ∈ Σ and µF < ∞, so the subspace measure on F is compact. ˇ µ (b) Let (X, Σ, µ) be a measure space, and write (X, Σ, ˇ) for either its completion or its c.l.d. version. (i) Suppose that µ is locally compact, and that µ ˇF < ∞. Then there is an E ∈ Σ such that E ⊆ F and µE = µ ˇF . Let µE be the subspace measure on E induced by the measure µ; then we are assuming that µE is compact. Let K ⊆ PE be a compact class such that µE is inner regular with respect to K. Then, as in the proof of 342Gb, the subspace measure µ ˇF on F induced by µ ˇ is also inner regular with respect to K, so µ ˇF is compact; as F is arbitrary, µ ˇ is locally compact. (ii) Now suppose that µ ˇ is locally compact, and that µE < ∞. Then the subspace measure µ ˇE is compact. But this is just the completion of the subspace measure µE , so µE is compact, by 342Gc; as E is arbitrary, µ is locally compact. (c) Put (a) and 342Hc together. (d) Let (X, Σ, µ) and (Y, T, ν) be locally compact measure spaces, with product (X × Y, Λ, λ). If W ∈ Λ and λW > 0, there are E ∈ Σ, F ∈ T such that µE < ∞, νF < ∞ and λ(W ∩ (E × F )) > 0. Now the subspace measure λE×F induced by λ on E × F is just the product of the subspace measures (251P(ii-α), so is compact, and the subspace measure λW ∩(E×F ) is therefore again compact, by 342Ga. By 342Hc, this is enough to show that λ is locally compact. 342J Examples It is time I listed some examples of compact measure spaces. (a) Lebesgue measure on Rr is compact. (Let K be the family of subsets of Rr which are compact for the usual topology. By 134Fb, Lebesgue measure is inner regular with respect to K.) (b) Similarly, any Radon measure on Rr (256A) is compact. (c) If (A, µ ¯) is any semi-finite measure algebra, the standard measure ν on its Stone space Z is compact. (By 322Qa, ν is inner regular with respect to the family of open-and-closed subsets of Z, which are all compact for the standard topology of Z, so form a compact class.) (d) The usual measure on {0, 1}I is compact, for any set I. (It is obvious that the usual measure on {0, 1} is compact; now use 342Gf.) Remark (a)-(c) above are special cases of the fact that all Radon measures are compact; I will return to this in §416. 342K One of the most important properties of (locally) compact measure spaces has been studied under the following name. Definition Let (X, Σ, µ) be a measure space. Then (X, Σ, µ), or µ, is perfect if whenever f : X → R is measurable, E ∈ Σ and µE > 0, then there is a compact set K ⊆ f [E] such that µf −1 [K] > 0.

*342N

Compact measure spaces

179

342L Theorem A semi-finite locally compact measure space is perfect. proof Let (X, Σ, µ) be a semi-finite locally compact measure space, f : X → R a measurable function, and E ∈ Σ a set of non-zero measure. Because µ is semi-finite, there is an F ∈ Σ such that F ⊆ E and 0 < µF < ∞. Now the subspace measure µF is compact; let T be a topology on F such that F is compact and µF is inner regular with respect to the family K of closed sets for P T. Let h²q iq∈Q be a family of strictly positive real numbers such that q∈Q ²q < 12 µF . (For instance, you could set ²q(n) = 2−n−3 µF where hq(n)in∈N is an enumeration of Q.) For each q ∈ Q, set Eq = {x : x ∈ F, f (x) ≤ q}, Eq0 = {x : x ∈ F, f (x) > q}, and T choose Kq , Kq0 ∈ K ∩ Σ such that Kq ⊆ Eq , Kq0 ⊆ Eq0 , 0 0 µ(Eq \ Kq ) ≤ ²q and µ(Eq \ Kq ) ≤ ²q . Then K = q∈Q (Kq ∪ Kq0 ) ∈ K ∩ Σ, K ⊆ F and P µ(F \ K) ≤ q∈Q µ(Eq \ Kq ) + µ(Eq0 \ Kq0 ) < µF , so µK > 0. The point is that f ¹K is continuous. P P For any q ∈ Q, {x : x ∈ K, f (x) ≤ q} = K ∩ Kq and {x : x ∈ K, f (x) > q} = K ∩ Kq0 . If H ⊆ R is open and x ∈ K ∩ f −1 [H], take q, q 0 ∈ Q such that f (x) ∈ ]q, q 0 ] ⊆ H; then G = K \ (Kq ∪ Kq0 0 ) is a relatively open subset of K containing x and included in f −1 [H]. Thus K ∩ f −1 [H] is relatively open in K; as H is arbitrary, f is continuous. Q Q Accordingly f [K] is a continuous image of a compact set, therefore compact; it is a subset of f [E], and µf −1 [f [K]] ≥ µK > 0. As f and E are arbitrary, µ is perfect. 342M I ought to give examples to distinguish between the concepts introduced here, partly on general principles, but also because it is not obvious that the concept of ‘locally compact’ measure space is worth spending time on at all. It is easy to distinguish between ‘perfect’ and ‘(locally) compact’; ‘locally compact’ and ‘compact’ are harder to separate. Example Let X be an uncountable set and µ the countable-cocountable measure on X (211R). Then µ is perfect but not compact or locally compact. proof (a) If f : X → R is measurable and E ⊆ X is measurable, with measure greater than 0, set A = {α S : α ∈ R, {x : x ∈ X, f (x) ≤ r} is negligible}. Then α ∈ A whenever α ≤ β ∈ A. Since / A, in which case A is bounded above by X = n∈N {x : f (x) ≤ n}, there is some n such that n ∈ n. Also there is some m ∈ N such that {x : f (x) > −m} is non-negligible, in which case it must be conegligible, and −m ∈ A, so A is non-empty. Accordingly γ = sup A is defined in R. Now for any k ∈ N, {x : f (x) ≤ γ − 2−k } is negligible, so {x : f (x) < γ} is negligible. Also, for any k, {x : f (x) ≤ γ + 2−k } is non-negligible, so {x : f (x) > γ2−k } must be negligible; accordingly, {x : f (x) > γ} is negligible. But this means that {x : f (x) = γ} is conegligible and has measure 1. Thus we have a compact set K = {γ} such that µf −1 [K] = 1, and γ must belong to f [E]. As f and E are arbitrary, µ is perfect. (b) µ is not compact. P P?? Suppose, if possible, that K ⊆ PX is a compact class such that µ is inner regular with respect to K. Then for every x ∈ X there is a measurable set Kx ∈ K such that Kx ⊆ X \ {x} and µKx > 0, that is, Kx is conegligible. But this means that {Kx : x ∈ X} must have the finite intersection property; as it also has empty intersection, K cannot be a compact class. X XQ Q (c) Because µ is totally finite, it cannot be locally compact (342Hb). *342N Example There is a complete locally determined localizable locally compact measure space which is not compact. proof (a) I refer to the example of 216E. In that construction, we have a set I and a family hfγ iγ∈C in X = {0, 1}I such that for every D ⊆ C there is an i ∈ I such that D = {γ : fγ (i) = 1}; moreover, #(C) > c. The σ-algebra Σ is the family of sets E ⊆ X such that for every γ there is a countable set J ⊆ I such that {x : x¹J = fγ ¹J} is a subset of either E or X \ E; and for E ∈ Σ, µE is #({γ : fγ ∈ E}) if this is finite, ∞ otherwise. Note that any subset of X determined by a countable set of coordinates belongs to Σ. For each γ ∈ C, let iγ ∈ I be such that fγ (iγ ) = 1, fδ (iγ ) = 0 for δ 6= γ. (In 216E I took I to be PC, and iγ would be {γ}.) Set Y = {x : x ∈ X, {γ : γ ∈ C, x(iγ ) = 1} is finite}.

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*342N

Give Y its subspace measure µY with domain ΣY . Then µY is complete, locally determined and localizable (214Id). Note that fγ ∈ Y for every γ ∈ C. (b) µY is locally compact. P P Suppose that F ∈ ΣY and µY F < ∞. If µY F = 0 then surely the subspace measure µF is compact. Otherwise, we can express F as E ∩ Y where E ∈ Σ and µE = µY F . Then D = {γ : fγ ∈ E} = {γ : fγ ∈ F } is finite. For γ ∈ D set G0γ = {x : x ∈ X, x(iγ ) = 1, x(iδ ) = 0 for every δ ∈ D \ {γ}} ∈ Σ, Kγ = {K : fγ ∈ K ⊆ F ∩ G0γ }.

S Then each Kγ is a compact class, and members of different Kγ ’s are disjoint, so K = γ∈D Kγ is a compact class. Now suppose that H belongs to the subpsace σ-algebra ΣF and µF H > 0. Then there is a γ ∈ D such that fγ ∈ H, so that H ∩ G0γ ∈ K ∩ ΣF and µF (H ∩ G0γ ) > 0. By 342E, this is enough to show that µF is compact. As F is arbitrary, µY is locally compact. Q Q (c) µY is not compact. P P?? Suppose, if possible, that µY is inner regular with respect to a compact class K ⊆ PY . For each γ ∈ C set Gγ = {x : x ∈ X, x(iγ ) = 1}, so that fγ ∈ Gγ ∈ Σ and µY (Gγ ∩ Y ) = 1. There must therefore be a Kγ ∈ K such that Kγ ⊆ Gγ ∩ Y and µY Kγ = 1 (since µY takes no value in ]0, 1[). Express Kγ as Y ∩ Eγ , where Eγ ∈ Σ, and let Jγ ⊆ I be a countable set such that Eγ ⊇ {x : x ∈ X, x¹Jγ = fγ ¹Jγ }. At this point I call on the full strength of 2A1P. There is a set B ⊆ C, of cardinal greater than c, such that fγ ¹Jγ ∩ Jδ = fδ ¹Jγ ∩ Jδ for all γ, δ ∈ B. But this means that, for any finite set D ⊆ B, we can define x ∈ X by setting x(i) = fα (i) if α ∈ D, i ∈ Jα , [ = 0 if i ∈ I \ Jα . α∈D

It is easy to check that {γ : γ ∈ C, x(iγ ) = 1} = D, so that x ∈ Y ; but now T T x ∈ Y ∩ α∈D Eα = α∈D Kα . What this shows is that {Kα : α ∈ B} has the finite intersection property. It must therefore have non-empty intersection; say T T y ∈ α∈B Kα ⊆ α∈B Gα . But now we have a member y of Y such that {γ : y(iγ ) = 1} ⊇ B is infinite, contrary to the definition of Y . X XQ Q 342X Basic exercises > (a) Show that a measure space (X, Σ, µ) is semi-finite iff µ is inner regular with respect to {E : µE < ∞}. (b) Find a proof of 342B based on 215A. (c) Let (X, Σ, µ) be a locally compact semi-finite measure space in which all singleton sets are negligible. Show that it is atomless. (d) Let (X, Σ, µ) be a measure space, and ν an indefinite-integral measure over µ (234B). Show that ν is compact, or locally compact, if µ is. (Hint: if K satisfies the conditions of 342E with respect to µ, then it satisfies them for ν.) (e) Let f : R → R be any non-decreasing function, and νf the corresponding Lebesgue-Stieltjes measure. Show that νf is compact. (Hint: 256Xg.) (f ) Let µ be Lebesgue measure on [0, 1], ν the countable-cocountable measure on [0, 1], and λ their c.l.d. product. Show that λ is a compact measure. (Hint: let K be the family of sets K × A where A ⊆ [0, 1] is cocountable and K ⊆ A is compact.)

342 Notes

Compact measure spaces

181

(g) (i) Give an example of a compact probability space (X, Σ, µ), a set Y and a function f : X → Y such that the image measure µf −1 is not compact. (ii) Give an example of a compact probability space (X, Σ, µ) and a σ-subalgebra T of Σ such that (X, T, µ¹ T) is not compact. (Hint: 342Xf.) (h) Let (X, Σ, µ) be a perfect measure space, and f : X → R a measurable function. Show that the image measure µf −1 is inner regular with respect to the compact subsets of R, so is a compact measure. (i) Let (X, Σ, µ) be a σ-finite measure space. Show that it is perfect iff for every measurable f : X → R there is a Borel set H ⊆ f [X] such that f −1 [H] is conegligible in X. (Hint: 342Xh for ‘only if’, 256C for ‘if’.) (j) Let (X, Σ, µ) be a complete totally finite perfect measure space and f : X → R a measurable function. Show that the image measure µf −1 is a Radon measure, and is the only Radon measure on R for which f is inverse-measure-preserving. (Hint: 256G.) (k) Suppose that (X, Σ, µ) is a perfect measure space. (i) Show that if (Y, T, ν) is a measure space, and f : X → Y is a function such that f −1 [F ] ∈ Σ for every F ∈ T and f −1 [F ] is µ-negligible for every ν-negligible set F , then (Y, T, ν) is perfect. (ii) Show that if T is a σ-subalgebra of Σ then (X, T, µ¹ T) is perfect. (l) Let (X, Σ, µ) be a perfect measure space such that Σ is the σ-algebra generated P∞ by a sequence of sets. Show that µ is compact. (Hint: if Σ is generated by {En : n ∈ N}, set f = n=0 3−n χEn and consider {f −1 [K] : K ⊆ f [X] is compact}.) (m) Let (X, Σ, µ) be a semi-finite measure space. Show that µ is perfect iff µ¹ T is compact for every countably generated σ-subalgebra T of Σ. (n) Show that (i) a measurable subspace of a perfect measure space is perfect (ii) a semi-finite measure space is perfect iff all its totally finite subspaces are perfect (iii) the direct sum of any family of perfect measure spaces is perfect (iv) the c.l.d. product of two perfect measure spaces is perfect (hint: put 342Xm and 342Ge together) (v) the product of any family of perfect probability spaces is perfect (vi) a measure space is perfect iff its completion is perfect (vii) the c.l.d. version of a perfect measure space is perfect (viii) any purely atomic measure space is perfect (ix) an indefinite-integral measure over a perfect measure is perfect. (o) Let µ be Lebesgue measure on R, let A be a subset of R, and let µA be the subspace measure. Show that µA is compact iff it is perfect iff A is Lebesgue measurable. (Hint: if µA is perfect, consider the image measure hµ−1 A on R, where h(x) = x for x ∈ A.) 342Y Further exercises (a) Show that the space (X, Σ, µ) of 216E and 342N is a compact measure space. (Hint: use the usual topology on X = {0, 1}I .) (b) Give an example of a compact complete locally determined measure space which is not localizable. (Hint: in 216D, add a point to each horizontal and vertical section of X, so that all the sections become compact measure spaces.) 342 Notes and comments The terminology I find myself using in this section – ‘compact’, ‘locally compact’, ‘countably compact’, ‘perfect’ – is not entirely satisfactory, in that it risks collision with the same words applied to topological spaces. For the moment, this is not a serious problem; but when in Volume 4 we come to the systematic analysis of spaces which have both topologies and measures present, it will be necessary to watch our language carefully. Of course there are cases in which a ‘compact class’ of the sort discussed here can be taken to be the family of compact sets for some familiar topology, as in 342Ja-342Jd, but in others this is not so (see 342Xf); and even when we have a familiar compact class, the topology constructed from it by the method of 342Da need not be one we might expect. (Consider, for instance, the topology on R for which the closed sets are just the sets which are compact for the usual topology.)

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342 Notes

I suppose that ‘compact’ and ‘perfect’ measure spaces look reasonably natural objects to study; they offer to illuminate one of the basic properties of Radon measures, the fact that (at least for totally finite Radon measures on Euclidean space) the image measure of a Radon measure under a measurable function is again Radon (256G, 342Xj). Indeed this was the original impetus for the study of perfect measures (Gnedenko & Kolmogorov 54, Sazonov 66). It is not obvious that there is any need to examine ‘locally compact’ measure spaces, but actually they are the chief purpose of this section, since the main theorem of the next section is an alternative characterization of semi-finite locally compact measure spaces (343B). Of course you may feel that the fact that ‘locally compact’ and ‘compact’ coincide for strictly localizable spaces (342Hb) excuses you from troubling about the distinction at first reading. As with any new classification of measure spaces, it is worth finding out how the classes of ‘compact’ and ‘perfect’ measure spaces behave with respect to the standard constructions. I run through the basic facts in 342G-342I, 342Xd, 342Xk and 342Xn. We can also look for relationships between the new properties and those already studied. Here, in fact, there is not much to be said; 342N and 342Yb show that ‘compactness’ is largely independent of the classification in §211. However there are interactions with the concept of ‘atom’ (342Xc, 342Xn(viii)). I give examples to show that perfect measure spaces need not be locally compact, and that locally compact measure spaces need not be compact (342M, 342N). The standard examples of measure spaces which are not perfect are non-measurable subspaces (342Xo); I will return to these in the next section (343L-343M). Something which is not important to us at the moment, but is perhaps worth taking note of, is the following observation. To determine whether a measure space (X, Σ, µ) is compact, we need only the structure (X, Σ, N ), where N is the σ-ideal of negligible sets, since that is all that is referred to in the criterion of 342E. The same is true of local compactness, by 342Hc, and of perfectness, by the definition in 342K. Compare 342Xd, 342Xk, 342Xn(ix). Much of the material of this section will be repeated in Volume 4 as part of a more systematic analysis of inner regularity.

343 Realization of homomorphisms We are now in a position to make progress in one of the basic questions of abstract measure theory. In §324 I have already described the way in which a function between two measure spaces can give rise to a homomorphism between their measure algebras. In this section I discuss some conditions under which we can be sure that a homomorphism can be represented by a function. The principal theorem of the section is 343B. If a measure space (X, Σ, µ) is locally compact, then many homomorphisms from the measure algebra of µ to other measure algebras will be representable by functions into X; moreover, this characterizes locally compact spaces. In general, a homomorphism between measure algebras can be represented by widely different functions (343I, 343J). But in some of the most important cases (e.g., Lebesgue measure) representing functions are ‘almost’ uniquely defined; I introduce the concept of ‘countably separated’ measure space to describe these (343D-343H). 343A Preliminary remarks It will be helpful to establish some vocabulary and a couple of elementary facts. (a) If (X, Σ, µ) and (Y, T, ν) are measure spaces, with measure algebras A and B, I will say that a function f : X → Y represents a homomorphism π : B → A if f −1 [F ] ∈ Σ and (f −1 [F ])• = π(F • ) for every F ∈ T. (Perhaps I should emphasize here that some homomorphisms are representable in this sense, and some are not; see 343M below for examples of non-representable homomorphisms.) (b) If (X, Σ, µ) and (Y, T, ν) are measure spaces, with measure algebras A and B, f : X → Y is a function, and π : B → A is a sequentially order-continuous Boolean homomorphism, then {F : F ∈ T, f −1 [F ] ∈ Σ and f −1 [F ]• = πF • } is a σ-subalgebra of T. (The verification is elementary.)

343B

Realization of homomorphisms

183

(c) Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with measure algebras A and B, and π : B → A a ˆ µ ˆ νˆ) be the Boolean homomorphism which is represented by a function f : X → Y . Let (X, Σ, ˆ), (Y, T, completions of (X, Σ, µ), (Y, T, ν); then A and B can be identified with the measure algebras of µ ˆ and ˆ µ ˆ νˆ). P νˆ (322Da). Now f still represents π when regarded as a function from (X, Σ, ˆ) to (Y, T, P If G is ν-negligible, there is a negligible F ∈ T such that G ⊆ F ; since f −1 [F ]• = πF • = 0, ˆ If G is any element of T, ˆ there is an f −1 [F ] is µ-negligible, so f −1 [E] is negligible, therefore belongs to Σ. F ∈ T such that G4F is negligible, so that ˆ f −1 [G] = f −1 [F ]4f −1 [G4F ] ∈ Σ, and f −1 [G]• = f −1 [F ]• = πF • = πG• . Q Q (d) In particular, if (X, Σ, µ) and (Y, T, ν) are measure spaces, and f : X → Y is inverse-measurepreserving, then f is still inverse-measure-preserving with respect to the completed measures µ ˆ and νˆ (apply (c) with π : B → A the homomorphism induced by f ). (See 235Hc.) 343B Theorem Let (X, Σ, µ) be a non-empty semi-finite measure space, and (A, µ ¯) its measure algebra. Let (Z, Λ, λ) be the Stone space of (A, µ ¯); for E ∈ Σ write E ∗ for the open-and-closed subset of Z corresponding to the image E • of E in A. Then the following are equiveridical: (i) (X, Σ, µ) is locally compact in the sense of 342Ad. (ii) There is a family K ⊆ Σ such that (α) whenever TE ∈ Σ and µE > 0 there is a K ∈ K such that 0 0 K⊆E T and µK > 0 (β) whenever K ⊆ K is such that µ( K0 ) > 0 for every non-empty finite set K0 ⊆ K , then K0 6= ∅. (iii) ThereTis a family K ⊆ Σ such that (α)0 µ is inner regular withTrespect to K (β) whenever K0 ⊆ K is such that µ( K0 ) > 0 for every non-empty finite set K0 ⊆ K0 , then K0 6= ∅. (iv) There is a function f : Z → X such that f −1 [E]4E ∗ is negligible for every E ∈ Σ. (v) Whenever (Y, T, ν) is a complete strictly localizable measure space, with measure algebra B, and π : A → B is an order-continuous Boolean homomorphism, then there is a g : Y → X representing π. (vi) Whenever (Y, T, ν) is a complete strictly localizable measure space, with measure algebra B, and π : A → B is an order-continuous measure-preserving Boolean homomorphism, then there is a g : Y → X representing π. proof (a)(i)⇒(ii) Because µ is semi-finite, there is a partition of unity hai ii∈I in A such that µ ¯ai < ∞ for each i. For each i ∈ I, let Ei ∈ Σ be such that Ei• = ai . Then the subspace measure µEi on Ei is S compact; let Ki ⊆ PEi be a compact class such that µ is inner regular with respect to K . Set K = Ei i i∈I Ki . If T 0 K0 ⊆ K and µ( K0 ) > 0 for every non-empty finite K ⊆ K, then K ⊆ K for some i, and surely has the 0 i T finite intersection property, so K0 6= ∅; thus K0 satisfies (β) of condition (ii). And if E ∈ Σ, µE > 0 then there must be some i ∈ I such that Ei• ∩ ai 6= 0, that is, µ(E ∩ Ei ) > 0, in which case there is a K ∈ Ki ⊆ K such that K ⊆ E ∩ Ei and µK > 0; so that K satisfies condition (α). (b)(ii)⇒(iii) Suppose that K ⊆ Σ witnesses that (ii) is true. If µX = 0 then K already witnesses that (iii) is true, so we need consider only the case µX > 0. Set L = {K0 ∪ . . . ∪ Kn : K0 , . . . , Kn ∈ K}. Then LT witnesses that (iii) is true. P P By 342Ba, µ is inner regular with respect to L. Let L0 ⊆ L be such that µ( L0 ) > 0 for every non-empty finite L0 ⊆ L0 . Then T F0 = {A : A ⊆ X, there is a finite L0 ⊆ L0 such that X ∩ L0 \ A is negligible} is a filter on X, so there is an ultrafilter F on X including F0 . T Note that every conegligible set belongs to F0 , so no negligible set can belong to F. Set K0 = K ∩ F; then T K0 belongs to F, so is not negligible, for every non-empty finite K0 ⊆ K0 . Accordingly there is some x ∈ K0 . But any member of L0 is of the form L = K0 ∪ . . . ∪ Kn where each Ki ∈ K; because F is an T ultrafilter and L ∈ F, there must be some i ≤ n such that Ki ∈ F, in which case x ∈ Ki ⊆ L. Thus x ∈ L0 . As L0 is arbitrary, L satisfies the condition (β). Q Q

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The lifting theorem

343B

(c)(iii)⇒(iv) Let K ⊆ Σ witness that (iii) ForTany z ∈ Z, set Kz = T {K : K ∈ K, z ∈ K ∗ }. If T T is true. ∗ ∗ ∗ K0 , . . . , Kn ∈ Kz , then z ∈ i≤n Ki = ( i≤n Ki ) , so ( i≤n Ki ) 6= ∅ and µ( i≤n Ki ) > 0. By (β) of T T condition (iii), Kz 6= ∅; and even if Kz = ∅, X ∩ Kz 6= ∅ because X is non-empty. So we may choose T f (z) ∈ X ∩ Kz . This defines a function f : Z → X. Observe that, for K ∈ K and z ∈ Z, z ∈ K ∗ =⇒ K ∈ Kz =⇒ f (z) ∈ K =⇒ z ∈ f −1 [K], so that K ∗ ⊆ f −1 [K]. Now take any E ∈ Σ. Consider S S U1 = {K ∗ : K ∈ K, K ⊆ E} ⊆ {E ∗ ∩ f −1 [K] : K ∈ K, K ⊆ E} ⊆ E ∗ ∩ f −1 [E], S U2 = {K ∗ : K ∈ K, K ⊆ X \ E} ⊆ (X \ E)∗ ∩ f −1 [X \ E] = Z \ (f −1 [E] ∪ E ∗ ), so that f −1 [E]4E ∗ ⊆ Z \ (U1 ∪ U2 ). Now U1 and U2 are open subsets of Z, so M = Z \ (U1 ∪ U2 ) is closed, and in fact M is nowhere dense. P P?? Otherwise, there is a non-zero a ∈ A such that the corresponding open-and-closed set b a is included in M , and an F ∈ Σ of non-zero measure such that a = F • . At least one of F ∩ E, F \ E is non-negligible and therefore includes a non-negligible member K of K. But in this case K ∗ is a non-empty open subset of M which is included in either U1 or U2 , which is impossible. X XQ Q By the definition of λ (321J-321K), M is λ-negligible, so f −1 [E]4E ∗ ⊆ M is negligible, as required. (d)(iv)⇒(v) Now assume that f : Z → X witnesses (iv), and let (Y, T, ν) be a complete strictly localizable measure space, with measure algebra B, and π : A → B an order-continuous Boolean homomorphism. If νY = 0 then any function from Y to X will represent π, so we may suppose that νY > 0. Write W for the Stone space of B. Then we have a continuous function φ : W → Z such that φ−1 [b a] = π ca for every a ∈ A (312P), and φ−1 [M ] is nowhere dense in W for every nowhere dense M ⊆ Z (313R). It follows that φ−1 [M ] is meager for every meager M ⊆ Z, that is, φ−1 [M ] is negligible in W for every negligible M ⊆ Z. By 341Q, there is an inverse-measure-preserving function h : Y → W such that h−1 [bb]• = b for every b ∈ B. Consider g = f φh : Y → X. If E ∈ Σ, set a = E • ∈ A, so that E ∗ = b a ⊆ Z, and M = f −1 [E]4E ∗ is λ-negligible; consequently −1 φ [M ] is negligible in W . Because h is inverse-measure-preserving, g −1 [E]4h−1 [φ−1 [E ∗ ]] = h−1 [φ−1 [f −1 [E]]]4h−1 [φ−1 [E ∗ ]] = h−1 [φ−1 [M ]] is negligible. But φ−1 [E ∗ ] = π ca, so g −1 [E]• = h−1 [φ−1 [E ∗ ]]• = πa. As E is arbitrary, g induces the homomorphism π. (e)(v)⇒(vi) is trivial. (f )(vi)⇒(iv) Assume (vi). Let ν be the c.l.d. version of λ, T its domain, and B its measure algebra; then ν is strictly localizable (322Qb). The embedding Λ ⊆ T corresponds to an order-continuous measurepreserving Boolean homomorphism from A to B (322Db). By (vi), there is a function f : Z → X such that f −1 [E] ∈ T and f −1 [E]• = (E ∗ )• in B for every E ∈ Σ. But as ν and λ have the same negligible sets (322Qb), f −1 [E]4E ∗ is λ-negligible for every E ∈ Σ, as required by (iv). α) To begin with (down to the end of (γ) below) I suppose that µ is totally finite. In this (g)(iv)⇒(i)(α case we have a function g : X → Z such that E4g −1 [E ∗ ] is negligible for every E ∈ Σ (341Q again). We are supposing also that there is a function f : Z → X such that f −1 [E]4E ∗ is negligible for every E ∈ Σ. Write K for the family of sets K ⊆ E such that K ∈ Σ and there is a compact set L ⊆ Z such that f [L] ⊆ K ⊆ g −1 [L]. β ) µ is inner regular with respect to K. P (β P Take F ∈ Σ and γ < µF . Choose hVn in∈N , hFn in∈N as follows. F0 = F . Given that µFn > γ, then λ(f −1 [Fn ] ∩ Fn∗ ) = λFn∗ = µFn > γ, ∗ so there is an open-and-closed set Vn ⊆ f −1 [Fn ] ∩ Fn∗ with λVn > γ. Express Vn as Fn+1 where Fn+1 ∈ Σ; −1 ∗ ∗ −1 ∗ since Fn 4g [Fn ] is negligible, and Vn ⊆ T Fn , we may take it that T Fn+1 ⊆ g [Fn ]. Continue. At the end of the induction, set K = n∈N Fn ∈ Σ and L = n∈N Fn∗ . Because Fn+1 \ Fn ⊆ g −1 [Fn∗ ] \ Fn is negligible for each n, µK = limn→∞ µFn ≥ γ, while K ⊆ F and L is surely compact. We have

343D

Realization of homomorphisms

L⊆ so f [L] ⊆ K. Also K⊆

T n∈N

T n∈N

Vn ⊆

T

Fn+1 ⊆

n∈N

185

f −1 [Fn ] = f −1 [K],

T n∈N

g −1 [Fn∗ ] = g −1 [L].

So K ∈ K. As F and γ are arbitrary, µ is inner regular with respect to K. Q Q (γγ ) Next, P Suppose that K0 ⊆ K has the finite intersection property. If K0 = ∅, T 0 K is a compact class. 0P of course K 6= ∅; suppose that K is non-empty. Let L be the family of closed sets L ⊆ Z such that 0 g −1 [L] includes some property, and Z is compact, so T member of K . Then L has the finite intersection there is some z ∈ L; also Z ∈ L, so z ∈ Z. For any K ∈ K0 , there is some T closed set L ⊆ Z such that f [L] ⊆ K ⊆ g −1 [L], so that L ∈ L and z ∈ L and f (z) ∈ K. Thus f (z) ∈ K0 . As K0 is arbitrary, K is a compact class. Q Q So K witnesses that µ is a compact measure. (δδ ) Now consider the general case. Take any E ∈ Σ of finite measure. If E = ∅ then surely the subspace measure µE is compact. Otherwise, we can identify the measure algebra of µE with the principal ideal AE • of A generated by E • (322Ja), and E ∗ ⊆ Z with the Stone space of AE • (312S). Take any x0 ∈ E and define f˜ : E ∗ → E by setting f˜(z) = f (z) if z ∈ E ∗ ∩ f −1 [E], x0 if z ∈ E ∗ \ f −1 [E]. Then f and f˜ agree almost everywhere on E ∗ , so f˜−1 [F ]4F ∗ is negligible for every F ∈ ΣE , that is, f˜ represents the canonical isomorphism between the measure algebras of µE and the subspace measure λE ∗ on E ∗ . But this means that condition (iv) is true of µE , so µE is compact, by (α)-(γ) above. As E is arbitrary, µ is locally compact. This completes the proof. 343C Examples (a) Let κ be an infinite cardinal. We know that the usual measure νκ on {0, 1}κ is compact (342Jd). It follows that if (X, Σ, µ) is any complete probability space such that the measure algebra Bκ of νκ can be embedded as a subalgebra of the measure algebra A of µ, there is an inversemeasure-preserving function from X to {0, 1}κ . By 332P, this is so iff every non-zero principal ideal of A has Maharam type at least κ. Of course this does not depend in any way on the results of the present chapter. If Bκ can be embedded in A, there must be a stochastically independent family hEξ iξ 0, apply (b) to the normalized measure (µX)−1 µ; or argue directly from 343B, using the fact that Lebesgue measure on [0, µX] is compact; or use the idea suggested in 343Xd.) (d) Throughout the work above – in §254 as well as in 343B – I have taken the measures involved to be complete. It does occasionally happen, in this context, that this restriction is inconvenient. Typical results not depending on completeness in the domain space X are in 343Xc-343Xd. Of course these depend not only on the very special nature of the codomain spaces {0, 1}I or [0, 1], but also on the measures on these spaces being taken to be incomplete. 343D Uniqueness of realizations The results of 342E-342J, together with 343B, give a respectable number of contexts in which homomorphisms between measure algebras can be represented by functions between measure spaces. They say nothing about whether such functions are unique, or whether we can distinguish, among the possible representations of a homomorphism, any canonical one. In fact the proof of 343B, using the Lifting Theorem as it does, strongly suggests that this is like looking for a canonical lifting, and I am sure that (outside a handful of very special cases) any such search is vain. Nevertheless, we do have a weak kind of uniqueness theorem, valid in a useful number of spaces, as follows.

186

The lifting theorem

343D

Definition A measure space (X, Σ, µ) is countably separated if there is a countable set A ⊆ Σ separating the points of X in the sense that for any distinct x, y ∈ X there is an E ∈ A containing one but not the other. (Of course this is a property of the structure (X, Σ) rather than of (X, Σ, µ).) 343E Lemma A measure space (X, Σ, µ) is countably separated iff there is an injective measurable function from X to R. proof If (X, Σ, µ) is countably separated, let A ⊆ Σ be a countable set separating the points of X. Let hEn in∈N be a sequence running over A ∪ {∅}. Set P∞ f = n=0 3−n χEn : X → R. Then f is measurable (because every En is measurable) and injective (because if x 6= y in X and n = min{i : #(Ei ∩ {x, y}) = 1} and x ∈ En , then P P P f (x) ≥ 3−n + i i>n 3−i + i 0 there is a K ∈ K such that K ⊆ E and µK > 0. And K is a compact class. P P If K0 ⊆ K has the finite intersection property, L = {f [K] : K ∈ K0 } is aTfamily of compact sets in R with the finite intersection property, and has non-empty intersection; so that K0 is also non-empty, because f is injective. Q Q By 342E, (X, Σ, µ) is compact. 343L The time has come to give examples of spaces which are not locally compact, so that we can expect to have measure-preserving homomorphisms not representable by inverse-measure-preserving functions. The most commonly arising ones are covered by the following result. Proposition Let (X, Σ, µ) be a complete locally determined countably separated measure space, and A ⊆ X a set such that the subspace measure µA is perfect. Then A is measurable. proof ?? Otherwise, there is a set E ∈ Σ such that µE < ∞ and B = A ∩ E ∈ / Σ. Let f : X → R be an injective measurable function (343E again). Then f ¹B is ΣB -measurable, where ΣB is the domain of the subspace measure µB on B. Set K = {f −1 [L] : L ⊆ f [B], L is compact in R}. Just as in the proof of 343K, K is a compact class and S µB is inner regular with respect to K. By 342Bb, there is a sequence hK i in K such that µ (B \ B n∈N K Sn ) = 0. But of course K ⊆ Σ, because f is S n n∈N X Σ-measurable, so n∈N Kn ∈ Σ. Because µ is complete, B \ n∈N Kn ∈ Σ and B ∈ Σ. X 343M Example 343L tells us that any non-measurable set X of Rr , or of {0, 1}N , with their usual measures, is not perfect, therefore not (locally) compact, when given its subspace measure. To find a non-representable homomorphism, we do not need to go through the whole apparatus of 343B. Take Y to be a measurable envelope of X (132Ed). Then the identity function from X to Y induces an isomorphism of their measure algebras. But there is no function from Y to X inducing the same isomorphism. P P?? Writing Z for Rr or {0, 1}N and µ for its measure, Z is countably separated; suppose hEn in∈N is a sequence of measurable sets in Z separating its points. For each n, (Y ∩ En )• in the measure algebra of µY corresponds to (X ∩ En )• in the measure algebra of µX . So if f : Y → X were a function representing the isomorphism of the measure algebras, (Y ∩ En )4f −1 [En ] would have to be negligible for each n, and S A = n∈N (Y ∩ En )4f −1 [En ] would be negligible. But for y ∈ Y \ A, f (y) belongs to just the same En as y does, so must be equal to y. Accordingly X ⊇ Y \ A and X is measurable. X XQ Q 343X Basic exercises (a) Let (X, Σ, µ) be a semi-finite measure space. (i) Suppose that there is a set A ⊆ X, of full outer measure, such that the subspace measure on A is compact. Show that µ is locally compact. (Hint: show that µ satisfies (ii) or (v) of 343B.) (ii) Suppose that for every non-negligible E ∈ Σ there is a non-negligible set A ⊆ E such that the subspace measure on A is compact. Show that µ is locally compact. (b) Let hXi ii∈I be a family of non-empty sets, with product X; write πi : X → Xi for the coordinate N map. Suppose we are given a σ-algebra Σi of subsets of Xi for each i; let Σ = c i∈I Σi be the corresponding σ-algebra of subsets of X generated by {πi−1 [E] : i ∈ I, E ∈ Σi }. Let µ be a totally finite measure with domain Σ, and for i ∈ I let µi be the image measure µπi−1 . Check that the domain of µi is Σi . Show that if every (Xi , Σi , µi ) is compact, then so is (X, Σ, µ). (Hint: either show that µ satisfies (v) of 343B or adapt the method of 342Gf.) (c) Let I be any set. Let T be the σ-algebra of subsets of {0, 1}I generated by the sets Fi = {z : z(i) = 1} for i ∈ I, and ν any probability measure with domain T; let B be the measure algebra of ν. Let (X, Σ, µ) be a measure space with measure algebra A, and φ : B → A an order-continuous Boolean homomorphism. Show that there is an inverse-measure-preserving function f : X → {0, 1}I representing φ. (Hint: for each i ∈ I, take Ei ∈ Σ such that Ei• = φFi• ; set f (x)(i) = 1 if x ∈ Ei , and use 343Ab.)

343Yc

Realization of homomorphisms

189

(d) Let (X, Σ, µ) be an atomless probability space. Let µB be the restriction of Lebesgue measure to the σ-algebra of Borel subsets of [0, 1]. Show that there is a function g : X → [0, 1] which is inverseN measure-preserving P∞ −n−1 for µ and µB . (Hint: find an f : X → {0, 1} as in 343Xc, and set g = hf where h(z) = n=0 2 g(n), as in 254K; or choose Eq ∈ Σ such that µEq = q, Eq ⊆ Eq0 whenever q ≤ q 0 in [0, 1] ∩ Q, and set f (x) = inf{q : x ∈ Eq } for x ∈ E1 .) (e) Let (X, Σ, µ) be a countably separated measure space, with measure algebra A. (i) Show that {x} ∈ Σ for every x ∈ X. (ii) Show that every atom of A is of the form {x}• for some x ∈ X. (f ) Let I k be the split interval, with its usual measure µ described in 343J, and h : I k → [0, 1] the canonical surjection. Show that the canonical isomorphism between the measure algebras of µ and Lebesgue measure on [0, 1] is given by the formula ‘E • 7→ h[E]• for every measurable E ⊆ I k ’. (g) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with measure algebras (A, µ ¯), (B, ν¯). Suppose that X ∩ Y = ∅ and that we have a measure-preserving isomorphism π : A → B. Set Λ = {W : W ⊆ X ∪ Y, W ∩ X ∈ Σ, W ∩ Y ∈ T, π(W ∩ X)• = (W ∩ Y )• }, and for W ∈ Λ set λW = µ(W ∩ X) = ν(W ∩ Y ). Show that (X ∪ Y, Λ, λ) is a measure space which is locally compact, or perfect, if (X, Σ, µ) is. (h) Let (X, Σ, µ) be a complete perfect totally finite measure space, (Y, T, ν) a complete countably separated measure space, and f : X → Y an inverse-measure-preserving function. Show that T = {F : F ⊆ Y, f −1 [F ] ∈ Σ}, so that a function h : Y → R is ν-integrable iff hf is µ-integrable. (Hint: if A ⊆ Y and E = f −1 [A] ∈ Σ, f ¹E is inverse-measure-preserving for the subspace measures µE , νA ; by 342Xk, νA is perfect, so by 343L A ∈ T. Now use 235L.) 343Y Further exercises (a) Let (X, Σ, µ) be a semi-finite measure space, and suppose that there is a compact class K ⊆ PX such that (α) whenever T E ∈ Σ and µE > 0 there is a non-negligible K ∈ K such that K ⊆ E (β) whenever K0 , . . . , Kn ∈ K and i≤n Ki = ∅ then there are measurable sets E0 , . . . , En T such that Ei ⊇ Ki for every i and i≤n Ei is negligible. Show that µ is locally compact. (b) (i) Show that a countably separated semi-finite measure space has magnitude at most c and Maharam type at most 2c . (ii) Show that the direct sum of c or fewer countably separated measure spaces is countably separated. (c) Let I k = {t+ : t ∈ [0, 1]} ∪ {t− : t ∈ [0, 1]} be the split interval (343J). (i) Show that the rules s− ≤ t− ⇐⇒ s+ ≤ t+ ⇐⇒ s ≤ t,

s+ ≤ t− ⇐⇒ s < t,

t− ≤ t+ for all t ∈ [0, 1] define a Dedekind complete total order on I k with greatest and least elements. (ii) Show that the intervals [0− , t− ], [t+ , 1+ ], interpreted for this ordering, generate a compact Hausdorff topology on I k for which the map h : I k → [0, 1] of 343J is continuous. (iii) Show that a subset E of I k is Borel for this topology iff the sets Er , El ⊆ [0, 1], as described in 343J, are Borel and Er 4El is countable. (iv) Show that if f : [0, 1] → R is of bounded variation then there is a continuous g : I k → R such that g = f h except perhaps at countably many points. (v) Show that the measure µ of 343J is inner regular with respect to the compact subsets of I k . (vi) Show that we have a lower density φ for µ defined by setting 1

φE = {t− : 0 < t ≤ 1, lim µ(E ∩ [(t − δ)+ , t− ]) = 1} δ↓0 δ

1

∪ {t+ : 0 ≤ t < 1, lim µ(E ∩ [t+ , (t + δ)− ]) = 1} δ↓0 δ

for measurable sets E ⊆ I k .

190

The lifting theorem

343Yd

(d) Set X = {0, 1}c , with its usual measure µ. Show that there is an inverse-measure-preserving function f : X → X such that f [X] is non-measurable but f induces the identity automorphism of the measure algebra of µ. (Hint: use the idea of 343I.) Show that under these conditions f [X], with its subspace measure, must be compact. (Hint: use 343B(iv).) (e) Let µHr be r-dimensional Hausdorff measure on Rs , where s ≥ 1 is an integer and r ≥ 0 (§264). (i) Show that µHr is countably separated. (ii) Show that the c.l.d. version of µHr is compact. (Hint: 264Yi.) (f ) Give an example of a countably separated probability space (X, Σ, µ) and a function f from X to a set Y such that the image measure µf −1 is not countably separated. (Hint: use 223B to show that if E ⊆ R is Lebesgue measurable and not negligible, then E + Q is conegligible; or use the zero-one law to show that if E ⊆ PN is measurable and not negligible for the usual measure on PN, then {a4b : a ∈ E, b ∈ [N] (c) Show that there is no lifting φ of Lebesgue measure on R which is ‘symmetric’ in the sense that φ(−E) = −φE for every measurable set E, writing −E = {−x : x ∈ E}. (Hint: can 0 belong to φ([0, ∞[)?) > (d) Let µ be Lebesgue measure on X = R\{0}. Show that there is a lifting φ of µ such that φ(xE) = xφE for every x ∈ X and every measurable E ⊆ X, writing xE = {xy : y ∈ E}. (e) Let µ be the usual measure on X = {0, 1}I , for some set I, Σ its domain, and (A, µ ¯) its measure algebra. (i) Show that we can define πx (a) = a + x, for a ∈ A and x ∈ X, by the formula E • + x = (E + x)• ; and that x 7→ πx is a group homomorphism from X to the group of measure-preserving automorphisms of A. (ii) Define Σξ as in the proof of 345C, and set Aξ = {E • : E ∈ Σξ }. Say that a partial lifting θ : Aξ → Σ is translation-invariant if θ(a + x) = θa + x for every a ∈ Aξ and x ∈ X. Show that any such partial lifting can be extended to a translation-invariant partial lifting on Aξ+1 . (iii) Write out a proof of 345C in the language of 341F-341H. > (f ) Let φ be a lower density for Lebesgue measure on Rr which is translation-invariant in the sense that φ(E + x) = φE + x for every x ∈ Rr and every measurable set E. Show that φG ⊆ G for every open set G ⊆ Rr .

208

Liftings

345Xg

(g) Let µ be 1-dimensional Hausdorff measure on S 1 , as in 345Xb. Show that there is no translationinvariant lifting φ of µ such that φE is a Borel set for every E ∈ dom µ. 345Y Further exercises (a) Let (X, Σ, µ) be a complete measure space, and suppose that X has a group operation (x, y) 7→ xy (not necessarily abelian!) such that µ is left-translation-invariant, in the sense that xE = {xy : y ∈ E} ∈ Σ and µ(xE) = µE whenever E ∈ Σ and x ∈ X. Suppose that φ : Σ → Σ is a lower density which is left-translation-invariant in the sense that φ(xE) = x(φE) for every E ∈ Σ, x ∈ X. Show that there is a left-translation-invariant lifting φ : Σ → Σ such that φE ⊆ φE for every E ∈ Σ. (b) Write Σ for the σ-algebra of Lebesgue measurable subsets of R, and L0 (Σ) for the linear space of Σ-measurable functions from R to itself. Show that there is a linear operator T : L0 (µ) → L0 (Σ) such that (α) (T u)• = u for every u ∈ L0 (µ) (β) supx∈R |(T u)(x)| = kuk∞ for every u ∈ L∞ (µ) (γ) T u ≥ 0 whenever u ∈ L∞ (µ) and u ≥ 0 (δ) T is translation-invariant in the sense that T (Sx f )• = Sx T f • for every x ∈ R and f ∈ L0 (Σ), where (Sx f )(y) = f (x + y) for f ∈ L0 (Σ) and x, y ∈ R (²) T is reflection-invariant in the sense that T (Rf )• = RT f • for every f ∈ L0 (Σ), where (Rf )(x) = f (−x) for f ∈ L0 (Σ) and x ∈ R. (Hint: for f ∈ L0 (Σ), set p(f • ) = inf{α : α ∈ [0, ∞], limδ↓0

1 µ{x 2δ

: |x| ≤ δ, |f (x)| > α} = 0}.

Set V = {u : u ∈ L0 (µ), p(u) < ∞} and show that V is a linear subspace of L0 (µ) and that p¹V is a seminorm. Let h0 : V → R be a linear functional such that h0 (χR)• = 1 and h0 (u) ≤ p(u) for every u ∈ V . Extend h0 arbitrarily to a linear functional h1 : L0 (µ) → R; set h(f • ) = 12 (h1 (f • ) + h1 (Rf )• ). Set (T f • )(x) = h(S−x f )• . You will need 231C.) Show that there must be a u ∈ L1 (µ) such that u ≥ 0 but T u 6≥ 0. (c) Show that there is no translation-invariant lifting φ of the usual measure on {0, 1}N such that φE is a Borel set for every measurable set E. 345 Notes and comments I have taken a great deal of care over the concept of ‘translation-invariance’. I hope that you are already a little impatient with some of the details as I have written them out; but while it is very easy to guess at the structure of such arguments as part (e) of the proof of 345B, or (b-iii) and (c-viii) in the proof of 345C, I am not sure that one can always be certain of guessing correctly. A fair test of your intuition will be how quickly you can generate the formulae appropriate to a non-abelian group operation, as in 345Ya. Part (b) of the proof of 345C is based on the same idea as the proof of 341F. There is a useful simplification because the set Eξ in 345C, corresponding to the set E of the proof of 341F, is independent of the algebra Σξ in a very strong sense, so that the expression of an element of Σξ+1 in the form (F ∩ Eξ ) ∪ (G \ Eξ ) is unique. Interpreted in the terms of 341F, we have w = v = 1, so that the formula ¡ ¢ ¡ ¢ θ1 ((a ∩ e) ∪ (b \ e)) = θ((a ∩ v) ∪ (b \ v)) ∩ E ∪ θ((a \ w) ∪ (b ∩ w)) \ E used there becomes θ1 ((a ∩ e) ∪ (b \ e)) = (θa ∩ E) ∪ (θb \ E), matching the formula for φ1 in the proof of 345C. The results of this section are satisfying and natural; they have obvious generalizations, many of which are true. The most important measure spaces come equipped with a variety of automorphisms, and we can always ask which of these can be preserved by a lifting. The answers are not always obvious; I offer 345Xc and 346Xb as warnings, and 345Xd as an encouragement. 345Yb is striking (I have made it as striking as I can), but slightly off the most natural target; the sting is in the last sentence (see 341Ye).

346D

Consistent liftings

209

346 Consistent liftings I turn now to a different type of condition Q which we should naturally prefer our liftings to satisfy. If we have a product measure µ on a product X = i∈I Xi of probability spaces, then we can look for liftings φ which ‘respect coordinates’, that is, are compatible with the product structure in the sense that they factor through subproducts (346A). There seem to be obstacles in the way of the natural conjecture (346Za), and I give the partial results which are known. For Maharam-type-homogeneous spaces Xi , there is always a lifting which respects coordinates (346E), and indeed the translation-invariant liftings of §345 on {0, 1}I already have this property (346C). There is always a lower density on the product which respects coordinates, and we can ask for a little more (346G); using the full strength of 346G, we can enlarge this lower density to a lifting which respects single coordinates and initial segments of a well-ordered product (346H). In the case in which all the factors are copies of each other, we can arrange for the induced liftings on the factors to be copies also (346I, 346J, 346Yd). I end the section with an important fact about Stone spaces which is relevant here (346K-346L). 346A Definition Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product (X, Σ, µ). I will say that a lifting φ : Σ → Σ respects coordinates if φE is determined by coordinates in J whenever E ∈ Σ is determined by coordinates in J ⊆ I. Remark Recall that a set E ⊆ X is ‘determined by coordinates in Q J’ if x0 ∈ E whenever x ∈ E, x0 ∈ X −1 0 and x ¹J = x¹J; that is, if E is expressible as πJ [F ] for some F ⊆ i∈J Xi , where πJ (x) = x¹J for every x ∈ X; that is, if E = πJ−1 [πJ [E]]. See 254M. Recall also that in this case, Q if E is measurable for the product measure on X, then πJ [E] is measurable for the product measure on i∈J Xi (254Ob). 346B Proposition Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product (Z, Λ, λ). For J ⊆ I let (ZJ , ΛJ , λJ ) be the product of h(Xi , Σi , µi )ii∈J , and πJ : Z → ZJ the canonical map. Let φ : Λ → Λ be a lifting. If J ⊆ I is such that φW is determined by coordinates in J whenever W ∈ Λ is determined by coordinates in J, then φ induces a lifting φJ : ΛJ → ΛJ defined by the formula πJ−1 [φJ E] = φ(πJ−1 [E]) for every E ∈ ΛJ . proof If E ∈ ΛJ , then π −1 [E] and φ(πJ−1 [E]) are determined by coordinates in J, so φ(πJ−1 [E]) is of the form πJ−1 [F ] for some F ∈ ΛJ ; this defines φJ : ΛJ → ΛJ . It is now easy to see that φJ is a lifting. Remark Of course we frequently wish to use this result with a singleton set J = {j}. In this case we must remember that (ZJ , ΣJ , λJ ) corresponds to the completion of the probability space (Xj , Σj , µj ). 346C Theorem Let I be any set, and µ the usual measure on X = {0, 1}I . Then any translationinvariant lifting of µ respects coordinates. proof Suppose that E ⊆ X is a measurable set determined by coordinates in J ⊆ I; take x ∈ φE and x0 ∈ X such that x0 ¹J = x¹J. Set y = x0 − x; then y(i) = 0 for i ∈ J, so that E + y = y. Now x0 = x + y ∈ φE + y = φ(E + y) = φE because φ is translation-invariant. As x, x0 are arbitrary, φE is determined by coordinates in J. As E and J are arbitrary, φ respects coordinates. 346D

I describe a standard method of constructing liftings from other liftings.

Lemma Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with measure algebras A, B; suppose that f : X → Y is a (Σ, T)-measurable function inducing an isomorphism F • 7→ f −1 [F ]• : B → A. Then if φ : T → T is a lifting for ν, there is a corresponding lifting φ0 : Σ → Σ given by the formula φ0 E = f −1 [φF ] whenever µ(E4f −1 [F ]) = 0. proof If we say that π : B → A is the isomorphism induced by f , then

210

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346D

φ0 E = f −1 [θ(π −1 E • )], where θ : B → T is the lifting corresponding to φ : T → T. Since θ, π −1 and F 7→ f −1 [F ] are all Boolean homomorphisms, so is φ0 , and it is easy to check that (φ0 E)• = E • for every E ∈ Σ and that φ0 E = ∅ if µE = 0. Remark Compare the construction in 341P. 346E Theorem Let h(Xi , Σi , µi )ii∈I be a family of Maharam-type-homogeneous probability spaces, with product (X, Σ, µ). Then there is a lifting of µ which respects coordinates. proof (a) Replacing each µi by its completion does not change µ (254I), so we may suppose that all the µi are complete. In this case there is for each i an isomorphism between the measure algebra (Ai , µ ¯i ) of µi and the measure algebra (Bi , ν¯i ) of some {0, 1}Ji with its usual measure νi (331L). We may suppose that the sets Ji are disjoint. Each νi is compact (342Jd), so the isomorphisms are represented by inverse-measurepreserving functions fi : Xi → {0, 1}Ji (343Ca). S Set K = i∈I Ji , and let ν be the usual measure on Y = {0, 1}K , T its domain. We have a natural Q bijection between i∈I {0, 1}Ji and Y , so we obtain a function f : X → Y ; literally speaking, f (x)(j) = fi (x(i))(j) for i ∈ I, j ∈ Ji , x ∈ X. (b) Now f is inverse-measure-preserving and induces an isomorphism between the measure algebras A, B of µ, ν. P P(i) If L ⊆ K is finite and z ∈ {0, 1}L , then, setting Li = LcapJi for i ∈ I, Y µ{x : x ∈ X, f (x)¹L = z} = µ( {w : w ∈ Xi , fi (w)¹Li = z¹Li }) i∈I

=

Y

µi {w : w ∈ Xi , fi (w)¹Li = z¹Li }

i∈I

=

Y

νi {v : v ∈ {0, 1}Ji , v¹Li = z¹Li }

i∈I

(because every fi is inverse-measure-preserving) Y = 2−#(Li ) = 2−#(L) = ν{y : y ∈ Y, y¹L = z}. i∈I

So µf −1 [C] = νC for every basic cylinder set C ⊆ Y . By 254G, f is inverse-measure-preserving. (ii) Accordingly f induces a measure-preserving homomorphism π : B → A. To see that π is surjective, consider Λ0 = {E : E is Σ-measurable, E • ∈ π[B]}. Because π[B] is a closed subalgebra of A (324Kb), Λ0 is a σ-subalgebra of the domain Λ of µ, and of course it contains all µ-negligible sets. If i ∈ J and G ∈ Σi , then there is an H ⊆ {0, 1}Ji such that G4fi−1 [H] is µi -negligible. Now if E = {x : x ∈ X, x(i) ∈ G} and F = {y : y ∈ Y, y¹Ji ∈ H}, E4f −1 [F ] = {x : x(i) ∈ G4fi−1 [H]} N is µ-negligible, and E ∈ Λ0 . But this means that Λ0 ⊇ c i∈I Σi , and must therefore be the whole of Λ (254Ff). Q Q (c) By 345C, there is a translation-invariant lifting φ for ν; by 346C, this respects coordinates. By 346D, we have a corresponding lifting φ0 for µ such that φ0 f −1 [F ] = f −1 [φF ] for every F ∈ T. Now suppose that E ∈ Λ is determined by coordinates in L ⊆ I. Then there is an E 0 belonging to the σ-algebra Λ0L generated by

346G

Consistent liftings

211

{{x : x(i) ∈ G} : i ∈ L, G ∈ Σi }

S such that µ(E4E ) = 0 (254Ob). Write TL for the family of sets in T determined by coordinates in i∈L Ji . Then, just as in (b-ii), every member of Λ0L differs by a negligible set from some set of the form f −1 [F ] with F ∈ TL . So there is an F ∈ TL such that E4f −1 [F ] is µ-negligible. Consequently 0

φ0 E = φ0 f −1 [F ] = f −1 [φF ]. S But φ respects coordinates, so φF is determined by coordinates in i∈L Ji . It follows at once that f −1 [φF ] is determined by coordinates in L; that is, that φ0 E is determined by coordinates in L. As E and L are arbitrary, φ0 respects coordinates, and witnesses the truth of the theorem. 346F It seems to be unknown whether 346E is true of arbitrary probability spaces (346Za); I give some partial results in this direction. The following general method of constructing lower densities will be useful. Lemma Let (X, Σ, µ) and (Y, T, ν) be complete probability spaces, with product (X × Y, Λ, λ). If φ : Λ → Λ is a lower density, then we have a lower density φ1 : Σ → Σ defined by saying that φ1 E = {x : x ∈ X, {y : (x, y) ∈ φ(E × Y )} is conegligible in Y } for every E ∈ Σ. proof For E ∈ Σ, (E × Y )4φ(E × Y ) is negligible, so that Hx = {y : (x, y) ∈ (E × Y )4φ(E × Y )} is ν-negligible for almost every x ∈ X (252D). Now E4φ1 E = {x : Hx is not negligible} is negligible, so φ1 E ∈ Σ. If E, F ∈ Σ, then φ((E ∩ F ) × Y ) = φ((E × Y ) ∩ (F × Y )) = φ(E × Y ) ∩ φ(F × Y ), so that {y : (x, y) ∈ φ((E ∩ F ) × Y )} = {y : (x, y) ∈ φ(E × Y )} ∩ {y : (x, y) ∈ φ(F × Y )} is conegligible iff both {y : (x, y) ∈ φ(E × Y )} and {y : (x, y) ∈ φ(F × Y )} are conegligible, and φ1 (E ∩ F ) = φ1 E ∩ φ1 F . The rest is easy. Of course φ(∅ × Y ) = ∅ so φ1 ∅ = ∅. If E, F ∈ Σ and E4F is negligible, then (E × Y )4(F × Y ) is negligible, φ(E × Y ) = φ(F × Y ) and φ1 E = φ1 F . So φ1 is a lower density, as claimed. 346G Theorem Let h(Xi , Σi , µi )ii∈I be a family of probability spaces with product (X, Σ, µ). For J ⊆ I let ΣJ be the set of members of Σ which are determined by coordinates in J. Then there is a lower density φ : Σ → Σ such that (i) whenever J ⊆ I and E ∈ ΣJ then φE ∈ ΣJ , (ii) whenever J, K ⊆ I are disjoint, E ∈ ΣJ and F ∈ ΣK then φ(E ∪ F ) = φE ∪ φF . Q proof For each i ∈ I, set Yi = XiN , with the product measure Q νi ; set Y = i∈I Yi , with its product measure ν; set Zi = Xi × Yi , with its product measure λi , Q and Z = Q i∈I Zi , with its product measure λ. Then the Q natural identification of Z = i∈I Xi × Yi with i∈I Xi × i∈I Yi = X × Y makes λ correspond to the product of µ and ν (254N). Each (Zi , λi ) can be identified with an infinite power of (Xi , µi ), and is therefore Maharam-type-homogeneous (334E). Consequently there is a lifting φ : Λ → Λ which respects coordinates (346E). Regarding (Z, λ) as the product of (X, µ) and (Y, ν), we see that φ induces a lower density φ : Σ → Σ by the formula of 346F. Q If J ⊆ I and E ∈ Σ is determined by coordinates in J, then E × Y (regarded as a subset of i∈I Zi ) is determined by coordinates in J, so φ(E × Y ) also is. Now suppose that x ∈ φE, x0 ∈ X and x¹J = x0 ¹J. Then for any y ∈ Y , (x¹J, y¹J) = (x0 ¹J, y¹J), so (x, y) ∈ φ(E × Y ) iff (x0 , y) ∈ φ(E × Y ). Thus {y : (x0 , y) ∈ φ(E × Y )} = {y : (x, y) ∈ φ(E × Y )} is conegligible in Y , and x0 ∈ φE. This shows that φE is determined by coordinates in J.

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Now suppose that J and K are disjoint subsets of I, that E, F ∈ Σ are determined by coordinates in J, K respectively, and that x ∈ / φE ∪ φF . Then A = {y : (x, y) ∈ / φ(E × Y )} and B = {y : (x, y) ∈ / φ(F × Y )} are non-negligible. As noted just above, φ(E × Y ) is determined by coordinates in J, so A is determined Q by coordinates in J, and can be expressed as {y : y¹J ∈ A0 }, where A0 ⊆ YJ = i∈J Yi . Because y 7→ y¹J : Y → YJ is inverse-measure-preserving, A0 cannot be negligible in YJ . Similarly, B can be expressed as {y : y¹K ∈ B 0 } for some non-negligible B 0 ⊆ YK . By 251R/251Wm, A0 × B 0 × YI\(J∪K) , regarded as a subset of Y , is non-negligible, that is, C = {y : y ∈ Y, y¹J ∈ A0 , y¹K ∈ B 0 } is non-negligible. But C = A ∩ B = {y : (x, y) ∈ / φ(E × Y ) ∪ φ(F × Y )} = {y : (x, y) ∈ / φ((E ∪ F ) × Y }. So x ∈ / φ(E ∪ F ). As x is arbitrary, φ(E ∪ F ) ⊆ φE ∪ φF ; but of course φE ∪ φF ⊆ φ(E ∪ F ), because φ is a lower density, so that φ(E ∪ F ) = φE ∪ φF , as required. Remark See Macheras Musial & Strauss p99 for an alternative proof. 346H Theorem Let ζ be an ordinal, and h(Xξ , Σξ , µξ )iξ #(Iψ ). Take I ⊆ Kψ such that #(I) = #(Iψ ). We may regard X κ as X I × X κ\I , and in this form we can use the method of 346F to obtain a lower density φ : ΛI → ΛI from φ0 : Λκ → Λκ . Now −1 −1 φ(πIξ [E]) = πIξ [ψE] for every E ∈ Σ, ξ ∈ I. −1 −1 −1 −1 P P The point is that πIξ [E] × X κ\I corresponds to πκξ [E] ⊆ X κ , while φ0 (πκξ [E]) = πκξ [ψE] can be −1 −1 −1 κ\I [ψE]. identified with πIξ [ψE]×X . Now the construction of 346F obviously makes φ(πIξ [E]) equal to πIξ Q Q By 341Jb, there is a lifting φ : ΛI → ΛI such that φW ⊇ φW for every W ∈ ΛI . But now we must have −1 −1 −1 πIξ [ψE] = φ(πIξ [E]) ⊆ φ(πIξ [E]) −1 −1 = X I \ φ(πIξ [X \ E]) ⊆ X I \ φ(πIξ [X \ E]) −1 −1 −1 = X I \ πIξ [ψ(X \ E)] = X I \ πIξ [X \ ψE] = πIξ [ψE] −1 −1 and φ(πIξ [E]) = πIξ [ψE] for every E ∈ Σ, ξ ∈ I. But since #(I) = #(Iψ ), this must be impossible, by the choice of Iψ . X X This contradiction proves the theorem.

346K

Consistent liftings

215

346J Consistent liftings Let (X, Σ, µ) be a measure space. A lifting ψ : Σ → Σ is consistent if for every n ≥ 1 there is a lifting φn of the product measure on X n such that φn (E1 ×. . .×En ) = ψE1 ×. . .×ψEn for all E1 , . . . , En ∈ Σ. Thus 346I tells us, in part, that every complete probability space has a consistent lifting; it follows that every non-trivial complete totally finite measure space has a consistent lifting. I do not suppose you will be surprised to be told that not all liftings on probability spaces are consistent. What may be surprising is the fact that one of the standard liftings already introduced is not consistent. This depends on a general fact about Stone spaces of measure algebras which has further important applications, so I present it as a lemma. 346K Lemma Let (Z, T, ν) be the Stone space of the measure algebra of Lebesgue measure on [0, 1], and let λ be the product measure on Z × Z, with Λ its domain. Then there is a set W ∈ Λ, with λW < 1, ˜ = 1, where such that λ∗ W ˜ = S{G × H : G, H ⊆ Z are open-and-closed, (G × H) \ W is negligible}. W Remark For the sake of anybody who has already become acquainted with the alternative measures which can be put on the product of topological measure spaces, I ought to insist here that the ‘product measure’ λ is, as always in this volume, the ordinary completed product measure as defined in Chapter 25. proof (a) Let hEn in∈N be a sequence of measurable subsets of [0, 1], stochastically independent for Lebesgue 1 measure µ on [0, 1], such that µEn = n+2 for each n. Set an = En• in the measure of algebra of µ, and S ∗ an the corresponding compact open subset of Z. Set W = n∈N En∗ × En∗ . Then En = b λW ≤

P∞

2 n=0 (νEn )

=

P∞

1

n=2 n2

< 1.

˜ < 1. Then there are sequences hGn in∈N , hHn in∈N in T such that ?? Suppose, if possible, that λ∗ W S S ˜ W ⊆ n∈N Gn × Hn and λ( n∈N Gn × Hn ) < 1. Recall from 322Qc that νF = inf{νG : G is compact and open, F ⊆ G} ˜n, H ˜ n such that Gn ⊆ G ˜ n , Hn ⊆ H ˜ n for for every F ∈ T. Accordingly we can find compact open sets G every n ∈ N and S P∞ P∞ ˜ ˜ n=0 ν(Hn \ Hn ) < 1 − λ( n∈N Gn × Hn ), n=0 ν(Gn \ Gn ) + S ˜n × H ˜ n ) < 1. so that λ( n∈N G Let U0 be the family ˜ n : n ∈ N} ∪ {Z \ H ˜ n : n ∈ N}, {Z} ∪ {En∗ : n ∈ N} ∪ {Z \ G so that U0 is a countable subset of T. Let U be the set of finite intersections U0 ∩ U1 ∩ . . . ∩ Un where U0 , . . . , Un ∈ U0 , so that U is also a countable subset of T, and U is closed under ∩. (b) For U ∈ U, define Q(U ) as follows. If νU = 0, then Q(U ) = U . Otherwise, S Q(U ) = Z \ {En∗ : n ∈ N, ν(En∗ ∩ U ) > 0}. Then νQ(U ) is always 0. P P Of course this is true if S νU = 0, so suppose that νU > 0. Set I = {n : ν(En∗ ∩ U ) = 0}. Then we have νU 0 > 0, where U 0 = U \ n∈I En∗ , and Z \ En∗ ⊇ U 0 for every n ∈ I. Because hEn in∈N is stochastically independent for µ, hEn∗ in∈N is stochastically independent for ν, while S ν( n∈I En∗ ) ≤ 1 − νU 0 < 1. P P P∞ 1 By the Borel-Cantelli lemma (273K), n∈I νEn∗ < ∞. Consequently n∈N\I νEn∗ = ∞, because n=0 n+2 is infinite, so S ν(Z \ Q(U )) = ν( n∈N\I En∗ ) = 1, and νQ(U ) = 0. Q Q S (c) Set Q0 = U ∈U Q(U ); because U is countable, Q0 is negligible. Accordingly (Z \ Q0 )2 has measure S ˜n × H ˜ n ; take (w, z) ∈ (Z \ Q0 )2 \ S ˜ ˜ 1 and cannot be included in n∈N G n∈N Gn × Hn .

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346K

(d) We can find sequences hCn in∈N , hDn in∈N , hUn in∈N and hVn in∈N in U such that ˜n × H ˜ n ) = ∅, w ∈ Un+1 ⊆ Un , z ∈ Vn+1 ⊆ Vn , (Un+1 × Vn+1 ) ∩ (G νCn > 0, νDn > 0, Cn ⊆ Un , Dn ⊆ Vn+1 , Cn × Vn+1 ⊆ W , Un+1 × Dn ⊆ W for every n ∈ N. P P Build the sequences inductively, as follows. Start with U0 = V0 = Z. Given that ˜n × H ˜ n . If w ∈ ˜ n , set Un0 = Un \ G ˜n, w ∈ Un ∈ U , z ∈ Vn ∈ U , then we know that (w, z) ∈ / G / G 0 0 0 ˜ n . In either case, we have w ∈ Un0 ∈ U, z ∈ Vn0 ∈ U and Vn = Vn ; otherwise set Un = Un , Vn = Vn \ H ˜n × H ˜ n ) = ∅. (Un0 × Vn0 ) ∩ (G 0 / Q(Un0 ), so ∈ U, w ∈ / Q(Un0 ). But w ∈ Un0 , so this must be because νUn0 > 0. Now z ∈ Because U n S z ∈ {Ek∗ : k ∈ N, ν(Ek∗ ∩ Un0 ) > 0}. Take some k ∈ N such that z ∈ Ek∗ and ν(Ek∗ ∩ Un0 ) > 0, and set Vn+1 = Vn0 ∩ Ek∗ ,

Cn = Ek∗ ∩ Un0 ,

so that z ∈ Vn+1 ∈ U ,

Cn ⊆ Un ,

Cn × Vn+1 ⊆ Ek∗ × Ek∗ ⊆ W , νCn > 0.

Next, z ∈ / Q(Vn+1 ) and νVn+1 > 0; also w ∈ / Q(Vn+1 ), so there is an l such that w ∈ El∗ and ν(El∗ ∩Vn+1 ) > 0. Set Un+1 = Un0 ∩ El∗ ,

Dn = El∗ ∩ Vn+1 ,

so that w ∈ Un+1 ∈ U , Dn ⊆ Vn+1 ,

Un+1 × Dn ⊆ El∗ × El∗ ⊆ W ,

νDn > 0,

˜n × H ˜ n ) = ∅, ˜n × H ˜ n ) ⊆ (Un0 × Vn0 ) ∩ (G (Un+1 × Vn+1 ) ∩ (G and continue the process. Q Q S S P If m ≤ n, Dn ⊆ Vn+1 ⊆ Vm+1 , so (e) Setting C = n∈N Cn , D = n∈N Dn we see that C × D ⊆ W . P Cm × Dn ⊆ W . If m > n, Cm ⊆ Um ⊆ Un+1 , so Cm × Dn ⊆ W . Q Q Recall from 322Qa that the measurable sets of Z are precisely those of the form G4M where M is nowhere dense and negligible and G is compact and open. There must therefore be compact open sets G, H ⊆ Z such that G4C and H4D are negligible. Consequently (G × H) \ W ⊆ ((G \ C) × Z) ∪ (Z × (H \ D)) is negligible, and ˜ ⊆ G×H ⊆W

S n∈N

˜n × H ˜ n. G

˜n × H ˜ n are open, there must be some n such that But because G × H is compact (3A3J), and all the G S ˜ ˜ ˜ ˜ k ) = ∅ for every k, so G × H ⊆ k≤n Gk × Hk = S say. Now (Uk+1 × Vk+1 ) ∩ (Gk × H (Cn+2 × Dn+2 ) ∩ (G × H) ⊆ (Un+1 × Vn+1 ) ∩ S = ∅, and either Cn+2 ∩ G = ∅ or Dn+2 ∩ H = ∅. Since Cn+2 \ G ⊆ C \ G,

Dn+2 \ H ⊆ D \ H

are both negligible, one of Cn+2 , Dn+2 is negligible. But the construction took care to ensure that all the Ck , Dk were non-negligible. X X ˜ = 1, as required. (f ) Thus λ∗ W 346L Proposition Let (Z, T, ν) be the Stone space of the measure algebra of Lebesgue measure on [0, 1]. Let ψ : T → T be the canonical lifting, defined by setting ψE = G whenever E ∈ T, G is open-and-closed and E4G is negligible (341O). Then ψ is not consistent. proof ?? Suppose, if possible, that φ is a lifting on Z × Z such that φ(E × F ) = ψE × ψF for every E, F ∈ T. Let W ⊆ Z × Z be a set as in 346K, and consider φW . If G, H ⊆ Z are open-and-closed and (G × H) \ W is negligible, then

346Yd

Consistent liftings

217

G × H = ψG × ψH = φ(G × H) ⊆ φW ; ˜ ⊆ φW . But this means that that is, in the language of 346K, we must have W ˜ = 1 > λW , λ(φW ) ≥ λ∗ W which is impossible. X X Thus ψ fails the first test and cannot be consistent. 346X Basic exercises (a) Let (X, Σ, µ) be a measure space and φ a lower density for µ. Take H ∈ Σ and set A = X \ (φH ∪ φ(Z \ H)), φ0 E = φE ∪ (A ∩ φ(H ∪ E)) for E ∈ Σ. Show that φ0 is a lower density. > (b) Show that there is no lifting φ of Lebesgue measure on [0, 1]2 which is ‘symmetric’ in the sense that φ(E −1 ) = (φE)−1 for every measurable set E, writing E −1 = {(y, x) : (x, y) ∈ E}. (c) Let (X, Σ, µ) be a measure space and hφn in∈N a sequence of lower densities for µ. (i) Show that S T T E 7→ n∈N φn E and E 7→ n∈N m≥n φm E are also lower densities for µ. (ii) Show that if µ is complete S T and F is any filter on N, then E 7→ F ∈F n∈F φn E is a lower density for µ. (d) Let (X, Σ, µ) be a strictly localizable measure space, and G a countable group of measure space automorphisms from X to itself. Show that there is a lower density φ : Σ → Σ which is G-invariant in the T sense that φ(g −1 [E]) = g −1 [φE] for every E ∈ Σ, g ∈ G. (Hint: set φE = g∈G g[φ0 (g −1 [E])].) > (e) Let φ be lower Lebesgue density on R, and φ any lifting of Lebesgue measure on R such that φE ⊇ φE for every measurable set E. Show that φ is consistent. (Hint: given n ≥ 1, let φn be lower Lebesgue density on Rn . For x ∈ Rn let Ix be the ideal generated by S {W : x ∈ φn (Rn \ W )} ∪ i (g) Suppose, in 341H, that (X, Σ, µ) is a product of probability spaces, and that in the proof, instead of taking haξ iξ 0 then u 7→ αu is an order-isomorphism, so we have sup(αA) = α sup A if either side is defined; similarly, inf(αA) = α inf A. 351E Linear subspaces If U is a partially ordered linear space, and V is any linear subspace of U , then V , with the induced linear and order structures, is a partially ordered linear space; this is obvious from the definition. 351F Positive linear operators Let U and V be partially ordered linear spaces, and write L(U ; V ) for the linear space of all linear operators from U to V . For S, T ∈ L(U ; V ) say that S ≤ T iff Su ≤ T u for every u ∈ U + . Under this ordering, L(U ; V ) is a partially ordered linear space; its positive cone is {T : T u ≥ 0 for every u ∈ U + }. P P This is an elementary verification. Q Q Note that, for T ∈ L(U ; V ), T ≥ 0 =⇒ T u ≤ T u + T (v − u) = T v whenever u ≤ v in U =⇒ 0 = T 0 ≤ T u for every u ∈ U + =⇒ T ≥ 0, so that T ≥ 0 iff T is order-preserving. In this case we say that T is a positive linear operator. Clearly ST is a positive linear operator whenever U , V and W are partially ordered linear spaces and T : U → V , S : V → W are positive linear operators (cf. 313Ia). 351G Order-continuous positive linear operators: Proposition Let U and V be partially ordered linear spaces and T : U → V a positive linear operator. (a) The following are equiveridical: (i) T is order-continuous; (ii) inf T [A] = 0 in V whenever A ⊆ U is a non-empty downwards-directed set with infimum 0 in U ; (iii) sup T [A] = T w in V whenever A ⊆ U + is a non-empty upwards-directed set with supremum w in U. (b) The following are equiveridical: (i) T is sequentially order-continuous; (ii) inf n∈N T un = 0 in V whenever hun in∈N is a non-increasing sequence in U with infimum 0 in U ; (iii) supn∈N T un = T w in V whenever hun in∈N is a non-decreasing sequence in U + with supremum w in U . proof (a)(i)⇒(iii) is trivial. (iii)⇒(ii) Assuming (iii), and given that A is non-empty, downwards-directed and has infimum 0, take any u0 ∈ A and consider A0 = {u : u ∈ A, u ≤ u0 }, B = u0 − A0 . Then A0 is non-empty, downwards-directed and has infimum 0, so B is non-empty, upwards-directed and has supremum u0 (using 351Db); by (iii), sup T [B] = T u0 and (inverting again) inf T [A0 ] = inf T [u0 − B] = inf T u0 − T [B] = T u0 − sup T [B] = 0. But (because T is positive) 0 is surely a lower bound for T [A], so it is also the infimum of T [A]. As A is arbitrary, (ii) is true. α) If A ⊆ U is non-empty, downwards-directed and has (ii)⇒(i) Suppose now that (ii) is true. (α infimum w, then A − w is non-empty, downwards-directed and has infimum 0, so

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inf T [A − w] = 0, inf T [A] = inf(T [A − w] + T w) = T w + inf T [A − w] = T w. β ) If A ⊆ U is non-empty, upwards-directed and has supremum w, then −A is non-empty, downwards(β directed and has infimum −w, so sup T [A] = − inf(−T [A]) = − inf T [−A] = −T (−w) = T w. Putting these together, T is order-continuous. (b) The arguments are identical, replacing each directed set by an appropriate sequence. 351H Riesz homomorphisms (a) For the sake of a representation theorem below (351Q), I introduce the following definition. Let U , V be partially ordered linear spaces. A Riesz homomorphism from U to V is a linear operator T : U → V such that whenever A ⊆ U is a finite non-empty set and inf A = 0 in U , then inf T [A] = 0 in V . The following facts are now nearly obvious. (b) Any Riesz homomorphism is a positive linear operator. (For if T is a Riesz homomorphism and u ≥ 0, then inf{0, u} = 0 so inf{0, T u} = 0 and T u ≥ 0.) (c) Let U and V be partially ordered linear spaces and T : U → V a Riesz homomorphism. Then inf T [A] exists = T (inf A),

sup T [A] exists = T (sup A)

whenever A ⊆ U is a finite non-empty set and inf A, sup A exist. (Apply the definition in (a) to A0 = {u − inf A : u ∈ A},

A00 = {sup A − u : u ∈ A}.)

(d) If U , V and W are partially ordered linear spaces and T : U → V , S : V → W are Riesz homomorphisms then ST : U → W is a Riesz homomorphism. 351I Solid sets Let U be a partially ordered linear space. I will say that a subset A of U is solid if S A = {v : v ∈ U, −u ≤ v ≤ u for some u ∈ A} = u∈U [−u, u] in the notation of 2A1Ab. (I should perhaps remark that while this definition is well established in the case of Riesz spaces (§352), the extension to general partially ordered linear spaces is not standard. See 351Yb for a warning.) 351J Proposition Let U be a partially ordered linear space and V a solid linear subspace of U . Then the quotient linear space U/V has a partially ordered linear space structure defined by either of the rules u• ≤ w• iff there is some v ∈ V such that u ≤ v + w, (U/V )+ = {u• : u ∈ U + }, and for this partial ordering on U/V the map u 7→ u• : U → U/V is a Riesz homomorphism. proof (a) I had better start by giving priority to one of the descriptions of the relation ≤ on U/V ; I choose the first. To see that this makes U/V a partially ordered linear space, we have to check the following. (i) 0 ∈ V and u ≤ u + 0, so u• ≤ u• for every u ∈ U . (ii) If u1 , u2 , u3 ∈ U and u•1 ≤ u•2 , u•2 ≤ u•3 then there are v1 , v2 ∈ V such that u1 ≤ u2 + v1 , u2 ≤ u3 + v2 ; in which case v1 + v2 ∈ V and u1 ≤ u3 + v1 + v2 , so u•1 ≤ u•3 . (iii) If u, w ∈ U and u• ≤ w• , w• ≤ u• then there are v, v 0 ∈ V such that u ≤ w + v, w ≤ u + v 0 . Now there are v0 , v00 ∈ V such that −v0 ≤ v ≤ v0 , −v00 ≤ v 0 ≤ v00 , and in this case v0 , v00 ≥ 0 (351Cd), so −v0 − v00 ≤ −v 0 ≤ u − w ≤ v ≤ v0 + v00 ∈ V , . Accordingly u − w ∈ V and u• = w• . Thus U/V is a partially ordered set. (iv) If u1 , u2 , w ∈ U and u•1 ≤ u•2 , then there is a v ∈ V such that u1 ≤ u2 + v, in which case u1 + w ≤ u2 + w + v and u•1 + w• ≤ u•2 + w• . (v) If u ∈ U , α ∈ R and u• ≥ 0, α ≥ 0 then there is a v ∈ V such that u + v ≥ 0; now αv ∈ V and αu + αv ≥ 0, so αu• = (αu)• ≥ 0.

351N

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223

Thus U/V is a partially ordered linear space. (b) Now (U/V )+ = {u• : u ≥ 0}. P P If u ≥ 0 then of course u• ≥ 0 because 0 ∈ V and u + 0 ≥ 0. On the other hand, if we have any element p of (U/V )+ , there are u ∈ U , v ∈ V such that u• = p and u + v ≥ 0; but now p = (u + v)• is of the required form. Q Q (c) Finally, u 7→ u• is a Riesz homomorphism. P P Suppose that A ⊆ U is a non-empty finite set and that inf A = 0 in U . Then u• ≥ 0 for every u ∈ A, that is, 0 is a lower bound for {u• : u ∈ A}. Let p ∈ U/V be any other lower bound for {u• : u ∈ A}. Express p as w• where w ∈ U . For each u ∈ A, w• ≤P u• so there is 0 0 0 ∗ a vu ∈ V such that w ≤ u + vu . Next, there is a vu ∈ V such that −vu ≤ vu ≤ vu . Set v = u∈A vu0 ∈ V . Then vu ≤ vu0 ≤ v ∗ so w ≤ u + v ∗ for every u ∈ A, and w − v ∗ is a lower bound for A in U . Accordingly w − v ∗ ≤ 0, w ≤ 0 + v ∗ and p = w• ≤ 0. As p is arbitrary, inf{u• : u ∈ A} = 0; as A is arbitrary, u 7→ u• is a Riesz homomorphism. Q Q 351K Lemma Suppose that U is a partially ordered linear space, and that W , V are solid linear subspaces of U such that W ⊆ V . Then V1 = {v • : v ∈ V } is a solid linear subspace of U/W . proof (i) Because the map u 7→ u• is linear, V1 is a linear subspace of U/W . (ii) If p ∈ V1 , there is a v ∈ V such that p = v • ; because V is solid in U , there is a v0 ∈ V such that −v0 ≤ v ≤ v0 ; now v0• ∈ V1 and −v0• ≤ p ≤ v0• . (iii) If p ∈ V1 , q ∈ U/W and −p ≤ q ≤ p, take v0 ∈ V , u ∈ U such that v0• = p, u• = q. Because −v0• ≤ u• ≤ v0• , there are w, w0 ∈ W such that −v0 − w ≤ u ≤ v0 + w0 . Now −v0 − w, v0 + w0 both belong to V , which is solid, so both are mapped to 0 by the canonical Riesz homomorphism from U to U/V , and u must also be, that is, u ∈ V and q = u• ∈ V1 . (iv) Putting (ii) and (iii) together, V1 is solid. 351L Products If hUi ii∈I is any family of partially ordered linear spaces, we have a product linear Q space U = i∈I Ui ; if we set u ≤ v in U iff u(i) ≤ v(i) for every i ∈ I, U becomes a partially ordered linear space, with positive cone {u : u(i) ≥ 0 for every i ∈ I}. For each i ∈ I the map u 7→ u(i) : U → Ui is an order-continuous Riesz homomorphism (in fact, it preserves arbitrary suprema and infima). 351M Reduced powers of R (a) Let X be any set. Then RX is a partially ordered linear space if we say that f ≤ g means that f (x) ≤ g(x) for every x ∈ X, as in 351L. If now F is a filter on X, we have a corresponding set V = {f : f ∈ RX , {x : f (x) = 0} ∈ F}; it is easy to see that V is a linear subspace of RX , and is solid because f ∈ V iff |f | ∈ V . By the reduced power RX |F I shall mean the quotient partially ordered linear space RX /V . (b) Note that for f ∈ RX , f • ≥ 0 in RX |F ⇐⇒ {x : f (x) ≥ 0} ∈ F. P P (i) If f • ≥ 0, there is a g ∈ V such that f + g ≥ 0; now {x : f (x) ≥ 0} ⊇ {x : g(x) = 0} ∈ F. (ii) If {x : f (x) ≥ 0} ∈ F, then {x : (|f | − f )(x) = 0} ∈ F, so f • = |f |• ≥ 0. Q Q 351N On the way to the next theorem, the main result (in terms of mathematical depth) of this section, we need a string of lemmas. Lemma Let U be a partially ordered linear space. If u, v0 , . . . , vn ∈ U are such that u 6= 0 and v0 , . . . , vn ≥ 0 then there is a linear functional f : U → R such that f (u) 6= 0 and f (vi ) ≥ 0 for every i. proof The point is that at most one of u, −u can belong to the convex cone C generated by {v0 , . . . , vn }, because this is included in the convex cone set U + , and since u 6= 0 at most one of u, −u can belong to U + . Now however the Hahn-Banach theorem, in the form 3A5D, tells us that if u ∈ / C there is a linear functional f : U → R such that f (u) < 0, f (vi ) ≥ 0 for every i; while if −u ∈ / C we can get f (−u) < 0 and f (vi ) ≥ 0 for every i. Thus in either case we have a functional of the required type.

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351O Lemma Let U be a partially ordered linear space, and u0 a non-zero member of U . Then there is a solid linear subspace V of U such that u0 ∈ / V and whenever A ⊆ U is finite, non-empty and has infimum 0 then A ∩ V 6= ∅. proof (a) Let W be the family of all solid S linear subspaces of U not containing u0 . Then any non-empty totally ordered V ⊆ W has an upper bound V in W. By Zorn’s Lemma, W has a maximal element V say. This is surely a solid linear subspace of U not containing u0 . (b) Now for any w ∈ U + \ V there are α ≥ 0, v ∈ V + such that −αw − v ≤ u0 ≤ αw + v. P P Let V1 be {u : u ∈ U , there are α ≥ 0, v ∈ V + such that −αw − v ≤ u ≤ αw + v}. Then it is easy to check that V1 is a solid linear subspace of U , including V , and containing w; because w∈ / V , V1 6= V , so V1 ∈ / W and u ∈ V1 , as claimed. Q Q (c) It follows that if A ⊆ U is finite and non-empty and inf A = 0 in U then A∩V 6= ∅. P P?? Otherwise, for P every w ∈ A there must be αw ≥ 0, vw ∈ V + such that −αw w −vw ≤ u0 ≤ αw w +vw . Set α = 1+ w∈A αw , P v = w∈A vw ∈ V ; then −αw − v ≤ u0 ≤ αw + v for every w ∈ A. Accordingly α1 (u0 − v) ≤ w for every w ∈ A and α1 (u0 − v) ≤ 0, so u0 ≤ v. Similarly, − α1 (v + u0 ) ≤ w for every w ∈ A and −v ≤ u0 . But (because V is solid) this means that u0 ∈ V , which is not so. X XQ Q Accordingly V has the required properties. 351P Lemma Let U be a partially ordered linear space and u a non-zero element of U , and suppose that A0 , . . . , An are finite non-empty subsets of U such that inf Aj = 0 for every j ≤ n. Then there is a linear functional f : U → R such that f (u) 6= 0 and min f [Aj ] = 0 for every j ≤ n. proof By 351O, there is a solid linear subspace V of U such that u ∈ / V and Aj ∩ V 6= 0 for every S j ≤ n. Give the quotient space U/V its standard partial ordering (351J), and in U/V set C = {v • : v ∈ j≤n Aj }. Then C is a finite subset of (U/V )+ , while u• 6= 0, so by 351N there is a linear functional g : U/V → R such that g(u• ) 6= 0 but g(p) ≥ 0 S for every p ∈ C. Set f (v) = g(v • ) for v ∈ U ; then f : U → R is linear, f (u) 6= 0 and f (v) ≥ 0 for every v ∈ j≤n Aj . But also, for each j ≤ n, there is a vj ∈ Aj ∩ V , so that f (vj ) = 0; and this means that min f [Aj ] must be 0, as required. 351Q

Now we are ready for the theorem.

Theorem Let U be any partially ordered linear space. Then we can find a set X, a filter F on X and an injective Riesz homomorphism from U to the reduced power RX |F described in 351M. proof Let X be the set of all linear functionals f : U → R; define φ : U → RX by setting φ(u)(f ) = f (u) for every f ∈ X, u ∈ U , so that φ is linear. Let A be the family of non-empty finite sets A ⊆ U such that inf A = 0. For A ∈ A let FA be the set of those f ∈ X such that min f [A] = 0. Since 0 ∈ FA for every A ∈ A, the set T F = {F : F ⊆ X, there are A0 , . . . , An ∈ A such that F ⊇ j≤n FAj } is a filter on X. Set ψ(u) = φ(u)• ∈ RX |F for u ∈ U . The ψ : U → RX |F is an injective Riesz homomorphism. P P (i) ψ is linear because φ and h 7→ h• : RX → Rx |F are. (ii) If A ∈ A, then FA ∈ F . So, first, if v ∈ A, then {f : φ(v)(f ) ≥ 0} ∈ F, so that ψ(v) = φ(v)• ≥ 0 in RX |F (351Mb). Next, if w ∈ RX |F • and w ≤ ψ(v) for every v ∈ A, we can express w as h• where hT ≤ φ(v)• for every v ∈ A, that is, Hv = {f : h(f ) ≤ φ(v)(f )} ∈ F for every v ∈ A. But now H = FA ∩ v∈A Hv ∈ F, and for f ∈ H we have h(f ) ≤ minv∈A f (v) = 0. This means that w = h• ≤ 0. As w is arbitrary, inf ψ[A] = 0. As A is arbitrary, ψ is a Riesz homomorphism. (iii) Finally, ?? suppose, if possible, that there is a non-zero T u ∈ U such that ψ(u) = 0. Then F = {f : f (u) = 0} ∈ F, and there are A0 , . . . , An ∈ A such that F ⊇ j≤n FAj . By 351P, T there is an f ∈ j≤n FAj such that f (u) 6= 0; which is impossible. X X Accordingly ψ is injective, as claimed. Q Q

351 Notes

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225

351R Archimedean spaces (a) For a partially ordered linear space U , the following are equiveridical: (i) if u, v ∈ U are such that nu ≤ v for every n ∈ N then u ≤ 0 (ii) if u ≥ 0 in U then inf δ>0 δu = 0. P P (i)⇒(ii) If (i) is true and u ≥ 0, then of course δu ≥ 0 for every δ > 0; on the other hand, if v ≤ δu for every δ > 0, then nv ≤ n · n1 u = u for every n ≥ 1, while of course 0v = 0 ≤ u, so v ≤ 0. Thus 0 is the greatest lower bound of {δu : δ > 0}. (ii)⇒(i) If (ii) is true and nu ≤ v for every n ∈ N, then 0 ≤ v and u ≤ n1 v for every n ≥ 1. If now δ > 0, then there is an n ≥ 1 such that n1 ≤ δ, so that u ≤ n1 v ≤ δv (351Bc). Accordingly u is a lower bound for {δv : δ > 0} and u ≤ 0. Q Q (b) I will say that partially ordered linear spaces satisfying the equiveridical conditions of (a) above are Archimedean. (c) Any linear subspace of an Archimedean partially ordered linear space, with the induced partially ordered linear space structure, is Archimedean. Q (d) Any product of Archimedean partially ordered linear spaces is Archimedean. P P If U = i∈I Ui is a product of Archimedean spaces, and nu ≤ v in U for every n ∈ N, then for each i ∈ I we must have nu(i) ≤ v(i) for every n, so that u(i) ≤ 0; accordingly u ≤ 0. Q Q In particular, RX is Archimedean for any set X. 351X Basic exercises > (a) Let ζ be any ordinal. The lexicographic ordering on Rζ is defined by saying that f ≤ g iff either f = g or there is a ξ < ζ such that f (η) = g(η) for η < ξ and f (ξ) < g(ξ). Show that this is a total order on Rζ which renders Rζ a partially ordered linear space. (b) Let U be a partially ordered linear space and V a linear subspace of U . Show that the formulae of 351J define a partially ordered linear space structure on the quotient U/V iff V is order-convex, that is, u ∈ V whenever v1 , v2 ∈ V and v1 ≤ u ≤ v2 . (c) Let hUi ii∈I be a family of partially ordered linear spaces with product U . Define Ti : Ui → U by setting Ti x = u where u(i) = x, u(j) = 0 for j 6= i. Show that Ti is an injective order-continuous Riesz homomorphism. > (d) Let U be a partially ordered linear space and hVi ii∈I a family of partially orderedQlinear spaces with product V . Show that L(U ; V ) can be identified, as partially ordered linear space, with i∈I L(U ; Vi ). > (e) Show that if U , V are partially ordered linear spaces and V is Archimedean, then L(U ; V ) is Archimedean. 351Y Further exercises (a) Give an example of two partially ordered linear spaces U and V and a bijective Riesz homomorphism T : U → V such that T −1 : V → U is not a Riesz homomorphism. (b) (i) Let U be a partially ordered linear space. Show that U is a solid subset of itself (on the definition 351I) iff U = U + − U + . (ii) Give an example of a partially ordered linear space U satisfying this condition with an element u ∈ U such that the intersection of the solid sets containing u is not solid. (c) Let U be a partially ordered linear space, and suppose that A, B ⊆ U are two non-empty finite sets such that (α) u ∨ v = sup{u, v} is defined for every u ∈ A, v ∈ B (β inf A and inf B and (inf A) ∨ (inf B) are defined. Show that inf{u ∨ v : u ∈ A, v ∈ B} = (inf A) ∨ (inf B). (Hint: show that this is true if U = R, if U = RX and if U = RX |F, and use 351Q.) (d) Show that a reduced power RX |F, as described in 351M, is totally ordered iff F is an ultrafilter, and in this case has a natural structure as a totally ordered field. 351 Notes and comments The idea of ‘partially ordered linear space’ is a very natural abstraction from the elementary examples of RX and its subspaces, and the only possible difficulty lies in guessing the exact boundary at which one’s standard manipulations with such familiar spaces cease to be valid in the general case. (For instance, most people’s favourite examples are Archimedean, in the sense of 351R, so it is prudent

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to check your intuitions against a non-Archimedean space like that of 351Xa.) There is really no room for any deep idea to appear in 351B-351F. When I come to what I call ‘Riesz homomorphisms’, however (351H), there are some more interesting possibilities in the background. I shall not discuss the applications of Theorem 351Q to general partially ordered linear spaces; it is here for the sake of its application to Riesz spaces in the next section. But I think it is a very striking fact that not only does any partially ordered linear space U appear as a linear subspace of some reduced power of R, but the embedding can be taken to preserve any suprema and infima of finite sets which exist in U . This is in a sense a result of the same kind as the Stone representation theorem for Boolean algebras; it gives us a chance to confirm that an intuition valid for R or RX may in fact apply to arbitrary partially ordered linear spaces. If you like, this provides a metamathematical foundation for such results as those in 351B. I have to say that for partially ordered linear spaces it is generally quicker to find a proof directly from the definition than to trace through an argument relying on 351Q; but this is not always the case for Riesz spaces. I offer 351Yc as an example of a result where a direct proof does at least call for a moment’s thought, while the argument through 351Q is straightforward. ‘Reduced powers’ are of course of great importance for other reasons; I mention 351Yd as a hint of what can be done.

352 Riesz spaces In this section I sketch those fragments of the theory we need which can be expressed as theorems about general Riesz spaces or vector lattices. I begin with the definition (352A) and most elementary properties (352C-352F). In 352G-352J I discuss Riesz homomorphisms and the associated subspaces (Riesz subspaces, solid linear subspaces); I mention product spaces (352K, 352T) and quotient spaces (352Jb, 352U) and the form the representation theorem 351Q takes in the present context (352L-352M). Most of the second half of the section concerns the theory of ‘bands’ in Riesz spaces, with the algebras of complemented bands (352Q) and projection bands (352S) and a description of bands generated by upwards-directed sets (352V). I conclude with a description of ‘f -algebras’ (352W). 352A

I repeat a definition from 241E.

Definition A Riesz space or vector lattice is a partially ordered linear space which is a lattice. 352B Lemma If U is a partially ordered linear space, then it is a Riesz space iff sup{0, u} is defined for every u ∈ U . proof If U is a lattice, then of course sup{u, 0} is defined for every u. If sup{u, 0} is defined for every u, and v1 , v2 are any two members of U , consider w = v1 + sup{0, v2 − v1 }; by 351Db, w = sup{v1 , v2 }. Next, inf{v1 , v2 } = − sup{−v1 , −v2 } must also be defined in U ; as v1 and v2 are arbitrary, U is a lattice. 352C Notation In any Riesz space U I will write u+ = u ∨ 0,

u− = (−u) ∨ 0 = (−u)+ ,

|u| = u ∨ (−u)

where (as in any lattice) u ∨ v = sup{u, v} (and u ∧ v = inf{u, v}). I mention immediately a term which will be useful: a family hui ii∈I in U is disjoint if |ui | ∧ |uj | = 0 for all distinct i, j ∈ I. Similarly, a set C ⊆ U is disjoint if |u| ∧ |v| = 0 for all distinct u, v ∈ C. 352D Elementary identities Let U be a Riesz space. The translation-invariance of the order, and its invariance under positive scalar multiplication, reversal under negative multiplication, lead directly to the following, which are in effect special cases of 351D: u + (v ∨ w) = (u + v) ∨ (u + w),

u + (v ∧ w) = (u + v) ∧ (u + w),

352Ec

Riesz spaces

α(u ∨ v) = αu ∨ αv,

227

α(u ∧ v) = αu ∧ αv if α ≥ 0,

−(u ∨ v) = (−u) ∧ (−v). Combining and elaborating on these facts, we get u+ − u− = (u ∨ 0) − ((−u) ∨ 0) = u + (0 ∨ (−u)) − ((−u) ∨ 0) = u, u+ + u− = 2u+ − u = (2u ∨ 0) − u = u ∨ (−u) = |u|, u ≥ 0 ⇐⇒ −u ≤ 0 ⇐⇒ u− = 0 ⇐⇒ u = u+ ⇐⇒ u = |u|, | − u| = |u|,

| |u| | = |u|,

|αu| = |α||u| (looking at the cases α ≥ 0, α ≤ 0 separately),

u ∨ v + u ∧ v = u + (0 ∨ (v − u)) + v + ((u − v) ∧ 0) = u + (0 ∨ (v − u)) + v − ((v − u) ∨ 0) = u + v, u ∨ v = u + (0 ∨ (v − u)) = u + (v − u)+ , u ∧ v = u + (0 ∧ (v − u)) = u − (−0 ∨ (u − v)) = u − (u − v)+ , 1 2

1 2

1 2

u ∨ v = (2u ∨ 2v) = (u + v + (u − v) ∨ (v − u)) = (u + v + |u − v|), 1 2

u ∧ v = u + v − u ∨ v = (u + v − |u − v|), u+ ∨ u− = u ∨ (−u) ∨ 0 = |u|,

u+ ∧ u− = u+ + u− − (u+ ∨ u− ) = 0,

|u + v| = (u + v) ∧ ((−u) + (−v)) ≤ (|u| + |v|) ∧ (|u| + |v|) =|u| + |v|, ||u| − |v|| = (|u| − |v|) ∧ (|v| − |u|) ≤ |u − v| + |v − u| = |u − v|, |u ∨ v| ≤ |u| + |v| (because −|u| ≤ u ∨ v ≤ |u| ∨ |v| ≤ |u| + |v|) for u, v, w ∈ U and α ∈ R. 352E Distributive laws Let U be a Riesz space. (a) If A, B ⊆ U have suprema a, b in U , then C = {u ∧ v : u ∈ A, v ∈ B} has supremum a ∧ b. P P Of course u ∧ v ≤ a ∧ b for all u ∈ A, v ∈ B, so a ∧ b is an upper bound for C. Now suppose that c is any upper bound for C. If u ∈ A, v ∈ B then u − (u − v)+ = u ∧ v ≤ c,

u ≤ c + (u − v)+ ≤ c + (a − v)+

(because (a − v)+ = sup{a − v, 0} ≥ sup{u − v, 0} = (u − v)+ ). As u is arbitrary, a ≤ c + (a − v)+ and a ∧ v ≤ c. Now turn the argument round: v = (a ∧ v) + (v − a)+ ≤ c + (v − a)+ ≤ c + (b − a)+ , and this is true for every v ∈ B, so b ≤ c + (b − a)+ and a ∧ b ≤ c. As c is arbitrary, a ∧ b = sup C, as claimed. Q Q (b) Similarly, or applying (a) to −A and −B, inf{u ∨ v : u ∈ A, v ∈ B} = inf A ∨ inf B whenever A, B ⊆ U and the right-hand-side is defined. (c) In particular, U is a distributive lattice (definition: 3A1Ic).

228

Riesz spaces

352F

352F Further identities and inequalities At a slightly deeper level we have the following facts. Proposition Let U be a Riesz space. (a) If u, v, w ≥ 0 in U then u ∧ (v + w) ≤ (u ∧ v) + (u ∧ P w). Pn n (b) If u0 , . . . , un ∈ U and |ui | ∧ |uj | = 0 for i 6= j, then | i=0 αi ui | = i=0 |αi ||ui | for any α0 , . . . , αn . (c) If u, v ∈ U then u+ ∧ v + ≤ (u + v)+ ≤ u+ + v + . Pm Pn (d) If u0 , . . . , um , v0 , . . . , vn ∈ U + and i=0 ui = j=0 vj , then there is a family hwij ii≤m,j≤n in U + Pm Pn such that i=0 wij = vj for every j ≤ n and j=0 wij = ui for every i ≤ m. proof (a) u ∧ (v + w) ≤ [(u + w) ∧ (v + w)] ∧ u ≤ [(u ∧ v) + w] ∧ [(u ∧ v) + u] = (u ∧ v) + (u ∧ w). α) A simple (b)(i)(α the inductive step, shows that if v0 , . . . , vm , w0 , . . . , wn are Pm induction, Pn using (a) Pmfor P n β ) Next, if u ∧ v = 0 then non-negative then i=0 vi ∧ j=0 wj ≤ i=0 j=0 vi ∧ wj . (β (u − v)+ = u − (u ∧ v) = u,

(u − v)− = (v − u)+ = v − (v ∧ u) = v,

|u − v| = (u − v)+ + (u − v)− = u + v = |u + v|, so if |u| ∧ |v| = 0 then (u+ + v + ) ∧ (u− + v − ) ≤ (u+ ∧ u− ) + (u+ ∧ v − ) + (v + ∧ u− ) + (v + ∧ v − ) ≤ 0 + (|u| ∧ |v|) + (|v| ∧ |u|) + 0 = 0 and |u + v| = |(u+ + v + ) − (u− + v − )| = u+ + v + + u− + v − = |u| + |v|. (γγ ) Finally, if |u| ∧ |v| = 0 and α, β ∈ R, |αu| ∧ |βv| = |α||u| ∧ |β||v| ≤ (|α| + |β|)|u| ∧ (|α| + |β|)|v| = (|α| + |β|)(|u| ∧ |v|) = 0. (ii) We may therefore proceed by induction. The case n = 0 is trivial. For the inductive step to n + 1, setting u0i = αi ui we have |u0i | ∧ |u0j | = 0 for all i 6= j, by (i-γ). By (i-α), Pn Pn Pn |u0n+1 | ∧ | i=0 u0i | ≤ |u0n+1 | ∧ i=0 |u0i | ≤ i=0 |u0n+1 | ∧ |u0i | = 0, so by (i-β) and the inductive hypothesis Pn+1 Pn Pn+1 | i=0 u0i | = |u0n+1 | + | i=0 u0i | = i=0 |u0i | as required. (c) By 352E, u+ ∧ v + = (u ∨ 0) ∧ (v ∨ 0) = (u ∧ v) ∨ 0. Now 1 2

1 2

u ∧ v = (u + v − |u − v|) ≤ (u + v + |u + v|) = (u + v)+ , and of course 0 ≤ (u + v)+ , so u+ ∧ v + ≤ (u + v)+ . For the other inequality we need only note that u + v ≤ u+ + v + (because u ≤ u+ , v ≤ v + ) and 0 ≤ u+ + v + . Pm Pn (d) Write w for the common value of i=0 ui and j=0 vj . Induce on k = #({(i, j) : i ≤ m, j ≤ n, ui ∧ vj > 0}). If k = 0, that is, ui ∧ vj = 0 for all i, j, then (by (a), used repeatedly) we must have w ∧ w = 0, that is, w = 0, and we can take wij = 0 for all i, j. For the inductive step to k ≥ 1, take i∗ , j ∗ such that w ˜ = ui∗ ∧ vj ∗ > 0. Set

352Ic

Riesz spaces

229

u ˜i∗ = ui∗ − w, ˜ u ˜i = ui for i 6= i∗ ,

Then

Pm i=0

u ˜i =

v˜j ∗ = vj ∗ − w, ˜ v˜j = vj for j 6= j ∗ .

Pn j=0

v˜j = w − w ˜ and u ˜i ∧ v˜j ≤ ui ∧ vj for all i, j, while u ˜i∗ ∧ v˜j ∗ = 0; so that #({(i, j) : u ˜i ∧ vj > 0}) < k.

Pn By the inductive hypothesis, there are w ˜ij ≥ 0, for i ≤ m and j ≤ n, such that u ˜i = j=0 w ˜ij for each Pm Pn ∗ ∗ i, v˜j = i=0 w ˜ij for each j. Set wi∗ j ∗ = w ˜i∗ j ∗ + w, ˜ wij = w ˜ij for (i, j) 6= (i , j ); then ui = j=0 wij , Pm vj = i=0 wij so the induction proceeds. 352G Riesz homomorphisms: Proposition Let U be a Riesz space, V a partially ordered linear space and T : U → V a linear operator. Then the following are equiveridical: (i) T is a Riesz homomorphism in the sense of 351H; (ii) (T u)+ = sup{T u, 0} is defined and equal to T (u+ ) for every u ∈ U ; (iii) sup{T u, −T u} is defined and equal to T |u| for every u ∈ U ; (iv) inf{T u, T v} = 0 in V whenever u ∧ v = 0 in U . proof (i)⇒(iii) and (i)⇒(iv) are special cases of 351Hc. For (iii)⇒(ii) we have 1 2

1 2

1 2

1 2

1 2

sup{T u, 0} = T u + sup{ T u, − T u} = T u + T |u| = T (u+ ). For (ii)⇒(i), argue as follows. If (ii) is true and u, v ∈ U , then T u ∧ T v = inf{T u, T v} = T u + inf{0, T v − T u} = T u − sup{0, T (u − v)} is defined and equal to T u − T ((u − v)+ ) = T (u − (u − v)+ ) = T (u ∧ v). Inducing on n, inf i≤n T ui = T (inf i≤n ui ) for all u0 , . . . , un ∈ U ; in particular, if inf i≤n ui = 0 then inf i≤n T ui = 0; which is the definition I gave of Riesz homomorphism. Finally, for (iv)⇒(ii), we know from (iv) that 0 = inf{T (u+ ), T (u− )}, so −T (u+ ) = inf{0, −T u} and T (u+ ) = sup{0, T u}. 352H Proposition If U and V are Riesz spaces and T : U → V is a bijective Riesz homomorphism, then T is a partially-ordered-linear-space isomorphism, and T −1 : V → U is a Riesz homomorphism. proof Use 352G(ii). If v ∈ V , set u = T −1 v; then T (u+ ) = v + so T −1 (v + ) = u+ = (T −1 v)+ . Thus T −1 is a Riesz homomorphism; in particular, it is order-preserving, so T is an isomorphism for the order structures as well as for the linear structures. 352I Riesz subspaces (a) If U is a partially ordered linear space, a Riesz subspace of U is a linear subspace V such that sup{u, v} and inf{u, v} are defined in U and belong to V for every u, v ∈ V . In this case they are the supremum and infimum of {u, v} in V , so V , with the induced order and linear structure, is a Riesz space in its own right, and the embedding map u 7→ u : V → U is a Riesz homomorphism. (b) Generally, if U is a Riesz space, V is a partially ordered linear space and T : U → V is a Riesz homomorphism, then T [U ] is a Riesz subspace of V (because, by 351Hc, T u ∨ T u0 = T (u ∨ u0 ), T (u ∧ u0 ) = T u ∧ T u0 belong to T [U ] for all u, u0 ∈ U ). (c) If U is a Riesz space and V is a linear subspace of U , then V is a Riesz subspace of U iff |u| ∈ V for every u ∈ V . P P In this case, 1 2

u ∨ v = (u + v + |u − v|), belong to V for all u, v ∈ V . Q Q

1 2

u ∧ v = (u + v − |u − v|

230

Riesz spaces

352J

352J Solid subsets (a) If U is a Riesz space, a subset A of U is solid (in the sense of 351I) iff v ∈ A whenever u ∈ A and |v| ≤ |u|. P P (α) If A is solid, u ∈ V and |v| ≤ |u|, then there is some w ∈ A such that −w ≤ u ≤ w; in this case |v| ≤ |u| ≤ w and −w ≤ v ≤ w and v ∈ A. (β) Suppose that A satisfies the condition. If u ∈ A, then |u| ∈ A and −|u| ≤ u ≤ |u|. If w ∈ A and −w ≤ u ≤ w then −u ≤ w, |u| ≤ w = |w| and u ∈ A. Thus A is solid. Q Q In particular, if A is solid, then v ∈ A iff |v| ∈ A. For any set A ⊆ U , the set {u : there is some v ∈ A such that |u| ≤ |v|} is a solid subset of U , the smallest solid set including A; we call it the solid hull of A in U . Any solid linear subspace of U is a Riesz subspace (use 352Fc). If V ⊆ U is a Riesz subspace, then the solid hull of V in U is {u : there is some v ∈ V such that |u| ≤ v} and is a solid linear subspace of U . (b) If T is a Riesz homomorphism from a Riesz space U to a partially ordered linear space V , then its kernel W is a solid linear subspace of U . P P If u ∈ W and |v| ≤ |u|, then T |u| = sup{T u, T (−u)} = 0, while −|u| ≤ v ≤ |u|, so that −0 ≤ T v ≤ 0 and v ∈ W . Q Q Now the quotient space U/W , as defined in 351J, is a Riesz space, and is isomorphic, as partially ordered linear space, to the Riesz space T [U ]. P P Because U/W is the linear space quotient of V by the kernel of the linear operator T , we have an induced linear space isomorphism S : U/W → T [U ] given by setting Su• = T u for every u ∈ U . If p ≥ 0 in U/W there is a u ∈ U + such that u• = p (351J), so that Sp = T u ≥ 0. On the other hand, if p ∈ U/W and Sp ≥ 0, take u ∈ V such that u• = p. We have T (u+ ) = (T u)+ = (Sp)+ = Sp = T u, so that T (u− ) = T u+ − T u = 0 and u− ∈ W , p = (u+ )• ≥ 0. Thus Sp ≥ 0 iff p ≥ 0, and S is a partially-ordered-linear-space isomorphism. Q Q (c) Because a subset of a Riesz space is a solid linear subspace iff it is the kernel of a Riesz homomorphism, such subspaces are sometimes called ideals. 352K Q Products If hUi ii∈I is any family of Riesz spaces, then the product partially ordered linear space U = i∈I Ui (351L) is a Riesz space, with u ∨ v = hu(i) ∨ v(i)ii∈I ,

u ∧ v = hu(i) ∧ v(i)ii∈I ,

|u| = h|u(i)|ii∈I

for all u, v ∈ U . 352L Theorem Let U be any Riesz space. Then there are a set X, a filter F on X and a Riesz subspace of the Riesz space RX |F (351M) which is isomorphic, as Riesz space, to U . proof By 351Q, we can find such X and F and an injective Riesz homomorphism T : U → RX |F. By 352K, or otherwise, RX is a Riesz space; by 352Ib, RX |F is a Riesz space (recall that it is a quotient of RX by a solid linear subspace, as explained in 351M); by 352I, T [U ] is a Riesz subspace of RX |F; and by 352H it is isomorphic to U . 352M Corollary Any identity involving the operations +, −, ∨, ∧, + , and the relation ≤, which is valid in R, is valid in all Riesz spaces.

−

, | | and scalar multiplication,

Remark I suppose some would say that a strict proof of this must begin with a formal description of what the phrase ‘any identity involving the operations. . . ’ means. However I think it is clear in practice what is involved. Given a proposed identity like Pn Pn P 0 ≤ i=0 |αi ||ui | − | i=0 αi ui | ≤ i6=j (|αi | + |αj |)(|ui | ∧ |uj |), (compare 352Fb), then to check that it is valid in all Riesz spaces you need only check (i) that it is true in R (ii) that it is true in RX (iii) that it is true in any RX |F (iv) that it is true in any Riesz subspace of RX |F; and you can hope that the arguments for (ii)-(iv) will be nearly trivial, since (ii) is generally nothing but a coordinate-by-coordinate repetition of (i), and (iii) and (iv) involve only transformations of the formula by Riesz homomorphisms which preserve its structure.

352Oc

Riesz spaces

231

352N Order-density and order-continuity Let U be a Riesz space. (a) Definition A Riesz subspace V of U is quasi-order-dense if for every u > 0 in U there is a v ∈ V such that 0 < v ≤ u; it is order-dense if u = sup{v : v ∈ V, 0 ≤ v ≤ u} for every u ∈ U + . (b) If U is a Riesz space and V is a quasi-order-dense Riesz subspace of U , then the embedding V ⊆ U is order-continuous. P P Let A ⊆ V be a non-empty set such that inf A = 0 in V . ?? If 0 is not the infimum of A in U , then there is a u > 0 such that u is a lower bound for A in U ; now there is a v ∈ V such that 0 < v ≤ u, and v is a lower bound for A in V which is strictly greater than 0. X X Thus 0 = inf A in U . As A is arbitrary, the embedding is order-continuous. Q Q (c) (i) If V ⊆ U is an order-dense Riesz subspace, it is quasi-order-dense. (ii) If V is a quasi-order-dense Riesz subspace of U and W is a quasi-order-dense Riesz subspace of V , then W is a quasi-order-dense Riesz subspace of U . (iii) If V is an order-dense Riesz subspace of U and W is an order-dense Riesz subspace of V , then W is an order-dense Riesz subspace of U . (Use (b).) (iv) If V is a quasi-order-dense solid linear subspace of U and W is a quasi-order-dense Riesz subspace of U then V ∩ W is quasi-order-dense in V , therefore in U . (d) I ought somewhere to remark that a Riesz homomorphism, being a lattice homomorphism, is ordercontinuous iff it preserves arbitrary suprema and infima; compare 313L(b-iv) and (b-v). (e) If V is a Riesz subspace of U , we say that it is regularly embedded in U if the identity map from V to U is order-continuous, that is, whenever A ⊆ V is non-empty and has infimum 0 in V , then 0 is still its greatest lower bound in U . Thus quasi-order-dense Riesz subspaces and solid linear subspaces are regularly embedded. 352O Bands Let U be a Riesz space. (a) Definition A band or normal subspace of U is an order-closed solid linear subspace. (b) If V ⊆ U is a solid linear subspace then it is a band iff sup A ∈ V whenever A ⊆ V + is a non-empty, upwards-directed subset of V with a supremum in U . P P Of course the condition is necessary; I have to show that it is sufficient. (i) Let A ⊆ V be any non-empty upwards-directed set with a supremum in V . Take any u0 ∈ A and set A1 = {u − u0 : u ∈ A, u ≥ u0 }. Then A1 is a non-empty upwards-directed subset of V + , and u0 + A1 = {u : u ∈ A, u ≥ u0 } has the same upper bounds as A, so sup A1 = sup A − u0 is defined in U and belongs to V . Now sup A = u0 + sup A1 also belongs to V . (ii) If A ⊆ V is non-empty, downwards-directed and has an infimum in U , then −A ⊆ V is upwards-directed, so inf A = sup(−A) belongs to V . Thus V is order-closed. Q Q (c) For any set A ⊆ U set A⊥ = {v : v ∈ U, |u| ∧ |v| = 0 for every u ∈ A}. Then A⊥ is a band. P P (i) Of course 0 ∈ A⊥ . (ii) If v, w ∈ A⊥ and u ∈ A, then |u| ∧ |v + w| ≤ (|u| ∧ |v|) + (|u| ∧ |w|) = 0, ⊥

⊥

so v + w ∈ A . (iii) If v ∈ A

and |w| ≤ |v| then 0 ≤ |u| ∧ |w| ≤ |u| ∧ |v| = 0

⊥

for every u ∈ A, so w ∈ A . (iv) If v ∈ A⊥ then nv ∈ A⊥ for every n, by (ii). So if α ∈ R, take n ∈ N such that |α| ≤ n; then |αv| = |α||v| ≤ n|v| ∈ A⊥ and αv ∈ A⊥ . Thus A⊥ is a solid linear subspace of U . (v) If B ⊆ (A⊥ )+ is non-empty and upwards-directed and has a supremum w in U , then |u| ∧ |w| = |u| ∧ w = supv∈B |u| ∧ v = 0 by 352Ea, so w ∈ A⊥ . Thus A⊥ is a band. Q Q

232

Riesz spaces

352Od

(d) For any A ⊆ U , A ⊆ (A⊥ )⊥ . Also B ⊥ ⊆ A⊥ whenever A ⊆ B. So A⊥⊥⊥ ⊆ A⊥ ⊆ A⊥⊥⊥ and A⊥ = A⊥⊥⊥ . (e) If W is another Riesz space and T : U → W is an order-continuous Riesz homomorphism then its kernel is a band. (For {0} is order-closed in W and the inverse image of an order-closed set by an order-continuous order-preserving function is order-closed.) 352P Complemented bands Let U be a Riesz space. A band V ⊆ U is complemented if V ⊥⊥ = V , that is, if V is of the form A⊥ for some A ⊆ U (352Od). In this case its complement is the complemented band V ⊥ . 352Q Theorem In any Riesz space U , the set C of complemented bands forms a Dedekind complete Boolean algebra, with V ∩C W = V ∩ W , 1C = U ,

V ∪C W = (V + W )⊥⊥ ,

0C = {0},

1C \C V = V ⊥ ,

V ⊆ C W ⇐⇒ V ⊆ W for V , W ∈ C. proof To show that C is a Boolean algebra, I use the identification of Boolean algebras with complemented distributive lattices (311L). (a) Of course C is partially ordered by ⊆. If V , W ∈ C then V ∩ W = V ⊥⊥ ∩ W ⊥⊥ = (V ⊥ ∪ W ⊥ )⊥ ∈ C, and V ∩ W must be inf{V, W } in C. The map V 7→ V ⊥ : C → C is an order-reversing bijection, so that V ⊆ W iff W ⊥ ⊆ V ⊥ and V ∨ W = sup{V, W } will be (V ⊥ ∩ W ⊥ )⊥ ; thus C is a lattice. Note also that V ∨ W must be the smallest complemented band including V + W , that is, it is (V + W )⊥⊥ . (b) If V1 , V2 , W ∈ C then (V1 ∨ V2 ) ∧ W = (V1 ∧ W ) ∨ (V2 ∧ W ). P P Of course (V1 ∨ V2 ) ∧ W ⊇ (V1 ∧ W ) ∨ (V2 ∧ W ). ?? Suppose, if possible, that there is a u ∈ (V1 ∨ V2 ) ∩ W \ ((V1 ∩ W ) ∨ (V2 ∩ W )). Then u∈ / ((V1 ∩ W )⊥ ∩ (V2 ∩ W )⊥ )⊥ , so there is a v ∈ (V1 ∩ W )⊥ ∩ (V2 ∩ W )⊥ such that u1 = |u| ∧ |v| > 0. Now / Vj⊥ , and there is a vj ∈ Vj such that u2 = u1 ∧|vj | > 0. / V1⊥ ∩V2⊥ ; say u1 ∈ u1 ∈ V1 ∨V2 = (V1⊥ ∩V2⊥ )⊥ so u1 ∈ ⊥ In this case we still have u2 ∈ (Vj ∩ W ) , because u2 ≤ |v|, but also u2 ∈ Vj and u2 ∈ W because u2 ≤ |u|; but this means that u2 = u2 ∧ u2 = 0, which is absurd. X X Thus (V1 ∨ V2 ) ∧ W ⊆ (V1 ∧ W ) ∨ (V2 ∧ W ) and the two are equal. Q Q (c) Now if V ∈ C, V ∧ V ⊥ = {0} is the least member of C, because if v ∈ V ∩ V ⊥ then |v| = |v| ∧ |v| = 0. By 311L, C has a Boolean algebra structure, with the Boolean relations described; by 312L, this structure is uniquely defined. (d) Finally, if V ⊆ C is non-empty, then T

V=(

S V ∈V

V ⊥ )⊥ ∈ C

and is inf V in C. So C is Dedekind complete. 352R Projection bands Let U be a Riesz space. (a) A projection band in U is a set V ⊆ U such that V + V ⊥ = U . In this case V is a complemented band. P P If v ∈ V ⊥⊥ then v is expressible as v1 + v2 where v1 ∈ V , v2 ∈ V ⊥ . Now |v| = |v1 | + |v2 | ≥ |v2 | (352Fb), so

352S

Riesz spaces

233

|v2 | = |v2 | ∧ |v2 | ≤ |v2 | ∧ |v| = 0 and v = v1 ∈ V . Thus V = V projection band.

⊥⊥

is a complemented band. Q Q Observe that U = V ⊥ + V ⊥⊥ so V ⊥ is also a

(b) Because V ∩ V ⊥ is always {0}, we must have U = V ⊕ V ⊥ for any projection band V ⊆ U ; accordingly there is a corresponding band projection PV : U → U defined by setting P (v + w) = v whenever v ∈ V , w ∈ V ⊥ . In this context I will say that v is the component of v + w in V . The kernel of P is V ⊥ , the set of values is V , and P 2 = P . Moreover, P is an order-continuous Riesz homomorphism. P P (i) P is a linear operator because V and V ⊥ are linear subspaces. (ii) If v ∈ V , w ∈ V ⊥ then |v + w| = |v| + |w|, by 352Fb, so P |v + w| = |v| = |P (v + w)|; consequently P is a Riesz homomorphism (352G). (iii) If A ⊆ U is downwardsdirected and has infimum 0, then P u ≤ u for every u ∈ A, so inf P [A] = 0; thus P is order-continuous. Q Q (c) Note that for any band projection P , and any u ∈ U , we have |P u| ∧ |u − P u| = 0, so that |u| = |P u| + |u − P u| and (in particular) |P u| ≤ |u|; consequently P [W ] ⊆ W for any solid linear subspace W of U. (d) A linear operator P : U → U is a band projection iff P u ∧ (u − P u) = 0 for every u ∈ U + . P P I remarked in (c) that the condition is satisfied for any band projection. Now suppose that P has the property. (i) For any u ∈ U + , P u ≥ 0 and u − P u ≥ 0; in particular, P is a positive linear operator. (ii) If u, v ∈ U + then u − P u ≤ (u + v) − P (u + v), so P v ∧ (u − P u) ≤ P (u + v) ∧ ((u + v) − P (u + v)) = 0 and P v ∧ (u − P u) = 0. (iii) If u, v ∈ U then |P v| ≤ P |v|, |u − P u| ≤ |u| − P |u| (because w 7→ w − P w is a positive linear operator), so |P v| ∧ |u − P u| ≤ P |v| ∧ (|u| − P |u|) = 0. (iv) Setting V = P [U ], we see that u − P u ∈ V ⊥ for every u ∈ U , so that u = u + (u − P u) ∈ V + V ⊥ for every u, and U = V + V ⊥ ; thus V is a projection band. (v) Since P u ∈ V and u − P u ∈ V ⊥ for every u ∈ U , P is the band projection onto V . Q Q 352S Proposition Let U be any Riesz space. (a) The family B of projection bands in U is a subalgebra of the Boolean algebra C of complemented bands in U . (b) For V ∈ B let PV : U → V be the corresponding projection. Then for any e ∈ U + , PV ∩W e = PV e ∧ PW e = PV PW e,

PV ∨W e = PV e ∨ PW e

for all V , W ∈ B. In particular, band projections commute. (c) If V ∈ B then the algebra of projection bands of V is just the principal ideal of B generated by V . proof (a) Of course 0C = {0} ∈ B. If V ∈ B then V ⊥ = 1C \ V belongs to B. If now W is another member of B, then (V ∩ W ) + (V ∩ W )⊥ ⊇ (V ∩ W ) + V ⊥ + W ⊥ . But if u ∈ U then we can express u as v + v 0 , where v ∈ V and v 0 ∈ V ⊥ , and v as w + w0 , where w ∈ W and w0 ∈ W ⊥ ; and as |w| ≤ |v|, we also have w ∈ V , so that u = w + v 0 + w0 ∈ (V ∩ W ) + V ⊥ + W ⊥ . This shows that V ∩ W ∈ B. Thus B is closed under intersection and complements and is a subalgebra of C. (b) If V , W ∈ B and e ∈ U + , we have e = e1 + e2 + e3 + e4 where e1 = PW PV e ∈ V ∩ W ,

e2 = P W ⊥ P V e ∈ V ∩ W ⊥ ,

234

Riesz spaces

e3 = PW PV ⊥ e ∈ V ⊥ ∩ W , e1 + e2 = PV e,

352S

e4 = PW ⊥ PV ⊥ e ∈ V ⊥ ∩ W ⊥ , e1 + e3 = PW e.

Now e2 + e3 + e4 belongs to (V ∩ W )⊥ , so e1 must be the component of e in V ∩ W ; similarly e4 is the component of e in V ⊥ ∩ W ⊥ , and e1 + e2 + e3 is the component of e in V ∨ W . But as e2 ∧ e3 = 0, we have PV ∩W e = e1 = (e1 + e2 ) ∧ (e1 + e3 ) = PV e ∧ PW e, PV ∨W e = e1 + e2 + e3 = (e1 + e2 ) ∨ (e1 + e3 ) = PV e ∨ PW e, as required. It follows that PV PW = PV ∩W = PW ∩V = PW PV . (c) If V , W ∈ B and W ⊆ V , then of course W is a band in the Riesz space V (because V is order-closed in U , so that for any set A ⊆ W its supremum in U will be its supremum in V ). For any v ∈ V , we have an expression of it as w + w0 , where w ∈ W and w0 ∈ W ⊥ , taken in U ; but as |w| + |w0 | = |w + w0 | = |v| ∈ V , w0 belongs to V , and is in WV⊥ , the band in V orthogonal to W . Thus W + WV⊥ = V and W is a projection band in V . Conversely, if W is a projection band in V , then W ⊥ (taken in U ) includes WV⊥ + V ⊥ , so that W + W ⊥ ⊇ W + WV⊥ + V ⊥ = V + V ⊥ = U and W ∈ B. Thus the algebra of projection bands in V is, as a set, equal to the principal ideal BV ; because their orderings agree, or otherwise, their Boolean algebra structures coincide. 352T Products again (a) If U = subspace

Q i∈I

Ui is a product of Riesz spaces, then for any J ⊆ I we have a

VJ = {u : u ∈ U, u(i) = 0 for all i ∈ I \ J} Q of U , canonically isomorphic to i∈J Ui . Each VJ is a projection band, its complement being VI\J ; the map J 7→ VJ is a Boolean homomorphism from PI to the algebra B of projection bands in U , and hV{i} ii∈I is a partition of unity in B. (b) Conversely, if U is a Riesz space and (V0 , . . . , Vn ) is a finite partition Pn of unity in the algebra B of projection bands in U , then every element of U is uniquely expressible as i=0 ui where ui ∈ V Qi for each i. (Induce on n.) This decomposition corresponds to a Riesz space isomorphism between U and i≤n Vi . 352U Quotient spaces (a) If U is a Riesz space and V is a solid linear subspace, then the quotient partially ordered linear space U/V (351J) is a Riesz space; if U and W are Riesz spaces and T : U → W a Riesz homomorphism, then the kernel V of T is a solid linear subspace of U and the Riesz subspace T [U ] of W is isomorphic to U/V (352Jb). (b) Suppose that U is a Riesz space and V a solid linear space. Then the canonical map from U to U/V is order-continuous iff V is a band. P P (i) If u 7→ u• is order-continuous, its kernel V is a band, by 352Oe. (ii) If V is a band, and A ⊆ U is non-empty and downwards-directed and has infimum 0, let p ∈ U/V be any lower bound for {u• : u ∈ A}. Express p as w• . Then ((u − w)− )• = (u• − w• )− = 0, that is, (w − u)+ = (u − w)− ∈ V for every u ∈ A. But this means that w+ = supu∈A (w − u)+ ∈ V ,

p+ = (w+ )• = 0,

that is, p ≤ 0. As p is arbitrary, inf u∈A u• = 0; as A is arbitrary, u 7→ u• is order-continuous. Q Q 352V Principal bands Let U be a Riesz space. Evidently the intersection of any family of Riesz subspaces of U is a Riesz subspace, the intersection of any family of solid linear subspaces is a solid linear subspace, the intersection of any family of bands is a band; we may therefore speak of the band generated by a subset A of U , the intersection of all the bands including A. Now we have the following description of the band generated by a single element.

352W

Riesz spaces

235

Lemma Let U be a Riesz space. (a) If A ⊆ U + is upwards-directed and 2w ∈ A for every w ∈ A, then an element u of U belongs to the band generated by A iff |u| = supw∈A |u| ∧ w. (b) If u ∈ U and w ∈ U + , then u belongs to the band of U generated by w iff |u| = supn∈N |u| ∧ nw. proof (a) Let W be the band generated by A and W 0 the set of elements of U satisfying the condition. (i) If u ∈ W 0 then |u| ∧ w ∈ W for every w ∈ A, because W is a solid linear subspace; because W is also order-closed, |u| and u belong to W . Thus W 0 ⊆ W . (ii) Now W 0 is a band. α) If u ∈ W 0 and |v| ≤ |u| then P P(α supw∈A |v| ∧ w = supw∈A |v| ∧ |u| ∧ w = |v| ∧ supw∈A |u| ∧ w = |v| ∧ |u| = |v| by 352Ea, so v ∈ W 0 . β ) If u, v ∈ W 0 then, for any w1 , w2 ∈ A there is a w ∈ A such that w ≥ w1 ∨ w2 . Now (β w1 + w2 ≤ 2w ∈ A, and (|u| + |v|) ∧ 2w ≥ (|u| ∧ w1 ) + (|v| ∧ w2 ). So any upper bound for {(|u|+|v|)∧w : w ∈ A} must also be an upper bound for {|u|∧w : w ∈ A}+{|v|∧w : w ∈ A} and therefore greater than or equal to sup({|u| ∧ w : w ∈ A} + {|v| ∧ w : w ∈ A}) = supw∈A |u| ∧ w + supw∈A |v| ∧ w = |u| + |v| (351Dc). But this means that supw∈A (|u| + |v|) ∧ w must be |u| + |v|, and |u| + |v| belongs to W 0 ; it follows from (i) that u + v belongs to W 0 . (γγ ) Just as in 352Oc, we now have nu ∈ W 0 for every n ∈ N, u ∈ W 0 , and therefore αu ∈ W 0 for every α ∈ R, u ∈ W 0 , since |αu| ≤ |nu| if |α| ≤ n. Thus W 0 is a solid linear subspace of U . (δδ ) Now suppose that C ⊆ (W 0 )+ has a supremum v in U . Then any upper bound of {v ∧ w : w ∈ A} must also be an upper bound of {u ∧ w : u ∈ C, w ∈ A} and greater than or equal to u = supw∈A u ∧ w for every u ∈ C, therefore greater than or equal to v = sup C. Thus v = supw∈A v ∧ w and v ∈ W 0 . As C is arbitrary, W 0 is a band (352Ob). Q Q (iii) Since A is obviously included in W 0 , W 0 must include W ; putting this together with (i), W = W 0 , as claimed. (b) Apply (a) with A = {nw : n ∈ N}. 352W f -algebras Some of the most important Riesz spaces have multiplicative structures as well as their order and linear structures. A particular class of these structures appears sufficiently often for it to be useful to develop a little of its theory. The following definition is a common approach. (a) Definition An f -algebra is a Riesz space U with a multiplication × : U × U → U such that u × (v × w) = (u × v) × w, (u + v) × w = (u × w) + (v × w),

u × (v + w) = (u × v) + (u × w),

α(u × v) = (αu) × v = u × (αv) for all u, v, w ∈ U and α ∈ R, and u × v ≥ 0 whenever u, v ≥ 0, if u ∧ v = 0 then (u × w) ∧ v = (w × u) ∧ v = 0 for every w ≥ 0. An f -algebra is commutative if u × v = v × u for all u, v.

236

Riesz spaces

352Wb

(b) Let U be an f -algebra. (i) If u ∧ v = 0 in U , then u × v = 0. P P u ∧ (u × v) = 0 so (u × v) ∧ (u × v) = 0. Q Q (ii) u × u ≥ 0 for every u ∈ U . P P (u+ − u− ) × (u+ − u− ) = u+ × u+ − u+ × u− − u− × u+ + u− × u− = u+ × u+ + u− × u− ≥ 0. Q Q (iii) If u, v ∈ U then |u × v| = |u| × |v|. P P u+ × v + , u+ × v − , u− × v + and u+ × u− are disjoint, so |u × v| = |u+ × v + − u+ × v − − u− × v + + u− × v − | = u+ × v + + u+ × v − + u− × v + + u− × v − = |u| × |v| by 352Fb. Q Q (iv) If v ∈ U + the maps u 7→ u × v, u 7→ v × u : U → U are Riesz homomorphisms. P P The first four clauses of the definition in (a) ensure that they are linear operators. If u ∈ U , then |u| × v = |u × v|,

v × |u| = |v × u|

by (iii), so we have Riesz homomorphisms, by 352G(iii). Q Q (c) Let hUi ii∈I be a family of f -algebras, with Riesz space product U (352K). If we set u × v = hu(i) × v(i)ii∈I for all u, v ∈ U , then U becomes an f -algebra. 352X Basic exercises > (a) Let U be any Riesz space. Show that |u+ − v + | ≤ |u − v| for all u, v ∈ U . > (b) Let U , V be a Riesz spaces and T : U → V a linear operator. Show that the following are equiveridical: (i) T is a Riesz homomorphism; (ii) T (u∨v) = T u∨T v for all u, v ∈ U ; (iii) T (u∧v) = T u∧T v for all u, v ∈ U ; (iv) |T u| = T |u| for every u ∈ U . (c) Let U be a Riesz space and V a solid linear subspace; for u ∈ U write u• for the corresponding element of U/V . Show that if A ⊆ U is solid then {u• : u ∈ A} is solid in U/W . (d) Let U and V be Riesz spaces and T : U → V a Riesz homomorphism with kernel W . Show that if W is a band in U and T [U ] is regularly embedded in V then T is order-continuous. (e) Give U = R2 its lexicographic ordering (351Xa). Show that it has a band V which is not complemented. (f ) Let U be a Riesz space, C the algebra of complemented bands in U . Show that for any V ∈ C the algebra of complemented bands of V is just the principal ideal of C generated by V . > (g) Let U = C([0, 1]) be the space of continuous functions from [0, 1] to R, with its usual linear and order structures, so that it is a Riesz subspace of R[0,1] . Set V = {u : u ∈ U, u(t) = 0 if t ≤ 12 }. Show that V is a band in U and that V ⊥ = {u : u(t) = 0 if t ≥ 21 }, so that V is complemented but is not a projection band. (h) Show that the Boolean homomorphism J 7→ VJ : PI → B of 352Ta is order-continuous. (i) Let U be a Riesz space and A ⊆ U + an upwards-directed set. Show that the band generated by A is {u : |u| = supn∈N,w∈A |u| ∧ nw}. >(j) (i) Let X be any set. Setting (u × v)(x) = u(x)v(x) for u, v ∈ RX , x ∈ X, show that RX is a commutative f -algebra. (ii) With the same definition of ×, show that `∞ (X) is an f -algebra. (iii) If X is a topological space, show that C(X), Cb (X) are f -algebras. (iv) If (X, Σ, µ) is a measure space, show that L0 (µ), L∞ (µ) (§241, §243) are f -algebras.

353A

Archimedean and Dedekind complete Riesz spaces

237

(k) Let U ⊆ RZ be the set of sequences u such that {n : u(n) 6= 0} is bounded above in Z. For u, v ∈ U (i) say that u ≤ v if either P∞ u = v or there is an n ∈ Z such that u(n) < v(n), u(i) = v(i) for every i > n (ii) say that (u ∗ v)(n) = i=−∞ u(i)v(n − i) for every n ∈ Z. Show that U is an f -algebra under this ordering and multiplication. (l) Let U be an f -algebra. (i) Show that any complemented band of U is an ideal in the ring (U, +, ×). (ii) Show that if P : U → U is a band projection, then P (u × v) = P u × P v for every u, v ∈ U . 1 γ

(m) Let U be an f -algebra with multiplicative identity e. Show that u − γe ≤ u2 for every u ∈ U , γ > 0. (Hint: (u+ − γe)2 ≥ 0.) 352 Notes and comments In this section we begin to see a striking characteristic of the theory of Riesz spaces: repeated reflections of results in Boolean algebra. Without spelling out a complete list, I mention the distributive laws (313Bc, 352Ea) and the behaviour of order-continuous homomorphisms (313Pa, 352N, 352Oe, 352Ub, 352Xd). Riesz subspaces correspond to subalgebras, solid linear subspaces to ideals and Riesz homomorphisms to Boolean homomorphisms. We even have a correspondence, though a weaker one, between the representation theorems available; every Boolean algebra is isomorphic to a subalgebra of a power of Z2 (311D-311E), while every Riesz space is isomorphic to a Riesz subspace of a quotient of a power of R (352L). It would be a closer parallel if every Riesz space were embeddable in some RX ; I must emphasize that the differences are as important as the agreements. Subspaces of RX are of great importance, but are by no means adequate for our needs. And of course the details – for instance, the identities in 352D-352F, or 352V – frequently involve new techniques in the case of Riesz spaces. Elsewhere, as in 352G, I find myself arguing rather from the opposite side, when applying results from the theory of general partially ordered linear spaces, which has little to do with Boolean algebra. In the theory of bands in Riesz spaces – corresponding to order-closed ideals in Boolean algebras – we have a new complication in the form of bands which are not complemented, which does not arise in the Boolean algebra context; but it disappears again when we come to specialize to Archimedean Riesz spaces (353B). (Similarly, order-density and quasi-order-density coincide in both Boolean algebras (313K) and Archimedean Riesz spaces (353A).) Otherwise the algebra of complemented bands in a Riesz space looks very like the algebra of order-closed ideals in a Boolean algebra (314Yh, 352Q). The algebra of projection bands in a Riesz space (352S) would correspond, in a Boolean algebra, to the algebra itself. I draw your attention to 352H. The result is nearly trivial, but it amounts to saying that the theory of Riesz spaces will be ‘algebraic’, like the theories of groups or linear spaces, rather than ‘analytic’, like the theories of partially ordered linear spaces or topological spaces, in which we can have bijective morphisms which are not isomorphisms.

353 Archimedean and Dedekind complete Riesz spaces I take a few pages over elementary properties of Archimedean and Dedekind (σ)-complete Riesz spaces. 353A Proposition Let U be an Archimedean Riesz space. Then every quasi-order-dense Riesz subspace of U is order-dense. proof Let V ⊆ U be a quasi-order-dense Riesz subspace, and u ≥ 0 in U . Set A = {v : v ∈ V, v ≤ u}. ?? Suppose, if possible, that u is not the least upper bound of A. Then there is a u1 < u such that v ≤ u1 for every v ∈ A. Because 0 ∈ A, u1 ≥ 0. Because V is quasi-order-dense, there is a v > 0 in V such that v ≤ u − u1 . Now nv ≤ u1 for every n ∈ N. P P Induce on n. For n = 0 this is trivial. For the inductive step, given nv ≤ u1 , then (n + 1)v ≤ u1 + v ≤ u, so (n + 1)v ∈ A and (n + 1)v ≤ u1 . Thus the induction proceeds. Q Q But this is impossible, because v > 0 and U is supposed to be Archimedean. X X So u = sup A. As u is arbitrary, V is order-dense.

238

Riesz spaces

353B

353B Proposition Let U be an Archimedean Riesz space. Then (a) for every A ⊆ U , the band generated by A is A⊥⊥ , (b) every band in U is complemented. proof (a) Let V be the band generated by A. Then V is surely included in A⊥⊥ , because this is a band including A (352O). ?? Suppose, if possible, that V 6= A⊥⊥ . Then there is a w ∈ A⊥⊥ \ V , so that |w| ∈ / V. Set B = {v : v ∈ V, v ≤ |w|}; then B is upwards-directed and non-empty. Because V is order-closed, |w| cannot be the supremum of A, and there is a u0 > 0 such that |w|−u0 ≥ v for every v ∈ B. Now u0 ∧|w| 6= 0, so u0 ∈ / A⊥ , and there is a u1 ∈ A such that v = u0 ∧ |u1 | > 0. In this case nv ∈ B for every n ∈ N. P P Induce on n. For n = 0 this is trivial. For the inductive step, given that nv ∈ B, then nv ≤ |w| − u0 so (n + 1)v ≤ nv + u0 ≤ |w|; but also (n + 1)v ≤ nv + |u1 | ∈ V , so (n + 1)v ∈ B. Q Q But this means that nv ≤ |w| for every n, which is impossible, because U is Archimedean. X X (b) Now if V ⊆ U is any band, it is surely the band generated by itself, so is equal to V ⊥⊥ , and is complemented. Remark We may therefore speak of the band algebra of an Archimedean Riesz space, rather than the ‘complemented band algebra’ (352Q). 353C Corollary Let U be an Archimedean Riesz space and v ∈ U . Let V be the band in U generated by v. If u ∈ U , then u ∈ V iff there is no w such that 0 < w ≤ |u| and w ∧ |v| = 0. proof By 353B, V = {v}⊥⊥ . Now, for u ∈ U , u∈ / V ⇐⇒ ∃ w ∈ {v}⊥ , |u| ∧ |w| > 0 ⇐⇒ ∃ w ∈ {v}⊥ , 0 < w ≤ |u|. Turning this round, we have the condition announced. 353D Proposition Let U be an Archimedean Riesz space and V an order-dense Riesz subspace of U . Then the map W 7→ W ∩ V is an isomorphism between the band algebras of U and V . proof If W ⊆ U is a band, then W ∩ V is surely a band in V (it is order-closed in V because it is the inverse image of the order-closed set W under the embedding V ⊆ U , which is order-continuous by 352Nc and 352Nb). If W , W 0 are distinct bands in U , say W 0 6⊆ W , then W 0 6⊆ W ⊥⊥ , by 353B, so W 0 ∩ W ⊥ 6= {0}; because V is order-dense, V ∩ W 0 ∩ W ⊥ 6= {0}, and V ∩ W 0 6= V ∩ W . Thus W 7→ W ∩ V is injective. If Q ⊆ V is a band in V , then its complementary band in V is just Q⊥ ∩ V , where Q⊥ is taken in U . So (because V , like U , is Archimedean, by 351Rc) Q = (Q⊥ ∩ V )⊥ ∩ V = W ∩ V , where W = (Q⊥ ∩ V )⊥ is a band in U . Thus the map W 7→ W ∩ V is an order-preserving bijection between the two band algebras. By 312L, it is a Boolean isomorphism, as claimed. 353E Lemma Let U be an Archimedean Riesz space and V ⊆ U a band such that sup{v : v ∈ V, 0 ≤ v ≤ u} is defined for every u ∈ U + . Then V is a projection band. proof Take any u ∈ U + and set v = sup{v 0 : v 0 ∈ V + , v 0 ≤ u}, w = u − v. v ∈ V because V is a band. Also w ∈ V ⊥ . P P?? If not, there is some v0 ∈ V + such that w ∧ v0 > 0. Now for any n ∈ N we see that nv0 ≤ u =⇒ nv0 ≤ v =⇒ (n + 1)v0 ≤ v + w = u, so an induction on n shows that nv0 ≤ u for every n; which is impossible, because U is supposed to be Archimedean. X XQ Q Accordingly u = v +w ∈ V +V ⊥ . As u is arbitrary, U + ⊆ V +V ⊥ , and V is a projection band (352R). 353F Lemma Let U be an Archimedean Riesz space. If A ⊆ U is non-empty and bounded above and B is the set of its upper bounds, then inf(B − A) = 0. proof ?? If not, let w > 0 be a lower bound for B − A. If u ∈ A and v ∈ B, then v − u ≥ w, that is, u ≤ v − w; as u is arbitrary, v − w ∈ B. Take any u0 ∈ A, v0 ∈ B. Inducing on n, we see that v0 − nw ∈ B for every n ∈ N, so that v0 − nw ≥ u0 , nw ≤ v0 − u0 for every n; but this is impossible, because U is supposed to be Archimedean. X X

353J

Archimedean and Dedekind complete Riesz spaces

239

353G Dedekind completeness Recall from 314A that a partially ordered set P is Dedekind (σ)complete if (countable) non-empty sets with upper and lower bounds have suprema and infima in P . For a Riesz space U , U is Dedekind complete iff every non-empty upwards-directed subset of U + with an upper bound has a least upper bound, and is Dedekind σ-complete iff every non-decreasing sequence in U + with an upper bound has a least upper bound. P P (Compare 314Bc.) (i) Suppose that any non-empty upwardsdirected order-bounded subset of U + has an upper bound, and that A ⊆ U is any non-empty set with an upper bound. Take u0 ∈ A and set B = {u0 ∨ u1 ∨ . . . ∨ un − u0 : u1 , . . . , un ∈ A}. Then B is an upwards-directed subset of U + , and if w is an upper bound of A then w − u0 is an upper bound of B. So sup B is defined in U , and in this case u0 + sup B = sup A. As A is arbitrary, U is Dedekind complete. (ii) Suppose that order-bounded non-decreasing sequences in U + have suprema, and that A ⊆ U is any countable non-empty set with an upper bound. Let hun in∈N be a sequence running over A, and set vn = supi≤n ui − u0 for each n. Then hvn in∈N is a non-decreasing order-bounded sequence in U + , and u0 + supn∈N vn = sup A. (iii) Finally, still supposing that order-bounded non-decreasing sequences in U + have suprema, if A ⊆ U is non-empty, countable and bounded below, inf A will be defined and equal to − sup(−A). Q Q 353H Proposition Let U be a Dedekind σ-complete Riesz space. (a) U is Archimedean. (b) For any v ∈ U the band generated by v is a projection band. (c) If u, v ∈ U , then u is uniquely expressible as u1 + u2 , where u1 belongs to the band generated by v and |u2 | ∧ |v| = 0. proof (a) Suppose that u, v ∈ U are such that nu ≤ v for every n ∈ N. Then nu+ ≤ v + for every n, and A = {nu+ : n ∈ N} is a countable non-empty upwards-directed set with an upper bound; say w = sup A. Since A + u+ ⊆ A, w + u+ = sup(A + u+ ) ≤ w, and u ≤ u+ ≤ 0. As u, v are arbitrary, U is Archimedean. (b) Let V be the band generated by v. Take any u ∈ U + and set A = {v 0 : v 0 ∈ V, 0 ≤ v 0 ≤ u}. Then {u ∧ n|v| : n ∈ N} is a countable set with an upper bound, so has a supremum u1 say in U . Now u1 is an upper bound for A. P P If v 0 ∈ A, then v 0 = supn∈N v 0 ∧ n|v| ≤ u1 by 352V. Q Q Since u ∧ n|v| ∈ A ⊆ V for every n, u1 ∈ V and u1 = sup A. As u is arbitrary, 353E tells us that V is a projection band. (c) Again let V be the band generated by v. Then {v}⊥⊥ is a band containing v, so {v} ⊆ V ⊆ {v}⊥⊥ ,

{v}⊥ ⊇ V ⊥ ⊇ {v}⊥⊥⊥ = {v}⊥

(352Od), and V ⊥ = {v}⊥ . Now, if u ∈ U , u is uniquely expressible in the form u1 + u2 where u1 ∈ V and u2 ∈ V ⊥ , by (b). But u2 ∈ V ⊥ ⇐⇒ u2 ∈ {v}⊥ ⇐⇒ |u2 | ∧ |v| = 0. So we have the result. 353I Proposition In a Dedekind complete Riesz space, all bands are projection bands. proof Use 353E, noting that the sets {v : v ∈ V, 0 ≤ v ≤ u} there are always non-empty, upwards-directed and bounded above, so always have suprema. 353J Proposition (a) Let U be a Dedekind σ-complete Riesz space. (i) If V is a solid linear subspace of U , then V is (in itself) Dedekind σ-complete. (ii) If W is a sequentially order-closed Riesz subspace of U then W is Dedekind σ-complete. (iii) If V is a sequentially order-closed solid linear subspace of U , the canonical map from U to V is sequentially order-continuous, and the quotient Riesz space U/V is also Dedekind σ-complete. (b) Let U be a Dedekind complete Riesz space.

240

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353J

(i) If V is a solid linear subspace of U , then V is (in itself) Dedekind complete. (ii) If W ⊆ U is an order-closed Riesz subspace then W is Dedekind complete. proof (a)(i) If hun in∈N is a non-decreasing sequence in V + with an upper bound v ∈ V , then w = supn∈N un is defined in U ; but as 0 ≤ w ≤ v, w ∈ V and w = supn∈N un in V . Thus V is Dedekind σ-complete. (ii) If hun in∈N is a non-decreasing order-bounded sequence in W , then u = supn∈N un is defined in U ; but because W is sequentially order-closed, u ∈ W and u = supn∈N un in W . (iii) Let hun in∈N be a non-decreasing sequence in U with supremum u. Then of course u• is an upper bound for A = {u•n : n ∈ N} in U/V . Now let p be any other upper bound for A. Express p as v • . Then for each n ∈ N we have u•n ≤ p, so that (un − v)+ ∈ V . Because V is sequentially order-closed, (u − v)+ = supn∈N (un − v)+ ∈ V and u• ≤ p. Thus u• is the least upper bound of A. Similarly, if hun in∈N is a non-increasing sequence in U with infimum u, then u• = inf n∈N u•n in U/V . Thus u 7→ u• is sequentially order-continuous. Now suppose that hpn in∈N is a non-decreasing sequence in (U/V )+ with an upper bound p ∈ (U/V )+ . Let u ∈ U + be such that u• = p, and for each n ∈ N let un ∈ U + be such that u•n = pn . Set vn = u ∧ supi≤n ui for each n; then vn• = pn for each n, and hvn in∈N is a non-decreasing order-bounded sequence in U . Set v = supn∈N vn ; by the last paragraph, v • = supn∈N pn in U/V . As hpn in∈N is arbitrary, U/V is Dedekind σ-complete, as claimed. (b) The argument is the same as parts (i) and (ii) of the proof of (a). 353K Proposition Let U be a Riesz space and V a quasi-order-dense Riesz subspace of U which is (in itself) Dedekind complete. Then V is a solid linear subspace of U . proof Suppose that v ∈ V , u ∈ U and that |u| ≤ |v|. Consider A = {w : w ∈ V, 0 ≤ w ≤ u+ }. Then A is a non-empty subset of V with an upper bound in V (viz., |v|). So A has a supremum v0 in V . Because the embedding V ⊆ U is order-continuous (352N), v0 is the supremum of A in U . But as V is order-dense (353A), v0 = u+ and u+ ∈ V . Similarly, u− ∈ V and u ∈ V . As u, v are arbitrary, V is solid. 353L Order units Let U be a Riesz space. (a) An element e of U + is an order unit in U if U is the solid linear subspace of itself generated by e; that is, if for S every u ∈ U there is an n ∈ N such that |u| ≤ ne. (For the solid linear subspace generated by v ∈ U + is n∈N [−nv, nv].) (b) An element e of U + is a weak order unit in U if U is the principal band generated by e; that is, if u = supn∈N u ∧ ne for every u ∈ U + (352Vb). Of course an order unit is a weak order unit. (c) If U is Archimedean, then an element e of U + is a weak order unit iff {e}⊥⊥ = U (353B), that is, iff {e}⊥ = {0} (because {e}⊥ = {0} =⇒ {e}⊥⊥ = {0}⊥ = U =⇒ {e}⊥ = {e}⊥⊥⊥ = U ⊥ = {0},) that is, iff u ∧ e > 0 whenever u > 0. 353M Theorem Let U be an Archimedean Riesz space with order unit e. Then it can be embedded as an order-dense and norm-dense Riesz subspace of C(X), where X is a compact Hausdorff space, in such a way that e corresponds to χX; moreover, this embedding is essentially unique. Remark Here C(X) is the space of all continuous functions from X to R; because X is compact, they are all bounded, so that χX is an order unit in C(X) = Cb (X). proof (a) Let X be the set of Riesz homomorphisms x from U to R such that x(e) = 1. Define T : U → RX by setting (T u)(x) = x(u) for x ∈ X, u ∈ U ; then it is easy to check that T is a Riesz homomorphism, just because every member of X is a Riesz homomorphism, and of course T e = χX.

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(b) The key to the proof is the fact that X separates the points of U , that is, that T is injective. I choose the following method to show this. Suppose that w ∈ U and w > 0. Because U is Archimedean, there is a δ > 0 such that (w − δe)+ 6= 0. Now there is an x ∈ X such that x(w) ≥ δ. P P (i) By 351O, there is a solid linear subspace V of U such that (w − δe)+ ∈ / V and whenever u ∧ v = 0 in U then one of u, v belongs to V . (ii) Because V 6= U , e ∈ / V , so no non-zero multiple of e can belong to V . Also observe that if u, v ∈ U \ V , then one of (u − v)+ , (v − u)+ must belong to V , while neither u = u ∧ v + (u − v)+ nor v = u ∧ v + (v − u)+ does; so u ∧ v ∈ / V . (iii) For each u ∈ U set Au = {α : α ∈ R, (u − αe)+ ∈ V }. Then α ≥ β ∈ Au =⇒ 0 ≤ (u − αe)+ ≤ (u − βe)+ ∈ V =⇒ α ∈ Au . Also Au is non-empty and bounded below, because if α ≥ 0 is such that −αe ≤ u ≤ αe then α ∈ Au and −α − 1 ∈ / Au (since (u − (−α − 1)e)+ ≥ e ∈ / V ). (iv) Set x(u) = inf Au for every u ∈ U ; then α ∈ Au for every α > x(u), α ∈ / Au for every α < x(u). (v) If u, v ∈ U and α > x(u), β > x(v) then ((u + v) − (α + β)e)+ ≤ (u − αe)+ + (v − βe)+ ∈ V (352Fc), so α + β ∈ Au+v ; as α and β are arbitrary, x(u + v) ≤ x(u) + x(v). (vi) If u, v ∈ U and α < x(u), β < x(v) then ((u + v) − (α + β)e)+ ≥ (u − αe)+ ∧ (v − βe)+ ∈ / V, using (ii) of this argument and 352Fc, so α + β ∈ / Au+v . As α and β are arbitrary, x(u + v) ≥ x(u) + x(v). (vii) Thus x : U → R is additive. (viii) If u ∈ U , γ > 0 then α ∈ Au =⇒ (γu − αγe)+ = γ(u − αe)+ ∈ V =⇒ γα ∈ Aγu ; thus Aγu ⊇ γAu ; similarly, Au ⊇ γ −1 Aγu so Aγu = γAu and x(γu) = γx(u). (ix) Consequently x is linear, since we know already from (vii) that x(0u) = 0.x(u), x(−u) = −x(u). (x) If u ≥ 0 then u + αe ≥ αe ∈ /V for every α > 0, that is, −α ∈ / Au for every α > 0, and x(u) ≥ 0; thus x is a positive linear functional. (xi) If u ∧ v = 0, then one of u, v belongs to V , so min(x(u), x(v)) ≤ 0 and (using (x)) min(x(u), x(v)) = 0; thus x is a Riesz homomorphism (352G(iv)). (xii) Ae = [1, ∞[ so x(e) = 1. Thus x ∈ X. (xiii) δ ∈ / Aw so x(w) ≥ δ. Q Q (c) Thus T w 6= 0 whenever w > 0; consequently |T w| = T |w| 6= 0 whenever w 6= 0, and T is injective. I now have to define the topology of X. This is just the subspace topology on X if we regard X as a subset of RU with its product topology. To see that X is compact, observe that if for each u ∈ U we choose an Q αu such that |u| ≤ αu e, then X is a subspace of Q = u∈U [−αu , αu ]. Because Q is a product of compact spaces, it is compact, by Tychonoff’s theorem (3A3J). Now X is a closed subset of Q. P P X is just the intersection of the sets {x : x(u + v) = x(u) + x(v)}, {x : x(u+ ) = max(x(u), 0)},

{x : x(αu) = αx(u)}, {x : x(e) = 1}

as u, v run over U and α over R; and each of these is closed, so X is an intersection of closed sets and therefore itself closed. Q Q Consequently X also is compact. Moreover, the coordinate functionals x 7→ x(u) are continuous on Q, therefore on X also, that is, T u : X → R is a continuous function for every u ∈ U . Note also that because Q is a product of Hausdorff spaces, Q and X are Hausdorff (3A3Id). (d) So T is a Riesz homomorphism from U to C(X). Now T [U ] is a Riesz subspace of C(X), containing χX, and such that if x, y ∈ X are distinct there is an f ∈ T [U ] such that f (x) 6= f (y) (because there is surely a u ∈ U such that x(u) 6= y(u)). By the Stone-Weierstrass theorem (281A), T [U ] is k k∞ -dense in C(X). Consequently it is also order-dense. P P If f > 0 in C(X), set ² = 13 kf k∞ , and let u ∈ U be such that + kf − T uk∞ ≤ ²; set v = (u − ²e) . Since 0 < (f − 2²χX)+ ≤ (T u − ²χX)+ ≤ f + = f , 0 < T v ≤ f . As f is arbitrary, T [U ] is quasi-order-dense, therefore order-dense (353A). Q Q (e) I have still to show that the representation is (essentially) unique. Suppose, then, that we have another representation of U as a norm-dense Riesz subspace of C(Z), with e this time corresponding to χZ; to simplify the notation, let us suppose that U is actually a subspace of C(Z). Then for each z ∈ Z,

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we have a functional zˆ : U → R defined by setting zˆ(u) = u(z) for every u ∈ U ; of course zˆ is a Riesz homomorphism such that zˆ(e) = 1, that is, zˆ ∈ X. Thus we have a function z 7→ zˆ : Z → X. For any u ∈ U , the function z 7→ zˆ(u) = u(z) is continuous, so the function z 7→ zˆ is continuous (3A3Ib). If z1 , z2 are distinct members of Z, there is an f ∈ C(Z) such that f (z1 ) 6= f (z2 ) (3A3Bf); now there is a u ∈ U such that kf − uk∞ ≤ 31 |f (z1 ) − f (z2 )|, so that u(z1 ) 6= u(z2 ) and zˆ1 6= zˆ2 . Thus z 7→ zˆ is injective. Finally, it is also surjective. P P Suppose that x ∈ X. Set V = {u : u ∈ U, x(u) = 0}; then V is a solid linear subspace of U (352Jb), not containing e. For z ∈ V + set Gv = {z : v(z) > 1}. Because e ∈ / V , Gv 6= Z. G = {Gv : v ∈ V + } is an upwards-directed family of open setsSin Z, not containing Z; consequently, because Z is compact, G cannot be an open cover of Z. Take z ∈ Z \ G. Then v(z) ≤ 1 for every v ∈ V + ; because α|v| ∈ V + whenever v ∈ V , α ≥ 0, we must have v(z) = 0 for every v ∈ V . Now, given any u ∈ U , consider v = u − x(u)e. Then x(v) = 0 so v ∈ V and v(z) = 0, that is, u(z) = (v + x(u)e)(z) = v(z) + x(u)e(z) = x(u). As u is arbitrary, zˆ = x; as x is arbitrary, we have the result. Q Q Thus z 7→ zˆ is a continuous bijection from the compact Hausdorff space Z to the compact Hausdorff space X; it must therefore be a homeomorphism (3A3Dd). This argument shows that if U is embedded as a norm-dense Riesz subspace of C(Z), where Z is compact and Hausdorff, then Z must be homeomorphic to X. But it shows also that a homeomorphism is canonically defined by the embedding; z ∈ Z corresponds to the Riesz homomorphism u 7→ u(z) in X. 353N Lemma Let U be a Riesz space, V an Archimedean Riesz space and S, T : U → V Riesz homomorphisms such that Su ∧ T u0 = 0 in V whenever u ∧ u0 = 0 in U . Set W = {u : Su = T u}. Then W is a solid linear subspace of U ; if S and T are order-continuous, W is a band. proof (a) It is easy to check that, because S and T are Riesz homomorphisms, W is a Riesz subspace of U . (b) If w ∈ W and 0 ≤ u ≤ w in U , then Su ≤ T u. P P?? Otherwise, set e = Sw = T w, and let Ve be the solid linear subspace of V generated by e, so that Ve is an Archimedean Riesz space with order unit, containing both Su and T u. By 353M (or its proof), there is a Riesz homomorphism x : Ve → R such that x(e) = 1 and x(Su) > x(T u). Take α such that x(Su) > α > x(T u), and consider u0 = (u − αw)+ , u00 = (αw − u)+ . Then x(Su0 ) = max(0, x(Su) − αx(Sw)) = max(0, x(Su) − α) > 0, x(T u00 ) = max(0, αx(T w) − x(T u)) = max(0, α − x(T u)) > 0, so x(Su0 ∧ T u00 ) = min(x(Su0 ), x(T u00 )) > 0 and Su0 ∧ T u00 > 0, while u0 ∧ u00 = 0. X XQ Q Similarly, T u ≤ Su and u ∈ W . As u and w are arbitrary, W is a solid linear subspace. (c) Finally, suppose that S and T are order-continuous, and that A ⊆ W is a non-empty upwards-directed set with supremum u in U . Then Su = sup S[A] = sup T [A] = T u and u ∈ W . As u and A are arbitrary, W is a band (352Ob). 353O f -algebras I give two results on f -algebras, intended to clarify the connexions between the multiplicative and lattice structures of the Riesz spaces in Chapter 36. Proposition Let U be an Archimedean f -algebra (352W). Then (a) the multiplication is separately order-continuous in the sense that the maps u 7→ u × w, u 7→ w × u are order-continuous for every w ∈ U + ; (b) the multiplication is commutative. proof (a) Let A ⊆ U be a non-empty set with infimum 0, and v0 ∈ U + a lower bound for {u × w : u ∈ A}. Fix u0 ∈ A. If u ∈ A and δ > 0, then v0 ∧ (u0 − 1δ u)+ ≤ δu0 × w. P P Set v = v0 ∧ (u0 − 1δ u)+ . Then

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δv ∧ (u − δu0 )+ ≤ (δu0 − u)+ ∧ (u − δu0 )+ = 0, so v ∧ (u − δu0 )+ = 0 and v ∧ ((u − δu0 )+ × w) = 0. But v ≤ v0 ≤ u × w ≤ (u − δu0 )+ × w + δu0 × w, so v ≤ ((u − δu0 )+ × w) ∧ v + (δu0 × w) ∧ v ≤ δu0 × w, by 352Fa. Q Q Taking the infimum over u, and using the distributive laws (352E), we get v0 ∧ u0 ≤ δu0 × w. Taking the infimum over δ, and using the hypothesis that U is Archimedean, v0 ∧ u0 = 0. But this means that v0 ∧ (u0 × w) = 0, while v0 ≤ u0 × w, so v0 = 0. As v0 is arbitrary, inf u∈A u × w = 0; as A is arbitrary, u 7→ u × w is order-continuous. Similarly, u 7→ w × u is order-continuous. (b)(i) Fix v ∈ U + , and for u ∈ U set Su = u × v,

T u = v × u.

Then S and T are both order-continuous Riesz homomorphisms from U to itself (352W(b-iv) and (a) above). Also, Su ∧ T u0 = 0 whenever u ∧ u0 = 0. P P 0 = (u × v) ∧ u0 = (u × v) ∧ (v × u0 ). Q Q So W = {u : u × v = v × u} is a band in U (353N). Of course v ∈ W (because Sv = T v = v 2 ). If u ∈ W ⊥ , then v ∧ |u| = 0 so Su = T u = 0 (352W(b-i)), and u ∈ W ; but this means that W ⊥ = {0} and W = W ⊥⊥ = U (353Bb). Thus v × u = u × v for every u ∈ U . (ii) This is true for every v ∈ U + . Of course it follows that v × u = u × v for every u, v ∈ U , so that multiplication is commutative. 353P Proposition Let U be an Archimedean f -algebra with multiplicative identity e. (a) e is a weak order unit in U . (b) If u, v ∈ U then u × v = 0 iff |u| ∧ |v| = 0. (c) If u ∈ U has a multiplicative inverse u−1 then |u| also has a multiplicative inverse; if u ≥ 0 then u−1 ≥ 0 and u is a weak order unit. (d) If V is another Archimedean f -algebra with multiplicative identity e0 , and T : U → V is a positive linear operator such that T e = e0 , then T is a Riesz homomorphism iff T (u × v) = T u × T v for all u, v ∈ U . proof (a) e = e2 ≥ 0 by 352W(b-iii). If u ∈ U and e ∧ |u| = 0 then |u| = (e × |u|) ∧ |u| = 0; by 353Lc, e is a weak order unit. (b) If |u| ∧ |v| = 0 then u × v = 0, by 352W(b-ii). If w = |u| ∧ |v| > 0, then w2 ≤ |u| × |v|. Let n ∈ N be such that nw 6≤ e, and set w1 = (nw − e)+ , w2 = (e − nw)+ . Then 0 6= w1 = w1 × e = w1 × w2 + w1 × (e ∧ nw) = w1 × (e ∧ nw) ≤ (nw)2 ≤ n2 |u| × |v| = n2 |u × v|, so u × v 6= 0. (c) u × u−1 = e so |u| × |u−1 | = |e| = e (352W(b-iii)), and |u−1 | = |u|−1 . (Recall that inverses in any semigroup with identity are unique, so that we need have no inhibitions in using the formulae u−1 , |u|−1 .) Now suppose that u ≥ 0. Then u−1 = |u−1 | ≥ 0. If u ∧ |v| = 0 then e ∧ |v| = (u × u−1 ) ∧ |v| = 0, so v = 0; accordingly u is a weak order unit. (d)(i) If T is multiplicative, and u ∧ v = 0 in U , then T u × T v = T (u × v) = 0 and T u ∧ T v = 0, by (b). So T is a Riesz homomorphism, by 352G.

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(ii) Accordingly I shall henceforth assume that T is a Riesz homomorphism and seek to show that it is multiplicative. If u, v ∈ U + , then T (u × v) and T u × T v both belong to the band generated by T u. P P Write W for this band. (α) For any n ≥ 1 we have (v − ne)2 ≥ 0, that is, 2nv ≤ v 2 + n2 e, so n(v − ne) ≤ 2nv − n2 e ≤ v 2 . Consequently 1 n

T (u × v) − nT u = T (u × v) − nT (u × e) = T (u × (v − ne)) ≤ T (u × v 2 ) because v 0 7→ T (u × v 0 ) is a positive linear operator; as V is Archimedean, inf n∈N (T (u × v) − nT u)+ = 0 and T (u × v) = supn∈N T (u × v) ∧ nT u belongs to W . (β) If w ∧ |T u| = 0 then w ∧ |T u × T v| = w ∧ (|T u| × |T v|) = 0; so T u × T v ∈ W ⊥⊥ = W . Q Q (iii) Fix v ∈ U + . For u ∈ U , set S1 u = T u × T v and S2 u = T (u × v). Then S1 and S2 are both Riesz homomorphisms from U to V . If u ∧ u0 = 0 in U , then S1 u ∧ S2 u0 = 0 in V , because (by (ii) just above) S1 u belongs to the band generated by T u, while S2 u0 belongs to the band generated by T u0 , and T u ∧ T u0 = T (u ∧ u0 ) = 0. By 353N, W = {u : S1 u = S2 u} is a solid linear subspace of U . Of course it contains e, since S1 e = T e × T v = e0 × T v = T v = T (e × v) = S2 e. In fact u ∈ W for every u ∈ U + . P P As noted in (ii) just above, u − ne ≤

1 2 nu

for every n ≥ 1. So

|S1 u − S2 u| = |S1 (u − ne)+ + S1 (u ∧ ne) − S2 (u − ne)+ − S2 (u ∧ ne)| 1 n

≤ S1 (u − ne)+ + S2 (u − ne)+ ≤ (S1 u2 + S2 u2 ) for every n ≥ 1, and |S1 u − S2 u| = 0, that is, S1 u = S2 u. Q Q So W = U , that is, T u × T v = T (u × v) for every u ∈ U . And this is true for every v ∈ U + . It follows at once that it is true for every v ∈ U , so that T is multiplicative, as claimed. 353X Basic exercises > (a) Let U be a Riesz space in which every band is complemented. Show that U is Archimedean. (b) A Riesz space U has the principal projection property iff the band generated by any single member of U is a projection band. Show that any Dedekind σ-complete Riesz space has the principal projection property, and that any Riesz space with the principal projection property is Archimedean. > (c) Fill in the missing part (b-iii) of 353J. (d) Let U be an Archimedean f -algebra with an order-unit which is a multiplicative identity. Show that U can be identified, as f -algebra, with a subspace of C(X) for some compact Hausdorff space X. 353Y Further exercises (a) Let U be a Riesz space in which every quasi-order-dense solid linear subspace is order-dense. Show that U is Archimedean. (b) Let X be a completely regular Hausdorff space. Show that C(X) is Dedekind complete iff Cb (X) is Dedekind complete iff X is extremally disconnected. (c) Let X be a compact Hausdorff space. Show that C(X) is Dedekind σ-complete iff G is open for every cozero set G ⊆ X. (Cf. 314Yf.) Show that in this case X is zero-dimensional. (d) Let U be an Archimedean Riesz space such that {un : n ∈ N} has a supremum in U whenever hun in∈N is a sequence in U such that um ∧un = 0 whenever m 6= n. Show that U has the principal projection property, but need not be Dedekind σ-complete.

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(e) Let U be an Archimedean Riesz space. Show that the following are equiveridical: (i) U has the countable sup property (241Yd) (ii) for every A ⊆ U there is a countable B ⊆ A such that A and B have the same upper bounds; (iii) every disjoint subset of U + is countable. (f ) Let U be an Archimedean Riesz space with order unit e, and k ke the corresponding norm. Let Z be the unit ball of U ∗ . Show that for a linear functional f : U → R the following are equiveridical: (i) f is an extreme point of Z, that is, f ∈ Z and Z \ {f } is convex (ii) |f (e)| = 1 and one of f , −f is a Riesz homomorphism. (g) Let U be an Archimedean f -algebra. Show that an element e of U is a multiplicative identity iff e2 = e and e is a weak order unit. (Hint: start by showing that under these conditions, e × u = 0 ⇒ u = 0.) (h) Let U be an Archimedean f -algebra with a multiplicative identity. Show that if u ∈ U then u is invertible iff |u| is invertible. 353 Notes and comments As in the last section, many of the results above have parallels in the theory of Boolean algebras; thus 353A corresponds to 313K, 353G corresponds in part to remarks in 314Bc and 314Xa, and 353J corresponds to 314C-314E. Riesz spaces are more complicated; for instance, principal ideals in Boolean algebras are straightforward, while in Riesz spaces we have to distinguish between the solid linear subspace generated by an element and the band generated by the same element. Thus an ‘order unit’ in a Boolean ring would just be an identity, while in a Riesz space we must distinguish between ‘order unit’ and ‘weak order unit’. As this remark may suggest to you, (Archimedean) Riesz spaces are actually closer in spirit to arbitrary Boolean rings than to the Boolean algebras we have been concentrating on so far; to the point that in §361 below I will return briefly to general Boolean rings. Note that the standard definition of ‘order-dense’ in Boolean algebras, as given in 313J, corresponds to the definition of ‘quasi-order-dense’ in Riesz spaces (352Na); the point here being that Boolean algebras behave like Archimedean Riesz spaces, in which there is no need to make a distinction. I give the representation theorem 353M more for completeness than because we need it in any formal sense. In 351Q and 352L I have given representation theorems for general partially ordered linear spaces, and general Riesz spaces, as quotients of spaces of functions; in 368F below I give a theorem for Archimedean Riesz spaces corresponding rather more closely to the expressions of the Lp spaces as quotients of spaces of measurable functions. In 353M, by contrast, we have a theorem expressing Archimedean Riesz spaces with order units as true spaces of functions, rather than as spaces of equivalence classes of functions. All these theorems are important in forming an appropriate mental picture of ordered linear spaces, as in 352M. I give a bare-handed proof of 353M, using only the Riesz space structure of C(X); if you know a little about extreme points of dual unit balls you can approach from that direction instead, using 353Yf. The point is that (as part (d) of the proof makes clear) the space X can be regarded as a subset of the normed space dual U ∗ of U with its weak* topology. In this treatise generally, and in the present chapter in particular, I allow myself to be slightly prejudiced against normed-space methods; you can find them in any book on functional analysis, and I prefer here to develop techniques like those in part (b) of the proof of 353M, which will be a useful preparation for such theorems as 368E. There is a very close analogy between 353M and the Stone representation of Boolean algebras (311E, 311I-311K). Just as the proof of 311E looked at the set of ring homomorphisms from A to the elementary Boolean algebra Z2 , so the proof of 353M looks at Riesz homomorphisms from U to the elementary M -space R. Later on, the most important M -spaces, from the point of view of this treatise, will be the L∞ spaces of §363, explicitly defined in terms of Stone representations (363A). Of the two parts of 353O, it is (a) which is most important for the purposes of this book. The f -algebras we shall encounter in Chapter 36 can be seen to be commutative for different, and more elementary, reasons. The (separate) order-continuity of multiplication, however, is not always immediately obvious. Similarly, the uniferent Riesz homomorphisms we shall encounter can generally be seen to be multiplicative without relying on the arguments of 353Pd.

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Riesz Spaces

§354 intro.

354 Banach lattices The next step is a brief discussion of norms on Riesz spaces. I start with the essential definitions (354A, 354D) with the principal properties of general Riesz norms (354B-354C) and order-continuous norms (354E). I then describe two of the most important classes of Banach lattice: M -spaces (354F-354L) and L-spaces (354M-354R), with their elementary properties. For M -spaces I give the basic representation theorem (354K-354L), and for L-spaces I give a note on uniform integrability (354P-354R). 354A Definitions (a) If U is a Riesz space, a Riesz norm or lattice norm on U is a norm k k such that kuk ≤ kvk whenever |u| ≤ |v|; that is, a norm such that k|u|k = kuk for every u and kuk ≤ kvk whenever 0 ≤ u ≤ v. (b) A Banach lattice is a Riesz space with a Riesz norm under which it is complete. Remark We have already seen many examples of Banach lattices; I list some in 354Xa below. 354B Lemma Let U be a Riesz space with a Riesz norm k k. (a) U is Archimedean. (b) The maps u 7→ |u| and u 7→ u+ are uniformly continuous. (c) For any u ∈ U , the sets {v : v ≤ u} and {v : v ≥ u} are closed; in particular, the positive cone of U is closed. (d) Any band in U is closed. (e) If V is a norm-dense Riesz subspace of U , then V + = {v : v ∈ V, v ≥ 0} is norm-dense in the positive cone U + of U . proof (a) If u, v ∈ U are such that nu ≤ v for every n ∈ N, then nu+ ≤ v + so nku+ k ≤ kv + k for every n, and ku+ k = 0, that is, u+ = 0 and u ≤ 0. As u, v are arbitrary, U is Archimedean. (b) For any u, v ∈ U , ||u| − |v|| ≤ |u − v| (352D), so k|u| − |v|k ≤ ku − vk; thus u 7→ |u| is uniformly continuous. Consequently u 7→ 12 (u + |u|) = u+ is uniformly continuous. (c) Now {v : v ≤ u} = {v : (v − u)+ = 0} is closed because the function v 7→ (v − u)+ is continuous and {0} is closed. Similarly {v : v ≥ u} = {v : (u − v)+ = 0} is closed. (d) If V ⊆ U is a band, then V = V ⊥⊥ (353Bb), that is, V = {v : |v| ∧ |w| = 0 for every w ∈ V ⊥ }. Because the function v 7→ |v| ∧ |w| = 21 (|v| + |w| − ||v| − |w||) is continuous, all the sets {v : |v| ∧ |w| = 0} are closed, and so is their intersection V . (e) Observe that V + = {v + : v ∈ V } and U + = {u+ : u ∈ U }; recall that u 7→ u+ is continuous, and apply 3A3Eb. 354C Lemma If U is a Banach lattice and hu Pn∞in∈N is a sequence in U such that supn∈N un is defined in U , with k supn∈N un k ≤ n=0 kun k.

P∞ n=0

kun k < ∞, then

proof Set vn = supi≤n ui for each n. Then 0 ≤ vn+1 − vn ≤ (un+1 − un )+ ≤ |un+1 − un | for each n ∈ N, so

P∞ n=0

kvn+1 − vn k ≤

P∞ n=0

kun+1 − un k ≤

P∞ n=0

kun+1 k + kun k

is finite, and hvn in∈N is Cauchy. Let u be its limit; because hvn in∈N is non-decreasing, and the sets {v : v ≥ vn } are all closed, u ≥ vn for each n ∈ N. On the other hand, if v ≥ vn for every n, then (u − v)+ = limn→∞ (vn − v)+ = 0, and u ≤ v. So u = supn∈N vn = supn∈N un is the required supremum.

354E

Banach lattices

To estimate its norm, observe that |vn | ≤ for the inductive step), so that

Pn i=0

|ui | for each n (induce on n, using the last item in 352D

kuk = limn→∞ kvn k ≤

354D

247

P∞ i=0

k|ui |k =

P∞ i=0

kui k.

I come now to the basic properties according to which we classify Riesz norms.

Definitions (a) A Fatou norm on a Riesz space U is a Riesz norm on U such that whenever A ⊆ U + is non-empty and upwards-directed and has a least upper bound in U , then k sup Ak = supu∈A kuk. (Observe that, once we know that k k is a Riesz norm, we can be sure that kuk ≤ k sup Ak for every u ∈ A, so that all we shall need to check is that k sup Ak ≤ supu∈A kuk.) (b) A Riesz norm on a Riesz space U has the Levi property if every upwards-directed norm-bounded set is bounded above. (c) A Riesz norm on a Riesz space U is order-continuous if inf u∈A kuk = 0 whenever A ⊆ U is a non-empty downwards-directed set with infimum 0. 354E Proposition Let U be a Riesz space with an order-continuous Riesz norm k k. (a) If A ⊆ U is non-empty and upwards-directed and has a supremum, then sup A ∈ A. (b) k k is Fatou. (c) If A ⊆ U is non-empty and upwards-directed and bounded above, then for every ² > 0 there is a u ∈ A such that k(v − u)+ k ≤ ² for every v ∈ A. (d) Any non-decreasing order-bounded sequence in U is Cauchy. (e) If U is a Banach lattice it is Dedekind complete. (f) Every order-dense Riesz subspace of U is norm-dense. proof (a) Suppose that A ⊆ U is non-empty and upwards-directed and has a least upper bound u0 . Then B = {u0 − u : u ∈ A} is downwards-directed and has infimum 0. So inf u∈A ku0 − uk = 0, and u0 ∈ A. (b) If, in (a), A ⊆ U + , then we must have ku0 k ≤ inf u∈A kuk + ku − u0 k ≤ supu∈A kuk. As A is arbitary, k k is a Fatou norm. (c) Let B be the set of upper bounds for A. Then B is downwards-directed; because A is upwardsdirected, B − A = {v − u : v ∈ B, u ∈ A} is downwards-directed. By 353F, inf(B − A) = 0. So there are w ∈ B, u ∈ A such that kw − uk ≤ ². Now if v ∈ A, (v − u)+ = (v ∨ u) − u ≤ w − u, so k(v − u)+ k ≤ ². (d) If hun in∈N is a non-decreasing order-bounded sequence, and ² > 0, then, applying (c) to {un : n ∈ N}, we find that there is an m ∈ N such that kum − un k ≤ ² whenever m ≥ n. (e) Now suppose that U is a Banach lattice. Let A ⊆ U be any non-empty set with an upper bound. Set A0 = {u0 ∨ . . . ∨ un : u0 , . . . , un ∈ A}, so that A0 is upwards-directed and has the same upper bounds as A. For each n, choose un ∈ A0 such that k(u − un )+ k ≤ 2−n for every u ∈ A0 . Set vn = supi≤n ui for each n; then vn ∈ A0 and kvm − vn k ≤ k(vm − un )+ k ≤ 2−n for all m ≥ n. So hvn in∈N is Cauchy and has a limit v say. If u ∈ A, then k(u − v)+ k = limn→∞ k(u − vn )+ k = 0, so u ≤ v; while if w is any upper bound for A, then k(v − w)+ k = limn→∞ k(vn − w)+ k = 0 and v ≤ w. Thus v = sup A and A has a supremum. (f ) If V is an order-dense Riesz subspace of U and u ∈ U + , set A = {v : v ∈ V, v ≤ u}. Then A is upwards-directed and has supremum u, so u ∈ A ⊆ V , by (a). Thus U + ⊆ V ; it follows at once that U = U+ − U+ ⊆ V .

248

Riesz Spaces

354F

354F Lemma If U is an Archimedean Riesz space with an order unit e (definition: 353L), there is a Riesz norm k ke defined on U by the formula kuke = min{α : |u| ≤ αe} for every u ∈ U . proof This is a routine verification. Because e is so always has an infimum α0 say; now |u| − α0 e ≤ |u| − α0 e ≤ 0 and |u| ≤ α0 e, so that the minimum subadditivity and homogeneity of k ke are immediate

an order-unit, {α : |u| ≤ αe} is always non-empty, δe for every δ > 0, so (because U is Archimedean) is attained. In particular, kuke = 0 iff u = 0. The from the facts that |u + v| ≤ |u| + |v|, |αu| = |α||u|.

354G Definitions (a) If U is an Archimedean Riesz space and e an order unit in U , the norm k ke as defined in 354F is the order-unit norm on U associated with e. (b) An M -space is a Banach lattice in which the norm is an order-unit norm. (c) If U is an M -space, its standard order unit is the order unit e such that k ke is the norm of U . (To see that e is uniquely defined, observe that it is sup{u : u ∈ U, kuk ≤ 1}.) 354H Examples (a) For any set X, `∞ (X) is an M -space with standard order unit χX. (As remarked in 243Xl, the completeness of `∞ (X) can be regarded as the special case of 243E in which X is given counting measure.) (b) For any topological space X, the space Cb (X) of bounded continuous real-valued functions on X is an M -space with standard order unit χX. (It is a Riesz subspace of `∞ (X) containing the order unit of `∞ (X), therefore in its own right an Archimedean Riesz space with order unit. To see that it is complete, it is enough to observe that it is closed in `∞ (X) because a uniform limit of continuous functions is continuous (3A3Nb).) (c) For any measure space (X, Σ, µ), the space L∞ (µ) is an M -space with standard order unit χX • . 354I Lemma Let U be an Archimedean Riesz space with order unit e, and V a subset of U which is dense for the order-unit norm k ke . Then for any u ∈ U there are sequences hvn in∈N , hwn in∈N in V such that vn ≤ vn+1 ≤ u ≤ wn+1 ≤ wn and kwn − vn ke ≤ 2−n for every n; so that u = supn∈N vn = inf n∈N wn in U. If V is a Riesz subspace of U , and u ≥ 0, we may suppose that vn ≥ 0 for every n. Consequently V is order-dense in U . proof For each n ∈ N, take vn , wn ∈ V such that ku −

3 e − vn ke 2n+3

≤

1 , 2n+3

ku +

3 e − wn ke 2n+3

≤

1 . 2n+3

Then u−

1 e 2n+1

≤ vn ≤ u −

1 e 2n+2

≤u≤u+

1 e 2n+2

≤ wn ≤ u +

1 e. 2n+1

Accordingly hvn in∈N is non-decreasing, hwn in∈N is non-increasing and kwn −vn ke ≤ 2−n for every n. Because U is Archimedean, supn∈N vn = inf n∈N wn = u. If V is a Riesz subspace of U , then replacing vn by vn+ if necessary we may suppose that every vn is non-negative; and V is order-dense by the definition in 352N. 354J Proposition Let U be an Archimedean Riesz space with an order unit e. Then k ke is Fatou and has the Levi property. proof This is elementary. If A ⊆ U + is non-empty, upwards-directed and norm-bounded, then it is bounded above by αe, where α = supu∈A kuke . This is all that is called for in the Levi property. If moreover sup A is defined, then sup A ≤ αe so k sup Ak ≤ α, as required in the Fatou property.

354O

Banach lattices

249

354K Theorem Let U be an Archimedean Riesz space with order unit e. Then it can be embedded as an order-dense and norm-dense Riesz subspace of C(X), where X is a compact Hausdorff space, in such a way that e corresponds to χX and k ke corresponds to k k∞ ; moreover, this embedding is essentially unique. proof This is nearly word-for-word a repetition of 353M. The only addition is the mention of the norms. But let X and T : U → C(X) be as in 353M. Then, for any u ∈ U , |u| ≤ kuke e, so that |T u| = T |u| ≤ kuke T e = kuke χX, and kT uk∞ ≤ kuke . On the other hand, if 0 < δ < kuke then u1 = (|u| − δe)+ > 0, so that T u1 = (|T u| − δχX)+ > 0 and kT uk∞ ≥ δ; as δ is arbitrary, kT uk∞ ≥ kuke . 354L Corollary Any M -space U is isomorphic, as Banach lattice, to C(X) for some compact Hausdorff X, and the isomorphism is essentially unique. X can be identified with the set of Riesz homomorphisms x : U → R such that x(e) = 1, where e is the standard order unit of U , with the topology induced by the product topology on RU . proof By 354K, there are a compact Hausdorff space X and an embedding of U as a norm-dense Riesz subspace of C(X) matching k ke to k k∞ . Since U is complete under k ke , its image is closed in C(X) (3A4Ff), and must be the whole of C(X). The expression is unique just in so far as the expression of 353M/354K is unique. In particular, we may, if we wish, take X to be the set of normalized Riesz homomorphisms from U to R, as in the proof of 353M. Remark If U is an M -space, then the construction of 353M represents U as C(X), where X is the set of uniferent Riesz homomorphisms from U to R; this is sometimes called the spectrum of U . 354M I come now to a second fundamental class of Banach lattices, in a strong sense ‘dual’ to the class of M -spaces, as will appear in §356. Definition An L-space is a Banach lattice U such that ku + vk = kuk + kvk whenever u, v ∈ U + . Example If (X, Σ, µ) is any measure space, then L1 (µ), with its norm k k1 , is an L-space (242D, 242F). In particular, taking µ to be counting measure on N, `1 is an L-space (242Xa). 354N Theorem If U is an L-space, then its norm is order-continuous and has the Levi property. proof (a) Both of these are consequences of the following fact: if A ⊆ U is norm-bounded and non-empty and upwards-directed, then sup A is defined in U and belongs to the norm-closure of A in U . P P For u ∈ A, set γ(u) = sup{kv − uk : v ∈ A, v ≥ u}. We surely have γ(u) ≤ kuk + supv∈A kvk < ∞. Choose a sequence hun in∈N in A such that un+1 ≥ un and kun+1 − un k ≥ 12 γ(un ) for each n. Then kun+1 − u0 k =

Pn i=0

kui+1 − ui k ≥

1 2

Pn i=0

γ(ui )

P∞

for every n, using the definition of ‘L-space’. Because A is bounded, i=0 γ(ui ) < ∞ and limn→∞ γ(un ) = 0. But kum − un k ≤ γ(un ) whenever m ≥ n, so hun in∈N is Cauchy and has a limit u∗ in U . For each n ∈ N, u∗ ≥ un because um ≥ un for every m ≥ n (see 354Bc). If u ∈ A, n ∈ N then there is a u0 ∈ A such that u0 ≥ u ∨ un ; now (u − u∗ )+ ≤ u0 − un so k(u − u∗ )+ k ≤ ku0 − un k ≤ γ(un ); as n is arbitrary, k(u − u∗ )+ k = 0 and u ≤ u∗ . Thus u∗ is an upper bound for A. But if v is any upper bound for A, then un ≤ v for every n so u∗ ≤ v. Thus u∗ is the least upper bound of A; and u∗ ∈ A because it is the norm limit of hun in∈N . Q Q (b) This shows immediately that the norm has the Levi property. But also it must be order-continuous. P P If A ⊆ U is non-empty and downwards-directed and has infimum 0, take any u0 ∈ A and consider B = {u0 − u : u ∈ A, u ≤ u0 }. Then B is upwards-directed and has supremum u0 , so u0 ∈ B and inf u∈A kuk ≤ inf v∈B ku0 − vk = 0. Q Q 354O Proposition If U is an L-space and V is a norm-closed Riesz subspace of U , then V is an L-space in its own right. In particular, any band of U is an L-space.

250

Riesz Spaces

354O

proof For any Riesz subspace V of U , we surely have ku + vk = ku| + kvk whenever u, v ∈ V + ; so if V is norm-closed, therefore a Banach lattice, it must be an L-space. But in any Banach lattice, a band is norm-closed (354Bd), so a band in an L-space is again an L-space. 354P Uniform integrability in L-spaces Some of the ideas of §246 can be readily expressed in this abstract context. Definition Let U be an L-space. A set A ⊆ U is uniformly integrable if for every ² > 0 there is a w ∈ U + such that k(|u| − w)+ k ≤ ² for every u ∈ A. 354Q Since I have already used the phrase ‘uniformly integrable’ based on a different formula, I had better check instantly that the two definitions are consistent. Proposition If (X, Σ, µ) is any measure space, then a subset of L1 = L1 (µ) is uniformly integrable in the sense of 354P iff it is uniformly integrable in the sense of 246A. proof (a) If A ⊆ L1 is uniformly integrable in the sense of 246A, then for any ² > 0 there are M ≥ 0, R E ∈ Σ such that µE < ∞ and (|u| − M χE • )+ ≤ ² for every u ∈ A; now w = M χE • belongs to (L1 )+ and k(|u| − w)+ k ≤ ² for every u ∈ A. As ² is arbitrary, A is uniformly integrable in the sense of 354P. (b) Now suppose that A is uniformly integrable in the sense of 354P. Let ² > 0. Then there is a w ∈ (L1 )+ such that k(|u| − w)+ k ≤ 21 ² for every u ∈ A. There is a simple function h : X → R such that kw − h• k ≤ 21 ² (242M); now take E = {x : h(x) 6= 0}, M = supx∈X |h(x)| (I pass over the trivial case X = ∅), so that h ≤ M χE and (|u| − M χE • )+ ≤ (|u| − w)+ + (w − M χE • )+ ≤ (|u| − w)+ + (w − h• )+ ,

R

(|u| − M χE • )+ ≤ k(|u| − w)+ k + kw − h• k ≤ ²

for every u ∈ A. As ² is arbitrary, A is uniformly integrable in the sense of 354P. 354R

I give abstract versions of the easiest results from §246.

Theorem Let U be an L-space. (a) If A ⊆ U is uniformly integrable, then (i) A is norm-bounded; (ii) every subset of A is uniformly integrable; (iii) for any α ∈ R, αA is uniformly integrable; (iv) there is a uniformly integrable, solid, convex, norm-closed set C ⊇ A; (v) for any other uniformly integrable set B ⊆ U , A ∪ B and A + B are uniformly integrable. (b) For any set A ⊆ U , the following are equiveridical: (i) A is uniformly integrable; (ii) limn→∞ (|un | − supi 0 there are u0 , . . . , un ∈ A such that k(|u| − supi≤n |ui |)+ k ≤ ² for every u ∈ A; (iv) A is norm-bounded and any disjoint sequence in the solid hull of A is norm-convergent to 0. (c) If V ⊆ U is a closed Riesz subspace then a subset of V is uniformly integrable when regarded as a subset of V iff it is uniformly integrable when regarded as a subset of U . R proof (a)(i) There must be a w ∈ U + such that (|u| − w)+ ≤ 1 for every u ∈ A; now |u| ≤ |u| − w + |w| ≤ (|u| − w)+ + |w|,

kuk ≤ k(|u| − w)+ k + kwk ≤ 1 + kwk

for every u ∈ A, so A is norm-bounded. (ii) This is immediate from the definition. (iii) Given ² > 0, we can find w ∈ U + such that |α|k(|u| − w)+ k ≤ ² for every u ∈ A; now k(|v| − |α|w)+ k ≤ ² for every v ∈ αA. (iv) If A is empty, take C = A. Otherwise, try

354R

Banach lattices

251

C = {v : v ∈ U, k(|v| − w)+ k ≤ supu∈A k(|u| − w)+ k for every w ∈ U + }. Evidently A ⊆ C, and C satisfies the definition 354M because A does. The functionals v 7→ k(|v| − w)+ k : U → R are all continuous for k k (because the operators v 7→ |v|, v 7→ v − w, v 7→ v + , v 7→ kvk are continuous), so C is closed. If |v 0 | ≤ |v| and v ∈ C, then k(|v 0 | − w)+ k ≤ k(|v| − w)+ k ≤ supu∈A k(|u| − w)+ k for every w, and v 0 ∈ C. If v = αv1 + βv2 where v1 , v2 ∈ C, α ∈ [0, 1] and β = 1 − α, then |v| ≤ α|v1 | + β|v2 |, so |v| − w ≤ (α|v1 | − αw) + (β|v2 | − βw) ≤ (α|v1 | − αw)+ + (β|v2 | − βw)+ and (|v| − w)+ ≤ α(|v1 | − w)+ + β(|v2 | − w)+ for every w; accordingly k(|v| − w)+ k ≤ αk(|v1 | − w)+ k + βk(|v2 | − w)+ k ≤ (α + β) sup k(|u| − w)+ k = sup k(|u| − w)+ k u∈A

u∈A

for every w, and v ∈ C. Thus C has all the required properties. (v) I show first that A ∪ B is uniformly integrable. P P Given ² > 0, let w1 , w2 ∈ U + be such that k(|u| − w1 )+ k ≤ ² for every u ∈ A,

k(|u| − w2 )+ k ≤ ² for every u ∈ B.

Set w = w1 ∨ w2 ; then k(|u| − w)+ k ≤ ² for every u ∈ A ∪ B. As ² is arbitrary, A ∪ B is uniformly integrable. Q Q Now (iv) tells us that there is a convex uniformly integrable set C including A ∪ B, and in this case A + B ⊆ 2C, so A + B is also uniformly integrable, using (ii) and (iii). (b)(i)⇒(ii)&(iv) Suppose that A is uniformly integrable and that hun in∈N is any sequence in the solid hull of A. Set vn = supi≤n |ui | for n ∈ N and v00 = v0 = |u0 |,

vn0 = vn − vn−1 = (|un | − supi 0, there is a w ∈ U + such that k(|u| − w)+ k ≤ ² for every u ∈ A, and therefore for every u in the solid hull of A. Of course supn∈N kvn ∧ wk ≤ kwk is finite, so there is an n ∈ N such that kvi ∧ wk ≤ ² + kvn ∧ wk for every i ∈ N. But now, for m > n, 0 vm ≤ (|um | − vn )+ ≤ (|um | − |um | ∧ w)+ + ((|um | ∧ w) − vn )+

≤ (|um | − w)+ + (vm ∧ w) − (vn ∧ w), so that 0 kvm k ≤ k(|um | − w)+ k + k(vm ∧ w) − (vn ∧ w)k

= k(|um | − w)+ k + kvm ∧ wk − kvn ∧ wk ≤ 2², using the L-space property of the norm for the equality in the middle. As ² is arbitrary, limn→∞ vn0 = 0. As hun in∈N is arbitrary, condition (ii) is satisfied; but so is condition (iv), because we know from (a-i) that A is norm-bounded, and if hun in∈N is disjoint then vn0 = |un | for every n, so that in this case limn→∞ un = 0. (ii)⇒(iii)⇒(i) are elementary. not-(i)⇒not-(iv) Now suppose that A is not uniformly integrable. If it is not norm-bounded, we can stop. Otherwise, there is some ² > 0 such that supu∈A k(|u| − w)+ k > ² for every w ∈ U + . Consequently we shall be able to choose inductively a sequence hun in∈N in A such that k(|un | − 2n supi ² for P∞ −i every n ≥ 1. Because A is norm-bounded, i=0 2 kui k is finite, and we can set

252

Riesz Spaces

354R

P∞ vn = (|un | − 2n supi (d) Let U be a Riesz space with a Riesz norm. (i) Show that any order-bounded set in U is normbounded. (ii) Show that in Rr , with any of the standard Riesz norms (354Xa(i)), norm-bounded sets are order-bounded. (iii) Show that in `1 (N) there is a sequence converging to 0 (for the norm) which is not orderbounded. (iv) Show that in c 0 any sequence converging to 0 is order-bounded, but there is a norm-bounded set which is not order-bounded. (e) Let U be a Riesz space with a Riesz norm. Show that it is a Banach lattice iff non-decreasing Cauchy sequences are convergent. (Hint: if kun+1 − un k ≤ 2−n for every n, show that hsupi≤n ui in∈N is Cauchy, and that hun in∈N converges to inf n∈N supm≥n um .) (f ) Let U be a Riesz space with a Riesz norm. Show that U is a Banach lattice iff every non-decreasing Cauchy sequence hun in∈N in U + has a least upper bound u with kuk = limn→∞ kun k. (g) Let U be a Banach lattice. Suppose that B ⊆ U is solid and supn∈N un ∈ B whenever hun in∈N is a non-decreasing sequence in B with a supremum in U . Show that B is closed. (Hint: show first that u ∈ B whenever there is a sequence hun in∈N in B ∩ U + such that ku −un k ≤ 2−n for every n; do this by considering vm = inf n≥m un .) (h) Let U be any Riesz space with a Riesz norm. Show that the Banach space completion of U (3A5Ib) has a unique partial ordering under which it is a Banach lattice. >(i) Show that the space c0 of sequences convergent to 0, with k k∞ , is a Banach lattice with an ordercontinuous norm which does not have the Levi property.

354Yc

Banach lattices

253

> (j) Show that `∞ , with k k∞ , is a Banach lattice with a Fatou norm which has the Levi property but is not order-continuous. (k) Let U be a Riesz space with a Fatou norm. Show that if V ⊆ U is a regularly embedded Riesz subspace (definition: 352Ne) then the induced norm on V is a Fatou norm. (l) Let U be a Riesz space and k k a Riesz norm on U which is order-continuous in the sense of 354Dc. Show that it is order-continuous in the sense of 313H when regarded as a function from U + to [0, ∞[. (m) Let U be a Riesz space with an order-continuous norm. Show that if V ⊆ U is a regularly embedded Riesz subspace then the induced norm on V is order-continuous. (n) Let U be a Dedekind σ-complete Riesz space with a Fatou norm which has the Levi property. Show that it is a Banach lattice. (Hint: 354Xf.) Q (o) Let hUi ii∈I be any family of Banach lattices and let V1 , V∞ be the subspaces of U = i∈I Ui as described in 354Xb. Show that V1 , V∞ have norms which are Fatou, or have the Levi property, iff every Ui has. Show that the norm of V1 is order-continuous iff the norm of every Ui is. (p) Let U be a Banach lattice with an order-continuous norm. Show that a Riesz subspace of U (indeed, any sublattice of U ) is norm-closed iff it is order-closed in the sense of 313D, and in this case is itself a Banach lattice with an order-continuous norm. > (q) Let U be an M -space and V a norm-closed Riesz subspace of U containing the standard order unit of U . (i) Show that V , with the induced norm, is an M -space. (ii) Deduce that the space c of convergent sequences is an M -space if given the norm k k∞ inherited from `∞ . (r) Show that a Banach lattice U is an M -space iff (i) its norm is a Fatou norm with the Levi property (ii) ku ∨ vk = max(kuk, kvk) for all u, v ∈ U + . >(s) Describe a topological space X such that the space c of convergent sequences (354Xq) can be identified with C(X). (t) Let D ⊆ R be any non-empty set, and PnV the space of functions f : D → R of bounded variation (§224). For f ∈ V set kf k = sup{|f (t0 )| + i=1 |f (ti ) − f (ti−1 )| : t0 ≤ t1 ≤ . . . ≤ tn in D} (224Yb). Let V + be the set of non-negative, non-decreasing functions in V . Show that V + is the positive cone of V for a Riesz space ordering under which V is an L-space. 354Y Further exercises (a) Let U be a Riesz space with a Riesz norm, and V a norm-dense Riesz subspace of U . Suppose that the induced norm on V is Fatou, when regarded as a norm on the Riesz space V . Show (i) that V is order-dense in U (ii) that the norm of U is Fatou. (Hint: for (i), show that if u ∈ U + , vn ∈ V + and ku − vn k ≤ 2−n−2 kuk for every n, then kv0 − inf i≤n vi k ≥ 41 kuk for every n, so that 0 cannot be inf n∈N vn in V .) (b) Let U be a Riesz space with a Riesz norm. Show that the following are equiveridical: (i) limn→∞ un = 0 whenever hun in∈N is a disjoint order-bounded sequence in U + (ii) limn→∞ un+1 − un = 0 for every orderbounded non-decreasing sequence hun in∈N in U (iii) whenever A ⊆ U + is a non-empty downwards-directed set in U + with infimum 0, inf u∈A supv∈A,v≤u ku − vk = 0. (Hint: for (i)⇒(ii), show by induction that limn→∞ un = 0 whenever hun in∈N is an order-bounded sequence such that, for some fixed k, inf i∈K ui = 0 for every K ⊆ N of size k; now show that if hun in∈N is non-decreasing and 0 ≤ un ≤ u for every n, then inf i∈K (ui+1 − ui − k1 u)+ = 0 whenever K ⊆ N, #(K) = k ≥ 1. For (iii)⇒(i), set A = {u : ∃ n, u ≥ ui ∀ i ≥ n}. See Fremlin 74a, 24H.) (c) Show that any Riesz space with an order-continuous norm has the countable sup property (definition: 241Yd).

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354Yd

(d) Let U be a Banach lattice. Show that the following are equiveridical: (i) the norm on U is ordercontinuous; (ii) U satisfies the conditions of 354Yb; (iii) every order-bounded monotonic sequence in U is Cauchy. (e) Let U be a Riesz space with a Fatou norm. Show that the norm on U is order-continuous iff it satisfies the conditions of 354Yb. R (f ) For f ∈ C([0, 1]), set kf k1 = |f (x)|dx. Show that k k1 is a Riesz norm on C([0, 1]) satisfying the conditions of 354Yb, but is not order-continuous. (g) Let U be a Riesz space with a Riesz norm k k. Show that (U, k k) satisfies the conditions of 354Yb iff the norm of its completion is order-continuous. (h) Let U be a Riesz space with a Riesz norm, and V ⊆ U a norm-dense Riesz subspace such that the induced norm on V is order-continuous. Show that the norm of U is order-continuous. (Hint: use 354Ya.) (i) Let U be an Archimedean Riesz space. For any e ∈ U + , let Ue be the solid linear subspace of U generated by e, so that e is an order unit in Ue , and let k ke be the corresponding order-unit norm on Ue . We say that U is uniformly complete if Ue is complete under k ke for every e ∈ U + . (i) Show that any Banach lattice is uniformly complete. (ii) Show that any Dedekind σ-complete Riesz space is uniformly complete (cf. 354Xn). (iii) Show that if U is a uniformly complete Riesz space with a Riesz norm which has the Levi property, then U is a Banach lattice. (iv) Show that if U is a Banach lattice then a set A ⊆ U is closed, for the norm topology, iff A ∩ Ue is k ke -closed for every e ∈ U + . (v) Let V be a solid linear subspace of U . Show that the quotient Riesz space U/V (352U) is Archimedean iff V ∩ Ue is k ke -closed for every e ∈ U + . (vi) Show that if U is uniformly complete and V ⊆ U is a solid linear subspace such that U/V is Archimedean, then U/V is uniformly complete. (vii) Show that U is Dedekind σ-complete iff it is uniformly complete and has the principal projection property (353Xb). (Hint: for (vii), use 353Yc.) (j) Let U be a Banach lattice such that ku + vk = kuk + kvk whenever u ∧ v = 0. Show that U is an L-space. (Hint: by 354Yd, the norm is order-continuous, so U is Dedekind complete. If u, v ≥ 0, set e = u + v, and represent Ue as C(X) where X is extremally disconnected (353Yb); now approximate u and v by functions taking only finitely many values to show that ku + vk = kuk + kvk.) (k) Let U be a uniformly complete Archimedean Riesz space (354Yi). Set UC = U × U with the complex linear structure defined by identifying (u, v) ∈ U × U as u + iv ∈ UC , so that u = Re(u + iv), v = Im(u + iv) and (α + iβ)(u + iv) = (αu − βv) + i(αv + βu). (i) Show that for w ∈ UC we can define |w| ∈ U by setting |w| = sup|ζ|=1 Re(ζw). (ii) Show that if U is a uniformly complete Riesz subspace of RX for some set X, then we can identify UC with the linear subspace of CX generated by U . (iii) Show that |w + w0 | ≤ |w| + |w0 |, |γw| = |γ||w| for all w ∈ UC , γ ∈ C. (iv) Show that if w ∈ UC and |w| ≤ u1 + u2 , where u1 , u2 ∈ U + , then w is expressible as w1 + w2 where |wj | ≤ uj for both j. (Hint: set e = u1 + u2 and represent Ue as C(X).) (v) Show that if U0 is a solid linear subspace of U , then, for w ∈ UC , |w| ∈ U0 iff Re w, Im w both belong to U0 . (vi) Show that if U has a Riesz norm then we have a norm on UC defined by setting kwk = k|w|k, and that if U is a Banach lattice then UC is a (complex) Banach space. (vii) Show that if U = Lp (µ), where (X, Σ, µ) is a measure space and p ∈ [1, ∞], then UC can be identified with LpC (µ) as defined in 242P, 243K, 244O. (We may call UC the complexification of the Riesz space U .) (l) Let (X, Σ, µ) be a measure space and V a Banach lattice. Write L1V for the space of Bochner integrable functions from conegligible subsets of X to V , and L1V for the corresponding set of equivalence classes (253Yf). (i) Show that L1V is a Banach lattice under the ordering defined by saying that f • ≤ g • iff f (x) ≤ g(x) in V for µ-almost every x ∈ X. (ii) Show that when V = L1 (ν), for some other measure space (Y, T, ν), then this ordering on L1V agrees with the ordering of L1 (λ) where λ is the (c.l.d.) product measure on X × Y and we identify L1V with L1 (λ), as in 253Yi. (iii) Show that if V has an order-continuous norm, so has L1V . (Hint: 354Yd(ii).) (iv) Show that if µ is Lebesgue measure on [0, 1] and V = `∞ , then L1V is not Dedekind σ-complete.

355A

Spaces of linear operators

255

354 Notes and comments Apart from some of the exercises, the material of this section is pretty strictly confined to ideas which will be useful later in this volume. The basic Banach lattices of measure theory are the Lp spaces of Chapter 24; these all have Fatou norms with the Levi property (244Ye-244Yf), and for p < ∞ their norms are order-continuous (244Yd). In Chapter 36 I will return to these spaces in a more abstract context. Here I am mostly concerned to establish a vocabulary in which their various properties, and the relationships between these properties, can be expressed. In normed Riesz spaces we have a very rich mixture of structures, and must take particular care over the concepts of ‘boundedness’, ‘convergence’ and ‘density’, which have more than one possible interpretation. In particular, we must scrupulously distinguish between ‘order-bounded’ and ‘norm-bounded’ sets. I have not yet formally introduced any of the various concepts of order-convergence (see §367), but I think that even so it is best to get into the habit of reminding oneself, when a convergent sequence appears, that it is convergent for the norm topology, rather than in any sense related directly to the order structure. I should perhaps warn you that for the study of M -spaces 354L is not as helpful as it may look. The trouble is that apart from a few special cases (as in 354Xs) the topological space used in the representation is actually more complicated and mysterious than the M -space it is representing. After the introduction of M -spaces, this section becomes a natural place for ‘uniformly complete’ spaces (354Yi). For the moment I leave these in the exercises. But I mention them now because they offer a straightforward route towards a theory of ‘complex Riesz spaces’ (354Yk). In large parts of functional analysis it is natural, and in some parts it is necessary, to work with normed spaces over C rather than over R, and for L2 spaces in particular it is useful to have a proper grasp of the complex case. And while the insights offered by the theory of Riesz spaces are not especially important in such areas, I think we should always seek connexions between different topics. So it is worth remembering that uniformly complete Riesz spaces have complexifications. I shall have a great deal more to say about L-spaces when I come to spaces of additive functionals (§362) and to L1 spaces again (§365) and to linear operators on them (§371); and before that, there will be something in the next section on their duals, and on L-spaces which are themselves dual spaces. For the moment I just give some easy results, direct translations of the corresponding facts in §246, which have natural expressions in the language of this section, holding deeper ideas over. In particular, the characterization of uniformly integrable sets as relatively weakly compact sets (247C) is valid in general L-spaces (356Q). For an extensive treatment of Banach lattices, going very much deeper than I have space for in this volume, see Lindenstrauss & Tzafriri 79. For a careful exposition of a great deal of useful information, see Schaefer 74.

355 Spaces of linear operators We come now to a discussion of linear operators between Riesz spaces. Of course linear operators are central to any kind of functional analysis, and a feature of the theory of Riesz spaces is the way the order structure picks out certain classes of operators for special consideration. Here I concentrate on positive and order-continuous operators, with a brief mention of sequential order-continuity. It turns out, in fact, that we need to work with operators which are differences of positive operators or of order-continuous positive operators. I define the basic spaces L∼ , L× and L∼ c (355A, 355G), with their most important properties (355B, 355E, 355H-355I) and some remarks on the special case of Banach lattices (355C, 355K). At the same time I give an important theorem on extension of operators (355F) and a corollary (355J). The most important case is of course that in which the codomain is R, so that our operators become real-valued functionals; I shall come to these in the next section. 355A Definition Let U and V be Riesz spaces. A linear operator T : U → V is order-bounded if T [A] is order-bounded in V for every order-bounded A ⊆ U . I will write L∼ (U ; V ) for the set of order-bounded linear operators from U to V .

256

Riesz spaces

355B

355B Lemma If U and V are Riesz spaces, (a) a linear operator T : U → V is order-bounded iff {T u : 0 ≤ u ≤ w} is bounded above in V for every w ∈ U +; (b) in particular, any positive linear operator from U to V belongs to L∼ = L∼ (U ; V ); (c) L∼ is a linear space; (d) if W is another Riesz space and T : U → V and S : V → W are order-bounded linear operators, then ST : U → W is order-bounded. proof (a) This is elementary. If T ∈ L∼ and w ∈ U + , [0, w] is order-bounded, so its image must be orderbounded in V , and in particular bounded above. On the other hand, if T satisfies the condition, and A is order-bounded, then A ⊆ [u1 , u2 ] for some u1 ≤ u2 , and T [A] ⊆ T [u1 + [0, u2 − u1 ]] = T u1 + T [[0, u2 − u1 ]] is bounded above; similarly, T [−A] is bounded above, so T [A] is bounded below; as A is arbitrary, T is order-bounded. (b) If T is positive then {T u : 0 ≤ u ≤ w} is bounded above by T w for every w ≥ 0, so T ∈ L∼ . (c) If T1 , T2 ∈ L∼ , α ∈ R and A ⊆ U is order-bounded, then there are v1 , v2 ∈ V such that Ti [A] ⊆ [−vi , vi ] for both i. Setting v = (1 + |α|)v1 + v2 , (αT1 + T2 )[A] ⊆ [−v, v]; as A is arbitrary, αT1 + T2 belongs to L∼ ; as α, T1 , T2 are arbitrary, and since the zero operator surely belongs to L∼ , L∼ is a linear subspace of the space of all linear operators from U to V . (d) This is immediate from the definition; if A ⊆ U is order-bounded, then T [A] ⊆ V and (ST )[A] = S[T [A]] ⊆ W are order-bounded. 355C Theorem If U and V are Banach lattices then every order-bounded linear operator (in particular, every positive linear operator) from U to V is continuous. proof ?? Suppose, if possible, that T : U → V is an order-bounded linear operator which is not continuous. Then for each n ∈ N we can find a un ∈ U such that kun k ≤ 2−n but kT un k ≥ n. Now u = supn∈N |un | is defined in U (354C), and there is a v ∈ V such that −v ≤ T w ≤ v whenever −u ≤ w ≤ u; but this means that kvk ≥ kT un k ≥ n for every n, which is impossible. X X 355D Lemma Let U be a Riesz space and V any linear space over R. Then a function T : U + → V extends to a linear operator from U to V iff T (u + u0 ) = T u + T u0 ,

T (αu) = αT u

for all u, u0 ∈ U + and every α > 0, and in this case the extension is unique. proof For in this case we can, and must, set T1 u = T u1 − T u2 whenever u1 , u2 ∈ U + and u = u1 − u2 ; it is elementary to check that this defines T1 u uniquely for every u ∈ U , and that T1 is a linear operator extending T . 355E Theorem Let U be a Riesz space and V a Dedekind complete Riesz space. (a) The space L∼ of order-bounded linear operators from U to V is a Dedekind complete Riesz space; its positive cone is the set of positive linear operators from U to V . In particular, every order-bounded linear operator from U to V is expressible as the difference of positive linear operators. (b) For T ∈ L∼ , T + and |T | are defined in the Riesz space L∼ by the formulae T + (w) = sup{T u : 0 ≤ u ≤ w}, Pn Pn |T |(w) = sup{T u : |u| ≤ w} = sup{ i=0 |T ui | : i=0 |ui | = w} for every w ∈ U + . (c) If S, T ∈ L∼ then

355E

Spaces of linear operators

(S ∨ T )(w) = sup0≤u≤w Su + T (w − u),

257

(S ∧ T )(w) = inf 0≤u≤w Su + T (w − u)

for every w ∈ U + . (d) Suppose that A ⊆ L∼ is non-empty and upwards-directed. Then A is bounded above in L∼ iff {T u : T ∈ A} is bounded above in V for every u ∈ U + , and in this case (sup A)(u) = supT ∈A T u for every u ≥ 0. (e) Suppose that A ⊆ (L∼ )+ is non-empty and downwards-directed. Then inf A = 0 in L∼ iff inf T ∈A T u = 0 in V for every u ∈ U + . proof (a)(i) Suppose that T ∈ L∼ . For w ∈ U + set RT (w) = sup{T u : 0 ≤ u ≤ w}; this is defined because V is Dedekind complete and {T u : 0 ≤ u ≤ w} is bounded above in V . Then RT (w1 + w2 ) = RT w1 + RT w2 for all w1 , w2 ∈ U + . P P Setting Ai = [0, wi ] for each i, and A = [0, w], then of course A1 + A2 ⊆ A; but also A ⊆ A1 + A2 , because if u ∈ A then u = (u ∧ w1 ) + (u − w1 )+ , and 0 ≤ (u − w1 )+ ≤ (w − w1 )+ = w2 , so u ∈ A1 + A2 . Consequently RT w = sup T [A] = sup T [A1 + A2 ] = sup(T [A1 ] + T [A2 ]) = sup T [A1 ] + sup T [A2 ] = RT w1 + RT w2 by 351Dc. Q Q Next, it is easy to see that RT (αw) = αRT w for w ∈ U + , α > 0, just because u 7→ αu, v 7→ αv are isomorphisms of the partially ordered linear spaces U and V . It follows from 355D that we can extend RT to a linear operator from U to V . Because RT u ≥ T 0 = 0 for every u ∈ U + , RT is a positive linear operator. But also RT u ≥ T u for every u ∈ U + , so RT − T is also positive, and T = RT − (RT − T ) is the difference of two positive linear operators. (ii) This shows that every order-bounded operator is a difference of positive operators. But of course if T1 and T2 are positive, then (T1 − T2 )u ≤ T1 w whenever 0 ≤ u ≤ w in U , so that T1 − T2 is order-bounded, by the criterion in 355Ba. Thus L∼ is precisely the set of differences of positive operators. (iii) Just as in 351F, L∼ is a partially ordered linear space if we say that S ≤ T iff Su ≤ T u for every u ∈ U + . Now it is a Riesz space. P P Take any T ∈ L∼ . Then RT , as defined in (i), is an upper ∼ ∼ bound for {0, T } in L . If S ∈ L is any other upper bound for {0, T }, then for any w ∈ U + we must have Sw ≥ Su ≥ T u whenever u ∈ [0, w], so that Sw ≥ RT w; as w is arbitrary, S ≥ RT ; as S is arbitrary, RT = sup{0, T } in L∼ . Thus sup{0, T } is defined in L∼ for every T ∈ L∼ ; by 352B, L∼ is a Riesz space. Q Q (I defer the proof that it is Dedekind complete to (d-ii) below.) (b) As remarked in (a-iii), RT = T + for each T ∈ L∼ ; but this is just the formula given for T + . Now, if T ∈ L∼ and w ∈ U + , |T |(w) = 2T + w − T w = 2 sup T u − T w u∈[0,w]

= sup T (2u − w) = u∈[0,w]

sup

T u,

u∈[−w,w]

which is the first formula offered for |T |. In particular, if |u| ≤ w then T u,P−T u = T (−u) are both less than n or equal to |T |(w), so that |T u| ≤ |T |(w). So if u0 , . . . , un are such that i=0 |ui | = w, then Pn Pn i=0 |T ui | ≤ i=0 |T |(|ui |) = |T |(w). Pn Pn Thus B = { i=0 |T ui | : i=0 |ui | = w} is bounded above by |T |(w). On the other hand, if v is an upper bound for B and |u| ≤ w, then T u ≤ |T u| + |T (w − |u|)| ≤ v; as u is arbitrary, |T |(w) ≤ v; thus |T |(w) is the least upper bound for B. This completes the proof of part (ii) of the theorem. (c) We know that S ∨ T = T + (S − T )+ (352D), so that (S ∨ T )(w) = T w + (S − T )+ (w) = T w + sup (S − T )(u) 0≤u≤w

= sup T w + (S − T )(u) = sup Su + T (w − u) 0≤u≤w

0≤u≤w

258

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355E

for every w ∈ U + , by the formula in (b). Also from 352D we have S ∧ T = S + T − T ∨ S, so that

(S ∧ T )(w) = Sw + T w − sup T u + S(w − u) 0≤u≤w

=

inf Sw + T w − T u − S(w − u)

0≤u≤w

(351Db) =

inf Su + T (w − u)

0≤u≤w

for w ∈ U + . (d)(i) Now suppose that A ⊆ L∼ is non-empty and upwards-directed and that {T u : T ∈ A} is bounded above in V for every u ∈ U + . In this case, because V is Dedekind complete, we may set Ru = supT ∈A T u for every u ∈ U + . Now R(u1 + u2 ) = Ru1 + Ru2 for all u1 , u2 ∈ U + . P P Set Bi = {T ui : T ∈ A} for each i, B = {T (u1 + u2 ) : T ∈ A}. Then B ⊆ B1 + B2 , so R(u1 + u2 ) = sup B ≤ sup(B1 + B2 ) = sup B1 + sup B2 = Ru1 + Ru2 . On the other hand, if vi ∈ Bi for both i, there are Ti ∈ A such that vi = Ti ui for each i; because A is upwards-directed, there is a T ∈ A such that T ≥ Ti for both i, and now R(u1 + u2 ) ≥ T (u1 + u2 ) = T u1 + T u2 ≥ T1 u1 + t2 u2 = v1 + v2 . As v1 , v2 are arbitrary, R(u1 + u2 ) ≥ sup(B1 + B2 ) = sup B1 + sup B2 = Ru1 + Ru2 . Q Q It is also easy to see that R(αu) = αRu for every u ∈ U + , α > 0. So, using 355D again, R has an extension to a linear operator from U to V . Now if we fix any T0 ∈ A, we have T0 u ≤ Ru for every u ∈ U + , so R − T0 is a positive linear operator, and R = (R − T0 ) + T0 belongs to L∼ . Again, T u ≤ Ru for every T ∈ A, u ∈ U + , so R is an upper bound for A in L∼ ; and, finally, if S is any upper bound for A in L∼ , then Su is an upper bound for {T u : T ∈ A}, and must be greater than or equal to Ru, for every u ∈ U + ; so that R ≤ S and R = sup A in L∼ . (ii) Now L∼ is Dedekind complete. P P If A ⊆ L∼ is non-empty and bounded above by S say, then A = {T0 ∨ T1 ∨ . . . ∨ Tn : T0 , . . . , Tn ∈ A} is upwards-directed and bounded above by S, so {T u : T ∈ A0 } is bounded above by Su for every u ∈ U + ; by (i) just above, A0 has a supremum in L∼ , which will also be a supremum for A. Q Q 0

(e) Suppose that A ⊆ (L∼ )+ is non-empty and downwards-directed. Then −A = {−T : T ∈ A} is non-empty and upwards-directed, so inf A = 0 ⇐⇒ sup(−A) = 0 ⇐⇒ sup (−T w) = 0 for every w ∈ U + T ∈A

⇐⇒ inf T w = 0 for every w ∈ U + . T ∈A

355F Theorem Let U and V be Riesz spaces and U0 ⊆ U a Riesz subspace which is either order-dense or a solid linear subspace. Suppose that T0 : U0 → V is an order-continuous positive linear operator such that Su = sup{T0 w : w ∈ U0 , 0 ≤ w ≤ u} is defined in V for every u ∈ U + . Then (i) T0 has an extension to an order-continuous positive linear operator T : U → V . (ii) If T0 is a Riesz homomorphism so is T . (iii) If U0 is order-dense then T is unique. (iv) If U0 is order-dense and T0 is an injective Riesz homomorphism, then so is T . proof (a) I check first that S(u + u0 ) = Su + Su0 ,

S(γu) = γSu

355F

Spaces of linear operators

259

for all u, u0 ∈ U + and γ > 0. P P For every u ∈ U set A(u) = {v : v ∈ U0 , 0 ≤ v ≤ u}. Note that for any u ∈ U + , v ∈ U0+ v ∧ u = supw∈A(u) v ∧ w = sup A(v ∧ u); this is because sup A(u) = u if U0 is order-dense in U , while v ∧ u ∈ A(u) if U0 is solid. (i) If v ∈ A(u + u0 ) then sup(A(v ∧ u) + A(v − u)+ ) = sup A(v ∧ u) + sup A(v − u)+ = v ∧ u + (v − u)+ = v. Now T0 v = sup T0 [A(v ∧ u) + A(v − u)+ ] = sup(T0 [A(v ∧ u)] + T0 [A(v − u)+ ]) ≤ Su + Su0 because (v − u)+ ≤ u0 . As v is arbitrary, S(u + u0 ) ≤ Su + Su0 . (ii) Next, Su + Su0 = sup T0 [A(u)] + sup T0 [A(u0 )] = sup T0 [A(u) + A(u0 )] ≤ sup T0 [A(u + u0 )] = S(u + u0 ) because A(u) + A(u0 ) ⊆ A(u + u0 ), so S(u + u0 ) = Su + Su0 . (iii) Of course A(γu) = γA(u) so S(γu) = γSu. Q Q By 355D, S has an extension to a linear operator from U to V ; call this operator T . Of course T u = Su ≥ 0 whenever u ≥ 0, so T is positive. If u ∈ U0+ then T u = Su = T0 u, so T extends T0 . (b) If B ⊆ U + is non-empty and upwards-directed and has a supremum u0 ∈ U , then of course T u ≤ T u0 for every u ∈ B, so sup T [B] ≤ T u0 . On the other hand, for any v ∈ A(u0 ) we have also

v = supu∈B v ∧ u = supu∈B supw∈A(u) v ∧ w;

S u∈B

A(u) is upwards-directed, so T0 v = sup{T0 (v ∧ w) : w ∈

S u∈B

A(u)} ≤ supu∈B T u.

As v is arbitrary, T u0 = Su0 ≤ supu∈B T u. As B is arbitrary, T is order-continuous (351Ga). (c) Now suppose that T0 is a Riesz homomorphism. If u ∈ U then, in the language of (a) above,

T u+ ∧ T u− = sup T0 [A(u+ )] ∧ T0 [A(u− )] = sup{w ∧ w0 : w ∈ T0 [A(u+ )], w0 ∈ T0 [A(u− )] (352Ea) = sup{T0 (v ∧ v 0 ) : v ∈ A(u+ ), v 0 ∈ A(u− )} (because T0 is a Riesz homomorphism) = 0. So T is a Riesz homomorphism (352G(iv)). (d) If U0 is order-dense, any order-continuous linear operator extending T0 agrees with S and T on U + , so is equal to T . (e) Finally, if U0 is order-dense and T0 is an injective Riesz homomorphism, then for any non-zero u ∈ U there is a non-zero v ∈ U0 such that |v| ≤ |u|; so that |T u| = T |u| ≥ T0 |v| > 0 because T is a Riesz homomorphism, by (c). As u is arbitrary, T is injective.

260

Riesz spaces

355G

355G Definition Let U be a Riesz space and V a Dedekind complete Riesz space. Then L× (U ; V ) will be the set of those T ∈ L∼ (U ; V ) expressible as the difference of order-continuous positive linear operators, and ∼ L∼ c (U ; V ) will be the set of those T ∈ L (U ; V ) expressible as the difference of sequentially order-continuous positive linear operators. Because a composition of (sequentially) order-continuous functions is (sequentially) order-continuous, we shall have ST ∈ L× (U ; W ) whenever S ∈ L× (V ; W ), T ∈ L× (U ; V ), ∼ ∼ ST ∈ L∼ c (U ; W ) whenever S ∈ Lc (V ; W ), T ∈ Lc (U ; V ),

for all Riesz spaces U and all Dedekind complete Riesz spaces V , W . 355H Theorem Let U be a Riesz space and V a Dedekind complete Riesz space. Then (i) L× = L× (U ; V ) is a band in L∼ = L∼ (U ; V ), therefore a Dedekind complete Riesz space in its own right (ii) a member T of L∼ belongs to L× iff |T | is order-continuous. proof There is a fair bit to check, but each individual step is easy enough. (a) Suppose that S, T are order-continuous positive linear operators from U to V . Then S + T is ordercontinuous. P P If A ⊆ U is non-empty, downwards-directed and has infimum 0, then for any u1 , u2 ∈ A there is a u ∈ A such that u ≤ u1 , u ≤ u2 , and now (S + T )(u) ≤ Su1 + T u2 . Consequently any lower bound for (S + T )[A] must also be a lower bound for S[A] + T [A]. But since inf(S[A] + T [A]) = inf S[A] + inf T [A] = 0 (351Dc), inf(S + T )[A] must also be 0; as A is arbitrary, S + T is order-continuous, by 351Ga. Q Q (b) Consequently S + T ∈ L× for all S, T ∈ L× . Since −T and αT belong to L× for every T ∈ L× and α ≥ 0, we see that L× is a linear subspace of L∼ . (c) If T : U → V is an order-continuous linear operator, S : U → V is linear and 0 ≤ S ≤ T , then S is order-continuous. P P If A ⊆ U is non-empty, downwards-directed and has infimum 0, then any lower bound of S[A] must also be a lower bound of T [A], so inf S[A] = 0; as A is arbitrary, S is order-continuous. Q Q It follows that L× is a solid linear subspace of L∼ . P P If T ∈ L× and |S| ≤ |T | in L∼ , express T as T1 − T2 where T1 , T2 are order-continuous positive linear operators. Then S + , S − ≤ |S| ≤ |T | ≤ T1 + T2 , so S + and S − are order-continuous and S = S + − S − ∈ L× . Q Q Accordingly L× is a Dedekind complete Riesz space in its own right (353J(b-i)). (d) The argument of (c) also shows that if T ∈ L× then |T | is order-continuous; so that for T ∈ L∼ , T ∈ L× ⇐⇒ |T | ∈ L× ⇐⇒ |T | is order-continuous. (e) If C ⊆ (L× )+ is non-empty, upwards-directed and has a supremum T ∈ L∼ , then T is ordercontinuous, so belongs to L× . P P Suppose that A ⊆ U + is non-empty, upwards-directed and has supremum w. Then T w = supS∈C Sw = supS∈C supu∈A Su = supu∈A T u, putting 355Ed and 351G(a-iii) together. So (using 351Ga again) T is order-continuous. Q Q Consequently L× is a band in L∼ (352Ob). This completes the proof. 355I Theorem Let U be a Riesz space and V a Dedekind complete Riesz space. Then L∼ c (U ; V ) is a band in L∼ (U ; V ), and a member T of L∼ (U ; V ) belongs to L∼ c (U ; V ) iff |T | is sequentially order-continuous. proof Copy the arguments of 355H.

355Xe

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355J Proposition Let U be a Riesz space and V a Dedekind complete Riesz space. Let U0 ⊆ U be an order-dense Riesz subspace; then T 7→ T ¹U0 is an embedding of L× (U ; V ) as a solid linear subspace of L× (U0 ; V ). In particular, any T0 ∈ L× (U0 ; V ) has at most one extension in L× (U ; V ). proof (a) Because the embedding U0 ⊆ U is positive and order-continuous (352Nb), T ¹U0 is positive and order-continuous whenever T is; so T ¹U0 ∈ L× (U0 ; V ) whenever T ∈ L× (U ; V ). Because the map T 7→ T ¹U0 is linear, the image W of L× (U ; V ) is a linear subspace of L× (U0 ; V ). (b) If T ∈ L× (U ; V ) and T ¹U0 ≥ 0, then T ≥ 0. P P?? Suppose, if possible, that there is a u ∈ U + such × that T u 6≥ 0. Because |T | ∈ L (U ; V ) is order-continuous and A = {v : v ∈ U0 , v ≤ u} is an upwardsdirected set with supremum u, inf{|T |(u−v) : v ∈ A} = 0 and there is a v ∈ A such that T u+|T |(u−v) 6≥ 0. But T v = T u + T (v − u) ≤ T u + |T |(u − v) so T v 6≥ 0 and T ¹U0 6≥ 0. X XQ Q This shows that the map T 7→ T ¹U0 is an order-isomorphism between L× (U ; V ) and W , and in particular is injective. (c) Now suppose that S0 ∈ W and that |S| ≤ |S0 | in L× (U0 ; V ). Then S ∈ W . P P Take T0 ∈ L× (U ; V ) such that T0 ¹U0 = S0 . Then S1 = |T0 |¹U0 is a positive member of W such that S0 ≤ S1 , −S0 ≤ S1 , so S + ≤ S1 . Consequently, for any u ∈ U + , sup{S + v : v ∈ U0 , 0 ≤ v ≤ u} ≤ sup{S1 v : v ∈ U0 , 0 ≤ v ≤ u} ≤ |T0 |(u) is defined in V (recall that we are assuming that V is Dedekind complete). But this means that S + has an extension to an order-continuous positive linear operator from U to V (355F), and belongs to W . Similarly, S − ∈ W , so S ∈ W . Q Q This shows that W is a solid linear subspace of L× (U0 ; V ), as claimed. 355K Proposition Let U be a Banach lattice with an order-continuous norm. (a) If V is any Archimedean Riesz space and T : U → V is a positive linear operator, then T is ordercontinuous. (b) If V is a Dedekind complete Riesz space then L× (U ; V ) = L∼ (U ; V ). proof (a) Suppose that A ⊆ U + is non-empty and downwards-directed and has infimum 0. Then for each n ∈ N there is a un ∈ A such that kun k ≤ 4−n . By 354C, u = supn∈N 2n un is defined in U . Now T un ≤ 2−n T u for every n, so any lower bound for T [A] must also be a lower bound for {2−n T u : n ∈ N} and therefore (because V is Archimedean) less than or equal to 0. Thus inf T [A] = 0; as A is arbitrary, T is order-continuous. (b) This is now immediate from 355Ea and the definition of L× . 355X Basic exercises >(a) Let U and V be arbitrary Riesz spaces. (i) Show that the set L(U ; V ) of all linear operators from U to V is a partially ordered linear space if we say that S ≤ T whenever Su ≤ T u for every u ∈ U + . (ii) Show that if U and V are Banach lattices then the set of positive operators is closed in the normed space B(U ; V ) of bounded linear operators from U to V . > (b) If U is a Riesz space and k k, k k0 are two norms on U both rendering it a Banach lattice, show that they are equivalent, that is, give rise to the same topology. (c) Let U be a Riesz space with a Riesz norm, V an Archimedean Riesz space with an order unit, and T : U → V a linear operator which is continuous for the given norm on U and the order-unit norm on V . Show that T is order-bounded. (d) Let U be a Riesz space, V an Archimedean Riesz space, and T : U + → V + a map such that T (u1 + u2 ) = T u1 + T u2 for all u1 , u2 ∈ U + . Show that T has an extension to a linear operator from U to V. >(e) Show that if r, s ≥ 1 are integers then the Riesz space L∼ (Rr ; Rs ) can be identified with the space of real s × r matrices, saying that a matrix is positive iff every coefficient is positive, so that if T = hτij i1≤i≤s,1≤j≤r then |T |, taken in L∼ (Rr ; Rs ), is h|τij |i1≤i≤s,1≤j≤r . Show that a matrix represents a Riesz homomorphism iff each row has at most one non-zero coefficient.

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355Xf

> (f ) Let U be a Riesz space and V a Dedekind complete Riesz space. Show that if T0 , . . . , Tn ∈ L∼ (U ; V ) then Pn Pn (T0 ∨ . . . ∨ Tn )(w) = sup{ i=0 Ti ui : ui ≥ 0 ∀ i ≤ n, i=0 ui = w} for every w ∈ U + . ∼ > (g) Let U be a Riesz space, V a Dedekind complete Riesz Pnspace, and A ⊆ L (U ; V ) a non-empty ∼ set. P Show that A is bounded above in L (U ; V ) iff Cw = { i=0 Ti ui : T0 , . . . , Tn ∈ A, u0 , . . . , un ∈ n U + , i=0 ui = w} is bounded above in V for every w ∈ U + , and in this case (sup A)(w) = sup Cw for every + w∈U .

355Y Further exercises (a) Let U and V be Banach lattices. For T ∈ L∼ = L∼ (U ; V ), set kT k∼ = supw∈U + ,kwk≤1 inf{kvk : |T u| ≤ v whenever |u| ≤ w}. Show that k k∼ is a norm on L∼ under which L∼ is a Banach space, and that the set of positive linear operators is closed in L∼ . (b) Give an example of a continuous linear operator from `2 to itself which is not order-bounded. + (c) Let PnU and V be Riesz spaces+ and Pn T : U → V a linear operator. (i) Show that for any w ∈ U , Cw = { i=0 |T ui | : u0 , . . . , un ∈ U , i=0 ui = w} is upwards-directed, and has the same upper bounds as {TP u : |u| ≤ P w}. (Hint: induce on m and n to see that if u0 , . . . , P un , u00 , . . . , u0m ∈ U + are such m n m + 0 that i=0 ui = j=0 uj , there is a family huij ii≤n,j≤m in U such that j=0 uij = ui for every i ≤ n, Pn + 0 i=0 uij = uj for every j ≤ m.) (ii) Show that if sup Cw is defined for every w ∈ U , then S = T ∨ (−T ) ∼ is defined in the partially ordered linear space L (U ; V ) and Sw = sup Cw for every w ∈ U + .

(d) Let U , V and W be Riesz spaces, of which V and W are Dedekind complete. (i) Show that for any S ∈ L× (V ; W ), the map T 7→ ST : L∼ (U ; V ) → L∼ (U ; W ) belongs to L× (L∼ (U ; V ); L∼ (U ; W )), and is a Riesz homomorphism if S is. (Hint: 355Yc.) (ii) Show that for any T ∈ L∼ (U ; V ), the map S 7→ ST : L∼ (V ; W ) → L∼ (U ; W ) belongs to L× (L∼ (V ; W ); L∼ (U ; W )). (e) Let µ be the usual measure on {0, 1}N and c the Banach lattice of convergent sequences. Find a linear operator T : L2 (µ) → c which is norm-continuous, therefore order-bounded, such that 0 and T have no common upper bound in the partially ordered linear space of all linear operators from L2 (µ) to c . (f ) Let U and V be Banach lattices. Let Lreg be the linear space of operators from U to V expressible as the difference of positive operators. For T ∈ Lreg let kT kreg be inf{kT1 + T2 k : T1 , T2 : U → V are positive, T = T1 − T2 }. Show that k kreg is a norm under which Lreg is complete. (g) Let U and V be Riesz spaces. For this exercise only, say that L× (U ; V ) is to be the set of linear operators T : U → V such that whenever A ⊆ U is non-empty, downwards-directed and has infimum 0 then {v : v ∈ V + , ∃ w ∈ A, |T u| ≤ v whenever |u| ≤ w} has infimum 0 in V . (i) Show that L× (U ; V ) is a linear space. (ii) Show that if U is Archimedean then L× (U ; V ) ⊆ L∼ (U ; V ). (iii) Show that if U is Archimedean and V is Dedekind complete then this definition agrees with that of 355G. (iv) Show that for any Riesz spaces U , V and W , ST ∈ L× (U ; W ) for every S ∈ L× (V ; W ), T ∈ L× (U ; V ). (v) Show that if U and V are Banach lattices, then L× (U ; V ) is closed in L∼ (U ; V ) for the norm k k∼ of 355Ya. (vi) Show that if V is Archimedean and U is a Banach lattice with an order-continuous norm, then L× (U ; V ) = L∼ (U ; V ). (h) Let U be a Riesz space and V a Dedekind complete Riesz space. Show that the band projection P : L∼ (U ; V ) → L× (U ; V ) is given by the formula (P T )(w) = inf{sup T u : A ⊆ U + is non-empty, upwards-directed u∈A

and has supremum w} for every w ∈ U + , T ∈ (L∼ )+ . (Cf. 362Bd.)

355 Notes

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263

× (i) Show that if U is a Riesz space with the countable sup property (241Yd), then L∼ c (U ; V ) = L (U ; V ) for every Dedekind complete Riesz space V .

(j) Let U and V be Riesz spaces, of which V is Dedekind complete, and U0 a solid linear subspace of U . Show that the map T 7→ T ¹U0 is an order-continuous Riesz homomorphism from L× (U ; V ) onto a solid linear subspace of L× (U0 ; V ). (k) Let U be a uniformly complete Riesz space (354Yi) and V a Dedekind complete Riesz space. Let UC , VC be their complexifications (354Yk). Show that the complexification of L∼ (U ; V ) can be identified with the complex linear space of linear operators T : UC → VC such that BT (w) = {|T u| : |u| ≤ w} is bounded above in V for every w ∈ U + , and that now |T |(w) = sup BT (w) for every T ∈ L∼ (U ; V )C , w ∈ U + . (Hint: if u, v ∈ U and |u + iv| = w, then u and v can be simultaneously Pn approximated Pn for the order-unit norm k kw on the solid linear subspace generated by w by finite sums j=0 (cos θj )wj , j=0 (sin θj )wj where wj ∈ U + , Pn ∼ j=0 wj = w. Consequently |T (u + iv)| ≤ |T |(w) for every T ∈ LC .) 355 Notes and comments I have had to make some choices in the basic definitions of this chapter (355A, 355G). For Dedekind complete codomains V , there is no doubt what L∼ (U ; V ) should be, since the order-bounded operators (in the sense of 355A) are just the differences of positive operators (335Ea). (These are sometimes called ‘regular’ operators.) When V is not Dedekind complete, we have to choose between the two notions, as not every order-bounded operator need be regular (355Ye). In my previous book (Fremlin 74a) I chose the regular operators; I have still not encountered any really persuasive reason to settle definitively on either class. In 355G the technical complications in dealing with any natural equivalent of the larger space (see 355Yg) are such that I have settled for the narrower class, but explicitly restricting the definition to the case in which V is Dedekind complete. In the applications in this book, the codomains are nearly always Dedekind complete, so we can pass these questions by. The elementary extension technique in 355D may recall the definition of the Lebesgue integral (122L122M). In the same way, 351G may remind you of the theorem that a linear operator between normed spaces is continuous everywhere if it is continuous anywhere, or of the corresponding results about Boolean homomorphisms and additive functionals on Boolean algebras (313L, 326Ga, 326N). Of course 355Ea is the central fact about the space L∼ (U ; V ) for Dedekind complete V ; because we get a new Riesz space from old ones, the prospect of indefinite recursion immediately presents itself. For Banach lattices, L∼ (U ; V ) is a linear subspace of the space B(U ; V ) of bounded linear operators (355C); the question of when the two are equal will be of great importance to us. I give only the vaguest hints on how to show that they can be different (355Yb, 355Ye), but these should be enough to make it plain that equality is the exception rather than the rule. It is also very useful that we have effective formulae to describe the Riesz space operations on L∼ (U ; V ) (355E, 355Xf-355Xg, 355Yc). You may wish to compare these with the corresponding formulae for additive functionals on Boolean algebras in 326Yj and 362B. If we think of L∼ as somehow corresponding to the space of bounded additive functionals on a Boolean × correspond to the spaces of countably additive and completely additive algebra, the bands L∼ c and L functionals. In fact (as will appear in §362) this correspondence is very close indeed. For the moment, all × I have sought to establish is that L∼ are indeed bands. Of course any case in which L∼ (U ; V ) = c and L ∼ × L∼ (U ; V ) or L (U ; V ) = L (U ; V ) is of interest (355Kb, 355Yi). c c Between Banach lattices, positive linear operators are continuous (355C); it follows at once that the Riesz space structure determines the topology (355Xb), so that it is not to be wondered at that there are further connexions between the norm and the spaces L∼ and L× , as in 355K. 355F will be a basic tool in the theory of representations of Riesz spaces; if we can represent an orderdense Riesz subspace of U as a subspace of a Dedekind complete space V , we have at least some chance of expressing U also as a subspace of V . Of course it has other applications, starting with analysis of the dual spaces.

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§356 intro.

356 Dual spaces As always in functional analysis, large parts of the theory of Riesz spaces are based on the study of linear functionals. Following the scheme of the last section, I define spaces U ∼ , Uc∼ and U × , the ‘order-bounded’, ‘sequentially order-continuous’ and ‘order-continuous’ duals of a Riesz space U (356A). These are Dedekind complete Riesz spaces (356B). If U carries a Riesz norm they are closely connected with the normed space dual U ∗ , which is itself a Banach lattice (356D). For each of them, we have a canonical Riesz homomorphism from U to the corresponding bidual. The map from U to U ×× is particularly important (356I); when this map is an isomorphism we call U ‘perfect’ (356J). The last third of the section deals with L- and M -spaces and the duality between them (356N, 356P), with two important theorems on uniform integrability (356O, 356Q). 356A Definition Let U be a Riesz space. (a) I write U ∼ for the space L∼ (U ; R) of order-bounded real-valued linear functionals on U , the orderbounded dual of U . (b) Uc∼ will be the space L∼ c (U ; R) of differences of sequentially order-continuous positive real-valued linear functionals on U , the sequentially order-continuous dual of U . (c) U × will be the space L× (U ; R) of differences of order-continuous positive real-valued linear functionals on U , the order-continuous dual of U . Remark It is easy to check that the three spaces U ∼ , Uc∼ and U × are in general different (356Xa-356Xc). But the examples there leave open the question: can we find a Riesz space U , for which Uc∼ 6= U × , and which is actually Dedekind complete, rather than just Dedekind σ-complete, as in 356Xc? This leads to unexpectedly deep water; it is yet another form of the Banach-Ulam problem. Really this is a question for Volume 5, but in 363S below I collect the relevant ideas which are within the scope of the present volume. 356B Theorem For any Riesz space U , its order-bounded dual U ∼ is a Dedekind complete Riesz space in which Uc∼ and U × are bands, therefore Dedekind complete Riesz spaces in their own right. For f ∈ U ∼ , f + and |f | ∈ U ∼ are defined by the formulae f + (w) = sup{f (u) : 0 ≤ u ≤ w},

|f |(w) = sup{f (u) : |u| ≤ w}

+

for every w ∈ U . A non-empty upwards-directed set A ⊆ U ∼ is bounded above iff supf ∈A f (u) is finite for every u ∈ U , and in this case (sup A)(u) = supf ∈A f (u) for every u ∈ U + . proof 355E, 355H, 355I. 356C Proposition Let U be any Riesz space and P a band projection on U . Then its adjoint P 0 : U ∼ → U ∼ , defined by setting P 0 (f ) = f P for every f ∈ U ∼ , is a band projection on U ∼ . proof (a) Because P : U → U is a positive linear operator, P 0 f ∈ U ∼ for every f ∈ U ∼ (355Bd), and P 0 is a positive linear operator from U ∼ to itself. Set Q = I − P , the complementary band projection on U ; then Q0 is another positive linear operator on U ∼ , and P 0 f + Q0 f = f for every f . Now P 0 f ∧ Q0 f = 0 for every f ≥ 0. P P For any w ∈ U + , (P 0 f − Q0 f )+ (w) = sup (P 0 f − Q0 f )(u) = sup f (P u − Qu) 0≤u≤w

0≤u≤w 0

= f (P w) = (P f )(w), so (P 0 f − Q0 f )+ = P 0 f , that is, P 0 f ∧ Q0 f = 0. Q Q By 352Rd, P 0 is a band projection. 356D Proposition Let U be a Riesz space with a Riesz norm. (a) The normed space dual U ∗ of U is a solid linear subspace of U ∼ , and in itself is a Banach lattice with a Fatou norm and has the Levi property.

356E

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265

(b) The norm of U is order-continuous iff U ∗ ⊆ U × . (c) If U is a Banach lattice, then U ∗ = U ∼ , so that U ∼ , U × and Uc∼ are all Banach lattices. (d) If U is a Banach lattice with order-continuous norm then U ∗ = U × = U ∼ . proof (a)(i) If f ∈ U ∗ then sup|u|≤w f (u) ≤ sup|u|≤w kf kkuk = kf kkwk < ∞ for every w ∈ U + , so f ∈ U ∼ (355Ba). Thus U ∗ ⊆ U ∼ . (ii) If f , g ∈ U ∗ and |f | ≤ |g|, then for any w ∈ U |f (w)| ≤ |f |(|w|) ≤ |g|(|w|) = sup|u|≤|w| g(u) ≤ sup|u|≤|w| kgkkuk ≤ kgkkwk. As w is arbitrary, f ∈ U ∗ and kf k ≤ kgk; as f and g are arbitrary, U ∗ is a solid linear subspace of U ∼ and the norm of U ∗ is a Riesz norm. Because U ∗ is a Banach space it is also a Banach lattice. (iii) If A ⊆ (U ∗ )+ is non-empty, upwards-directed and M = supf ∈A kf k is finite, then supf ∈A f (u) ≤ M kuk is finite for every u ∈ U + , so g = sup A is defined in U ∼ (355Ed). Now g(u) = supf ∈A f (u) for every u ∈ U + , as also noted in 355Ed, so |g(u)| ≤ g(|u|) ≤ M k|u|k = M kuk for every u ∈ U , and kgk ≤ M . But as A is arbitrary, this simultaneously proves that the norm of U ∼ is Fatou and has the Levi property. (b)(i) Suppose that the norm is order-continuous. If f ∈ U ∗ and A ⊆ U is a non-empty downwardsdirected set with infimum 0, then inf u∈A |f |(u) ≤ inf u∈A kf kkuk = 0, so |f | ∈ U

×

×

∗

and f ∈ U . Thus U ⊆ U × .

(ii) Now suppose that the norm is not order-continuous. Then there is a non-empty downwards-directed set A ⊆ U , with infimum 0, such that inf u∈A kuk = δ > 0. Set B = {v : v ≥ u for some u ∈ A}. Then B is convex. P P If v1 , v2 ∈ B and α ∈ [0, 1], there are u1 , u2 ∈ A such that vi ≥ ui for both i; now there is a u ∈ A such that u ≤ u1 ∧ u2 , so that u = αu + (1 − α)u ≤ αv1 + (1 − α)v2 , and αv1 + (1 − α)v2 ∈ B. Q Q Also inf v∈B kvk = δ > 0. By the Hahn-Banach theorem (3A5Cb), there is an f ∈ U ∗ such that inf v∈B f (v) > 0. But now inf u∈A |f |(u) ≥ inf u∈A f (u) > 0 and |f | is not order-continuous; so U ∗ 6⊆ U × . (c) By 355C, U ∼ ⊆ U ∗ , so U ∼ = U ∗ . Now U × and Uc∼ , being bands, are closed linear subspaces (354Bd), so are Banach lattices in their own right. (d) Put (b) and (c) together. 356E Biduals If you have studied any functional analysis at all, it will come as no surprise that duals-of-duals are important in the theory of Riesz spaces. I start with a simple lemma. Lemma Let U be a Riesz space and f : U → R a positive linear functional. Then for any u ∈ U + there is a positive linear functional g : U → R such that 0 ≤ g ≤ f , g(u) = f (u) and g(v) = 0 whenever u ∧ v = 0. proof Set g(v) = supα≥0 f (v ∧ αu) for every v ∈ U + . Then it is easy to see that g(βv) = βg(v) for every v ∈ U + , β ∈ [0, ∞[. If v, w ∈ U + then (v ∧ αu) + (w ∧ αu) ≤ (v + w) ∧ 2αu ≤ (v ∧ 2αu) + (w ∧ 2αu) for every α ≥ 0 (352Fa), so g(v + w) = g(v) + g(w). Accordingly g has an extension to a linear functional from U to R (355D). Of course 0 ≤ g(v) ≤ f (v) for v ≥ 0, so 0 ≤ g ≤ f in U ∼ . We have g(u) = f (u), while if u ∧ v = 0 then αu ∧ v = 0 for every α ≥ 0, so g(v) = 0.

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Riesz spaces

356F

356F Theorem Let U be a Riesz space and V a solid linear subspace of U ∼ . For u ∈ U define u ˆ:V →R by setting u ˆ(f ) = f (u) for every f ∈ V . Then u 7→ u ˆ is a Riesz homomorphism from U to V × . proof (a) By the definition of addition and scalar multiplication in V , u ˆ is linear for every u; also αu c = αˆ u, (u1 + u2 )b= u ˆ1 + u ˆ2 for all u, u1 , u2 ∈ U and α ∈ R. If u ≥ 0 then u ˆ(f ) = f (u) ≥ 0 for every f ∈ V + , so u ˆ ≥ 0; accordingly every u ˆ is the difference of two positive functionals, and u 7→ u ˆ is a linear operator from U to V ∼ . (b) If B ⊆ V is a non-empty downwards-directed set with infimum 0, then inf f ∈B f (u) = 0 for every u ∈ U + , by 355Ee. But this means that u ˆ is order-continuous for every u ∈ U + , so that u ˆ ∈ V × for every u ∈ U. (c) If u ∧ v = 0 in U , then for any f ∈ V + there is a g ∈ [0, f ] such that g(u) = f (u), g(v) = 0 (356D). So (ˆ u ∧ vˆ)(f ) ≤ u ˆ(f − g) + vˆ(g) = f (u) − g(u) + g(v) = 0. As f is arbitrary, u ˆ ∧ vˆ = 0. As u and v are arbitrary, u 7→ u ˆ is a Riesz homomorphism (352G). 356G Lemma Suppose that U is a Riesz space such that U ∼ separates the points of U . Then U is Archimedean. proof ?? Otherwise, there are u, v ∈ U such that v > 0 and nv ≤ u for every n ∈ N. Now there is an f ∈ U ∼ such that f (v) 6= 0; but f (v) ≤ |f |(v) ≤ n1 |f |(u) for every n, so this is impossible. X X 356H Lemma Let U be an Archimedean Riesz space and f > 0 in U × . Then there is a u ∈ U such that (i) u > 0 (ii) f (v) > 0 whenever 0 < v ≤ u (iii) g(u) = 0 whenever g ∧ f = 0 in U × . Moreover, if u0 ∈ U + is such that f (u0 ) > 0, we can arrange that u ≤ u0 . proof (Because f > 0 there certainly is some u0 ∈ U such that f (u0 ) > 0.) (a) Set A = {v : 0 ≤ v ≤ u0 , f (v) = 0}. Then (v1 + v2 ) ∧ u0 ∈ A for all v1 , v2 ∈ A, so A is upwardsdirected. Because f (u0 ) > 0 = sup f [A] and f is order-continuous, u0 cannot be the least upper bound of A, and there is another upper bound u1 of A strictly less than u0 . Set u = u0 − u1 > 0. If 0 ≤ v ≤ u and f (v) = 0, then w ∈ A =⇒ w ≤ u1 =⇒ w + v ≤ u0 =⇒ w + v ∈ A; consequently nv ∈ A and nv ≤ u0 for every n ∈ N, so v = 0. Thus u has properties (i) and (ii). (b) Now suppose that g ∧ f = 0 in U × . Let ² > 0. Then for each n ∈ N there is a vn ∈ [0, u] such that f (vn ) + g(u − vn ) ≤ 2−n ² (355Ec). If v ≤ vn for every n ∈ N then f (v) = 0 so v = 0; thus inf n∈N vn = 0. Set wn = inf i≤n vi for each n ∈ N; then hwn in∈N is non-increasing and has infimum 0 so (because g is order-continuous) inf n∈N g(wn ) = 0. But Pn u − wn = supi≤n u − vi ≤ i=0 u − vi , so g(u − wn ) ≤

Pn i=0

g(u − vi ) ≤ 2²

for every n, and g(u) ≤ 2² + inf n∈N g(wn ) = 2². As ² is arbitrary, g(u) = 0; as g is arbitrary, u has the third required property. 356I Theorem Let U be any Archimedean Riesz space. Then the canonical map from U to U ×× (356F) is an order-continuous Riesz homomorphism from U onto an order-dense Riesz subspace of U ×× . If U is Dedekind complete, its image in U ×× is solid. proof (a) By 356F, u 7→ u ˆ : U → U ×× is a Riesz homomorphism. To see that it is order-continuous, take any non-empty downwards-directed set A ⊆ U with infimum 0. Then C = {ˆ u : u ∈ A} is downwards-directed, and for any f ∈ (U × )+

356K

Dual spaces

267

inf φ∈C φ(f ) = inf u∈A f (u) = 0 because f is order-continuous. As f is arbitrary, inf C = 0 (355Ee); as A is arbitrary, u 7→ u ˆ is ordercontinuous (351Ga). (b) Now suppose that φ > 0 in U ×× . By 356H, there is an f > 0 in U × such that φ(f ) > 0 and φ(g) = 0 whenever g ∧ f = 0. Next, there is a u > 0 in U such that f (u) > 0. Since U ≥ 0, u ˆ ≥ 0; since u ˆ(f ) > 0, u ˆ ∧ φ > 0. Because U ×× (being Dedekind complete) is Archimedean, inf α>0 αˆ u = 0, and there is an α > 0 such that ψ = (ˆ u ∧ φ − αˆ u)+ > 0. Let g ∈ (U × )+ be such that ψ(g) > 0 and θ(g) = 0 whenever θ ∧ ψ = 0 in U ×× . Let v ∈ U + be such that g(v) > 0 and h(v) = 0 whenever h ∧ g = 0 in U × . Because vˆ(g) = g(v) > 0, vˆ ∧ ψ > 0. As ψ ≤ u ˆ, vˆ ∧ u ˆ > 0 and vˆ ∧ αˆ u > 0. Set w = v ∧ αu; then w ˆ = vˆ ∧ αˆ u, by 356F, so w ˆ > 0. ?? Suppose, if possible, that w ˆ 6≤ φ. Then θ = (w ˆ − φ)+ > 0, so there is an h ∈ (U × )+ such that θ(h) > 0 0 0 and θ(h ) > 0 whenever 0 < h ≤ h (356H, for the fourth and last time). Now examine θ(h ∧ g) ≤ (αˆ u−φ∧u ˆ)+ (g) (because w ˆ ≤ αˆ u, φ ∧ u ˆ ≤ φ, h ∧ g ≤ g) =0 because (αˆ u−φ∧u ˆ)+ ∧ ψ = 0. So h ∧ g = 0 and h(v) = 0. But this means that θ(h) ≤ w(h) ˆ ≤ vˆ(h) = 0, which is impossible. X X ˆ of U is quasi-order-dense in U ×× , therefore order-dense Thus 0 < w ˆ ≤ φ. As φ is arbitrary, the image U (353A). ˆ . Express ψ as u (c) Now suppose that U is Dedekind complete and that 0 ≤ φ ≤ ψ ∈ U ˆ where u ∈ U , + and set A = {v : v ∈ U, v ≤ u , vˆ ≤ φ}. If v ∈ U and 0 ≤ vˆ ≤ φ, then w = v + ∧ u+ ∈ A and w ˆ = vˆ; thus ˆ . As φ and ψ are arbitrary, U ˆ is solid in U ×× . φ = sup{ˆ v : v ∈ A} = vˆ0 , where v0 = sup A. So φ ∈ U 356J Definition A Riesz space U is perfect if the canonical map from U to U ×× is an isomorphism. 356K Proposition A Riesz space U is perfect iff (i) it is Dedekind complete (ii) U × separates the points of U (iii) whenever A ⊆ U is non-empty and upwards-directed and {f (u) : u ∈ A} is bounded for every f ∈ U × , then A is bounded above in U . proof (a) Suppose that U is perfect. Because it is isomorphic to U ×× , which is surely Dedekind complete, U also is Dedekind complete. Because the map u 7→ u ˆ : U → U ×× is injective, U × separates the points of U . If A ⊆ U is non-empty and upwards-directed ad {f (u) : u ∈ A} is bounded above for every f ∈ U × , then B = {ˆ u : u ∈ A} is non-empty and upwards-directed and supφ∈B φ(f ) < ∞ for every f ∈ U × , so sup B is defined in U ×∼ (355Ed); but U ×× is a band in U ×∼ , so sup B ∈ U ×× and is of the form w ˆ for some w ∈ U . Because u 7→ u ˆ is a Riesz space isomorphism, w = sup A in U . Thus U satisfies the three conditions. (b) Suppose that U satisfies the three conditions. We know that u 7→ u ˆ is a Riesz homomorphism onto an order-dense Riesz subspace of U ×× (356I). It is injective because U × separates the points of U . If φ ≥ 0 in U ×× , set A = {u : u ∈ U + , u ˆ ≤ φ}. Then A is non-empty and upwards-directed and for any f ∈ U × supu∈A f (u) ≤ supu∈A |f |(u) ≤ supu∈A u ˆ(|f |) ≤ φ(|f |) < ∞, so by condition (iii) A has an upper bound in U . Since U is Dedekind complete, w = sup A is defined in U . Now w ˆ = supu∈A u ˆ = φ. As φ is arbitrary, the image of U includes (U ×× )+ , therefore is the whole of U ×× , and u 7→ u ˆ is a bijective Riesz homomorphism, that is, a Riesz space isomorphism.

268

Riesz spaces

356L

356L Proposition (a) Any band in a perfect Riesz space is a perfect Riesz space in its own right. (b) For any Riesz space U , U ∼ is perfect; consequently Uc∼ and U × are perfect. proof (a) I use the criterion of 356K. Let U be a perfect Riesz space and V a band in U . Then V is Dedekind complete because U is (353Jb). If v ∈ V \ {0} there is an f ∈ U × such that f (v) 6= 0; but the embedding V ⊆ U is order-continuous (352N), so g = f ¹ V belongs to V × , and g(v) 6= 0. Thus V × separates the points of V . If A ⊆ V is non-empty and upwards-directed and supv∈A g(v) is finite for every g ∈ V × , then supv∈A f (v) < ∞ for every f ∈ U × (again because f ¹ V ∈ V × ), so A has an upper bound in U ; because U is Dedekind complete, sup A is defined in U ; because V is a band, sup A ∈ V and is an upper bound for A in V . Thus V satisfies the conditions of 356K and is perfect. (b) U ∼ is Dedekind complete, by 355Ea. If f ∈ U ∼ \ {0}, there is a u ∈ U such that f (u) 6= 0; now u ˆ(f ) 6= 0, where u ˆ ∈ U ∼× (356F). Thus U ∼× separates the points of U ∼ . If A ⊆ U ∼ is non-empty and upwards-directed and supf ∈A φ(f ) is finite for every φ ∈ U ∼× , then, in particular, supf ∈A f (u) = supf ∈A u ˆ(f ) < ∞ for every u ∈ U , so A is bounded above in U ∼ , by 355Ed. Thus U ∼ satisfies the conditions of 356K and is perfect. By (a), it follows at once that U × and Uc∼ are perfect. 356M Proposition If U is a Banach lattice in which the norm is order-continuous and has the Levi property, then U is perfect. proof By 356Db, U ∗ = U × ; since U ∗ surely separates the points of U , so does U × . By 354Ee, U is Dedekind complete. If A ⊆ U is non-empty and upwards-directed and f [A] is bounded for every f ∈ U × , then A is norm-bounded, by the Uniform Boundedness Theorem (3A5Hb). Because the norm is supposed to have the Levi property, A is bounded above in U . Thus U satisfies all the conditions of 356K and is perfect. 356N L- and M -spaces I come now to the duality between L-spaces and M -spaces which I hinted at in §354. Proposition Let U be an Archimedean Riesz space with an order-unit norm. (a) U ∗ = U ∼ is an L-space. (b) If e is the standard order unit of U , then kf k = |f |(e) for every f ∈ U ∗ . (c) A linear functional f : U → R is positive iff it belongs to U ∗ and kf k = f (e). (d) If e 6= 0 there is a positive linear functional f on U such that f (e) = 1. proof (a)-(b) We know already that U ∗ ⊆ U ∼ is a Banach lattice (356Da). If f ∈ U ∼ then sup{|f (u)| : kuk ≤ 1} = sup{|f (u)| : |u| ≤ e} = |f |(e), ∗

so f ∈ U and kf k = |f |(e); thus U ∼ = U ∗ . If f , g ≥ 0 in U ∗ , then kf + gk = (f + g)(e) = f (e) + g(e) = kf k + kgk; thus U ∗ is an L-space. (c) As already remarked, if f is positive then f ∈ U ∗ and kf k = f (e). On the other hand, if f ∈ U ∗ and kf k = f (e), take any u ≥ 0. Set v = (1 + kuk)−1 u. Then 0 ≤ v ≤ e and ke − vk ≤ 1 and f (e − v) ≤ |f (e − v)| ≤ kf k = f (e). But this means that f (v) ≥ 0 so f (u) ≥ 0. As u is arbitrary, f ≥ 0. (d) By the Hahn-Banach theorem (3A5Ac), there is an f ∈ U ∗ such that f (e) = kf k = 1; by (c), f is positive. 356O Theorem Let U be an Archimedean Riesz space with order-unit norm. Then a set A ⊆ U ∗ = U ∼ is uniformly integrable iff it is norm-bounded and limn→∞ supf ∈A |f (un )| = 0 for every order-bounded disjoint sequence hun in∈N in U + .

356O

Dual spaces

269

proof (a) Suppose that A is uniformly integrable. Then it is surely norm-bounded (354Ra). If hun in∈N is ∗ a disjoint sequence in U + bounded above Pnby w, then for any ² > 0 we can find an h ≥ 0 in U such that + k(|f | − h) k ≤ ² for every f ∈ A. Now i=0 h(ui ) ≤ h(w) for every n, and limn→0 h(un ) = 0; since at the same time |f (un )| ≤ |f |(un ) ≤ h(un ) + (|f | − h)+ (un ) ≤ h(un ) + ²kun k ≤ h(un ) + ²kwk for every f ∈ A, n ∈ N, lim supn→∞ supf ∈A |f |(un ) ≤ ²kwk. As ² is arbitrary, limn→∞ supf ∈A |f |(un ) = 0, and the conditions are satisfied. (b)(i) Now suppose that A is norm-bounded but not uniformly integrable. Write B for the solid hull of A, M for supf ∈A kf k = supf ∈B kf k; then there is a disjoint sequence hgn in∈N in B ∩ (U ∗ )+ which is not norm-convergent to 0 (354R(b-iv)), that is, δ=

1 2

lim supn→∞ gn (e) =

1 2

lim supn→∞ kgn k > 0,

where e is the standard order unit of U . (ii) Set C = {v : 0 ≤ v ≤ e, supg∈B g(v) ≥ δ}, D = {w : 0 ≤ w ≤ e, lim supn→∞ gn (w) > δ}. Then for any u ∈ D we can find v ∈ C, w ∈ D such that v ∧ w = 0. P P Set δ 0 = lim supn→∞ gn (u), 0 η = (δ − δ)/(3 + M ) > 0; take k ∈ N so large that kη ≥ M . Because gn (u) ≥ δ 0 − η for infinitely many n, we can find P a set K ⊆ N, of size k, such that gi (u) ≥ δ 0 − η for every i ∈ K. Now we know that, for each i ∈ K, gi ∧ k j∈K,j6=i gj = 0, so there is a vi ≤ u such that P gi (u − vi ) + k j∈K,j6=i gj (vi ) ≤ η (355Ec). Now gi (vi ) ≥ gi (u) − η ≥ δ 0 − 2η, Set vi0 = (vi −

P j∈K,j6=i

gi (vj ) ≤

η k

for i, j ∈ K, i 6= j.

vj )+ for each i ∈ K; then P gi (vi0 ) ≥ gi (vi ) − j∈K,j6=i gi (vj ) ≥ δ 0 − 3η

for every i ∈ K, while vj0 ∧ vi0 = 0 for distinct i, j ∈ K. For each n ∈ N, P 1 0 i∈K gn (u ∧ vi ) ≤ gn (u) ≤ kgn k ≤ ηk, η

so there is some i(n) ∈ K such that 1 η

0 gn (u ∧ vi(n) ) ≤ η,

1 η

0 gn (u − vi(n) )+ ≥ gn (u) − η.

Since {n : gn (u) ≥ δ + 2η} is infinite, there is some m ∈ K such that J = {n : gn (u) ≥ δ + 2η, i(n) = m} is infinite. Try 0 v = (vm − ηu)+ ,

1 η

0 + ) . w = (u − vm

Then v, w ∈ [0, u] and v ∧ w = 0. Next, 0 gm (v) ≥ gm (vm ) − ηM ≥ δ 0 − 3η − ηM = δ,

so v ∈ C, while for any n ∈ J 1 η

0 )+ ≥ gn (u) − η ≥ δ + η; gn (w) = gn (u − vi(n)

since J is infinite, lim supn→∞ gn (w) ≥ δ + η > δ

270

Riesz spaces

356O

and w ∈ D. Q Q (iii) Since e ∈ D, we can choose inductively sequences hwn in∈N in D, hvn in∈N in C such that w0 = e, vn ∧ wn+1 = 0, vn ∨ wn+1 ≤ wn for every n ∈ N. But in this case hvn in∈N is a disjoint order-bounded sequence in [0, u], while for each n ∈ N, we can find fn ∈ A such that |fn |(vn ) > 23 δ. Now there is a un ∈ [0, vn ] such that |fn (un )| ≥ 31 δ. P P Set γ = sup0≤v≤vn |fn (v)|. Then fn+ (vn ), fn− (vn ) are both less than or equal to γ, so |fn |(vn ) ≤ 2γ and γ > 31 δ; so there is a un ∈ [0, vn ] such that |fn (un )| ≥ 31 δ. Q Q Accordingly we have a disjoint sequence hun in∈N in [0, e] such that supf ∈A |f (un )| ≥ 13 δ for every n ∈ N. (iv) All this is on the assumption that A is norm-bounded and not uniformly integrable. So, turning it round, we see that if A is norm-bounded and limn→∞ supf ∈A |f (un )| = 0 for every order-bounded disjoint sequence hun in∈N , A must be uniformly integrable. This completes the proof. 356P Proposition Let U be an L-space. (a) U is perfect. R ∗ ∼ × defined by setting R (b) U + = U − = U is an M -space; its standard order unit is the functional u = ku k − ku k for every u ∈ U . R R (c) If A ⊆ U is Rnon-empty and upwards-directed and supu∈A u is finite, then sup A is defined in U and sup A = supu∈A u. proof (a) By 354N we know that the norm on U is order-continuous and has the Levi property, so 356M tells us that U is perfect. (b) 356Dd tells us that U ∗ = U ∼ = U × . The L-space property tells us that the functional u 7→ kuk : U R+ → R is additive; of course it is also homogeneous, so by 355D it has an R R extension to a linear functional : U → R satisfying the given formula. Because u = kuk ≥ 0 for u ≥ 0, ∈ (U ∼ )+ . For f ∈ U ∼ , Z Z |f | ≤ ⇐⇒ |f |(u) ≤ u for every u ∈ U + ⇐⇒ |f (v)| ≤ kuk whenever |v| ≤ u ∈ U ⇐⇒ |f (v)| ≤ kvk for every v ∈ U ⇐⇒ kf k ≤ 1, so the norm on U ∗ = U ∼ is the order-unit norm defined from

R

, and U ∼ is an M -space, as claimed.

(c) Fix u0 ∈ A, and set B = {u+ : u ∈ A, u ≥ u0 }. Then B ⊆ U + is upwards-directed, and Z Z Z sup kvk = sup u+ = sup u + u− v∈B u∈A,u≥u0 u∈A,u≥u0 Z Z ≤ sup u + u− 0 < ∞. u∈A,u≥u0

Because k k has the Levi property, B is bounded above. But (because A is upwards-directed) every member of A is dominated by some member of B, so A also R is bounded R above. Because U is Dedekind complete, R sup A is defined in U . Finally, sup A = supu∈A u because , being a positive member of U × , is ordercontinuous. 356Q Theorem Let U be any L-space. Then a subset of U is uniformly integrable iff it is relatively weakly compact. proof (a) Let A ⊆ U be a uniformly integrable set. (i) Suppose that F is an ultrafilter on X containing A. Then A 6= ∅. Because A is norm-bounded, supu∈A |f (u)| < ∞ and φ(f ) = limu→F f (u) is defined in R for every f ∈ U ∗ (2A3Se). If f , g ∈ U ∗ then

356X

Dual spaces

271

φ(f + g) = limu→F f (u) + g(u) = limu→F f (u) + limu→F g(u) = φ(f ) + φ(g) (2A3Sf). Similarly, φ(αf ) = limu→F αf (u) = αφ(f ) ∗

∗

for every f ∈ U , α ∈ R. Thus φ : U → R is linear. Also |φ(f )| ≤ supu∈A |f (u)| ≤ kf k supu∈A kuk, so φ ∈ U

∗∗

=U

∗∼

.

(ii) Now the point of this argument is that φ ∈ U ∗× . P P Suppose that B ⊆ U ∗ is non-empty and downwards-directed and has infimum 0. Fix f0 ∈ B. Let ² > 0. Then there is a w ∈ U + such that k(|u| − w)+ k ≤ ² for every u ∈ A, which means that |f (u)| ≤ |f |(|u|) ≤ |f |(w) + |f |(|u| − w)+ ≤ |f |(w) + ²kf k for every f ∈ U ∗ and every u ∈ A. Accordingly |φ(f )| ≤ |f |(w)+²kf k for every f ∈ U ∗ . Now inf f ∈B f (w) = 0 (using 355Ee, as usual), so there is an f1 ∈ B such that f1 ≤ f0 and f1 (w) ≤ ². In this case |φ|(f1 ) = sup|f |≤f1 |φ(f )| ≤ sup|f |≤f1 |f |(w) + ²kf k ≤ f1 (w) + ²kf1 k ≤ ²(1 + kf0 k). As ² is arbitrary, inf f ∈B |φ|(f ) = 0; as B is arbitrary, |φ| is order-continuous and φ ∈ U ∗× . Q Q (iii) At this point, we recall that U ∗ = U × and that the canonical map from U to U ×× is surjective (356P). So there is a u0 ∈ U such that u ˆ0 = φ. But now we see that f (u0 ) = φ(f ) = limu→F f (u) for every f ∈ U ∗ ; which is just what is meant by saying that F → u0 for the weak topology on U (2A3Sd). Accordingly every ultrafilter on U containing A has a limit in U . But because the weak topology on U is regular (3A3Be), it follows that the closure of A for the weak topology is compact (3A3De), so that A is relatively weakly compact. (b) For the converse I use the criterion of 354R(b-iv). Suppose that A ⊆ U is relatively weakly compact. Then A is norm-bounded, by the Uniform Boundedness Theorem. Now let hun in∈N be any disjoint sequence in the solid hull of A. For each n, let Un be the band of U generated by un . Let Pn be the band projection from U onto Un (353H). Let vn ∈ A be such that |un | ≤ |vn |; then |un | = Pn |un | ≤ Pn |vn | = |Pn vn |, so kun k ≤ kPn vn k for each n. Let gn ∈ U ∗ be such that kgn k = 1 and gn (Pn vn ) = kPn vn k. Define T : U → RN by setting T u = hgn (Pn u)in∈N for each u ∈ U . Then T is a continuous linear operator from U to `1 . P P For m 6= n, Um ∩ Un = {0}, because |um | ∧ |un | = 0. So, for any u ∈ U , hPn uin∈N is a disjoint sequence in U , and Pn Pn i=0 |Pi u|k = k supi≤n |Pi u|k ≤ kuk i=0 kPi uk = k for every n; accordingly kT uk1 =

P∞ i=0

|gi Pi u| ≤

P∞ i=0

kPi uk ≤ kuk.

Since T is certainly a linear operator (because every coordinate functional gi Pi is linear), we have the result. Q Q Consequently T [A] is relatively weakly compact in `1 , because T is continuous for the weak topologies (2A5If). But `1 can be identified with L1 (µ), where µ is counting measure on N. So T [A] is uniformly integrable in `1 , by 247C, and in particular limn→∞ supw∈T [A] |w(n)| = 0. But this means that limn→∞ kun k ≤ limn→∞ |gn (Pn vn )| = limn→∞ |(T vn )(n)| = 0. As hun in∈N is arbitrary, A satisfies the conditions of 354R(b-iv) and is uniformly integrable. 356X Basic exercises (a) Show that if U = `∞ then U × = Uc∼ can be identified with `1 , and is P∞ ∼ ∼ properly included in U . (Hint: show that if f ∈ Uc then f (u) = n=0 u(n)f (en ), where en (n) = 1, en (i) = 0 for i 6= n.)

272

Riesz spaces

356Xb

(b) Show that if U = C([0, 1]) then U × = Uc∼ = {0}. (Hint: show that if f ∈ (Uc∼ )+ and hqn in∈N enumerates Q ∩ [0, 1], then for each n ∈ N there is a un ∈ U + such that un (qn ) = 1 and f (un ) ≤ 2−n .) (c) Let X be an uncountable set and µ the countable-cocountable measure on X, Σ its domain (211R). Let U be the space of bounded Σ-measurable real-valued functions on X. Show that U is a Dedekind σ-complete Banach lattice Rif given the supremum norm k k∞ . Show that U × can be identified with `1 (X) (cf. 356Xa), and that u 7→ u dµ belongs to Uc∼ \ U × . (d) Let U be a Dedekind σ-complete Riesz space and f ∈ Uc∼ . Let hun in∈N be an order-bounded sequence in U which is order-convergent to u ∈ U in the sense that u = inf n∈N supm≥n um = supn∈N inf m≥n um . Show that limn→∞ f (un ) exists and is equal to f (u). (e) Let U be any Riesz space. Show that the band projection P : U ∼ → U × is defined by the formula (P f )(u) = inf{sup f (v) : A ⊆ U is non-empty, upwards-directed v∈A

and has supremumu} for every f ∈ (U ∼ )+ , u ∈ U + . (Hint: show that the formula for P f always defines an order-continuous linear functional. Compare 355Yh, 356Yb and 362Bd.) (f ) Let U be any Riesz space. Show that the band projection P : U ∼ → Uc∼ is defined by the formula (P f )(u) = inf{supn∈N f (vn ) : hvn in∈N is a non-decreasing sequence with supremum u} for every f ∈ (U ∼ )+ , u ∈ U + . (g) Let U be a Riesz space with a Riesz norm. Show that U ∗ is perfect. (h) Let U be a Riesz space with a Riesz norm. Show that the canonical map from U to U ∗∗ is a Riesz homomorphism. (i) Let V be a perfect Riesz space and U any Riesz space. Show that L∼ (U ; V ) is perfect. (Hint: show that if u ∈ U , g ∈ V × then T 7→ g(T u) belongs to L∼ (U ; V )× .) (j) Let U be an M -space. Show that it is perfect iff it is Dedekind complete and U × separates the points of U . (k) Let U be a Banach lattice which, as a Riesz space, is perfect. Show that its norm has the Levi property. (l) Write out a proof from first principles that if hun in∈N is a sequence in `1 such that |un (n)| ≥ δ > 0 for every n ∈ N, then {un : n ∈ N} is not relatively weakly compact. (m) Let U be an L-space and A ⊆ U a non-empty set. Show that the following are equiveridical: (i) A is uniformly integrable (ii) inf f ∈B supu∈A |f (u)| for every non-empty downwards-directed set B ⊆ U × with infimum 0 (iii) inf n∈N supu∈A |fn (u)| = 0 for every non-increasing sequence hfn in∈N in U × with infimum 0 (iv) A is norm-bounded and limn→∞ supu∈A |fn (u)| = 0 for every disjoint order-bounded sequence hfn in∈N in U × . 356Y Further exercises (a) Let U be a Riesz space with the countable sup property. Show that U × = Uc∼ . (b) Let U be a Riesz space, and A a family of non-empty downwards-directed subsets of U + all with ∼ infimum 0. (i) Show that UA = {f : f ∈ U ∼ , inf u∈A |f |(u) = 0 for every A ∈ A} is a band in U ∼ . (ii) Set ∗ ∼ ∼ ∼ + A = {A0 + . . . + An : A0 , . . . , An ∈ A}. Sbow that UA = UA ) , and let g, h be ∗ . (iii) Take any f ∈ (U ∼ ∼ ⊥ the components of f in UA , (UA ) respectively. Show that g(u) = inf A∈A∗ supv∈A f (u − v)+ , for every u ∈ U + . (Cf. 362Xi.)

h(u) = supA∈A∗ inf v∈A f (u ∧ v)

356 Notes

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273

(c) Let U be a Riesz space. For any band V ⊆ U write V ◦ for {f : f ∈ U × , f (v) = 0 for every v ∈ V }. Show that V 7→ (V ⊥ )◦ is a surjective order-continuous Boolean homomorphism from the algebra of complemented bands of U onto the band algebra of U × , and that it is injective iff U × separates the points of U . (d) Let U be a Riesz space such that U ∼ separates the points of U . For any band V ⊆ U ∼ write V ◦ for {x : x ∈ U, f (x) = 0 for every f ∈ V }. Show that V 7→ (V ⊥ )◦ is a surjective Boolean homomorphism from the algebra of bands of U ∼ onto the band algebra of U , and that it is injective iff U ∼ = U × . (e) Let U be a Dedekind complete Riesz space such that U × separates the points of U and U is the solid linear subspace of itself generated by a countable set. Show that U is perfect. (f ) Let U be an L-space and hun in∈N a sequence in U such that hf (un )in∈N is Cauchy for every f ∈ U ∗ . Show that hun in∈N is convergent for the weak topology of U . (Hint: use 356Xm(iv) to show that {un : n ∈ N} is relatively weakly compact.) (g) Let U be a perfect Banach lattice with order-continuous norm and hun in∈N a sequence in U such that hf (un )in∈N is Cauchy for every f ∈ U ∗ . Show that hun in∈N is convergent for the weak topology of U . (Hint: set φ(f ) = limn→∞ fn (u). For any g ∈ (U ∗ )+ let Vg be the solid linear subspace of U ∗ generated by g, Wg = {u : g(|u|) = 0}⊥ , kukg = g(|u|) for u ∈ Wg . Show that the completion of Wg under k kg is an L-space with dual isomorphic to Vg , and hence (using 356Yf) that φ¹ Vg belongs to Vg× ; as g is arbitrary, φ ∈ V × and may be identified with an element of U .) (h) Let U be a uniformly complete Archimedean Riesz space with complexification V (354Yk). (i) Show that the complexification of U ∼ can be identified with the space of linear functionals f : V → C such that sup|v|≤u |f (v)| is finite for every u ∈ U + . (ii) Show that if U is a Banach lattice, then the complexification of U ∼ = U ∗ can be identified (as normed space) with V ∗ . (See 355Yk.) 356 Notes and comments The section starts easily enough, with special cases of results in §355 (356B). When U has a Riesz norm, the identification of U ∗ as a subspace of U ∼ , and the characterization of ordercontinuous norms (356D) are pleasingly comprehensive and straightforward. Coming to biduals, we need to think a little (356F), but there is still no real difficulty at first. In 356H-356I, however, something more substantial is happening. I have written these arguments out in what seems to be the shortest route to the main theorem, at the cost perhaps of neglecting any intuitive foundation. What I think we are really doing is matching bands in U , U × and U ×× , as in 356Yc. From now on, almost the first thing we shall ask of any new Riesz space will be whether it is perfect, and if not, which of the three conditions of 356K it fails to satisfy. For reasons which will I hope appear in the next chapter, perfect Riesz spaces are particularly important in measure theory; in particular, all Lp spaces for p ∈ [1, ∞[ are perfect (366D), as are the L∞ spaces of localizable measure spaces (365K). Further examples will be discussed in §369 and §374. Of course we have to remember that there are also important Riesz spaces which are not perfect, of which C([0, 1]) and c 0 are two of the simplest examples. The duality between L- and M -spaces (356N, 356P) is natural and satisfying. We are now in a position to make a determined attempt to tidy up the notion of ‘uniform integrability’. I give two major theorems. The first is yet another ‘disjoint-sequence’ characterization of uniformly integrable sets, to go with 246G and 354R. The essential difference here is that we are looking at disjoint sequences in a predual; in a sense, this means that the result is a sharper one, because the M -space U need not be Dedekind complete (for instance, it could be C([0, 1]) – this indeed is the archetype for applications of the theorem) and therefore need not have as many disjoint sequences as its dual. (For instance, in the dual of C([0, 1]) we have all the point masses δt , where δt (u) = u(t); these form a disjoint family in C([0, 1])∼ not corresponding to any disjoint family in C([0, 1]).) The essence of the proof is a device to extract a disjoint sequence in U to match approximately a subsequence of a given disjoint sequence in U ∼ . In the example just suggested, this would correspond, given a sequence htn in∈N of distinct points in [0, 1], to finding a subsequence htn(i) ii∈N which is discrete, so that we can find disjoint ui ∈ C([0, 1]) with ui (tn(i) ) = 1 for each i. The second theorem, 356Q, is a new version of a result already given in §247: in any L-space, uniform integrability is the same as relative weak compactness. I hope you are not exasperated by having been

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asked, in Volume 2, to master a complex argument (one of the more difficult sections of that volume) which was going to be superseded. Actually it is worse than that. A theorem of Kakutani (369E) tells us that every L-space is isomorphic to an L1 space. So 356Q is itself a consequence of 247C. I do at least owe you an explanation for writing out two proofs. The first point is that the result is sufficiently important for it to be well worth while spending time in its neighbourhood, and the contrasts and similarities between the two arguments are instructive. The second is that the proof I have just given was not really accessible at the level of Volume 2. It does not rely on every single page of this chapter, but the key idea (that U is isomorphic to U ×× , so it will be enough if we can show that A is relatively compact in U ×× ) depends essentially on 356I, which lies pretty deep in the abstract theory of Riesz spaces. The third is an aesthetic one: a theorem about L-spaces ought to be proved in the category of normed Riesz spaces, without calling on a large body of theory outside. Of course this is a book on measure theory, so I did the measure theory first, but if you look at everything that went into it, the proof in §247 is I believe longer, in the formal sense, than the one here, even setting aside the labour of proving Kakutani’s theorem. Let us examine the ideas in the two proofs. First, concerning the proof that uniformly integrable sets are relatively compact, the method here is very smooth and natural; the definition I chose of ‘uniform integrability’ is exactly adapted to showing that uniformly integrable sets are relatively compact in the order-continuous bidual; all the effort goes into the proof that L-spaces are perfect. The previous argument depended on identifying the dual of L1 as L∞ – and was disagreeably complicated by the fact that the identification is not always valid, so that I needed to reduce the problem to the σ-finite case (part (b-ii) of the proof of 247C). After that, the Radon-Nikod´ ym theorem did the trick. Actually Kakutani’s theorem shows that the side-step to σ-finite spaces is irrelevant. It directly represents an abstract L-space as L1 (µ) for a localizable measure µ, in which case (L1 )∗ ∼ = L∞ exactly. In the other direction, both arguments depend on a disjoint-sequence criterion for uniform integrability (246G(iii) or 354R(b-iv)). These criteria belong to the ‘easy’ side of the topic; straightforward Riesz space arguments do the job, whether written out in that language or not. (Of course the new one in this section, 356O, lies a little deeper.) I go a bit faster this time because I feel that you ought by now to be happy with the Hahn-Banach theorem and the Uniform Boundedness Theorem, which I was avoiding in Volume 2. And then of course I quote the result for `1 . This looks like cheating. But `1 really is easier, as you will find if you just write out part (a) of the proof of 247C for this case. It is not exactly that you can dispense with any particular element of the R argument; rather it is that the formulae become much more direct when you can write u(i) in place of Fi u, and ‘cluster points for the weak topology’ become pointwise limits of subsequences, so that the key step (the ‘sliding hump’, in which uk(j) (n(k(j))) is the only significant coordinate of uk(j) ), is easier to find. We now have a wide enough variety of conditions equivalent to uniform integrability for it to be easy to find others; I give a couple in 356Xm, corresponding in a way to those in 246G. You may have noticed, in the proof of 247C, that in fact the full strength of the hypothesis ‘relatively weakly compact’ is never used; all that is demanded is that a couple of sequences should have cluster points for the weak topology. So we see that a set A is uniformly integrable iff every sequence in A has a weak cluster point. But this extra refinement is nothing to do with L-spaces; it is generally true, in any normed space U , that a set A ⊆ U is relatively weakly compact iff every sequence in A has a cluster point in U for the weak topology (‘Eberlein’s ¨ the 69, 24.2.1, or Dunford & Schwartz 57, V.6.1). theorem’; see 462D in Volume 4, Ko There is a very rich theory concerning weak compactness in perfect Riesz spaces, based on the ideas here; some of it is explored in Fremlin 74a. As a sample, I give one of the basic properties of perfect Banach lattices with order-continuous norms: they are ‘weakly sequentially complete’ (356Yg).

Chap. 36 intro.

Introduction

275

Chapter 36 Function Spaces Chapter 24 of Volume 2 was devoted to the elementary theory of the ‘function spaces’ L0 , L1 , L2 and L associated with a given measure space. In this chapter I return to these spaces to show how they can be related to the more abstract themes of the present volume. In particular, I develop constructions to demonstrate, as clearly as I can, the way in which all the function spaces associated with a measure space in fact depend only on its measure algebra; and how many of their features can (in my view) best be understood in terms of constructions involving measure algebras. The chapter is very long, not because there are many essentially new ideas, but because the intuitions I seek to develop depend, for their logical foundations, on technically complex arguments. This is perhaps best exemplified by §364. If two measure spaces (X, Σ, µ) and (Y, T, ν) have isomorphic measure algebras (A, µ ¯), (B, ν¯) then the spaces L0 (µ), L0 (ν) are isomorphic as topological f -algebras; and more: for any isomorphism between (A, µ ¯) and (B, ν¯) there is a unique corresponding isomorphism between the L0 spaces. The intuition involved is in a way very simple. If f , g are measurable real-valued functions on X and Y respectively, then f • ∈ L0 (µ) will correspond to g • ∈ L0 (ν) if and only if [[f • > α]] = {x : f (x) > α}• ∈ A corresponds to [[g • > α]] = {y : g(y) > α}• ∈ B for every α. But the check that this formula is consistent, and defines an isomorphism of the required kind, involves a good deal of detailed work. It turns out, in fact, that the measures µ and ν do not enter this part of the argument at all, except through their ideals of negligible sets (used in the construction of A and B). This is already evident, if you look for it, in the theory of L0 (µ); in §241, as written out, you will find that the measure of an individual set is not once mentioned, except in the exercises. Consequently there is an invitation to develop the theory with algebras A which are not necessarily measure algebras. Here is another reason for the length of the chapter; substantial parts of the work are being done in greater generality than the corresponding sections of Chapter 24, necessitating a degree of repetition. Of course this is not ‘measure theory’ in the strict sense; but for thirty years now measure theory has been coloured by the existence of these generalizations, and I think it is useful to understand which parts of the theory apply only to measure algebras, and which can be extended to other σ-complete Boolean algebras, like the algebraic theory of L0 , or even to all Boolean algebras, like the theory of L∞ . Here, then, are two of the objectives of this chapter: first, to express the ideas of Chapter 24 in ways making explicit their independence of particular measure spaces, by setting up constructions based exclusively on the measure algebras involved; second, to set out some natural generalizations to other algebras. But to justify the effort needed I ought to point to some mathematically significant idea which demands these constructions for its expression, and here I mention the categorical nature of the constructions. Between Boolean algebras we have a variety of natural and important classes of ‘morphism’; for instance, the Boolean homomorphisms and the order-continuous Boolean homomorphisms; while between measure algebras we have in addition the measure-preserving Boolean homomorphisms. Now it turns out that if we construct the Lp spaces in the natural ways then morphisms between the underlying algebras give rise to morphisms between their Lp spaces. For instance, any Boolean homomorphism from A to B produces a multiplicative norm-contractive Riesz homomorphism from L∞ (A) to L∞ (B); if A and B are Dedekind σ-complete, then any sequentially order-continuous Boolean homomorphism from A to B produces a sequentially order-continuous multiplicative Riesz homomorphism from L0 (A) to L0 (B); and if (A, µ ¯) and (B, ν¯) are measure algebras, then any measure-preserving Boolean homomorphism from A to B produces norm-preserving Riesz homomorphisms from Lp (A, µ ¯) to Lp (B, ν¯) for every p ∈ [1, ∞]. All of these are ‘functors’, that is, a composition of homomorphisms between algebras gives rise to a composition of the corresponding operators between their function spaces, and are ‘covariant’, that is, a homomorphism from A to B leads to an operator from Lp (A) to Lp (B). But the same constructions lead us to a functor which is ‘contravariant’: starting from an order-continuous Boolean homomorphism from a semi-finite measure algebra (A, µ ¯) to a measure algebra (B, ν¯), we have an operator from L1 (B, ν¯) to L1 (A, µ ¯). This last is in fact a kind of conditional expectation operator. In my view it is not possible to make sense of the theory of measure-preserving transformations without at least an intuitive grasp of these ideas. Another theme is the characterization of each construction in terms of universal mapping theorems: for instance, each Lp space, for 1 ≤ p ≤ ∞, can be characterized as Banach lattice in terms of factorizations of functions of an appropriate class from the underlying algebra to Banach lattices. ∞

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Now let me try to sketch a route-map for the journey ahead. I begin with two sections on the space S(A); this construction applies to any Boolean algebra (indeed, any Boolean ring), and corresponds to the space of ‘simple functions’ on a measure space. Just because it is especially close to the algebra (or ring) A, there is a particularly large number of universal mapping theorems corresponding to different aspects of its structure (§361). In §362 I seek to relate ideas on additive functionals on Boolean algebras from Chapter 23 and §§326-327 to the theory of Riesz space duals in §356. I then turn to the systematic discussion of the function spaces of Chapter 24: L∞ (§363), L0 (§364), L1 (§365) and other Lp (§366), followed by an account of convergence in measure (§367). While all these sections are dominated by the objectives sketched in the paragraphs above, I do include a few major theorems not covered by the ideas of Volume 2, such as the Kelley-Nachbin characterization of the Banach spaces L∞ (A) for Dedekind complete A (363R). In the last two sections of the chapter I turn to the use of L0 spaces in the representation of Archimedean Riesz spaces (§368) and of Banach lattices separated by their order-continuous duals (§369).

361 S This is the fundamental Riesz space associated with a Boolean ring A. When A is a ring of sets, S(A) can be regarded as the linear space of ‘simple functions’ generated by the characteristic functions of members of A (361L). Its most important property is the universal mapping theorem 361F, which establishes a one-toone correspondence between (finitely) additive functions on A (361B-361C) and linear operators on S(A). Simple universal mapping theorems of this type can be interesting, but do not by themselves lead to new insights; what makes this one important is the fact that S(A) has a canonical Riesz space structure, norm and multiplication (361E). From this we can deduce universal mapping theorems for many other classes of function (361G, 361H, 361I, 361Xb). (Particularly important are countably additive and completely additive real-valued functionals, which will be dealt with in the next section.) While the exact construction of S(A) (and the associated map from A to S(A)) can be varied (361D, 361L, 361M, 361Ya), its structure is uniquely defined, so homomorphisms between Boolean rings correspond to maps between their S( )-spaces (361J), and (when A is an algebra) A can be recovered from the Riesz space S(A) as the algebra of its projection bands (361K). 361A Boolean rings In this section I speak of Boolean rings rather than algebras; there are ideas in §365 below which are more naturally expressed in terms of the ring of elements of finite measure in a measure algebra than in terms of the whole algebra. I should perhaps therefore recall some of the ideas of §311, which is the last time when Boolean rings without identity were mentioned, and set out some simple facts. (a) Any Boolean ring A can be represented as the ring of compact open subsets of a zero-dimensional locally compact Hausdorff space X (311I); X is just the set of surjective ring homomorphisms from A onto Z2 (311E). (b) If A and B are Boolean rings and π : A → B is a function, then the following are equiveridical: (i) π is a ring homomorphism; (ii) π(a \ b) = πa \ πb for all a, b ∈ A; (iii) π0 = 0 and π(a ∪ b) = πa ∪ πb, π(a ∩ b) = πa ∩ πb for all a, b ∈ A. P P See 312H. To prove (ii)⇒(iii), observe that if a, b ∈ A then π(a ∩ b) = πa \ π(a \ b) = πa ∩ πb. Q Q (c) If A and B are Boolean rings and π : A → B is a ring homomorphism, then π is order-continuous iff inf π[A] = 0 whenever A ⊆ A is non-empty and downwards-directed and inf A = 0 in A; while π is sequentially order-continuous iff inf n∈N πan = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0. (See 313L.) (d) The following will be a particularly important type of Boolean ring for us. If (A, µ ¯) is a measure algebra, then the ideal Af = {a : a ∈ A, µ ¯a < ∞} is a Boolean ring in its own right. Now suppose that (B, ν¯) is another measure algebra and Bf ⊆ B the corresponding ring of elements of finite measure. We

361D

S

277

can say that a ring homomorphism π : Af → Bf is measure-preserving if ν¯πa = µ ¯a for every a ∈ Af . f Now in this case π is order-continuous. P P If A ⊆ A is non-empty, downwards-directed and has infimum 0, then inf a∈A µ ¯a = 0, by 321F; but this means that inf a∈A ν¯πa = 0, and inf π[A] = 0 in Bf . Q Q 361B Definition Let A be a Boolean ring and U a linear space. A function ν : A → U is finitely additive, or just additive, if ν(a ∪ b) = νa + νb whenever a, b ∈ A and a ∩ b = 0. 361C Elementary facts We have the following immediate consequences of this definition, corresponding to 326B and 313L. Let A be a Boolean ring, U a linear space and ν : A → U an additive function. (a) ν0 = 0 (because ν0 = ν0 + ν0). (b) If a0 , . . . , am are disjoint in A, then ν(supj≤m aj ) =

Pm j=0

νaj .

(c) If B is another Boolean ring and π : B → A is a ring homomorphism, then νπ : B → U is additive. In particular, if B is a subring of A, then ν¹ B : B → U is additive. (d) If V is another linear space and T : U → V is a linear operator, then T ν : A → V is additive. (e) If U is a partially ordered linear space, then ν is order-preserving iff it is non-negative, that is, νa ≥ 0 for every a ∈ A. P P (α) If ν is order-preserving, then of course 0 = ν0 ≤ νa for every a ∈ A. (β) If ν is non-negative, and a ⊆ b in A, then νa ≤ νa + ν(b \ a) = νb. Q Q (f ) If U is a partially ordered linear space and ν is non-negative, then (i) ν is order-continuous iff inf ν[A] = 0 whenever A ⊆ A is a non-empty downwards-directed set with infimum 0 (ii) ν is sequentially order-continuous iff inf n∈N νan = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0. P P (i) If ν is order-continuous, then of course inf ν[A] = ν0 = 0 whenever A ⊆ A is a non-empty downwardsdirected set with infimum 0. If ν satisfies the condition, and A ⊆ A is a non-empty upwards-directed set with supremum c, then {c \ a : a ∈ A} is downwards-directed with infimum 0 (313Aa), so that sup νa = sup νc − ν(c \ a) = νc − inf ν(c \ a) a∈A

a∈A

a∈A

(by 351Db) = νc. Similarly, if A ⊆ A is a non-empty downwards-directed set with infimum c, then inf a∈A νa = inf a∈A νc + ν(a \ c) = νc + inf a∈A ν(a \ c) = νc. Putting these together, ν is order-continuous. (ii) If ν is sequentially order-continuous, then of course inf n∈N νan = ν0 = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0. If ν satisfies the condition, and han in∈N is a non-decreasing sequence in A with supremum c, then hc \ an in∈N is non-increasing and has infimum 0, so that supn∈N νan = supn∈N νc − ν(c \ an ) = νc − inf n∈N ν(c \ an ) = νc. Similarly, if han in∈N is a non-increasing sequence in A with infimum c, then han \ cin∈N is non-increasing and has infimum 0, so that inf n∈N νan = inf n∈N νc + ν(c \ an ) = νc + inf n∈N ν(c \ an ) = νc. Thus ν is sequentially order-continuous. Q Q 361D Construction Let A be a Boolean ring, and Z its Stone space. For a ∈ A write χa for the characteristic function of the open-and-compact subset b a of Z corresponding to a. Let S(A) be the linear subspace of RZ generated by {χa : a ∈ A}. Because χa is a bounded function for every a, S(A) is a subspace of the space `∞ (Z) of all bounded real-valued functions on Z (354Ha), and k k∞ is a norm on S(A). Because χa × χb = χ(a ∩ b) for all a, b ∈ A (writing × for pointwise multiplication of functions, as in 281B), S(A) is closed under ×.

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361E of S(A).

Function spaces

361E

I give a portmanteau proposition running through the elementary, mostly algebraic, properties

Proposition Let A be a Boolean ring, with Stone space Z. Write S for S(A). (a) If a0 , . . . , an ∈ A, there are disjoint b0 , . . . , bm such that each ai is expressible as the supremum of some of the bj . Pm (b) If u ∈ S, it is expressible in the form j=0 βj χbj where b0 , . . . , bm are disjoint members of A and βj ∈ R for each j. If all the bj are non-zero then kuk∞ = supj≤m |βj |. Pm (c) If u ∈ S is non-negative, it is expressible in the form j=0 βj χbj where b0 , . . . , bm are disjoint members Pm of A and βj ≥ 0 for each j, and simultaneously in the form j=0 γj χcj where c0 ⊇ c1 ⊇ . . . ⊇ cm and γj ≥ 0 for every j. P m (d) If u = j=0 βj χbj where b0 , . . . , bm are disjoint members of A and βj ∈ R for each j, then |u| = Pm j=0 |βj |χbj ∈ S. (e) S is a Riesz subspace of RZ ; in its own right, it is an Archimedean Riesz space. If A is a Boolean algebra, then S has an order unit χ1 and kuk∞ = min{α : α ≥ 0, |u| ≤ αχ1} for every u ∈ S. (f) The map χ : A → S is injective, additive, non-negative, a lattice homomorphism and order-continuous. (g) Suppose that u ≥ 0 in S and δ ≥ 0 in R. Then [[u > δ]] = sup{a : a ∈ A, (δ + η)χa ≤ u for some η > 0} is defined in A, and δχ[[u > δ]] ≤ u ≤ δχ[[u > 0]] ∨ kuk∞ [[u > δ]]. In particular, u ≤ kuk∞ χ[[u > 0]] and there is an η > 0 such that ηχ[[u > 0]] ≤ u. If u, v ≥ 0 in S then u ∧ v = 0 iff [[u > 0]] ∩ [[v > 0]] = 0. (h) Under ×, S is an f -algebra (352W) and a commutative normed algebra (2A4J). (i) For any u ∈ S, u ≥ 0 iff u = v × v for some v ∈ S. proof Write b a for the open-and-compact subset of Z corresponding to a ∈ A. (a) Induce on n. If n = 0 take m = 0, b0 = a0 . For the inductive step to n ≥ 1, take disjoint b0 , . . . , bm such that ai is the supremum of some of the bj for each i < n; now replace b0 , . . . , bm with b0 ∩ an , . . . , bm ∩ an , b0 \ an , . . . , bm \ an , an \ supj≤m bj to obtain a suitable string for a0 , . . . , an . Pn (b) If u = 0 set m = 0, b0 = 0, β0 = 0. Otherwise, express u as i=0 αi χai where ai ∈ A and α0 , . . . , αn are real numbers. Let b0 , . . . , bm be disjoint and such that every ai is expressible as the supremum of some Pk of the bj . Set γij = 1 if bj ⊆ ai , 0 otherwise, so that, because the bj are disjoint, χai = j=0 γij χbj for each i. Then Pn Pm Pm Pn u = i=0 αi χai = i=0 j=0 αi γij χbj = j=0 βj χbj , Pn setting βj = i=0 αi γij for each j ≤ k. The expression for kuk∞ is now obvious. (c)(i) If u ≥ 0P in (b), we must have βj = u(z) ≥ 0 whenever z ∈ bbj , so that βj ≥ 0 whenever bj 6= 0; m consequently u = j=0 |βj |χbj is in the required form. (ii) If we suppose that every βj is non-negative, and rearrange the terms of the sum so that β0 ≤ . . . ≤ βm , then we may set γ0 = β0 , γj = βj − βj−1 for 1 ≤ j ≤ m, cj = supi≥j bi to get Pm Pm Pm Pm Pi Pm j=0 γj χcj = j=0 i=j γj χbi = i=0 j=0 γj χbi = i=0 βi χbi = u. (d) is trivial, because bb0 , . . . , bbn are disjoint. (e) By (d), |u| ∈ S for every u ∈ S, so S is a Riesz subspace of RZ , and in itself is an Archimedean Riesz space. If A is a Boolean algebra, then χ1, the constant function with value 1, belongs to S, and is an order unit of S; while kuk∞ = min{α : α ≥ 0, |u(z)| ≤ α ∀ z ∈ Z} = min{α : α ≥ 0, |u| ≤ αχ1} for every u ∈ S.

361F

S

279

(f ) χ is injective because b a 6= bb whenever a 6= b. χ is additive because b a ∩ bb = ∅ whenever a ∩ b = 0. Of course it is non-negative. It is a lattice homomorphism because a 7→ b a : A → PZ and E 7→ χE : PZ → RZ are. To see that ν is order-continuous, take a non-empty downwards-directed A ⊆ A with infimum 0. ?? Suppose, if possible, that {χa : a ∈ A} does not have P infimum 0 in S. Then there is a u > 0 in S such that m u ≤ χa for every a ∈ A. Now u can be expressed as j=0 βj bj where b0 , . . . , bm are disjoint. There must be some z0 ∈ Z such that u(z0 ) > 0; take j such that z0 ∈ bbj , so that bj 6= 0 and βj = u(z0 ) > 0. But now, for any z ∈ bbj , a ∈ A, (χa)(z) ≥ u(z) = βj > 0 and z ∈ b a. As z is arbitrary, bbj ⊆ b a and bj ⊆ a; as a is arbitrary, bj is a non-zero lower bound for A in A. X X So inf χ[A] = 0 in S. As A is arbitrary, χ is order-continuous, by the criterion of 361C(f-i). Pm (g) Express u as j=0 βj χbj where b0 , . . . , bm are disjoint and every βj ≥ 0. Then given δ ≥ 0, η > 0, a ∈ A we have (δ + η)χa ≤ u iff a ⊆ sup{bj : j ≤ m, βj ≥ δ + η}. So [[u > δ]] = sup{bj : j ≤ m, βj > δ}. Writing c = [[u > δ]], d = [[u > 0]] = sup{bj : βj > 0}, we have u(z) ≤ kuk∞ if z ∈ b c, ≤ δ if z ∈ db \ b c, b = 0 if z ∈ / d. So δχc ≤ u ≤ kuk∞ χc ∨ δχd, as claimed. Taking δ = 0 we get u ≤ kuk∞ χd. Set η = min({1} ∪ {βj : j ≤ m, βj > 0}); then η > 0 and ηχd ≤ u. If u, v ∈ S + take η, η 0 > 0 such that ηχ[[u > 0]] ≤ u,

η 0 χ[[v > 0]] ≤ v.

Then min(η, η 0 )χ([[u > 0]] ∩ [[v > 0]]) ≤ u ∧ v ≤ max(kuk∞ , kvk∞ )χ([[u > 0]] ∩ [[v > 0]]). So u ∧ v = 0 =⇒ [[u > 0]] ∩ [[v > 0]] = 0 =⇒ u ∧ v = 0. (h) S is a commutative f -algebra and normed algebra just because it is a Riesz subspace of the f -algebra and commutative normed algebra `∞ (Z) and is closed under multiplication. Pm (i) If u = j=0 βj χbj where b0 , . . . , bm are disjoint and βj ≥ 0 for every j, then u = v × v where Pm p v = j=0 βj χbj . 361F

I now turn to the universal mapping theorems which really define the construction.

Theorem Let A be a Boolean ring, and U any linear space. Then there is a one-to-one correspondence between additive functions ν : A → U and linear operators T : S(A) → U , given by the formula ν = T χ. proof (a) The core of the proof isP the following observation. Let ν : A U be additive. If a0 , . . . , an ∈ A P→ n n and α0 , . . . , αn ∈ R are such that i=0 αi χai = 0 in S = S(A), then i=0 αi νai = 0 in U . P P As in 361E, we can find disjoint b0 , . . P . , bm such that eachPai is the supremum of some of the b ; set γ = 1 if bj ⊆ ai , 0 j ij Pn m m otherwise, so that χai = j=0 γij χbj , νai = j=0 γij νbj for each i. Set βj = i=0 αi γij , so that Pn Pm 0 = i=0 αi χai = j=0 βj χbj . Now βj νbj = 0 in U for each j, because either bj = 0 and νbj = 0, or there is some z ∈ bbj so that βj must be 0. Accordingly

280

Function spaces

0=

Pm j=0

βj νbj =

Pm Pn j=0

i=0

361F

αi γij νbi =

Pn i=0

αi νai . Q Q

Pn Pm (b) It follows that if u ∈ S is expressible simultaneously as i=0 αi χai = j=0 βj χbj , then Pn Pm i=0 αi χai + j=0 (−βj )χbj = 0 in S, so that

Pn i=0

and

αi νai + Pn i=0

Pm

j=0 (−βj )νbj

αi νai =

Pm j=0

= 0 in U ,

βj νbj .

We can therefore define T : S → U by setting Pn Pn T ( i=0 αi χai ) = i=0 αi νai whenever a0 , . . . , an ∈ A and α0 , . . . , αn ∈ R. (c) It is now elementary to check that T is linear, and that T χa = νa for every a ∈ A. Of course this last condition uniquely defines T , because {χa : a ∈ A} spans the linear space S. 361G Theorem Let A be a Boolean ring, and U a partially ordered linear space. Let ν : A → U be an additive function, and T : S(A) → U the corresponding linear operator. (a) ν is non-negative iff T is positive. (b) In this case, (i) if T is order-continuous or sequentially order-continuous, so is ν; (ii) if U is Archimedean and ν is order-continuous or sequentially order-continuous, so is T . (c) If U is a Riesz space, then the following are equiveridical: (i) T is a Riesz homomorphism; (ii) νa ∧ νb = 0 in U whenever a ∩ b = 0 in A; (iii) ν is a lattice homomorphism. proof Write S for S(A). (a) If T is positive, then surely νa = T χa ≥ 0 for P every a ∈ A, so ν = T χ is non-negative. If ν is m non-negative, and u ≥ 0 in S, then u is expressible as j=0 βj χbj where b0 , . . . , bm ∈ A and βj ≥ 0 for every j (361Ec), so that Pm T u = j=0 βj νbj ≥ 0. Thus T is positive. (b)(i) If T is order-continuous (resp. sequentially order-continuous) then ν = T χ is the composition of two order-continuous (resp. sequentially order-continuous) functions (361Ef), so must be order-continuous (resp. sequentially order-continuous). (ii) Assume now that U is Archimedean. α) Suppose that ν is order-continuous and that A ⊆ S is non-empty, downwards-directed and has (α infimum 0. Fix u0 ∈ A, set α = kuk∞ and a0 = [[u > 0]] (in the language of 361Eg). If α = 0 then of course inf u∈A T u = T u0 = 0. Otherwise, take any w ∈ U such that w 6≤ 0. Then there is some δ > 0 such that w 6≤ δνa0 , because U is Archimedean. Set A0 = {u : u ∈ A, u ≤ u0 }; because A is downwards-directed, A0 has the same lower bounds as A, and inf A0 = 0, while A0 is still downwards-directed. For u ∈ A0 set cu = [[u > δ]], so that δχcu ≤ u ≤ αχcu + δχ[[u > 0]] ≤ αχcu + δχa0 0

(361Eg). If u, v ∈ A and u ≤ v, then cu ⊆ cv , so C = {cu : u ∈ A0 } is downwards-directed; but if c is any lower bound for C in A, δχc is a lower bound for A0 in S, so is zero, and c = 0 in A. Thus inf C = 0 in A, and inf u∈A0 νcu = 0 in U . But this means, in particular, that α1 (w − δνa0 ) is not a lower bound for ν[C], and there is some u ∈ A0 such that α1 (w − δνa0 ) 6≤ νcu , that is, w − δνa0 6≤ ανcu , that is, w 6≤ δνa0 + ανcu . As u ≤ αχcu + δχa0 ,

361H

S

281

T u ≤ T (αχcu + δχa0 ) = ανcu + δνa0 , and w 6≤ T u. Since w is arbitrary, this means that 0 = inf T [A]; as A is arbitrary, T is order-continuous. β ) The argument for sequential order-continuity is essentially the same. Suppose that ν is se(β quentially order-continuous and that hun in∈N is a non-increasing sequence in S with infimum 0. Again set α = ku0 k, a0 = [[u0 > 0]]; again we may suppose that α > 0; again take any w ∈ U such that w 6≤ 0. As before, there is some δ > 0 such that w 6≤ δνa0 . For n ∈ N set cn = [[un > δ]], so that δχcn ≤ un ≤ αχcn + δχa0 . The sequence hcn in∈N is non-increasing because hun in∈N is, and if c ⊆ cn for every n, then δχc ≤ un for every n, so is zero, and c = 0 in A. Thus inf n∈N cn = 0 in A, and inf n∈N νcn = 0 in U , because ν is sequentially order-continuous. Replacing A0 , C in the argument above by {un : n ∈ N}, {cn : n ∈ N} we find an n such that w 6≤ T un . Since w is arbitrary, this means that 0 = inf n∈N T un ; as hun in∈N is arbitrary, T is sequentially order-continuous. (c)(i)⇒(iii) If T : S(A) → U is a Riesz homomorphism, and ν = T χ, then surely ν is a lattice homomorphism because T and χ are. (iii)⇒(ii) is trivial. Pm (ii)⇒(i) If νa ∧ νb = 0 whenever a ∩ b = 0, then for any u ∈ S(A) we have an expression of u as j=0 βj χbj , where b0 , . . . , bm ∈ A are disjoint. Now Pm Pm Pm |T u| = | j=0 βj νbj | = j=0 |βj |νbj = T ( j=0 |βj |χbj ) = T (|u|) by 352Fb and 361Ed. As u is arbitrary, T is a Riesz homomorphism (352G). 361H Theorem Let A be a Boolean ring and U a Dedekind complete Riesz space. Suppose that ν : A → U is an additive function and T : S = S(A) → U the corresponding linear operator. Then T ∈ L∼ = L∼ (S; U ) iff {νb : b ⊆ a} is order-bounded in U for every a ∈ A, and in this case |T | ∈ L∼ corresponds to |ν| : A → U , defined by setting |ν|(a) = sup{

n X

|νai | : a0 , . . . , an ⊆ a are disjoint}

j=0

= sup{νb − ν(a \ b) : b ⊆ a} for every a ∈ A. proof (a) Suppose that T ∈ L∼ and a ∈ A. Then for any b ⊆ a, we have χb ≤ χa so |νb| = |T χb| ≤ |T |(χa). Accordingly {νb : b ⊆ a} is order-bounded in U . (b) Now suppose that {νb : b ⊆ a} is order-bounded in U for every a ∈ A. Then for any a ∈ A we can define wa = sup{|νb| : b ⊆ a}; in this case, νb − ν(a \ b) ≤ 2wa whenever b ⊆ a, so θa = supb ⊆ a νb − ν(a \ b) is also defined in U . Considering b = a, b = 0 we see that θa ≥ |νa|. Next, θ : A → U is additive. P P Take a1 , a2 ∈ A such that a1 ∩ a2 = 0; set a0 = a1 ∪ a2 . For each j ≤ 2 set Aj = {ν(aj ∩ b) − ν(aj \ b) : b ∈ A} ⊆ U . Then A0 ⊆ A1 + A2 , because ν(a0 ∩ b) − ν(a0 \ b) = ν(a1 ∩ b) − ν(a1 \ b) + ν(a2 ∩ b) − ν(a2 \ b) for every b ∈ A. But also A1 + A2 ⊆ A0 , because if b1 , b2 ∈ A then ν(a1 ∩ b1 ) − ν(a1 \ b1 ) + ν(a2 ∩ b2 ) − ν(a2 \ b2 ) = ν(a0 ∩ b) − ν(a0 \ b) where b = (a1 ∩ b1 ) ∪ (a2 ∩ b2 ). So A0 = A1 + A2 , and θa0 = sup A0 = sup A1 + sup A2 = θa1 + θa2 (351Dc). Q Q

282

Function spaces

361H

We therefore Pn have a corresponding positive operator T1 : S → U such that θ = T1 χ. But we also see that θa = sup{ i=0 |νai | : a0 , . . . , an ⊆ a are disjoint} for every a ∈ A. P P If a0 , . . . , an are disjoint and included in a, then Pn Pn i=0 |νai | ≤ i=0 θai = θ(supi≤n ai ) ≤ θa. On the other hand,

Pn θa ≤ supb ⊆ a |νb| + |ν(a \ b)| ≤ sup{ i=0 |νai | : a0 , . . . , an ⊆ a are disjoint}. Q Q

It follows that T ∈ L∼ P Take any u ≥ 0 in S. Set a = [[u > 0]] (361Eg) and α = kuk∞ . If 0 < |v| ≤ u, P. nP then v is expressible as i=0 αi χai where a0 , . . . , an are disjoint and no αi nor ai is zero. Since |v| ≤ αχa, we must have |αi | ≤ α, ai ⊆ a for each i. So Pn Pn Pn |T v| = | i=0 αi χai | ≤ i=0 |αi ||νai | ≤ α i=0 |νai | = αθa. Thus {|T v| : |v| ≤ u} is bounded above by αθa. As u is arbitrary, T ∈ L∼ . Q Q (c) Thus T ∈ L∼ iff ν is order-bounded on the sets {b : b ⊆ a}, and in this case the two formulae offered for |ν| are consistent and make |ν| = θ. Finally, θ = |T |χ. P P Take a ∈ A. If a0 , . . . , an ⊆ a are disjoint, then Pn Pn Pn i=0 |νai | = i=0 |T χai | ≤ i=0 |T |(χai ) ≤ |T |(χa); so θa ≤ |T |(χa). On the other hand, the argument at the end of (b) above shows that |T |(χa) ≤ θa for every a. Thus |T |(χa) = θa for every a ∈ A, as required. Q Q 361I Theorem Let A be a Boolean ring, U a normed space and ν : A → U an additive function. Give S = S(A) its norm k k∞ , and let T : S → U be the linear operator corresponding to ν. Then T is a bounded linear operator iff {νa : a ∈ A} is bounded, and in this case kT k = supa,b∈A kνa − νbk. proof (a) If T is bounded, then kνa − νbk = kT (χa − χb)k ≤ kT kkχa − χbk∞ ≤ kT k for every a ∈ A, so ν is bounded and supa,b∈A kνa − νbk ≤ kT k. (b)(i) For the converse, in 361Ec. If u ∈ S and u ≥ 0 and kuk∞ ≤ 1, Pidea Pmwe need a refinement of an m P If u = 0, take n = 0, c0 = 0, γ0 = 1. then u is expressible as i=0 γi χci wherePγi ≥ 0 and i=0 γi = 1. P n Otherwise, start from an expression u = j=0 γj χcj where c0 ⊇ . . . ⊇ cn and every γj is non-negative, as in 361Ec. We may suppose that cn 6= 0, in which case Pn j=0 γj = u(z) ≤ 1 Pn for every z ∈ b cn ⊆ Z, the Stone space of A. Set m = n + 1, cm = 0 and γm = 1 − j=0 γj to get the required form. Q Q 0 , γn0 ≥ 0 (ii) The next Pm numbers: if γ0 , . . . , γm , γ0 ,0 . . . P P Pn fact0 we need is an elementary property of real n m and i=0 γi = j=0 γj , then there are δij ≥ 0 such that γi = j=0 δij for every i ≤ m and γj = i=0 δij for every j ≤ n. P P This is just the case U = R of 352Fd. Q Q

(iii) Now suppose that ν is bounded; set α0 = supa∈A kνak < ∞. Then α = supa,b∈A kνa − νbk ≤ 2α0 is also finite. If u ∈ S and kuk∞ ≤ 1, then we can express u as u+ − u− where u+ , u− are non-negative and also of norm at most 1. By (i), we can express these as Pm Pn u+ = i=0 γi χci , u− = j=0 γj0 χc0j Pm Pn where all the γi , γj0 are non-negative and i=0 γi = j=0 γj0 = 1. Take hδij ii≤m,j≤n from (ii). Set cij = ci , c0ij = c0j for all i, j, so that Pm Pn Pm Pn u+ = i=0 j=0 δij χcij , u− = i=0 j=0 δij χc0ij , u=

Pm Pn i=0

j=0 δij (χcij

− χc0ij ),

361J

S

Tu = kT uk ≤

Pm Pn

Pm Pn i=0

j=0 δij (νcij

j=0 δij kνcij

i=0

− νc0ij k ≤

283

− νc0ij ), Pm Pn i=0

j=0 δij α

= α.

As u is arbitrary, T is a bounded linear operator and kT k ≤ α, as required. 361J The last few paragraphs describe the properties of S(A) in terms of universal mapping theorems. The next theorem looks at the construction as a functor which converts Boolean algebras into Riesz spaces and ring homomorphisms into Riesz homomorphisms. Theorem Let A and B be Boolean rings and π : A → B a ring homomorphism. (a) We have a Riesz homomorphism Tπ : S(A) → S(B) given by the formula Tπ (χa) = χ(πa) for every a ∈ A. For any u ∈ S(A), kTπ uk∞ = min{ku0 k∞ : u0 ∈ S(A), Tπ u0 = Tπ u}; in particular, kTπ uk∞ ≤ kuk∞ . Moreover, Tπ (u × u0 ) = Tπ u × Tπ u0 for all u, u0 ∈ S(A). (b) Tπ is surjective iff π is surjective, and in this case kvk∞ = min{kuk∞ : u ∈ S(A), Tπ u = v} for every v ∈ S(B). (c) The kernel of Tπ is just the set of those u ∈ S(A) such that π[[|u| > 0]] = 0, defining [[. . . > . . . ]] as in 361Eg. (d) Tπ is injective iff π is injective, and in this case kTπ uk∞ = kuk∞ for every u ∈ S(A). (e) Tπ is order-continuous iff π is order-continuous. (f) Tπ is sequentially order-continuous iff π is sequentially order-continuous. (g) If C is another Boolean ring and φ : B → C is another ring homomorphism, then Tφπ = Tφ Tπ : S(A) → S(C). proof (a) The map χπ : A → S(B) is additive (361Cc), so corresponds to a linear operator T = Tπ : S(A) → S(B), by 361F. P χ and π are both lattice homomorphisms, so χπ also is, and T is a Riesz homomorphism n (361Gc). If u = i=0 αi χai , where a0 , . . . , an are disjoint, then look at I = {i : i ≤ n, πai 6= 0}. We have Pn P T u = i=0 αi χ(πai ) = i∈I αi χ(πai ) and πa0 , . . . , πan are disjoint, so that where u0 =

kT uk∞ = supi∈I |αi | = ku0 k∞ ≤ supai 6=0 |αi | ≤ kuk∞ ,

P i∈I

αi χai , so that T u0 = T u. If a, a0 ∈ A, then T (χa × χa0 ) = T χ(a ∩ a0 ) = χπ(a ∩ a0 ) = χπa × χπa0 = T χa × T χa0 ,

so T is multiplicative. (b) If π is surjective, then T [S(A)] must be the linear span of {T (χa) : a ∈ A} = {χ(πa) : a ∈ A} = {χb : b ∈ B}, so is the whole ofP S(B). If T is surjective, and b ∈ B, then there must be a u ∈ A such that T u = χb. We n can express u as i=0 αi χai where a0 , . . . , an are disjoint; now Pn χb = T u = i=0 αi χ(πai ), and πa0 , . . . , πan are disjoint in B, so we must have b = supi∈I πai = π(supi∈I ai ) ∈ π[A], where I = {i : αi = 1}. As b is arbitrary, π is surjective. Of course the formula for kvk∞ is a consequence of the formula for kT uk∞ in (a). (c)(i) If π[[|u| > 0]] = 0 then |u| ≤ αχa, where α = kuk∞ , a = [[|u| > 0]], so |T u| = T |u| ≤ αT (χa) = αχ(πa) = 0, Pn and T u = 0. (ii) If u ∈ S(A) and T u = 0, express u as i=0 αi χai where a0 , . . . , an are disjoint and every αi is non-zero (361Eb). In this case Pn 0 = |T u| = T |u| = i=0 |αi |χ(πai ),

284

Function spaces

361J

so πai = 0 for every i, and π[[|u| > 0]] = π(supi≤n ai ) = supi≤n πai = 0. (d) If T is injective and a ∈ A \ {0}, then χ(πa) = T (χa) 6= 0, so πa 6= 0; as a is arbitrary, π is injective. If π is injective then π[[|u| > 0]] 6= 0 for every non-zero u ∈ S(A), so T is injective, by (c). Now the formula in (a) shows that T is norm-preserving. (e)(i) If T is order-continuous and A ⊆ A is a non-empty downwards-directed set with infimum 0 in A, let b be any lower bound for π[A] in B. Then χb ≤ χ(πa) = T (χa) for any a ∈ A. But T χ is order-continuous, by 361Ef, so inf a∈A T (χa) = 0, and b must be 0. As b is arbitrary, inf a∈A πa = 0; as A is arbitrary, π is order-continuous. (ii) If π is order-continuous, so is χπ : A → S(B), using 361Ef again; but now by 361G(b-ii) T must be order-continuous. (f )(i) If T is sequentially order-continuous, and han in∈N is a non-increasing sequence in A with infimum 0, let b be any lower bound for {πan : n ∈ N} in B. Then χb ≤ χ(πan ) = T (χan ) for any a ∈ A. But T χ is sequentially order-continuous so inf n∈N T (χan ) = 0, and b must be 0. As b is arbitrary, inf n∈N πan = 0; as A is arbitrary, π is sequentially order-continuous. (ii) If π is sequentially order-continuous, so is χπ : A → S(B); but now T must be sequentially order-continuous. (g) We need only check that Tφπ (χa) = χ(φ(πa)) = Tφ (χ(πa)) = Tφ T (χa) for every a ∈ A. 361K Proposition Let A be a Boolean algebra. For a ∈ A write Va for the solid linear subspace of S(A) generated by χa. Then a 7→ Va is a Boolean isomorphism between A and the algebra of projection bands in S(A). proof Write S for S(A). (a) The point is that, for any a ∈ A, (i) |u| ∧ |v| = 0 whenever u ∈ Va , v ∈ V1\a , (ii) Va + V1\a = S. P P (i) is just because χa ∧ χ(1 \ a) = 0. As for (ii), if w ∈ S then w = (w × χa) + (w × χ(1 \ a)) ∈ Va + V1\a . Q Q (b) Accordingly Va + Va⊥ ⊇ Va + V1\a = U and Va is a projection band (352R). Next, any projection band U ⊆ S is of the form Va . P P We know that χ1 = u + v where u ∈ U , v ∈ U ⊥ . Since |u| ∧ |v| = 0, u and v must be the characteristic functions of complementary subsets of Z, the Stone space of A. But {z : u(z) 6= 0} = {z : u(z) ≥ 1} must be of the form b a, where a = [[u > 0]], in which case u = χa and v = χ(1 \ a). Accordingly U ⊇ Va and U ⊥ ⊇ V1\a . But this means that U must be Va precisely. Q Q (c) Thus a 7→ Va is a surjective function from A onto the algebra of projection bands in S. Now a ⊆ b ⇐⇒ χa ∈ Vb ⇐⇒ Va ⊆ Vb , so a 7→ Va is order-preserving and bijective. By 312L it is a Boolean isomorphism. 361L Proposition Let X be a set, and Σ a ring of subsets of X, that is, a subring of the Boolean ring PX. Then S(Σ) can be identified, as ordered linear space, with the linear subspace of `∞ (X) generated by the characteristic functions of members of Σ, which is a Riesz subspace of `∞ (X). The norm of S(Σ) corresponds to the uniform norm on `∞ (X), and its multiplication to pointwise multiplication of functions. proof Let Z be the Stone space of Σ, and for E ∈ Σ write χE for the characteristic function of E as a subset of X, χE ˆ for the characteristic function of the open-and-compact subset of Z corresponding to E.

361Xb

S

285

Of course χ : Σ → `∞ (X) is additive, so by 361F there is a linear operator T : S → `∞ (X), writing S for S(Σ), such that T (χE) ˆ = χE for every E ∈ Σ. Pm If u ∈ S, T u ≥ 0 iff u ≥ 0. P P Express u as ˆ j where E0 , . . . , Em are disjoint. Then j=0 βj χE Pm T u = j=0 βj χEj , so u ≥ 0 ⇐⇒ βj ≥ 0 whenever Ej 6= ∅ ⇐⇒ T u ≥ 0. Q Q But this means (α) that T u = 0 ⇐⇒ T u ≥ 0 & T (−u) ≥ 0 ⇐⇒ u ≥ 0 & − u ≥ 0 ⇐⇒ u = 0, so that T is injective and is a linear space isomorphism between S and its image S, which must be the linear space spanned by {χE : E ∈ Σ} (β) that T is an order-isomorphism between S and S. Because χE ∧ χF = 0 whenever E, F ∈ Σ and E ∩ F = ∅, T is a Riesz homomorphism and S is a Riesz subspace of `∞ (X) (361Gc). Now kuk∞ = inf{α : |u| ≤ αχX} ˆ = inf{α : |T u| ≤ αχX} = kT uk∞ for every u ∈ S. Finally, T (χE ˆ × χF ˆ ) = T (χ(E ˆ ∩ F )) = χ(E ∩ F ) = T (χE) ˆ × T (χF ˆ ) for all E, F ∈ Σ, so S is closed under pointwise multiplication and the multiplications of S, S are identified under T . 361M Proposition Let X be a set, Σ a ring of subsets of X, and I an ideal of Σ; write A for the quotient ring Σ/I. Let S be the linear span of {χE : E ∈ Σ} in RX , and write V = {f : f ∈ S, {x : f (x) 6= 0} ∈ I}. Then V is a solid linear subspace of S. Now S(A) becomes identified with the quotient Riesz space S/V , if for every E ∈ Σ we identify χ(E • ) ∈ S(A) with (χE)• ∈ S/V . If we give S its uniform norm inherited from `∞ (X), V is a closed linear subspace of S, and the quotient norm on S/V corresponds to the norm of S(A): kf • k = min{α : {x : |f (x)| > α} ∈ I}. If we write × for pointwise multiplication on S, then V is an ideal of the ring (S, +, ×), and the multiplication induced on S/V corresponds to the multiplication of S(A). proof Use 361J and 361L. We can identify S with S(Σ). Now the canonical ring homomorphism E 7→ E • corresponds to a surjective Riesz homomorphism T from S(Σ) to S(A) which takes χE to χ(E • ). For f ∈ S, [[|f | > 0]] is just {x : f (x) 6= 0}, so the kernel of T is just the set of those f ∈ S such that {x : f (x) 6= 0} ∈ I, which is V . So S(A) = T [S] ∼ = S/V . As noted in 361Ja, T (f × g) = T f × T g for all f , g ∈ S, so the multiplications of S/V and S(A) match. As for the norms, the norm of S(A) corresponds to the norm of S/V by the formulae in 361Ja or 361Jb. (To see that V is closed in S, we need note only that if f ∈ V then kT f k∞ = inf g∈V kf + gk∞ = inf g∈V kf − gk∞ = 0, so that T f = 0 and f ∈ V .) To check the formula for kf • k, take any f ∈ S. Express it as E0 , . . . , En ∈ Σ are disjoint. Set I = {i : Ei ∈ / I}; then

Pn i=0

αi χEi where

kT f k∞ = maxi∈I |αi | = min{α : {x : |f (x)| > α} ∈ I}. 361X Basic exercises (a) Let A be a Boolean ring and U a linear space. Show that a function ν : A → U is additive iff ν0 = 0 and ν(a ∪ b) + ν(a ∩ b) = νa + νb for all a, b ∈ A. >(b) Let U be an algebra over R, that is, a real linear space endowed with a multiplication × such that (U, +, ×) is a ring and α(w × z) = (αw) × z = w × (αz) for all w, z ∈ U and all α ∈ R. Let A be a Boolean ring, ν : A → U an additive function and T : S(A) → U the corresponding linear operator. Show that T is multiplicative iff ν(a ∩ b) = νa × νb for all a, b ∈ A.

286

Function spaces

361Xc

> (c) Let A be a Boolean ring, and U a Dedekind complete Riesz space. Suppose that ν : A → U is an additive function such that the corresponding linear operator T : S(A) → U belongs to L∼ = L∼ (S(A); U ). Show that T + ∈ L∼ corresponds to ν + : A → U , where ν + a = supb ⊆ a νb for every a ∈ A. (d) Let A and B be Boolean algebras. Show that there is a natural one-to-one correspondence between Boolean homomorphisms π : A → B and Riesz homomorphisms T : S(A) → S(B) such that T (χ1A ) = χ1B , given by setting T (χa) = χ(πa) for every a ∈ A. (e) Let A, B be Boolean rings and T : S(A) → S(B) a linear operator such that T (u × v) = T u × T v for all u, v ∈ S(A). Show that there is a ring homomorphism π : A → B such that T (χa) = χ(πa) for every a ∈ A. (f ) Let A and B be Boolean rings. Show that any isomorphism of the algebras S(A) and S(B) (using the word ‘algebra’ in the sense of 361Xb) must be a Riesz space isomorphism, and therefore corresponds to an isomorphism between A and B. (g) Let A, B be Boolean algebras and T : S(A) → S(B) a Riesz homomorphism. Show that there are a ring homomorphism π : A → B and a non-negative v ∈ S(B) such that T (χa) = v × χ(πa) for every a ∈ A. (h) Let A be a Boolean ring. Show that for any u ∈ S(A) the solid linear subspace of S(A) generated by u is a projection band in S(A). Show that the set of such bands is an ideal in the algebra of all projection bands, and is isomorphic to A. > (i) Let X be a set, Σ a σ-algebra of subsets of X. Show that the linear span S in RX of {χE : E ∈ Σ} is just the set of Σ-measurable functions f : X → R which take only finitely many values. (j) For any Boolean ring A, we may define its ‘complex S-space’ SC (A) as the linear span in CX of the characteristic functions of open-and-compact subsets of the Stone space Z of A. State and prove results corresponding to 361Ea-361Ed, 361Eh, 361F, 361L and 361M. 361Y P Further exercises (a) Let A be a Boolean ring. Let V be the linear space of all formal sums of n the form i=0 αi ai where α0 , . . . , αn ∈ R and a0 , . . . , an ∈ A. Let W ⊆ V be the linear subspace spanned by members of V of the form (a ∪ b) − a − b where a, b ∈ A are disjoint. Define χ0 : A → V /W by taking χ0 a to be the image in V /W of a ∈ V . Show, without using the axiom of choice, that the pair (V /W, χ0 ) has the universal mapping property of (S(A), χ) as described in 361F and that V /W has a Riesz space structure, a norm and a multiplicative structure as described in 361D-361E. Prove results corresponding to 361E-361M. (b) Let A be a Boolean ring and U a Dedekind complete Riesz space. Let A ⊆ L∼ = L∼ (S(A); U ) be a non-empty set. Suppose that T˜ = sup A is defined in L∼ , and that ν˜ = T˜χ. Show that for any a ∈ A, Pn ν˜a = sup{ i=0 Ti (χai ) : T0 , . . . , Tn ∈ A, a0 , . . . , an ⊆ a are disjoint, supi≤n ai = a}. (c) Let A be a Boolean algebra. Show that the algebra of all bands of S(A) can be identified with the Dedekind completion of A (314U). (d) Let A be a Boolean ring, and U a complex normed space. Let ν : A → U be an additive function and T : SC (A) → U the corresponding linear operator (cf. 361Xj). Show that (giving SC (A) its usual norm k k∞ ) Pn kT k = sup{k j=0 ζj νaj k : a0 , . . . , an ∈ A are disjoint, |ζj | = 1 for every j} if either is finite. (e) Let U be a Riesz space. Show that it is isomorphic to S(A), for some Boolean algebra A, iff it has an order unit and every solid linear subspace of U is a projection band. (f ) Let hAi ii∈I be a non-empty family of Boolean algebras, with free product A; write εi : Ai → A for the canonical maps, and

§362 intro.

S∼

287

C = {inf j∈J εj (aj ) : J ⊆ I is finite, aj ∈ Aj for every j ∈ J}. Suppose that U is a linear space and θ : C → U is such that θc = θ(c ∩ εi (a)) + θ(c ∩ εi (1 \ a)) whenever c ∈ C, i ∈ I and a ∈ Ai . Show that there is a unique additive function ν : A → U extending θ. (Hint: 326Q.) 361 Notes and comments The space S(A) corresponds of course to the idea of ‘simple function’ which belongs to the very beginnings of the theory of integration (122A). All that 361D is trying to do is to set up a logically sound description of this obvious concept which can be derived from the Boolean ring A itself. To my eye, there is a defect in the construction there. It relies on the axiom of choice, since it uses the Stone space; but none of the elementary properties of S(A) have anything to do with the axiom of choice. In 361Ya I offer an alternative construction which is in a formal sense more ‘elementary’. If you work through the suggestion there you will find, however, that the technical details become significantly more complicated, and would be intolerable were it not for the intuition provided by the Stone space construction. Of course this intuition is chiefly valuable in the finitistic arguments used in 361E, 361F and 361I; and for these arguments we really need the Stone representation only for finite Boolean rings, which does not depend on the axiom of choice. It is quite true that in most of this volume (and in most of this chapter) I use the axiom of choice without scruple and without comment. I mention it here only because I find myself using arguments dependent on choice to prove theorems of a type to which the axiom cannot be relevant. The linear space structure of S(A), together with the map χ, are uniquely determined by the first universal mapping theorem here, 361F. This result says nothing about the order structure, which needs the further refinement in 361Ga. What is striking is that the partial order defined by 361Ga is actually a lattice ordering, so that we can have a universal mapping theorem for functions to Riesz spaces, as in 361Gc and 361Ja. Moreover, the same ordering provides a happy abundance of results concerning order-continuous functions (361Gb, 361Je-361Jf). When the codomain is a Dedekind complete Riesz space, so that we have a Riesz space L∼ (S; U ), and a corresponding modulus function T 7→ |T | for linear operators, there are reasonably natural formulae for |T |χ in terms of T χ (361H); see also 361Xc and 361Yb. The multiplicative structure of S(A) is defined by 361Xb, and the norm by 361I. The Boolean ring A cannot be recovered from the linear space structure of S(A) alone (since this tells us only the cardinality of A), but if we add either the ordering or the multiplication of S(A) then A is easy to identify (361K, 361Xf). The most important Boolean algebras of measure theory arise either as algebras of sets or as their quotients, so it is a welcome fact that in such cases the spaces S(A) have straightforward representations in terms of the construction of A (361L-361M). In Chapter 24 I offered a paragraph in each section to sketch a version of the theory based on the field of complex numbers rather than the field of real numbers. This was because so many of the most important applications of these ideas involve complex numbers, even though (in my view) the ideas themselves are most clearly and characteristically expressed in terms of real numbers. In the present chapter we are one step farther away from these applications, and I therefore relegate complex numbers to the exercises, as in 361Xj and 361Yd.

362 S ∼ The next stage in our journey is the systematic investigation of linear functionals on spaces S = S(A). We already know that these correspond to additive real-valued functionals on the algebra A (361F). My purpose here is to show how the structure of the Riesz space dual S ∼ and its bands is related to the classes of additive functionals introduced in §§326-327. The first step is just to check the identification of the linear and order structures of S ∼ and the space M of bounded finitely additive functionals (362A); all the ideas needed for this have already been set out, and the basic properties of S ∼ are covered by the general results in §356. Next, we need to be able to describe the operations on M corresponding to the Riesz space operations

288

Function spaces

§362 intro.

| |, ∨, ∧ on S ∼ , and the band projections from S ∼ onto Sc∼ and S × ; these are dealt with in 362B, with a supplementary remark in 362D. In the case of measure algebras, we have some further important bands which present themselves in M , rather than in S ∼ , and which are treated in 362C. Since all these spaces are L-spaces, it is worth taking a moment to identify their uniformly integrable subsets; I do this in 362F. While some of the ideas here have interesting extensions to the case in which A is a Boolean ring without identity, these can I think be left to one side; the work of this section will be done on the assumption that every A is a Boolean algebra. 362A Theorem Let A be a Boolean algebra. Write S for S(A). (a) The partially ordered linear space of all finitely additive real-valued functionals on A may be identified with the partially ordered linear space of all real-valued linear functionals on S. (b) The linear space of bounded finitely additive real-valued functionals on A may be identified with the L-space S ∼ of order-bounded linear functionals on S. If f ∈ S ∼ corresponds to ν : A → R, then f + ∈ S ∼ corresponds to ν + , where ν + a = supb ⊆ a νb for every a ∈ A, and kf k = supa∈A νa − ν(1 \ a). (c) The linear space of bounded countably additive real-valued functionals on A may be identified with the L-space Sc∼ . (d) The linear space of completely additive real-valued functionals on A may be identified with the L-space S×. proof By 361F, we have a canonical one-to-one correspondence between linear functionals f : S → R and additive functionals νf : A → R, given by setting νf = f χ. (a) Now it is clear that νf +g = νf + νg , ναf = ανf for all f , g and α, so this one-to-one correspondence is a linear space isomorphism. To see that it is also an order-isomorphism, we need note only that νf is non-negative iff f is, by 361Ga. (b) Recall from 356N that, because S is a Riesz space with order unit (361Ee), S ∼ has a corresponding norm under which it is an L-space. (i) If f ∈ S ∼ , then supb∈A |νf b| = supb∈A |f (χb)| ≤ sup{|f (u)| : u ∈ S, |u| ≤ χ1} is finite, and νf is bounded. (ii) Now suppose that νf is bounded and that v ∈PS + . Then there is an α ≥ 0 such that v ≤ αχ1 n (361Ee). If u ∈ S and |u| ≤ v, then we can express u as i=0 αi χai where a0 , . . . , an are disjoint (361Eb); now |αi | ≤ α whenever ai 6= 0, so Pn Pn |f (u)| = | i=0 αi νf ai | ≤ α i=0 |νf ai | = α(νf c1 − νf c2 ) ≤ 2α supb∈A |νf b|, setting c1 = sup{ai : i ≤ n, νf ai ≥ 0}, c2 = sup{ai : i ≤ n, νf ai < 0}. This shows that {f (u) : |u| ≤ v} is bounded. As v is arbitrary, f ∈ S ∼ (356Aa). (iii) To check the correspondence between f + and νf+ , refine the arguments of (i) and (ii) as follows. Take any f ∈ S ∼ . If a ∈ A, νf+ a = supb ⊆ a νf b = supb ⊆ a f (χb) ≤ sup{f (u) : u ∈ S, 0 ≤ u ≤ χa} = f + (χa). Pn On the other hand, if u ∈ S and 0 ≤ u ≤ χa, then we can express u as i=0 αi χai where a0 , . . . , an are disjoint; now 0 ≤ αi ≤ 1 whenever ai 6= 0, so Pn f (u) = i=0 αi νf ai ≤ νf c ≤ νf+ a, where c = sup{ai : i ≤ n, νf ai ≥ 0}. As u is arbitrary, f + (χa) ≤ νf+ a. This shows that νf+ = f + χ is finitely additive, and that νf+ = νf + , as claimed.

S∼

362B

289

(iv) Now, for any f ∈ S ∼ ,

kf k = |f |(χ1) (356N) = (2f + − f )(χ1) (352D) = 2νf+ 1 − νf 1 (by (iii) just above) = sup 2νf a − νf 1 = sup νf a − νf (1 \ a). a∈A

a∈A

(c) If f ≥ 0 in S ∼ , then f is sequentially order-continuous iff νf is sequentially order-continuous (361Gb), that is, iff νf is countably additive (326Gc). Generally, an order-bounded linear functional belongs to Sc∼ iff it is expressible as the difference of two sequentially order-continuous positive linear functionals (356Ab), while a bounded finitely additive functional is countably additive iff it is expressible as the difference of two non-negative countably additive functionals (326H); so in the present context f ∈ Sc∼ iff νf is bounded and countably additive. (d) If f ≥ 0 in S ∼ , then f is order-continuous iff νf is order-continuous (361Gb), that is, iff νf is completely additive (326Kc). Generally, an order-bounded linear functional belongs to S × iff it is expressible as the difference of two order-continuous positive linear functionals (356Ac), while a finitely additive functional is completely additive iff it is expressible as the difference of two non-negative completely additive functionals (326M); so in the present context f ∈ S × iff νf is completely additive. 362B Spaces of finitely additive functionals The identifications in the last theorem mean that we can relate the Riesz space structure of S(A)∼ to constructions involving finitely additive functionals. I have already set out the most useful facts as exercises (326Yj, 326Ym, 326Yn, 326Yp, 326Yq); it is now time to repeat them more formally. Theorem Let A be a Boolean algebra. Write S = S(A), and let M be the Riesz space of bounded finitely additive real-valued functionals on A, Mσ ⊆ M the space of bounded countably additive functionals, and Mτ ⊆ Mσ the space of completely additive functionals. (a) For any µ, ν ∈ M , µ ∨ ν, µ ∧ ν and |ν| are defined by the formulae (µ ∨ ν)(a) = supb ⊆ a µb + ν(a \ b), (µ ∧ ν)(a) = inf b ⊆ a µb + ν(a \ b), |ν|(a) = supb ⊆ a νb − ν(a \ b) = supb,c ⊆ a νb − νc for every a ∈ A. Setting kνk = |ν|(1) = supa∈A νa − ν(1 \ a), M becomes an L-space. (b) Mσ and Mτ are projection bands in M , therefore L-spaces in their own right. In particular, |ν| ∈ Mσ for every ν ∈ Mσ , and |ν| ∈ Mτ for every ν ∈ Mτ . (c) The band projection Pσ : M → Mσ is defined by the formula (Pσ ν)(c) = inf{supn∈N νan : han in∈N is a non-decreasing sequence with supremum c} whenever c ∈ A and ν ≥ 0 in M . (d) The band projection Pτ : M → Mτ is defined by the formula (Pτ ν)(c) = inf{supa∈A νa : A is a non-empty upwards-directed set with supremum c} whenever c ∈ A and ν ≥ 0 in M .

290

Function spaces

362B

(e) If A ⊆ M is upwards-directed, then A is bounded above in M iff {ν1 : ν ∈ A} is bounded above in R, and in this case (if A 6= ∅) sup A is defined by the formula (sup A)(a) = supν∈A νa for every a ∈ A. (f) Suppose that µ, ν ∈ M . (i) The following are equiveridical: (α) ν belongs to the band in M generated by µ; (β) for every ² > 0 there is a δ > 0 such that |νa| ≤ ² whenever |µ|a ≤ δ; (γ) limn→∞ νan = 0 whenever han in∈N is a non-increasing sequence in A such that limn→∞ |µ|(an ) = 0. (ii) Now suppose that µ, ν ≥ 0, and let ν1 , ν2 be the components of ν in the band generated by µ and its complement. Then ν1 c = supδ>0 inf µa≤δ ν(c \ a),

ν2 c = inf δ>0 supa ⊆ c,µa≤δ νa

for every c ∈ A. proof (a) Of course µ ∨ ν = ν + (µ − ν)+ , µ ∧ ν = ν − (ν − µ)+ , |ν| = ν ∨ (−ν) (352D), so the formula of 362Ab gives (µ ∨ ν)(a) = νa + sup µb − νb = sup µb + ν(a \ b), b⊆a

b⊆a

(µ ∧ ν)(a) = νa − sup νb − µb = inf µb + ν(a \ b), b⊆a

b⊆a

|ν|(a) = sup νb − ν(a \ b) ≤ sup νb − νc = sup ν(b \ c) − ν(c \ b) b⊆a

b,c ⊆ a

b,c ⊆ a

≤ sup |ν|(b \ c) + |ν|(c \ b) = sup |ν|(b 4 c) ≤ |ν|(a). b,c ⊆ a

b,c ⊆ a

The formula offered for kνk corresponds exactly to the formula in 362Ab for the norm of the associated member of S(A)∼ ; because S(A)∼ is an L-space under its norm, so is M . × (b) By 362Ac-362Ad, Mσ and Mτ may be identified with S(A)∼ c and S(A) , which are projection bands in S(A)∼ (356B); so that Mσ and Mτ are projection bands in M , and are L-spaces in their own right (354O).

(c) Take any ν ≥ 0 in M . Set νσ c = inf{supn∈N νan : han in∈N is a non-decreasing sequence with supremum c} for every c ∈ A. Then of course 0 ≤ νσ c ≤ νc for every c. The point is that νσ is countably additive. P P Let hci ii∈N be a disjoint sequence in A, with supremum c. Then for any ² > 0 we have non-decreasing sequences han in∈N , hain in∈N , for i ∈ N, such that supn∈N an = c,

supn∈N ain = ci for i ∈ N,

supn∈N νan ≤ νσ c + ², supn∈N νain ≤ νσ ci + 2−i ² for every i ∈ N. Set bn = supi≤n ain for each n; then hbn in∈N is non-decreasing, and supn∈N bn = supi,n∈N ain = supi∈N ci = c, so νσ c ≤ sup νbn = sup n∈N

=

∞ X

n X

νain

n∈N i=0 ∞ X

sup νain ≤

i=0 n∈N

i=0

νσ ci + 2−i ² =

∞ X

νσ ci + 2².

i=0

On the other hand, han ∩ ci in∈N is a non-decreasing sequence with supremum c ∩ ci = ci for each i, so νσ ci ≤ supn∈N ν(an ∩ ci ), and

S∼

362B

∞ X

νσ ci ≤

i=0

∞ X

291

sup ν(an ∩ ci ) = sup

i=0 n∈N

∞ X

n∈N i=0

ν(an ∩ ci )

(because han in∈N is non-decreasing) ≤ sup νan n∈N

(because hci ii∈N is disjoint) ≤ νσ c + ². P∞

As ² is arbitrary, νσ c = i=0 νσ ci ; as hci ii∈N is arbitrary, νσ is countably additive. Q Q Thus νσ ∈ Mσ . On the other hand, if ν 0 ∈ Mσ and 0 ≤ ν 0 ≤ ν, then whenever c ∈ A and han in∈N is a non-decreasing sequence with supremum c, ν 0 c = supn∈N ν 0 an ≤ supn∈N νan . So we must have ν 0 c ≤ νσ c. This means that νσ = sup{ν 0 : ν 0 ∈ Mσ , ν 0 ≤ ν} = Pσ ν, as claimed. (d) The same ideas, with essentially elementary modifications, deal with the completely additive part. Take any ν ≥ 0 in M . Set ντ c = inf{supa∈A νa : A is a non-empty upwards-directed set with supremum c} for every c ∈ A. Then of course 0 ≤ ντ c ≤ νc for every c. The point is that ντ is completely additive. P P Note first that if c ∈ A, ² > 0 there is a non-empty upwards-directed A, with supremum c, such that supa∈A νa ≤ ντ c + ²νc; for if νc = 0 we can take A = {c}. Now let hci ii∈I be a partition of unity in A. Then for any ² > 0 we have non-empty upwards-directed sets A, Ai , for i ∈ I, such that sup A = 1,

sup Ai = ci for i ∈ I,

supa∈A νa ≤ ντ 1 + ²ν1,

supa∈Ai νa ≤ ντ ci + ²νci for every i ∈ I. Set B = {supi∈J ai : J ⊆ I is finite, ai ∈ Ai for every i ∈ J}; then B is non-empty and upwards-directed, and sup B = sup(

S i∈I

Ai ) = 1,

so X ντ 1 ≤ sup νb = sup{ νai : J ⊆ I is finite, ai ∈ Ai ∀ i ∈ J} b∈B

≤

X

i∈J

ντ ci + ²νci ≤ ²ν1 +

i∈I

X

ντ ci .

i∈I

On the other hand, A0i = {a ∩ ci : a ∈ A} is a non-empty upwards-directed set with supremum ci for each i, so ντ ci ≤ supa∈A0i νa, and X i∈I

ντ ci ≤

X i∈I

sup ν(a ∩ ci ) = sup

a∈A

a∈A

X

ν(a ∩ ci )

i∈I

≤ sup νa ≤ ντ 1 + ²ν1. a∈A

P

As ² is arbitrary, ντ c = i∈I ντ ci ; as hci ii∈I is arbitrary, ντ is completely additive, by 326N. Q Q Thus ντ ∈ Mτ . On the other hand, if ν 0 ∈ Mτ and 0 ≤ ν 0 ≤ ν, then whenever c ∈ A and A is a non-empty upwards-directed set with supremum c,

292

Function spaces

362B

ν 0 c = supa∈A ν 0 a ≤ supa∈A νa (using 326Kc). So we must have ν 0 c ≤ ντ c. This means that ντ = sup{ν 0 : ν 0 ∈ Mτ , ν 0 ≤ ν} = Pτ ν, as claimed. (e) If A is empty, of course it is bounded above in M , and {ν1 : ν ∈ A} = ∅ is bounded above in R; so let us suppose that A is not empty. In this case, if λ0 ∈ M is an upper bound for A, then λ0 1 is an upper bound for {ν1 : ν ∈ A}. On the other hand, if supν∈A ν1 = γ is finite, γ ∗ = sup{νa : ν ∈ A, a ∈ A} is finite. P P Fix ν0 ∈ A. Set γ1 = supa∈A |ν0 a| < ∞. Then for any ν ∈ A, a ∈ A there is a ν 0 ∈ A such that ν0 ∨ ν ≤ ν 0 , so that νa ≤ ν 0 a = ν 0 1 − ν 0 (1 \ a) ≤ γ − ν0 (1 \ a) ≤ γ + γ1 . So γ ∗ ≤ γ + γ1 < ∞. Q Q Set λa = supν∈A νa for every a ∈ A. Then λ : A → R is additive. P P If a, b ∈ A are disjoint, then λ(a ∪ b) = sup ν(a ∪ b) = sup νa + νb = sup νa + sup νb ν∈A

ν∈A

ν∈A

ν∈A

(because A is upwards-directed) = λa + λb. Q Q Also λa ≤ γ ∗ for every a, so |λa| = max(λa, −λa) = max(λa, λ(1 \ a) − λ1) ≤ γ ∗ + |λ1| for every a ∈ A, and λ is bounded. This shows that λ ∈ M , so that A is bounded above in M . Of course λ must be actually the least upper bound of A in M . α)⇒(β β ) Suppose that ν belongs to the band in M generated by µ, that is, |ν| = supn∈N |ν| ∧ n|µ| (f )(i)(α (352Vb). Let ² > 0. Then there is an n ∈ N such that |ν|(1) ≤ 12 ² + (|ν| ∧ n|µ|)(1) ((e) above). Set 1 δ = 2n+1 ² > 0. If |µ|(a) ≤ δ, then |νa| ≤ |ν|(a) = (|ν| ∧ n|µ|)(a) + (|ν| − |ν| ∧ n|µ|)(a) 1 2

≤ n|µ|(a) + (|ν| − |ν| ∧ n|µ|)(1) ≤ nδ + ² ≤ ². So (β) is satisfied. β )⇒(α α) Suppose that ν does not belong to the band in M generated by |µ|. Then there is a ν1 > 0 (β such that ν1 ≤ |ν| and ν1 ∧ |µ| = 0 (353C). For any δ > 0, there is an a ∈ A such that ν1 (1 \ a) + |µ|(a) ≤ min(δ, 12 ν1 1) ((a) above); now |µ|(a) ≤ δ but |ν|(a) ≥ ν1 a = ν1 1 − ν1 (1 \ a) ≥ ν1 1 − 12 ν1 1 = 12 ν1 1. Thus µ, ν do not satisfy (β) (with ² = 21 ν1 1). β )⇒(γγ ) is trivial. (β α) Observe first that if hck ik∈N is a non-increasing sequence in A such that limk→∞ µck = 0, (γγ )⇒(α then limk→∞ ν + ck = 0. P P Let ² > 0. Because ν + ∧ ν − = 0, there is a b ∈ A such that ν + b + ν − (1 \ b) ≤ ², by part (a). Now hck \ bik∈N is non-increasing and limk→∞ µ(ck \ b) = 0, so limk→∞ ν(ck \ b) = 0 and lim sup ν + ck = lim sup ν + (ck ∩ b) + ν(ck \ b) + ν − (ck \ b) k→∞

k→∞

≤ ν + b + ν − (1 \ b) ≤ ². As ² is arbitrary, limk→∞ ν + ck = 0. Q Q

S∼

362C

293

?? Now suppose, if possible, that ν + does not belong to the band generated by µ. Then there is a ν1 > 0 such that ν1 ≤ ν + and ν1 ∧ |µ| = 0. Set ² = 14 ν1 1 > 0. For each n ∈ N, we can choose an ∈ A such that |µ|an + ν1 (1 \ an ) ≤ 2−n ², by part (a) again. For n ≥ k, set bkn = supk≤i≤n ai ; then Pn |µ|bkn ≤ i=k |µ|ai ≤ 2−k+1 ², and hbkn in≥k is non-decreasing. Set γk = supn≥k ν1 bkn and choose m(k) ≥ k such that ν1 bk,m(k) ≥ γk −2−k ². Setting bk = bk,m(k) , we see that bk ∪ bk+1 = bkn where n = max(m(k), m(k + 1)), so that ν1 (bk ∪ bk+1 ) ≤ γk ≤ ν1 bk + 2−k ² and ν1 (bk+1 \ bk ) ≤ 2−k ². Set ck = inf i≤k bi for each k; then ν1 (bk+1 \ ck+1 ) = ν1 (bk+1 \ ck ) ≤ ν1 (bk+1 \ bk ) + ν1 (bk \ ck ) ≤ 2−k ² + ν1 (bk \ ck ) for each k; inducing on k, we see that ν1 (bk \ ck ) ≤

Pk−1 i=0

2−i ² ≤ 2²

for every k. This means that ν + ck ≥ ν1 ck ≥ ν1 bk − 2² ≥ ν1 ak − 2² = ν1 1 − ν1 (1 \ ak ) − 2² ≥ 4² − ² − 2² = ² for every k ∈ N. On the other hand, hck ik∈N is a non-decreasing sequence and |µ|ck ≤ |µ|bk ≤ 2−k+1 ² for every k, which contradicts the paragraph just above. X X This means that ν + must belong to the band generated by µ. Similarly ν − = (−ν)+ belongs to the band generated by µ and ν = ν + + ν − also does. (ii) Take c ∈ A. Set β1 = supδ>0 inf µa≤δ ν(c \ a),

β2 = inf δ>0 supa ⊆ c,µa≤δ νa.

Then β1 = supδ>0 inf a ⊆ c,µa≤δ ν(c \ a) = νc − β2 . Take any ² > 0. Because ν1 belongs to the band generated by µ, part (i) tells us that there is a δ > 0 such that ν1 a ≤ ² whenever µa ≤ δ. In this case, if µa ≤ δ, ν(c \ a) = νc − ν(c ∩ a) ≥ νc − ² ≥ ν1 c − ²; thus β1 ≥ inf µa≤δ ν(c \ a) ≥ ν1 c − ². As ² is arbitrary, β1 ≥ ν1 c. On the other hand, given ², δ > 0, there is an a ⊆ c such that µa + ν2 (c \ a) ≤ min(δ, ²), because µ ∧ ν2 = 0 (using (a) again). In this case, of course, µa ≤ δ, while νa ≥ ν2 a = ν2 c − ν2 (c \ a) ≥ ν2 c − ². Thus supa ⊆ c,µa≤δ νa ≥ ν2 c − ². As δ is arbitrary, β2 ≥ ν2 c − ². As ² is arbitrary, β2 ≥ ν2 c; but as β1 + β2 = νc = ν1 c + ν2 c, βi = νi c for both i, as claimed. 362C The formula in 362B(f-i) has, I hope, already reminded you of the concept of ‘absolutely continuous’ additive functional from the Radon-Nikod´ ym theorem (Chapter 23, §327). The expressions in 362Bf are limited by the assumption that µ, like ν, is finite-valued. If we relax this we get an alternative version of some of the same ideas. Theorem Let (A, µ ¯) be a measure algebra. Write S = S(A), and let M be the Riesz space of bounded finitely additive real-valued functionals on A. Write Mac = {ν : ν ∈ M is absolutely continuous with respect to µ ¯} (see 327A),

294

Function spaces

362C

Mtc = {ν : ν ∈ M is continuous with respect to the measure-algebra topology on A}, Mt = {ν : ν ∈ M , |ν|1 = supµ¯a 0 there is a δ > 0 such that |νa| ≤ 12 ² whenever µ ¯a ≤ δ. Now |ν 0 a| ≤ |ν 0 |(a) ≤ |ν|(a) ≤ 2 supc ⊆ a |νc| ≤ ² (using the formula for |ν| in 362Ba) whenever µ ¯a ≤ δ. As ² is arbitrary, ν 0 is absolutely continuous. Q Q (iii) If A ⊆ Mac is non-empty and upwards-directed and ν = sup A in M , then ν ∈ Mac . P P Let ² > 0. Then there is a ν 0 ∈ A such that ν1 ≤ ν 0 1 + 21 ² (362Be). Now there is a δ > 0 such that |νa| ≤ 12 ² whenever µ ¯a ≤ δ. If now µ ¯a ≤ δ, |νa| ≤ |ν 0 a| + (ν − ν 0 )(a) ≤ 21 ² + (ν − ν 0 )(1) ≤ ². As ² is arbitrary, ν is absolutely continuous with respect to µ ¯. Q Q Putting these together, we see that Mac is a band. (b)(i) We know that Mtc consists just of those ν ∈ M which are continuous at 0 (327Bc). Of course this is a linear subspace of M . (ii) If ν ∈ Mtc , ν 0 ∈ M and |ν 0 | ≤ |ν| then |ν| ∈ Mtc . P P Write Af = {d : d ∈ A, µ ¯d < ∞}. Given ² > 0 1 f ¯(a ∩ d) ≤ δ. Now there are d ∈ A , δ > 0 such that |νa| ≤ 2 ² whenever µ |ν 0 a| ≤ |ν 0 |(a) ≤ |ν|(a) ≤ 2 supc ⊆ a |νc| ≤ ² whenever µ ¯(a ∩ d) ≤ δ. As ² is arbitrary, ν 0 is continuous at 0 and belongs to Mtc . Q Q (iii) If A ⊆ Mtc is non-empty and upwards-directed and ν = sup A in M , then ν ∈ Mtc . P P Let ² > 0. Then there is a ν 0 ∈ A such that ν1 ≤ ν 0 1 + 21 ² (362Be). There are d ∈ Af , δ > 0 such that |νa| ≤ 21 ² whenever µ ¯(a ∩ d) ≤ δ. If now µ ¯(a ∩ d) ≤ δ, |νa| ≤ |ν 0 a| + (ν − ν 0 )(a) ≤ 21 ² + (ν − ν 0 )(1) ≤ ². As ² is arbitrary, ν is continuous at 0, therefore belongs to Mtc . Q Q Putting these together, we see that Mtc is a band. (c)(i) Mt is a linear subspace of M . P P Suppose that ν1 , ν2 ∈ Mt and α ∈ R. Given ² > 0, there are a1 , ² a2 ∈ Af such that |ν1 |(1 \ a1 ) ≤ 1+|α| , |ν2 |(1 \ a2 ) ≤ ². Set a = a1 ∪ a2 ; then µ ¯a < ∞ and |ν1 + ν2 |(1 \ a) ≤ 2²,

|αν1 |(1 \ a) ≤ ².

As ² is arbitrary, ν1 + ν2 and αν1 belong to Mt ; as ν1 , ν2 and α are arbitrary, Mt is a linear subspace of M . Q Q (ii) If ν ∈ Mt , ν 0 ∈ M and |ν 0 | ≤ |ν| then inf µ¯a 0. Next, for each n, we can find an ⊆ cn such that |νn an | ≥ 21 |νn |(cn ), so that lim supn∈N supν∈C |νan | ≥ lim supn→∞ |νn an | > 0. Since han in∈N , like hcn in∈N , is disjoint, the condition is not satisfied. This completes the proof. 362X Basic exercises (a) Let A be a Dedekind σ-complete Boolean algebra and ν1 , ν2 two countably additive functionals on A. Show that |ν1 |∧|ν2 | = 0 in the Riesz space of bounded finitely additive functionals on A iff there is a c ∈ A such that ν1 a = ν1 (a ∩ c) and ν2 a = ν2 (a \ c) for every a ∈ A. (b) Let (A, µ ¯) be a measure algebra. Write M , Mac as in 362C. Show that for any non-negative ν ∈ M , the component νac of ν in Mac is given by the formula νac c = supδ>0 inf µ¯a≤δ ν(c \ a).

296

Function spaces

362Xc

(c) Let (A, µ ¯) be a measure algebra. Write M , Mt as in 362C. (i) Show that Mt is just the set of those ν ∈ M such that νa = limb→F ν(a ∩ b) for every a ∈ A, where F is the filter on A generated by the sets {b : b ∈ Af , b ⊇ b0 } as b0 runs through the elements of A of finite measure. (ii) Show that the complementary band Mt⊥ of Mt in M is just the set of ν ∈ M such that νa = 0 whenever µ ¯a < ∞. (iii) Show that for any ν ∈ M , its component νt in Mt is given by the formula νt a = limb→F ν(a ∩ b) for every a ∈ A. (d) Let (A, µ ¯) be a measure algebra. Write M , Mσ , Mτ , Mac , Mtc and Mt as in 362B-362C. Show that (i) Mσ ⊆ Mac (ii) Mac ∩ Mt = Mtc ⊆ Mτ (iii) if (A, µ ¯) is σ-finite, then Mσ = Mt ∩ Mac . (e) Let A be a Boolean algebra. Let us say that a non-zero finitely additive functional ν : A → R is an atom if whenever a, b ∈ A and a ∩ b = 0 then at least one of νa, νb is zero. Show that for a non-zero finitely additive functional ν the following are equiveridical: (i) ν is an atom; (ii) ν is bounded and |ν| is an atom; (iii) ν is bounded and the corresponding linear functional f|ν| = |fν | ∈ S(A)∼ is a Riesz homomorphism; (iv) there are a multiplicative linear functional f : S(A) → R and an α ∈ R such that νa = αf (χa) for every a ∈ A. Show that a completely additive functional ν : A → R is an atom iff there are a ∈ A, α ∈ R \ {0} such that a is an atom in A and νb = α when a ⊆ b, 0 when a ∩ b = 0. (f ) Let A be a Boolean algebra. Let us say that a bounded finitely additive functional ν : A → R is atomless if for every ² > 0 there is a finite partition C of unity in A such that |ν|c ≤ ² for every c ∈ C (cf. 326Ya). (i) Show that the atomless functionals form a band Mc in the Riesz space M of all bounded finitely additive functionals on A. (ii) Show that the complementary band Mc⊥ consists of just those ν ∈ M P expressible as a sum i∈I νi of countably many atoms νi ∈ M . (iii) Show that if A is purely atomic then an atomless completely additive functional on A must be 0. (g) Let X be a set and Σ an algebra of subsets of X. Let M be the Riesz space of bounded finitely additive functionals on Σ, Mτ P the space of completely additive functionals and Mp the space of functionals expressible in the form νE = x∈E αx for some absolutely summable family hαx ix∈X of real numbers. (i) Show that Mp is a band in M . (ii) Show that if all singleton subsets of X belong to Σ then Mp = Mτ . (iii) Show that if Σ is a σ-algebra then every member of Mp is countably additive. (iv) Show that if X is a compact zero-dimensional Hausdorff space and Σ is the algebra of open-and-closed subsets of X then the complementary band Mp⊥ of Mp in M is the band Mc of atomless functionals described in 362Xf. (h) Let (X, Σ, µ) be a measure space. Let M be the Riesz space of bounded finitely additive functionals on Σ and Mσ the space of bounded countably additive functionals. Let Mtc , Mac be the spaces of truly continuous and bounded absolutely continuous additive functionals as defined in 232A. Show that Mtc and Mac are bands in M and that Mtc ⊆ Mσ ∩ Mac . Show that if µ is σ-finite then Mtc = Mσ ∩ Mac . (i) Let A be a Boolean algebra and M the Riesz space of bounded finitely additive functionals on A. (i) For any non-empty downwards-directed set A ⊆ A set NA = {ν : ν ∈ M, inf a∈A |ν|a = 0}. Show that NA is a band in M . (ii) For any non-empty set A of non-empty downwards-directed sets in A set MA = {ν : ν ∈ M, inf a∈A |ν|a = 0 ∀ A ∈ A}. Show that MA is a band in M . (iii) Explain how to represent as such MA the bands Mσ , Mτ , Mt , Mac , Mtc described above, and also any band generated by a single element of M . (iv) Suppose, in (ii), that A has the property that for any A, A0 ∈ A there is a B ∈ A such that for every b ∈ B there are a ∈ A, a0 ∈ A0 such that a ∪ a0 ⊆ b. Show that for any non-negative ν ∈ M , the component ν1 of ν in MA is given by the formula ν1 c = inf A∈A supa∈A ν(c \ a), so that the component ⊥ ν2 of ν in MA is given by the formula ν2 c = supA∈A inf a∈A ν(c ∩ a). (Cf. 356Yb.) 362Y Further exercises (a) Let A be a Boolean algebra. Let C be the band algebra of the Riesz space M of bounded finitely additive functionals on A (353B). Show that the bands Mσ , Mτ , Mc (362B, 362Xe, 362Xf) generate a subalgebra C0 of C with at most six atoms. Give an example in which C0 has six atoms. How many atoms can it have if (i) A is atomless (ii) A is purely atomic (iii) A is Dedekind σ-complete? (b) Let (A, µ ¯) be a measure algebra. Let C be the band algebra of the Riesz space M of bounded finitely additive functionals on A. Show that the bands Mσ , Mτ , Mc , Mac , Mtc , Mt (362B, 362C, 362Xe, 362Xf) generate a subalgebra C0 of C with at most twelve atoms. Give an example in which C0 has twelve atoms. How many atoms can it have if (i) A is atomless (ii) A is purely atomic (iii) (A, µ ¯) is semi-finite (iv) (A, µ ¯) is localizable (v) (A, µ ¯) is σ-finite (vi) (A, µ ¯) is totally finite?

362 Notes

S∼

297

(c) Give an example of a set X, a σ-algebra Σ of subsets of X, and a functional in Mp (as defined in 362Xg) which is not completely additive. (d) Let U be a Riesz space and f , g ∈ U ∼ . Show that the following are equiveridical: (α) g is in the band of U ∼ generated by f ; (β) for every u ∈ U + , ² > 0 there is a δ > 0 such that |g(v)| ≤ ² whenever 0 ≤ v ≤ u and |f |(v) ≤ δ; (γ) limn→∞ g(un ) = 0 whenever hun in∈N is a non-increasing sequence in U + and limn→∞ f (un ) = 0. (Hint: 362B(f-i).) (e) Let A be a weakly σ-distributive Boolean algebra (316Yg). Show that the ‘inf’ in the formula for Pσ ν in 362Bc can be replaced by ‘min’. (f ) Let A be any Boolean algebra and M the space of bounded finitely additive functionals on A. Let C ⊆ M be such that supν∈C |νa| < ∞ for every a ∈ A. (i) Suppose that supn∈N supν∈C |νan | is finite for every disjoint sequence han in∈N in A. Show that C is norm-bounded. (ii) Suppose that limn→∞ supν∈C |νan | = 0 for every disjoint sequence han in∈N in A. Show that C is uniformly integrable. (g) Let A be a Boolean algebra and Mτ the space of completely additive functionals on A. Let C ⊆ Mτ be such that supν∈C |νa| < ∞ for every atom a ∈ A. (i) Suppose that supn∈N supν∈C |νan | is finite for every disjoint sequence han in∈N in A. Show that C is norm-bounded. (ii) Suppose that limn→∞ supν∈C |νan | = 0 for every disjoint sequence han in∈N in A. Show that C is uniformly integrable. (h) Let A be a Dedekind σ-complete Boolean algebra and hνn in∈N a sequence of countably additive realvalued functionals on A. Suppose that νa = limn→∞ νn a is defined in R for every a ∈ A. Show that ν is countably additive and that {νn : n ∈ N} is uniformly integrable. (Hint: 246Yg.) Show that if every νn is completely additive, so is ν. (i) Let A be a Boolean algebra, M the Riesz space of bounded finitely additive functionals on A, and Mc ⊆ M the space of atomless functionals (362Xf). Show that for a non-negative ν ∈ M the component νc of ν in Mc is given by the formula Pn νc a = inf δ>0 sup{ i=0 νai : a0 , . . . , an ⊆ a are disjoint, νai ≤ δ for every i} for each a ∈ A. (j) Let A be a Boolean algebra and M the L-space of bounded additive real-valued functionals on A. Show that the complexification of M , as defined in 354Yk, can be identified with the Banach space of bounded additive functionals ν : A → C, writing Pn kνk = sup{ i=0 |νai | : a0 , . . . , an are disjoint elements of A} for such ν. 362 Notes and comments The Boolean algebras most immediately important in measure theory are of course σ-algebras of measurable sets and their quotient measure algebras. It is therefore natural to begin any investigation by concentrating on Dedekind σ-complete algebras. Nevertheless, in this section and the last (and in §326), I have gone to some trouble not to specialize to σ-complete algebras except when necessary. Partly this is just force of habit, but partly it is because I wish to lay a foundation for a further step forward: the investigation of the ways in which additive functionals on general Boolean algebras reflect the concepts of measure theory, and indeed can generate them. Some of the results in this direction can be surprising. I do not think it obvious that the condition (γ) in 362B(f-i), for instance, is sufficient in the absence of any hypothesis of Dedekind σ-completeness or countable additivity. Given a Boolean algebra A with the associated Riesz space M ∼ = S(A)∼ of bounded additive functionals on A, we now have a substantial list of bands in M : Mσ , Mτ , Mc (362Xf), and for a measure algebra the further bands Mac , Mtc and Mt ; for an algebra of sets we also have Mp (362Xg). These bands can be used to generate finite subalgebras of the band algebra of M (362Ya-362Yb), and for any such finite subalgebra we have a corresponding decomposition of M as a direct sum of the bands which are the atoms of the subalgebra (352Tb). This decomposition of M can be regarded as a recipe for decomposing its members into finite sums of functionals with special properties. What I called the ‘Lebesgue decomposition’ in 232I

298

Function spaces

362 Notes

is just such a recipe. In that context I had a measure space (X, Σ, µ) and was looking at the countably additive functionals from Σ to R, that is, at Mσ in the language of this section, and the bands involved in the decomposition were Mp , Mac and Mtc . But I hope that it will be plain that these ideas can be refined indefinitely, as we refine the classification of additive functionals. At each stage, of course, the exact enumeration of the subalgebra of bands generated by the classification (as in 362Ya-362Yb) is a necessary check that we have understood the relationships between the classes we have described. These decompositions are of such importance that it is worth examining the corresponding band projections. I give formulae for the action of band projections on (non-negative) functionals in 362Bc, 362Bd, 362B(f-ii), 362Xb, 362Xc(iii), 362Xi(iv) and 362Yi. Of course these are readily adapted to give formulae for the projections onto the complementary bands, as in 362Bf and 362Xi. If we have an algebra of sets, the completely additive functionals are (usually) of relatively minor importance; in the standard examples, they correspond to functionals defined as weighted sums of point masses (362Xg(ii)). The point is that measure algebras A appear as quotients of σ-algebras Σ of sets by σ-ideals I; consequently the countably additive functionals on A correspond exactly to the countably additive functionals on Σ which are zero on I; but the canonical homomorphism from Σ to A is hardly ever order-continuous, so completely additive functionals on A rarely correspond to completely additive functionals on Σ. On the other hand, when we are looking at countably additive functionals on Σ, we have to consider the possibility that they are singular in the sense that they are carried on some member of I; in the measure algebra context this possibility disappears, and we can often be sure that every countably additive functional is absolutely continuous, as in 327Bb. For any Boolean algebra A, we can regard it as the algebra of open-and-closed subsets of its Stone space Z; the points of Z correspond to Boolean homomorphisms from A to {0, 1}, which are the fundamental ‘atoms’ in the space of additive functionals on A (362Xe, 362Xg(iv)). It is the case that all non-negative additive functionals on a Boolean algebra A can be represented by appropriate measures on its Stone space (see 416P in Volume 4), but I prefer to hold this result back until it can take its place among other theorems on representing functionals by measures and integrals. It is one of the leitmotivs of this chapter, that Boolean algebras and Riesz spaces are Siamese twins; again and again, matching results are proved by the application of identical ideas. A typical example is the pair 362B(f-i) and 362Yd. Many of us have been tempted to try to describe something which would provide a common generalization of Boolean algebras and Riesz spaces (and lattice-ordered groups). I have not yet seen any such structure which was worth the trouble. Most of the time, in this chapter, I shall be using ideas from the general theory of Riesz spaces to suggest and illuminate questions in measure theory; but if you pursue this subject you will surely find that intuitions often come to you first in the context of Boolean algebras, and the applications to Riesz spaces are secondary. In 362E I give a condition for uniform integrability in terms of disjoint sequences, following the pattern established in 246G and repeated in 354R and 356O. The condition of 362E assumes that the set is normbounded; but if you have 246G to hand, you will see that it can be done with weaker assumptions involving atoms, as in 362Yf-362Yg. I mention once again the Banach-Ulam problem: if A is Dedekind complete, can S(A)∼ c be different from S(A)× ? This is obviously equivalent to the form given in the notes to §326 above. See 363S below.

363 L∞ In this section I set out to describe an abstract construction for L∞ spaces on arbitrary Boolean algebras, corresponding to the L∞ (µ) spaces of §243. I begin with the definition of L∞ (A) (363A) and elementary facts concerning its own structure and the embedding S(A) ⊆ L∞ (A) (363B-363D). I give the basic universal mapping theorems which define the Banach lattice structure of L∞ (363E) and a description of the action of Boolean homomorphisms on L∞ spaces (363F-363G) before discussing the representation of L∞ (Σ) and L∞ (Σ/I) for σ-algebras Σ and ideals I of sets (363H). This leads at once to the identification of L∞ (µ), as defined in Volume 2, with L∞ (A), where A is the measure algebra of µ (363I). Like S(A), L∞ (A) determines the algebra A (363J). I briefly discuss the dual spaces of L∞ ; they correspond exactly to the duals of S described in §362 (363K). Linear functionals on L∞ can for some purposes be treated as ‘integrals’ (363L).

363E

L∞

299

In the second half of the section I present some of the theory of Dedekind complete and σ-complete algebras. First, L∞ (A) is Dedekind (σ-)complete iff A is (363M). The spaces L∞ (A), for Dedekind σcomplete A, are precisely the Dedekind σ-complete Riesz spaces with order unit (363N-363P). The spaces L∞ (A), for Dedekind complete A, are precisely the normed spaces which may be put in place of R in the Hahn-Banach theorem (363R). Finally, I mention some equivalent forms of the Banach-Ulam problem (363S). 363A Definition Let A be a Boolean algebra, with Stone space Z. I will write L∞ (A) for the space C(Z) = Cb (Z) of continuous real-valued functions from Z to R, endowed with the linear structure, order structure, norm and multiplication of C(Z) = Cb (Z). (Recall that because Z is compact (311I), {u(z) : z ∈ Z} is bounded for every u ∈ L∞ (A) = C(Z) (2A3N(b-iii)), that is, C(Z) = Cb (Z). Of course if A = {0}, so that Z = ∅, then C(Z) has just one member, the empty function.) 363B Theorem Let A be any Boolean algebra; write L∞ for L∞ (A). (a) L∞ is an M -space; its standard order unit is the constant function taking the value 1 at each point; in particular, it is a Banach lattice with a Fatou norm and the Levi property. (b) L∞ is a commutative Banach algebra and an f -algebra. (c) If u ∈ L∞ then u ≥ 0 iff there is a v ∈ L∞ such that u = v × v. proof (a) See 354Hb and 354J. (b)-(c) are obvious from the definitions of Banach algebra (2A4J) and f -algebra (352W) and the ordering of L∞ = C(Z). 363C Proposition Let A be any Boolean algebra. Then S(A) is a norm-dense, order-dense Riesz subspace of L∞ (A), closed under multiplication. proof Let Z be the Stone space of A. Using the definition of S = S(A) set out in 361D, it is obvious that S is a linear subspace of L∞ = L∞ (A) closed under multiplication. Because S, like L∞ , is a Riesz subspace of RX (361Ee), S is a Riesz subspace of L∞ . By the Stone-Weierstrass theorem (in either of the forms given in 281A and 281E), S is norm-dense in L∞ . Consequently it is order-dense (354I). 363D Proposition Let A be a Boolean algebra. If we regard χa ∈ S(A) (361D) as a member of L∞ (A) for each a ∈ A, then χ : A → L∞ (A) is additive, order-preserving, order-continuous and a lattice homomorphism. proof Because the embedding S = S(A) ⊆ L∞ (A) = L∞ is a Riesz homomorphism, χ : A → L∞ is additive and a lattice homomorphism (361F-361G). Because S is order-dense in L∞ (363C), the embedding S ⊆ L∞ is order-continuous (352Nb), so χ : A → L∞ is order-continuous (361Gb). 363E Theorem Let A be a Boolean algebra, and U a Banach space. Let ν : A → U be a bounded additive function. (a) There is a unique bounded linear operator T : L∞ (A) → U such that T χ = ν; in this case kT k = supa,b∈A kνa − νbk. (b) If U is a Banach lattice then T is positive iff ν is non-negative; and in this case T is order-continuous iff ν is order-continuous, and sequentially order-continuous iff ν is sequentially order-continuous. (c) If U is a Banach lattice then T is a Riesz homomorphism iff ν is a lattice homomorphism iff νa∧νb = 0 whenever a ∩ b = 0. proof Write S = S(A), L∞ = L∞ (A). (a) By 361I there is a unique bounded linear operator T0 : S → U such that T0 χ = ν, and kT0 k = sup{kνa − νbk : a, b ∈ A}. But because U is a Banach space and S is dense in L∞ , T0 has a unique extension to a bounded linear operator T : L∞ → U with the same norm (2A4I). (b)(i) If T is positive then T0 is positive so ν is non-negative, by 361Ga.

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Function spaces

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(ii) If ν is non-negative then T0 is positive, by 361Ga in the other direction. But if u ∈ L∞+ and ² > 0, then by 354I there is a v ∈ S + such that ku − vk∞ ≤ ²; now kT u − T vk ≤ ²kT k. But T v = T0 v belongs to the positive cone U + of U . As ² is arbitrary, T u belongs to the closure of U + , which is U + (354Bc). As u is arbitrary, T is positive. (iii) Now suppose that ν is order-continuous as well as non-negative, and that A ⊆ L∞ is a non-empty downwards-directed set with infimum 0. Set B = {v : v ∈ S, there is some u ∈ A such that v ≥ u}. Then B is downwards-directed (indeed, v1 ∧ v2 ∈ B for every v1 , v2 ∈ B), and u = inf{v : v ∈ B, u ≤ v} for every u ∈ A (354I again), so B has the same lower bounds as A and inf B = 0 in L∞ and in S. But we know from 361Gb that T0 is order-continuous, while any lower bound for {T u : u ∈ A} in U must also be a lower bound for {T v : v ∈ B} = {T0 v : v ∈ B}, so inf u∈A T u = inf v∈B T0 v = 0 in U . As A is arbitrary, T is order-continuous (351Ga). (iv) Suppose next that ν is only sequentially order-continuous, and that hun in∈N is a non-increasing sequence in L∞ with infimum 0. For each n, k choose wnk ∈ S such that un ≤ wnk and kwnk − un k∞ ≤ 2−k (354I once more), and set wn0 = inf j,k≤n wjk for each n. Then hwn0 in∈N is a non-increasing sequence in S, and any lower bound of {wn0 : n ∈ N} is also a lower bound of {un : n ∈ N}, so 0 = inf n∈N wn0 in S and L∞ . Since T0 : S → U is sequentially order-continuous (361Gb), inf n∈N T un ≤ inf n∈N T wn0 = inf n∈N T0 wn0 = 0 in U . As hun in∈N is arbitrary, T is sequentially order-continuous. (v) On the other hand, if T is order-continuous or sequentially order-continuous, so is ν = T χ, because χ is order-continuous (363D). (c) We know that T0 : S → U is a Riesz homomorphism iff ν is a lattice homomorphism iff νa ∧ νb = 0 whenever a ∩ b = 0, by 361Gc. But T0 is a Riesz homomorphism iff T is. P P If T is a Riesz homomorphism so is T0 , because the embedding S ⊆ L∞ is a Riesz homomorphism. On the other hand, if T0 is a Riesz homomorphism, then the functions u 7→ u+ 7→ T (u+ ), u 7→ T u 7→ (T u)+ are continuous (by 354Bb) and agree on S, so agree on L∞ , and T is a Riesz homomorphism, by 352G. Q Q 363F Theorem Let A and B be Boolean algebras, and π : A → B a Boolean homomorphism. (a) There is an associated multiplicative Riesz homomorphism Tπ : L∞ (A) → L∞ (B), of norm at most 1, defined by saying that Tπ (χa) = χ(πa) for every a ∈ A. (b) For any u ∈ L∞ (A), there is a u0 ∈ L∞ (A) such that Tπ u = Tπ u0 and ku0 k∞ = kTπ uk∞ ≤ kuk∞ . (c)(i) The kernel of Tπ is the closed linear subspace of L∞ (A) generated by {χa : a ∈ A, πa = 0}. (ii) The set of values of Tπ is the closed linear subspace of L∞ (B) generated by {χ(πa) : a ∈ A}. (d) Tπ is surjective iff π is surjective, and in this case kvk∞ = min{kuk∞ : Tπ u = v} for every v ∈ L∞ (B). (e) Tπ is injective iff π is injective, and in this case kTπ uk∞ = kuk∞ for every u ∈ L∞ (A). (f) Tπ is order-continuous, or sequentially order-continuous, iff π is. (g) If C is another Boolean algebra and θ : B → C is another Boolean homomorphism, then Tθπ = Tθ Tπ : L∞ (A) → L∞ (C). proof Let Z and W be the Stone spaces of A and B. By 312P there is a continuous function φ : W → Z such that π ca = φ−1 [b a] for every a ∈ A, where b a is the open-and-closed subset of Z corresponding to a ∈ A. Write T for Tπ . (a) For u ∈ L∞ (A) = C(Z), set T u = uφ : W → R. Then T u ∈ C(W ) = L∞ (B). It is obvious, or at any rate very easy to check, that T : L∞ (A) → L∞ (B) is linear, multiplicative, a Riesz homomorphism and of norm 1 unless B = {0}, W = ∅. If a ∈ A, then T (χa) = (χa)φ = (χb a)φ = χ(φ−1 [b a]) = χ(πa), identifying χa ∈ L∞ (A) with the characteristic function χb a : Z → {0, 1} of the set b a. Of course Tπ = T is the only continuous linear operator with these properties, by 363Ea. (b) Set α = kT uk∞ , u0 (z) = max(−α, min(u(z), α)) for z ∈ Z; that is, u0 = (−αe) ∨ (u ∧ αe) in L∞ (A), where e is the standard order unit of L∞ (A). Then T e is the standard order unit of L∞ (B), so

L∞

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301

T u0 = (−αT e) ∨ (T u ∧ αT e) = T u, while ku0 k∞ ≤ α = kT uk∞ = kT u0 k∞ ≤ ku0 k∞ ≤ kuk∞ . (c)(i) Let U be the closed linear subspace of L∞ (A) generated by {χa : πa = 0}, and U0 the kernel of T . Because T is continuous and linear, U0 is a closed linear subspace, and T (χa) = χ0 = 0 whenever πa = 0; so U ⊆ U0 . Now take any u ∈ U0 and ² > 0. Then T (u+ ) = (T u)+ = 0, so u+ ∈ U0 . By 354I there is a 0 + + 0 0 + 0 u0 ∈ S(A) Pn such that 0 ≤ u ≤ u and ku − u k∞ ≤ ². Now 0 ≤ T u ≤ T u = 0, so T u = 0. Express 0 u as i=0 αi χai where αi ≥ 0 for each i. For each i, αi χ(πai ) = T (αi χai ) = 0, so πai = 0 or αi = 0; in either case αi χai ∈ U . Consequently u0 ∈ U . As ² is arbitrary and U is closed, u+ ∈ U . Similarly, u− = (−u)+ ∈ U and u = u+ − u− ∈ U . As u is arbitrary, U0 ⊆ U and U0 = U . (ii) Let V be the closed linear subspace of L∞ (B) generated by {χ(πa) : a ∈ A}, and V0 = T [L∞ (A)]. Then T [S(A)] ⊆ V , so V0 = T [S(A)] ⊆ T [S(A)] ⊆ V = V . On the other hand, V0 is a closed linear subspace in L∞ (B). P P It is a linear subspace because T is a linear operator. To see that it is closed, take any v ∈ V 0 . Then there is a sequence hvn in∈N in V0 such that kv − vn k∞ ≤ 2−n for every n ∈ N. Choose un ∈ L∞ (A) such that T u0 = v0 , while T un = vn − vn−1 and kun k∞ = kvn − vn−1 k∞ for n ≥ 1 (using (b) above). Then P∞ P∞ n=1 kv − vn k∞ + kv − vn−1 k∞ n=1 kun k∞ ≤ Pn is finite, so u = limn→∞ i=0 ui is defined in the Banach space L∞ (A), and Pn T u = limn→∞ i=0 T ui = limn→∞ vn = v. As v is arbitrary, V0 is closed. Q Q Since χ(πa) = T (χa) ∈ V0 for every a ∈ A, V ⊆ V0 and V = V0 , as required. (d) If π is surjective, then T is surjective, by (c-ii). If T is surjective and b ∈ B, then there is a u ∈ L∞ (A) such that T u = χb. Now there is a u0 ∈ S(A) such that ku − u0 k∞ ≤ 31 , so that kT u0 − χbk∞ ≤ 31 . Taking a ∈ A such that {z : u0 (z) ≥ 21 } = b a, we must have πa = b, since bb = {w : (T u0 )(w) ≥ 1 } = φ−1 [b a] = π ca. 2 As b is arbitrary, π is surjective. Now (b) tells us that in this case kvk∞ = min{kuk∞ : T u = v} for every v ∈ L∞ (B). (e) By (c-i), T is injective iff π is injective. In this case, for any u ∈ L∞ (A), kT uk∞ = kT |u|k∞ (because T is a Riesz homomorphism) ≥ sup{kT u0 k∞ : u0 ∈ S(A), u0 ≤ |u|} = sup{ku0 k∞ : u0 ∈ S(A), u0 ≤ |u|} (by 361Jd) = kuk∞ (by 354I) ≥ kT uk∞ , and kT uk∞ = kuk∞ . (f ) If T is (sequentially) order-continuous then π = T χ is (sequentially) order-continuous, by 363D. If π is (sequentially) order-continuous then χπ : A → L∞ (B) is (sequentially) order-continuous, so T is (sequentially) order-continuous, by 363Eb. (g) This is elementary, in view of the uniqueness of Tθπ .

302

Function spaces

363G

363G Corollary Let A be a Boolean algebra. (a) If C is a subalgebra of A, then L∞ (C) can be identified, as Banach lattice and as Banach algebra, with the closed linear subspace of L∞ (A) generated by {χc : c ∈ C}. (b) If I is an ideal of A, then L∞ (A/I) can be identified, as Banach lattice and as Banach algebra, with the quotient space L∞ (A)/V , where V is the closed linear subspace of L∞ (A) generated by {χa : a ∈ I}. proof Apply 363Fc to the identity map from C to A and the canonical map from A onto A/I. 363H Representations of L∞ (A) Much of the importance of the concept of L∞ (A) arises from the way it is naturally represented in the contexts in which the most familiar Boolean algebras appear. Proposition Let X be a set and Σ a σ-algebra of subsets of X. (a) L∞ (Σ) can be identified, as Banach algebra and Banach lattice, with the space L∞ of bounded Σmeasurable real-valued functions on X, with the norm kf k∞ = supx∈X |f (x)| for f ∈ L∞ ; this identification matches χE ∈ L∞ (Σ) with the characteristic function of E as a subset of X, for every E ∈ Σ. In particular, for any set X, L∞ (PX) can be identified with `∞ (X). (b) If A is a Dedekind σ-complete Boolean algebra and π : Σ → A is a surjective sequentially ordercontinuous Boolean homomorphism with kernel I, then L∞ (A) can be identified, as Banach algebra and Banach lattice, with L∞ /W, where W = {f : f ∈ L∞ , {x : f (x) 6= 0} ∈ I} is a closed ideal and solid linear subspace of L∞ . For f ∈ L∞ , kf • k∞ = min{α : α ≥ 0, {x : |f (x)| > α} ∈ I}. (c) In particular, if I is any σ-ideal of Σ and E 7→ E • is the canonical homomorphism from Σ onto A = Σ/I, then we have an identification of L∞ (A) with a quotient of L∞ , and for any E ∈ Σ we can identify χ(E • ) ∈ L∞ (A) with the equivalence class (χE)• ∈ L∞ /W of the characteristic function χE. proof (a) For the elementary properties of the space of Σ-measurable functions, see §121. In particular, it is easy to check that L∞ is a Riesz space, with a Riesz norm, and a normed algebra. To check that it is a Banach space, observe that if hfn in∈N is a Cauchy sequence in L∞ , then |fm (x) − fn (x)| ≤ kfm − fn k∞ → 0 as m, n → ∞, so f (x) = limn→∞ fn (x) is defined for every x ∈ X; now f is Σ-measurable, by 121Fa. Of course kf k∞ ≤ supn∈N kfn k∞ < ∞, so f ∈ L∞ , while kf − fn k∞ ≤ supm≥n kfm − fn k∞ → 0 as n → ∞, so f = limn→∞ fn in L∞ . As hfn in∈N is arbitrary, L∞ is complete. By 361L we can identify S(Σ), as Riesz space and normed algebra, with the linear span S of {χE : E ∈ Σ}, which is a subspace of L∞ . Now the point is that it is dense. P P If f ∈ L∞ and ² > P0, then for each n ∈ Z set En = {x : n² ≤ f (x) < (n + 1)²} ∈ Σ; then En = ∅ if |n| > 1 + 1² kf k∞ , so g = n∈Z n²χEn belongs to S, and of course kf − gk∞ ≤ ². Q Q Consequently the canonical normed space isomorphism between S(Σ) and S extends (uniquely) to a normed space isomorphism between L∞ (Σ) and L∞ (use 2A4I). Because the operations ∨ and × are continuous in both L∞ (Σ) and L∞ , and their actions on S(Σ) and S are identified by our isomorphism, the isomorphism between L∞ (Σ) and L∞ identifies their lattice and algebra structures. (b)(i) By 363F, we have a multiplicative Riesz homomorphism T = Tπ from L∞ (Σ) to L∞ (A) which is surjective (363Fd) and has kernel the closed linear subspace W of L∞ (Σ) generated by {χE : E ∈ I}. Now under the isomorphism described in (a), W corresponds to W. P P (α) W is a linear subspace of L∞ because {x : (f + g)(x) 6= 0} ⊆ {x : f (x) 6= 0} ∪ {x : g(x) 6= 0} ∈ I, {x : (αf )(x) 6= 0} ⊆ {x : f (x) 6= 0} ∈ I whenever f , g ∈ W and α ∈ R. (β) If hfn in∈N is a sequence in W converging to f ∈ L∞ , then S {x : f (x) 6= 0} ⊆ n∈N {x : fn (x) 6= 0} ∈ I, so f ∈ W. Thus W is a closed linear subspace of L∞ . (γ) If E ∈ I, then χE, taken in S(Σ) or L∞ (Σ), corresponds to the function χE : X → {0, 1}, which belongs to W; so that W must correspond to the closed linear span in L∞ of such characteristic functions, which is a subspace of W. (δ) On the other hand, if f ∈ W and ² > 0, set

L∞

363K

303

En = {x : n² ≤ f (x) ≤ (n + 1)²}, En0 = {x : −(n + 1)² ≤ f (x) ≤ −n²} P∞ for n ≥ 1; all these belong to I, so g = n=1 n(χEn − χEn0 ) ∈ W corresponds to a member of W , while kf − gk∞ ≤ ². As W is closed, f also must correspond to some member of W . As f is arbitrary, W and W match exactly. Q Q (ii) Because T is a multiplicative Riesz homomorphism, L∞ (A) ∼ = L∞ (Σ)/W is matched canonically, ∞ in its linear, order and multiplicative structures, with L /W. We know also that kvk∞ = inf{kuk∞ : u ∈ L∞ (Σ), T u = v} for every v ∈ L∞ (A) (363Fd), that is, that the norm of L∞ (A) corresponds to the quotient norm on L∞ (Σ)/W . As for the given formula for the norm, take any f ∈ L∞ . There is a g ∈ L∞ such that T f = T g and kT f k∞ = kgk∞ . (Here I am treating T as an operator from L∞ onto L∞ (A).) In this case {x : |f (x)| > kT f k∞ } ⊆ {x : f (x) 6= g(x)} ∈ I. On the other hand, if {x : |f (x)| > α} ∈ I, and we set h = −α1 ∨ (f ∧ 1), then T h = T f , so kT f k∞ ≤ khk∞ ≤ α. (c) Put (a) and (b) together. 363I Corollary Let (X, Σ, µ) be a measure space, with measure algebra A. Then L∞ (µ) can be identified, as Banach lattice and Banach algebra, with L∞ (A); the identification matches (χE)• ∈ L∞ (µ) with χ(E • ) ∈ L∞ (A), for every E ∈ Σ. Remark The space I called L∞ (µ) in Chapter 24 is not strictly speaking the space L∞ ∼ = L∞ (Σ) of 363H; ∞ 0 I took L (µ) ⊆ L (µ) to be the set of essentially bounded, virtually measurable functions defined almost everywhere on X, and in general this is larger. But, as remarked in the notes to §243, L∞ (µ) can equally ∞ above, because every function well be regarded as a quotient of what I there called L∞ strict , which is the L ∞ ∞ in L (µ) is equal almost everywhere to some member of Lstrict . 363J Recovering the algebra A: Proposition Let A be a Boolean algebra. For a ∈ A write Va for the solid linear subspace of L∞ (A) generated by χa. Then a 7→ Va is a Boolean isomorphism between A and the algebra of projection bands in L∞ (A). proof The proof is nearly identical to that of 361K. If a ∈ A, u ∈ Va and v ∈ V1\a , then |u| ∧ |v| = 0 because χa ∧ χ(1 \ a) = 0; and if w ∈ L∞ (A) then w = (w × χa) + (w × χ(1 \ a)) ∈ Va + V1\a because |w×χa| ≤ kwk∞ χa and |w×χ(1 \ a)| ≤ kwk∞ χ(1 \ a). So Va and V1\a are complementary projection bands in L∞ = L∞ (A). Next, if U ⊆ L∞ is a projection band, then χ1 is expressible as u + v where u ∈ U , v ∈ U ⊥ ; thinking of L∞ as the space of continuous real-valued functions on the Stone space Z of A, u and v must be the characteristic functions of complementary subsets E, F of Z, which must be open-and-closed, so that E = b a, F = 1d \ a. In this case Va ⊆ U and V1\a ⊆ U ⊥ , so U must be Va precisely. Thus a 7→ Va is surjective. Finally, just as in 361K, a ⊆ b ⇐⇒ Va ⊆ Vb , so we have a Boolean isomorphism. 363K Dual spaces of L∞ The questions treated in §362 yield nothing new in the present context. I spell out the details. Proposition Let A be a Boolean algebra. Let M , Mσ and Mτ be the L-spaces of bounded finitely additive functionals, bounded countably additive functionals and completely additive functionals on A. Then the embedding S(A) ⊆ L∞ (A) induces Riesz space isomorphisms between S(A)∼ ∼ = M and L∞ (A)∼ = L∞ (A)∗ , ∼ ∼ ∞ ∼ × ∼ ∞ × S(A)c = Mσ and L (A)c , and S(A) = Mτ and L (A) . proof Write S = S(A), L∞ = L∞ (A). (a) For the identifications S ∼ ∼ = M , Sc∼ ∼ = Mσ and S × ∼ = Mτ see 362A.

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(b) L∞∗ = L∞∼ either because L∞ is a Banach lattice (356Dc) or because L∞ has an order-unit norm, so that a linear functional on L∞ is order-bounded iff it is bounded on the unit ball. (c) If f is a positive linear functional on L∞ , then f ¹S is a positive linear functional. Because S is orderdense in L∞ (363C), the embedding is order-continuous (352Nb); so if f is (sequentially) order-continuous, ∼ so is f ¹S. Accordingly the restriction operator f 7→ f ¹S gives maps from L∞∼ to S ∼ , (L∞ )∼ c to Sc and ∞× × ∞∼ + ∞ L to S . If f ∈ L and f ¹S ≥ 0, then f (u ) ≥ 0 for every u ∈ S and therefore for every u ∈ L , and f ≥ 0; so all these restriction maps are injective positive linear operators. (d) I need to show that they are surjective. (i) If g ∈ S ∼ , then g is bounded on the unit ball {u : u ∈ S, kuk∞ ≤ 1}, so has an extension to a continuous linear f : L∞ → R (2A4I); thus S ∼ = {f ¹S : f ∈ L∞∼ }. This means that f 7→ f ¹S is actually a Riesz space isomorphism between L∞∼ and S ∼ . In particular, |f |¹S = |f ¹S| for any f ∈ L∞∼ . (ii) If f : L∞ → R is a positive linear operator and f ¹S ∈ Sc∼ , let hun in∈N be a non-increasing sequence in L with infimum 0. For each n, k ∈ N there is a vnk ∈ S such that un ≤ vnk ≤ un + 2−k e, where e is the standard order unit of L∞ (354I, as usual); set wn = inf i,k≤n vik ; then hwn in∈N is a non-increasing sequence in S with infimum 0, so ∞

0 ≤ inf n∈N f (un ) ≤ inf n∈N f (wn ) = 0. As hun in∈N is arbitrary, f ∈

(L∞ )∼ c .

Consequently, for general f ∈ L∞∼ ,

∞ ∼ ∼ ∼ f ∈ (L∞ )∼ c ⇐⇒ |f | ∈ (L )c ⇐⇒ |f ¹S| ∈ Sc ⇐⇒ f ¹S ∈ Sc , ∼ and the map f 7→ f ¹S : (L∞ )∼ c → Sc is a Riesz space isomorphism.

(iii) Similarly, if f ∈ L∞∼ is non-negative and f ¹S ∈ S × , then whenever A ⊆ L∞ is non-empty, downwards-directed and has infimum 0, B = {w : w ∈ S, ∃ u ∈ A, w ≥ u} has infimum 0, so inf u∈A f (u) ≤ inf w∈B f (w) ≤ 0 and f ∈ L∞× . As in (ii), it follows that f 7→ f ¹S is a surjection from L∞× onto S × . *363L Integration with respect to a finitely additive functional (a) If A is a Boolean algebra and ν : A → R is a bounded additive functional, then by 363K we have a corresponding functional fν ∈ L∞ (A)∗ defined by saying that fν (χa) R = νa for every a ∈ A. There are contexts in which it is convenient, and even helpful, to use the formula u dν in place of fν (u) for u ∈ L∞ = L∞ (A). When doing so, we must of course remember that we may have lost some of the standard properties of ‘integration’. But enough of our intuitions (including, for instance, the idea of stochastic independence) remain valid to make the formula a guide to interesting ideas. (b) LetR M be the L-space of bounded finitely additive functionals on A (362B). Then we have a function (u, ν) 7→ u dν : L∞ × M → R. Now this map is bilinear. P P For µ, ν ∈ M , u, v ∈ L∞ and α ∈ R, R R R R R u + v dν = u dν + v dν, αu dν = α u dν just because fν is linear. On the other side, we have (fµ + fν )(χa) = fµ (χa) + fν (χa) = µa + νa = (µ + ν)(a) = fµ+ν (χa) ∞ Rfor every a ∈ A,R so thatRfµ + fν and fµ+νR must agree on R S(A) and therefore on L . But this means that u d(µ + ν) = u dµ + u dν. Similarly, u d(αµ) = α u dµ. Q Q R (c) If ν is non-negative, we have u dν ≥ 0 whenever u ≥ 0, as in part (c) of the proof of 363K. Consequently, for any ν ∈ M and u ∈ L∞ , Z Z Z Z Z | u dν| = | u+ dν + − u− dν + − u+ dν − + u− dν − | Z Z Z Z + + − + + − ≤ u dν + u dν + u dν + u− dν − Z Z = |u|d|ν| ≤ kuk∞ χ1 d|ν| = kuk∞ |ν|(1) = kuk∞ kνk.

L∞

363Lf

305

R So (u, ν) 7→ u dν has norm (as defined in 253Ab) at most 1. If A 6= 0, the norm is exactly 1. (For this we need to know that there is a ν ∈ M + such that ν1 = 1. Take any z in the Stone space of A and set νa = 1 if z ∈ b a, 0 otherwise.) (d) We do not have any result corresponding to B.Levi’s theorem in this language, because (even if ν is non-negative andR countably additive) there is no reason to suppose that supn∈N un is defined in L∞ just because supn∈N un dν is finite. But if ν is countably additive and A is Dedekind σ-complete, we have something corresponding to Lebesgue’s Dominated Convergence Theorem (363Yh). (e) One formula which we can imitate in the present context is that of 252O, where the ordinary integral is represented in the form

R

f dµ =

R∞ 0

µ{x : f (x) ≥ t}dt.

In the context of general Boolean algebras, we cannot directly represent the set [[f ≥ t]] = {x : f (x) ≥ t} (though in the next section I will show that in Dedekind σ-complete Boolean algebras there is an effective expression of this idea, and I will use it in the principal definition of §365). But what we can say is the following. If A is any Boolean algebra, and ν : A → [0, ∞[ is a non-negative additive functional, and u ∈ L∞ (A)+ , then

R

u dν =

R∞ 0

sup{νa : tχa ≤ u}dt,

where the right-hand integral is taken with respect to Lebesgue measure. RP P (i) For t ≥ 0 set h(t) = ∞ sup{νa : tχa ≤ u}. Then h is non-increasing and zero for t > kuk∞ , so 0 h(t)dt is defined in R. If we set hn (t) = h(2−n (k + 1)) whenever k, n ∈ N and 2−n k ≤ t < 2−n (k + 1), then hhn (t)in∈N is a nondecreasing sequence which h is continuous at t, which is almost everywhere R ∞converges to h(t) Rwhenever ∞ (222A, or otherwise); so 0 h(t)dt = limn→∞ 0 hn (t)dt. Next, given n ∈ N and ² > 0, we can choose for each k ≤ k ∗ = b2n kuk∞ c an ak such that 2−n (k + 1)χak ≤ u and νak ≥ h(2−n (k + 1)) − ². In this case Pk∗ −n χak ≤ u, so k=0 2 Z

∗

∞

hn (t)dt = 2

−n

0

k X

∗

h(2

−n

(k + 1)) ≤ kuk∞ ² + 2

−n

k=0

= kuk∞ ² +

k X

νak

k=0

Z X k∗

Z 2−n χak dν ≤ kuk∞ ² +

u dν.

k=0

R∞ R As n and ² are arbitrary, 0 h(t)dt ≤ u dν. (ii) In the other direction, there is for any ² > 0 a v ∈ S(A) Pm such that v ≤ u ≤ v + ²χ1. If we express v as j=0 γj χcj where c0 ⊇ . . . ⊇ cm and γj ≥ 0 for every j Pk (361Ec), then we shall have h(t) ≥ νck whenever t ≤ j=0 γj , so

R∞

As ² is arbitrary,

R∞ 0

0

h(t)dt ≥

R

h(t)dt ≥

Pm

k=0

γk νck =

R

v dν ≥

R

u dν − ²ν1.

u dν and the two ‘integrals’ are equal. Q Q

R (f ) The formula dν is especially natural when A is an algebra of sets, so that L∞ can be directly interpreted as a space of functions (363Yf); better still, when A is actually a σ-algebra of subsets of a set X, L∞ can be identified R with the space of bounded A-measurable functions on X, as in 363Ha. So in such contexts I may write g dν when g : X → R is bounded and A-measurable, and ν : A → R is an additive functional. But I will try to take care to signal any such deviation from the normal principle that the R symbol refers to the sequentially order-continuous integral defined in §122 with the minor modifications introduced in §§133 and 135. My purpose in this paragraph has been only to indicate something of what can be done with finite additivity alone.

306

363M the last.

Function spaces

363M

Now I come to a fundamental fact underlying a number of theorems in both this volume and

Theorem Let A be a Boolean algebra. (a) A is Dedekind σ-complete iff L∞ (A) is Dedekind σ-complete. (b) A is Dedekind complete iff L∞ (A) is Dedekind complete. proof (a)(i) Suppose that A is Dedekind σ-complete. By 314M, we may identify A with a quotient Σ/M, where M is the ideal of meager subsets of the Stone space Z of A, and Σ = {E4A : E ∈ E, A ∈ M}, writing E = {b a : a ∈ A} for the algebra of open-and-closed subsets of Z. By 363H, L∞ = L∞ (A) can be identified with L∞ /V, where L∞ is the space of bounded Σ-measurable functions from Z to R, and V is the space of functions zero except on a member of I. Now suppose that hun in∈N is a sequence in L∞ with an upper bound u ∈ L∞ . Express un , u as fn• , f • where fn , f ∈ L∞ . Set g(z) = supn∈N min(fn (z), f (z)) for every z ∈ Z; then g ∈ L∞ (121F), so we have a corresponding member v = g • of L∞ . For each n ∈ N, u ≥ un so (fn − f )+ ∈ V, {z : fn (z) > g(z)} ⊆ {z : fn (z) > f (z)} ∈ M and v ≥ un . If w ∈ L for every n, so

∞

and w ≥ un for every n, then express w as h• where h ∈ L∞ ; we have (fn − h)+ ∈ V {z : g(z) > h(z)} ⊆

S

n∈N {z

: fn (z) > h(z)} ∈ M

because M is a σ-ideal, and (g − h)+ ∈ V, w ≥ v. Thus v = supn∈N un in L∞ . As hun in∈N is arbitrary, L∞ is Dedekind σ-complete (using 353G). (ii) Now suppose that L∞ is Dedekind σ-complete, and that A is a countable non-empty set in A. In this case {χa : a ∈ A} has a least upper bound u in L∞ . Take a v ∈ S(A) such that 0 ≤ v ≤ u and ku − vk∞ ≤ 13 ; set b = [[v > 13 ]], as defined in 361Eg. If a ∈ A, then k(χa − v)+ k∞ ≤ ku − vk∞ ≤ 31 , so 2 3 χa ≤ v and a ⊆ b. If c ∈ A is any upper bound for A, then v ≤ u ≤ χc so b ⊆ c. Thus b = sup A in A. As A is arbitrary, A is Dedekind σ-complete. (b)(i) For the second half of this theorem I use an argument which depends on joining the representation described in (a-i) above with the original definition of L∞ in 363A. The point is that C(Z) ⊆ L∞ , and for any f ∈ C(Z) = L∞ (A) its equivalence class f • in L∞ /V corresponds to f itself. P P Perhaps it will help to give a name T to the canonical isomorphism from L∞ /V to L∞ . Then V = {f : T f • = f } is a closed linear subspace of C(Z), because f 7→ f • and T are continuous linear operators. But if a ∈ A, then (b a)• , the • ∞ equivalence class of b a ∈ Σ in Σ/M, corresponds to a (see the proof of 314M), so (χb a) ∈ L /V corresponds to χa; that is, T (χb a)• = χb a, if we identify χa ∈ L∞ with χb a : Z → {0, 1}. So V contains χb a for every a ∈ A; because V is a linear subspace, S(A) ⊆ V ; because V is closed, L∞ ⊆ V . Q Q For a general f ∈ L∞ , g = T f • must be the unique member of C(Z) such that g • = f • , that is, such that {z : g(z) 6= f (z)} is meager. (ii) Suppose now that A is actually Dedekind complete. In this case Z is extremally disconnected (314S). Consequently every open set belongs to Σ. P P If G is open, then G is open-and-closed; but A = G \ G is a closed set with empty interior, so is meager, and G = G4A ∈ Σ. Q Q Let A ⊆ L∞ = C(Z) be any non-empty set with an upper bound in C(Z). For each z ∈ Z set g(z) = supu∈A u(z). Then S Gα = {z : g(z) > α} = u∈A {z : u(z) > α} is open for every α ∈ R (that is, g is lower semi-continuous). Thus Gα ∈ Σ for every α, so g ∈ L∞ , and v = T g • is defined in C(Z). For any u ∈ A, g ≥ u in L∞ , so v = T g • ≥ T u• = u in L∞ ; thus v is an upper bound for A in L∞ . On the other hand, if w is any upper bound for A in L∞ = C(Z), then surely w(z) ≥ u(z) for every z ∈ Z, u ∈ A, so w ≥ g and w = T w• ≥ T g • = v. This means that v is the least upper bound of A. As A is arbitrary, L∞ is Dedekind complete.

363Q

L∞

307

(iii) Finally, if L∞ is Dedekind complete, then the argument of (b-ii), applied to arbitrary non-empty subsets A of A, shows that A is also Dedekind complete. 363N

Much of the importance of L∞ spaces in the theory of Riesz spaces arises from the next result.

Proposition Let U be a Dedekind σ-complete Riesz space with an order unit. Then U is isomorphic, as Riesz space, to L∞ (A), where A is the algebra of projection bands in U . proof (a) By 353M, U is isomorphic to a norm-dense Riesz subspace of C(X) for some compact Hausdorff space X; for the rest of this argument, therefore, we may suppose that U actually is such a subspace. (b) Now U = C(X). P P If g ∈ C(X) then by 354I there are sequences hfn in∈N , hfn0 in∈N in U such that fn ≤ g ≤ gn and kgn − fn k∞ ≤ 2−n for every n. Now {fn : n ∈ N} has a least upper bound f in U ; since we must have fn ≤ f ≤ gn for every n, f = g and g ∈ U . Q Q (c) Next, X is zero-dimensional. P P Suppose that G ⊆ X is open and x ∈ G. Then there is an open set G1 such that x ∈ G1 ⊆ G1 ⊆ G (3A3Bc). There is an f ∈ C(X) such that 0 ≤ f ≤ χG1 and f (x) > 0 (also by 3A3Bc); write H for {y : f (y) > 0}. Set g = supn∈N (nf ∧ χX), the supremum being taken in U = C(X). For each y ∈ H, we must have g(y) ≥ min(1, nf (y)) for every n, so that g(y) = 1. On the other hand, if y ∈ X \ H, there is an h ∈ C(X) such that h(y) > 0 and 0 ≤ h ≤ χ(X \ H); now h ∧ f = 0 so h ∧ g = 0 and g(y) = 0. Thus χH ≤ g ≤ χH. The set {y : g(y) ∈ {0, 1}} is closed and includes H ∪ (X \ H) so must be the whole of X; thus G2 = {y : g(y) > 12 } = {y : g(y) ≥ 12 } is open-and-closed, and we have x ∈ H ⊆ G2 ⊆ H ⊆ G1 ⊆ G. As x, G are arbitrary, the set of open-and-closed subsets of X is a base for the topology of X, and X is zero-dimensional. Q Q (d) We can therefore identify X with the Stone space of its algebra E of open-and-closed sets (311J). But in this case 363A immediately identifies U = C(X) with L∞ (E). By 363J, E is isomorphic to A, so U∼ = L∞ (A). Remark Note that in part (c) of the argument above, we have to take great care over the interpretation of ‘sup’. In the space of all real-valued functions on X, the supremum of {nf ∧ χX : n ∈ N} is just χH. But g is supposed to be the least continuous function greater than or equal to nf ∧ χX for every n, and is therefore likely to be strictly greater than χH, even though sandwiched between χH and χH. 363O Corollary Let U be a Dedekind σ-complete M -space. Then U is isomorphic, as Banach lattice, to L∞ (A), where A is the algebra of projection bands of U . proof This is merely the special case of 363N in which U is known from the start to be complete under an order-unit norm. 363P Corollary Let U be any Dedekind σ-complete Riesz space and e ∈ U + . Then the solid linear subspace Ue of U generated by e is isomorphic, as Riesz space, to L∞ (A) for some Dedekind σ-complete Boolean algebra A; and if U is Dedekind complete, so is A. proof Because U is Dedekind σ-complete, so is Ue (353J(a-i)). Apply 363N to Ue to see that Ue ∼ = L∞ (A) for some A. Because Ue is Dedekind σ-complete, so is A, by 363Ma; while if U is Dedekind complete, so are Ue and A, by 353J(b-i) and 363Mb. 363Q The next theorem will be a striking characterization of the Dedekind complete L∞ spaces as normed spaces. As a warming-up exercise I give a much simpler result concerning their nature as Banach lattices. Proposition Let A be a Dedekind complete Boolean algebra. Then for any Banach lattice U , a linear operator T : U → L∞ = L∞ (A) is continuous iff it is order-bounded, and in this case kT k = k|T |k, where the modulus |T | is taken in L∼ (U ; L∞ ).

308

Function spaces

363Q

proof It is generally true that order-bounded operators between Banach lattices are continuous (355C). If T : U → L∞ is continuous, then for any w ∈ U + |u| ≤ w =⇒ kuk ≤ kwk =⇒ kT uk∞ ≤ kT kkwk =⇒ |T u| ≤ kT kkwkχ1. So T is order-bounded. As L∞ is Dedekind complete (363Mb), |T | is defined in L∼ (U ; L∞ ) (355Ea). For any w ∈ U , |T ||w| = sup{|T u| : |u| ≤ |w|} ≤ kT kkwkχ1, so k|T |(w)k ≤ kT kkwk; accordingly k|T |k ≤ kT k. On the other hand, of course, |T w| ≤ |T ||w| ≤ k|T |kkwkχ1 for every w ∈ U , so kT k ≤ k|T |k and the two norms are equal. Remark Of course what is happening here is that the spaces L∞ (A), for Dedekind complete A, are just the Dedekind complete M -spaces; this is an elementary consequence of 363N and 363M. 363R

Now for something much deeper.

Theorem Let U be a normed space over R. Then the following are equiveridical: (i) there is a Dedekind complete Boolean algebra A such that U is isomorphic, as normed space, to L∞ (A); (ii) whenever V is a normed space, V0 a linear subspace of V , and T0 : V0 → U is a bounded linear operator, there is an extension of T0 to a bounded linear operator T : V → U with kT k = kT0 k. proof For the purposes of the argument below, let us say that a normed space U satisfying the condition (ii) has the ‘Hahn-Banach property’. Part A: (i)⇒(ii) I have to show that L∞ (A) has the Hahn-Banach property for every Dedekind complete Boolean algebra A. Let V be a normed space, V0 a linear subspace of V , and T0 : V0 → L∞ = L∞ (A) a bounded linear operator. Set γ = kT0 k. Let P be the set of all functions T such that dom T is a linear subspace of V including V0 and T : dom T → U is a bounded linear operator extending T0 and with norm at most γ. Order P by saying that T1 ≤ T2 if T2 extends T1 . Then any non-empty totally ordered subset Q of P has an upper bound in P. P P S Set dom T = {dom T1 : T1 ∈ Q}, T v = T1 v whenever T1 ∈ Q and v ∈ dom T1 ; it is elementary to check that T ∈ P, so that T is an upper bound for Q in P. Q Q By Zorn’s Lemma, P has a maximal element T˜. Now dom T˜ = V . P P?? Suppose, if possible, otherwise. Write V˜ = dom T˜ and take any v˜ ∈ V \ V˜ ; let V1 be the linear span of V˜ ∪{˜ v }, that is, {v+α˜ v : v ∈ V˜ , α ∈ R}. ∞ ˜ If v1 , v2 ∈ V then, writing e for the standard order unit of L , T˜v1 + T˜v2 = T˜(v1 + v2 ) ≤ kT˜(v1 + v2 )k∞ e ≤ γkv1 + v2 ke ≤ γkv1 − v˜ke + γkv2 + v˜ke, so T˜v1 − γkv1 − v˜ke ≤ γkv2 + v˜ke − T˜v2 . Because L∞ is Dedekind complete (363Mb), u ˜ = supv1 ∈V˜ T˜v1 − γkv1 − v˜ke is defined in L∞ and u ˜ ≤ γkv2 + v˜ke − T v2 for every v2 ∈ V˜ . Putting these together, we have T˜v − u ˜ ≤ γkv − v˜ke,

T˜v + u ˜ ≤ γkv + v˜ke

for all v ∈ V˜ . Consequently, if v ∈ V˜ , then for α > 0 T˜v + α˜ u = α(T˜( 1 v) + u ˜) ≤ αγk 1 v + v˜ke = γkv + α˜ v ke, α

α

while for α < 0 T˜v + α˜ u = |α|(T˜(− α1 v) − u ˜) ≤ |α|γk − α1 v − v˜ke = γkv + α˜ v ke, and of course

363R

L∞

309

T˜v ≤ kT˜vk∞ e ≤ γkvke. So we have T˜v + α˜ u ≤ γkv + α˜ v ke for every v ∈ V˜ , α ∈ R. Define T1 : V1 → L∞ by setting T1 (v + α˜ v ) = T˜v + α˜ u for every v ∈ V˜ , α ∈ R. (This is well-defined ˜ because v˜ ∈ / V , so any member of V1 is uniquely expressible as v + α˜ v where v ∈ V˜ and α ∈ R.) Then T1 ∞ is a linear operator, extending T0 , from a linear subspace of V to L . But from the calculations above we know that T1 v ≤ γkvke for every v ∈ V1 ; since we also have T1 v = −T1 (−v) ≥ −γk − vke = −γkvke, kT1 vk∞ ≤ γkvk for every v ∈ V1 , and T1 ∈ P. But now T1 is a member of P properly extending T˜, which is supposed to be impossible. X XQ Q Accordingly T˜ : V → L∞ is an extension of T0 to the whole of V , with the same norm as T0 . As V and T0 are arbitrary, L∞ has the Hahn-Banach property. Part B: (ii)⇒(i) Now let U be a normed space with the Hahn-Banach property. If U = {0} then of course it is isomorphic to L∞ (A), where A = {0}, so henceforth I will take it for granted that U 6= {0}. (a) Let Z be the unit ball of the dual U ∗ of U , with the weak* topology. Then Z is a compact Hausdorff space (3A5F). For u ∈ U set Zu = {f : f ∈ Z, |f (u)| = kuk}; then Zu is a closed subset of Z (because f 7→ f (u) is continuous), and is non-empty, by the Hahn-Banach theorem (3A5Ab, or Part A above!) Now let P be the set of those closed sets X ⊆ Z such that X ∩ Zu 6= ∅ for every u ∈ U . If Q ⊆ P is non-empty T and totally ordered, then Q ∈ P, because for any u ∈ U {X ∩ Zu : X ∈ Q} is a downwards-directed family of non-empty compact sets, so must have non-empty intersection. By Zorn’s Lemma, upside down, P has a minimal element X; with its relative topology, X is a compact Hausdorff space. (b) We have a linear operator R : U → C(X) given by setting (Ru)(x) = x(u) for every u ∈ U , x ∈ X; because X ⊆ Z, kRuk∞ ≤ kuk, and because X ∈ P, kRuk∞ = kuk, for every u ∈ U . Moreover, if G ⊆ X is a non-empty open set (in the relative topology of X) then X \ G cannot belong to P, because X is minimal, so there is a (non-zero) u ∈ U such that |x(u)| < kuk for every x ∈ X \ G. Replacing u by kuk−1 u if need be, we may suppose that kuk = 1. What this means is that W = R[U ] is a linear subspace of C(X) which is isomorphic, as normed space, to U , and has the property that whenever G ⊆ X is a non-empty relatively open set there is an f ∈ W such that kf k∞ = 1 and |f (x)| < 1 for every x ∈ X \ G. Observe that, because X \ G is compact, there is now some α < 1 such that |f (u)| ≤ α for every f ∈ X \ G. Because W is isomorphic to U , it has the Hahn-Banach property. (c) Now consider V = `∞ (X), V0 = W , T0 : V0 → W the identity map. Because W has the Hahn-Banach property, there is a linear operator T : `∞ (X) → W , extending T0 , and of norm kT0 k = 1. (d) If h ∈ `∞ (X) and x0 ∈ X \ {x : h(x) 6= 0}, then (T h)(x0 ) = 0. P P?? Otherwise, set G = {y : y ∈ X \ {x : h(x) 6= 0}, (T h)(y) 6= 0}. This is a non-empty open set in X, so there are f ∈ W , α < 1 such that kf k∞ = 1, |f (x)| ≤ α for every x ∈ X \ G. Because kf k∞ = 1, there must be some x1 ∈ X such that |f (x1 )| = 1, and of course x1 ∈ G, so that (T h)(x1 ) 6= 0. But let δ > 0 be such that δkhk∞ ≤ 1 − α. Then, because h(x) = 0 for x ∈ G, |f (x)| + |δh(x)| ≤ 1 for every x ∈ X, and kf + δhk∞ , kf − δhk∞ are both less than or equal to 1. As T f = f and kT k = 1, this means that kf − δT hk∞ ≤ 1,

kf + δT hk∞ ≤ 1;

consequently |f (x1 )| + δ|(T h)(x1 )| = max(|(f + δT h)(x1 )|, |(f − δT h)(x1 )|) ≤ 1. But |f (x1 )| = 1 and δ(T h)(x1 ) 6= 0, so this is impossible. X XQ Q

310

Function spaces

363R

(e) It follows that T h = h for every h ∈ C(X). P P?? Suppose, if possible, otherwise. Then there is a δ > 0 such that G = {x : |(T h)(x) − h(x)| > δ} is not empty. Let f ∈ W be such that kf k = 1 but |f (x)| < 1 for every x ∈ X \ G. Then there is an x0 ∈ X such that |f (x0 )| = 1; of course x0 must belong to G. Set f1 =

h(x0 ) f, f (x0 )

so that f1 ∈ W and f1 (x0 ) = h(x0 ). Set h1 (x) = max(h(x) − δ, min(h(x) + δ, f1 (x)))

for x ∈ X. Then h1 ∈ C(X). Setting H = {x : |h(x) − h(x0 )| + |f1 (x) − f1 (x0 )| < δ}, H is an open set containing x0 and |f1 (x) − h(x)| ≤ |f1 (x0 ) − h(x0 )| + δ = δ,

h1 (x) = f1 (x)

for every x ∈ H. Consequently x0 ∈ / {x : (f1 − h1 )(x) 6= 0}, and T (f1 − h1 )(x0 ) = 0, by (d). But this means that (T h1 )(x0 ) = (T f1 )(x0 ) = f1 (x0 ) = h(x0 ), so that |h(x0 ) − (T h)(x0 )| = |T (h1 − h)(x0 )| ≤ kT (h1 − h)k∞ ≤ kh1 − hk∞ ≤ δ, which is impossible, because x0 ∈ G. X XQ Q (f ) This tells us at once that W = C(X). But (d) also tells us that X is extremally disconnected. P P Let G ⊆ X be any open set. Then χX = χG + χ(X \ G), so χX = T (χX) = h1 + h2 , where h1 = T (χG), h2 = T (χ(X \ G)). Now from (d) we see that h1 must be zero on X \ G while h2 must be zero on G. Thus we have h1 (x) = 1 for x ∈ G; as h1 is continuous, h1 (x) = 1 for x ∈ G, and h1 = χG. Of course it follows that G is open. As G is arbitrary, X is extremally disconnected. Q Q (g) Being also compact and Hausdorff, therefore regular (3A3Bc), X is zero-dimensional (3A3Bd). We may therefore identify X with the Stone space of its regular open algebra A (314S), and W = C(X) with L∞ (A). Thus R : U → C(X) is a Banach space isomorphism between U and C(X) ∼ = L∞ (A); so U is of the type declared. 363S The Banach-Ulam problem At a couple of points already (232Hc, the notes to §326) I have remarked on a problem which was early recognised as a fundamental question in abstract measure theory. I now set out some formulations of the problem which arise naturally from the work done so far. I will do this by writing down a list of statements which are equiveridical in the sense that if one of them is true, so are all the others; the ‘Banach-Ulam problem’ asks whether they are indeed true. I should remark that this is not generally counted as an ‘open’ problem. It is in fact believed by most of us that these statements are independent of the usual axioms of Zermelo-Fraenkel set theory, including the axiom of choice and even the continuum hypothesis. As such, this problem belongs to Volume 5 rather than anywhere earlier, but its manifestations will become steadily more obtrusive as we continue through this volume and the next, and I think it will be helpful to begin collecting them now. The ideas needed to show that the statements here imply each other are already accessible; in particular, they involve no set theory beyond Zorn’s Lemma. These implications constitute the following theorem, derived from Luxemburg 67a. Theorem The following statements are equiveridical. (i) There are a set X and a probability measure ν, with domain PX, such that ν{x} = 0 for every x ∈ X. (ii) There are a localizable measure space (X, Σ, µ) and an absolutely continuous countably additive functional ν : Σ → R which is not truly continuous, so has no Radon-Nikod´ ym derivative (definitions: 232Ab, 232Hf). (iii) There are a Dedekind complete Boolean algebra A and a countably additive functional ν : A → R which is not completely additive. (iv) There is a Dedekind complete Riesz space U such that Uc∼ 6= U × .

363Xd

L∞

311

proof (a)(i)⇒(ii) Let X be a set with a probability measure ν, defined on PX, such that ν{x} = 0 for every x ∈ X. Let µ be counting measure on X. Then (X, PX, µ) is strictly localizable, and ν : PX → R is countably additive; also νE = 0 whenever µE is finite, so ν is absolutely continuous with respect to µ. But if µE < ∞ then E is finite and ν(X \ E) = 1, so ν is not truly continuous, and has no Radon-Nikod´ ym derivative (232D). (b)(ii)⇒(iii) Let (X, Σ, µ) be a localizable measure space and ν : Σ → R an absolutely continuous countably additive functional which is not truly continuous. Let (A, µ ¯) be the measure algebra of µ; then we have an absolutely continuous countably additive functional ν¯ : A → R defined by setting ν¯E • = νE for every E ∈ Σ (327C). Since ν is not truly continuous, ν¯ is not completely additive (327Ce). Also A is Dedekind complete, because µ is localizable, so A and ν¯ witness (iii). (c)(iii)⇒(i) Let A be a Dedekind complete Boolean algebra and ν : A → R a countably additive functional which is not completely additive. Because ν is bounded (326I), therefore expressible as the difference of non-negative countably additive functionals (326H), there must be a non-negative countably additive functional ν 0 on A which is not completely additive. P By 326N, there is a partition of unity hai ii∈I in A such that i∈I ν 0 ai < ν 0 1. Set K = {i : i ∈ I, ν 0 ai > 0}; then K must be countable, so P ν 0 (supi∈I\K ai ) = ν 0 1 − ν 0 (supi∈K ai ) = ν 0 1 − i∈K ν 0 ai > 0. For J ⊆ I set µJ = ν 0 (supi∈J\K ai ); the supremum is always defined because A is Dedekind complete. Because ν 0 is countably additive and non-negative, so is µ; because ν 0 ai = 0 for i ∈ J \ K, µ{i} = 0 for every i ∈ I. Multiplying µ by a suitable scalar, if need be, (I, PI, µ) witnesses that (i) is true. (d)(iii)⇒(iv) If A is a Dedekind complete Boolean algebra with a countably additive functional which is not completely additive, then U = L∞ (A) is a Dedekind complete Riesz space (363Mb) and Uc∼ 6= U × , by 363K (recalling, as in (c) above, that the functional must be bounded). (e)(iv)⇒(iii) Let U be a Dedekind complete Riesz space such that U × 6= Uc∼ . Take f ∈ Uc∼ \ U × ; replacing f by |f | if need be, we may suppose that f ≥ 0 is sequentially order-continuous but not ordercontinuous (355H, 355I). Let A be a non-empty downwards-directed set in U , with infimum 0, such that inf u∈A f (u) > 0 (351Ga). Take e ∈ A, and consider the solid linear subspace Ue of U generated by e; write g for the restriction of f to Ue . Because the embedding of Ue in U is order-continuous, g ∈ (Ue )∼ c ; because A ∩ Ue is downwards-directed and has infimum 0, and inf u∈A∩Ue g(u) = inf u∈A f (u) > 0, Ue× .

g∈ / But Ue is a Riesz space with order unit e, and is Dedekind complete because U is; so it can be identified with L∞ (A) for some Boolean algebra A (363N), and A is Dedekind complete, by 363M. ∞ × Accordingly we have a Dedekind complete Boolean algebra A such that L∞ (A)∼ c 6= L (A) . By 363K, there is a (bounded) countably additive functional on A which is not completely additive, and (iii) is true. 363X Basic exercises (a) Let A be a Boolean algebra and U a Banach algebra. Let ν : A → U be a bounded additive function and T : L∞ (A) → U the corresponding bounded linear operator. Show that T is multiplicative iff ν(a ∩ b) = νa × νb for all a, b ∈ A. > (b) Let A, B be Boolean algebras and T : L∞ (A) → L∞ (A) a linear operator. Show that the following are equiveridical: (i) there is a Boolean homomorphism π : A → A such that T = Tπ (ii) T (u × v) = T u × T v for all u, v ∈ L∞ (A) (iii) T is a Riesz homomorphism and T eA = eB , where eA is the standard order unit of L∞ (A). (c) Let A, B be Boolean algebras and T : L∞ (A) → L∞ (B) a Riesz homomorphism. Show that there are a Boolean homomorphism π : A → B and a v ≥ 0 in L∞ (B) such that T u = v × Tπ u for every u ∈ L∞ (A), where Tπ is the operator associated with π (363F). (d) Let A be a Boolean algebra and C a subalgebra of A. Show that L∞ (C), regarded as a subspace of L (A) (363Ga), is order-dense in L∞ (A) iff C is order-dense in A. ∞

312

Function spaces

363Xe

> (e) Let (X, Σ, µ) be a measure space with measure algebra A, and L∞ the space of bounded Σmeasurable real-valued functions on X. A linear lifting of µ is a positive linear operator T : L∞ (A) → L∞ such that T (χ1A ) = χX and (T u)• = u for every u ∈ L∞ (A), writing f 7→ f • for the canonical map from L∞ to L∞ (A) (363H-363I). (i) Show that if θ : A → Σ is a lifting in the sense of 341A then Tθ , as defined in 363F, is a linear lifting. (ii) Show that if T : L∞ (A) → L∞ is a linear lifting, then there is a corresponding lower density θ : A → Σ defined by setting θa = {x : T (χa)(x) = 1} for each a ∈ A. (iii) Show that θ, as defined in (ii), is a lifting iff T is a Riesz homomorphism. (f ) Let U be any commutative ring with multiplicative identity 1. Show that the set A of idempotents in U (that is, elements a ∈ U such that a2 = a) is a Boolean algebra with identity 1, writing a ∩ b = ab, 1 \ a = 1 − a for a, b ∈ A. (g) Let A be a Boolean algebra. Show that A is isomorphic to the Boolean algebras of idempotents of S(A) and L∞ (A). (h) Let A be a Dedekind σ-complete Boolean algebra. (i) Show that for any u ∈ L∞ (A), α ∈ R there are elements [[u ≥ α]], [[u > α]] ∈ A, where [[u ≥ α]] is the largest a ∈ A such that u × χa ≥ αχa, and [[u > α]] = supβ>α [[u ≥ β]]. (ii) Show that in the context of 363H, if u corresponds to f • for f ∈ L∞ , then [[u ≥ α]] = {x : f (x) ≥ α}• , [[u > α]] = {x : f (x) > α}• . (iii) Show that if A ⊆ L∞ is non-empty and v ∈ L∞ , then v = sup A iff [[v > α]] = supu∈A [[u > α]] for every α ∈ R; in particular, v = u iff [[v > α]] = [[u > α]] for every α. (iv) Show that a function φ : R → A is of the form φ(α) = [[u > α]] iff (α) φ(α) = supβ>α φ(β) for every α ∈ R (β) there is an M such that φ(M ) = 0, φ(−M ) = 1. (v) Put (iii) and (iv) together to give a proof that L∞ is Dedekind σ-complete if A is. (i) Let A be a Dedekind σ-complete Boolean algebra and U ⊆ L∞ (A) a (sequentially) order-closed Riesz subspace containing χ1. Show that U can be identified with L∞ (B) for some (sequentially) order-closed subalgebra B ⊆ A. (Hint: set B = {b : χb ∈ U } and use 363N.) (j) For a Boolean algebra A, with Stone space Z, write L∞ C (A) for the Banach algebra C(Z; C) of continuous complex-valued functions on Z. Prove results corresponding to 363C, 363Ea, 363F-363I for the complex case. 363Y Further exercises (a) Let A be a Boolean algebra. Given the linear structure, ordering, multiplication and norm of S(A) as described in §361, show that a norm completion of S(A) will serve for L∞ (A) in the sense that all the results of 363B-363Q can be proved with no use of the axiom of choice except an occasional appeal to countably many choices in sequential forms of the theorems. (b) Let A be a Boolean algebra. Show that A is ccc iff L∞ (A) has the countable sup property (241Yd, 353Ye). (c) Let X be an extremally disconnected topological space, and G its regular open algebra. Show that there is a natural isomorphism between L∞ (G) and Cb (X). (d) Let X be a compact Hausdorff space. Let us say that a linear subspace U of C(X) is `∞ -complemented in C(X) if there is a linear subspace V such that C(X) = U ⊕ V and ku + vk∞ = max(kuk∞ , kvk∞ ) for all u ∈ U , v ∈ V . Show that there is a one-to-one correspondence between such subspaces U and openand-closed subsets E of X, given by setting U = {f : f ∈ C(X), f (x) = 0 ∀ x ∈ X \ E}. Hence show that if A is any Boolean algebra, there is a canonical isomorphism between A and the partially ordered set of `∞ -complemented subspaces of L∞ (A). (e) Let A be a Boolean algebra. (i) If u ∈ L∞ = L∞ (A), show that |u| = e, the standard order unit of L , iff max(ku + vk∞ , ku − vk∞ ) > 1 whenever v ∈ L∞ \ {0}. (ii) Show that if u, v ∈ L∞ then |u| ∧ |v| = 0 iff kαu + v + wk∞ ≤ max(kαu + wk∞ , kv + wk∞ ) whenever α = ±1 and w ∈ L∞ . (iii) Show that if T : L∞ → L∞ is a normed space automorphism then there are a Boolean automorphism π : A → A and a w ∈ L∞ such that |w| = e and T u = w × Tπ u for every u ∈ L∞ . ∞

363 Notes

L∞

313

(f ) Let X be a set, Σ an algebra of subsets of X, and I an ideal in Σ. (i) Show that L∞ (Σ) can be identified, as Banach lattice and Banach algebra, with the space L∞ of bounded functions f : X → R such that whenever α < β in R there is an E ∈ Σ such that {x : f (x) ≤ α} ⊆ E ⊆ {x : f (x) ≤ β}. (ii) Show that L∞ = {gφ : g ∈ C(Z)}, where Z is the Stone space of Σ and φ : X → Z is a function (to be described). (iii) Show that L∞ (Σ/I) can be identified, as Banach lattice and Banach algebra, with L∞ /V, where V is the set of those functions f ∈ L∞ such that for every ² > 0 there is a member of I including {x : |f (x)| ≥ ²}. (g) Let (X, Σ, µ) be a complete probability space with measure algebra A. Let hBn in∈N be a nondecreasing sequence of closed subalgebras of A such that A is the closed subalgebra of itself generated by S • 1 1 n∈N Bn , and set Σn = {F : F ∈ Bn } for each n. Let Pn : L (µ) → L (µ¹Σn ) be the conditional ∞ expectation operator for each n, so that Pn ¹L (µ) is a positive linear operator from L∞ (µ) ∼ = L∞ (A) to ∞ ∞ ∼ L (µ¹Σn ) = L (Bn ). Suppose that we are given for each n a lifting θn : Bn → Σn and that θn+1 b = θn b whenever n ∈ N, b ∈ Bn . Let Tn : L∞ (Bn ) → L∞ be the corresponding linear liftings (363Xe), and F any non-principal ultrafilter on N. (i) Show that for any u ∈ L∞ (A), hTn Pn uin∈N converges almost everywhere. (ii) For u ∈ L∞ (A) set (T u)(x) = limn→F (Tn Pn u)(x) for x ∈ X, u ∈ L∞ (A). Show that T is a linear lifting of µ. (iii) Use 363Xe(ii) and 341J to show that there is a lifting θ of µ extending every θn . (iv) Use this as the countable-cofinality inductive step in a proof of the Lifting Theorem (using partial liftings rather than partial lower densities, as suggested in 341Li). (h) Let A be a Dedekind σ-complete Boolean algebra and ν : A → R a countably additive func∞ tional. Suppose that hun in∈N is an order-bounded R sequence in LR (A) such that inf n∈N supm≥n um and supn∈N inf m≥n um are equal to u say. Show that u dν = limn→∞ un dν. (i) Let Σ be the family of those sets E ⊆ [0, 1] such that µ(int E) = µE, where µ is Lebesgue measure. (i) Show that Σ is an algebra of subsets of [0, 1] and that every member of Σ is Lebesgue measurable. (ii) Show that if we identify L∞ (Σ) with a set of real-valued functions on [0, 1], as in 363Yf, R then we get just the space of Riemann integrable functions. (iii) Show that if we write ν for µ¹Σ, then dν, as defined in 363L, is just the Riemann integral. (j) Show that a normed space over C has the Hahn-Banach property of 363R for complex spaces iff it is isomorphic to L∞ C (A), as described in 363Xj, for some Dedekind complete Boolean algebra A. 363 Notes and comments As with S(A), I have chosen a definition of L∞ (A) in terms of the Stone space of A; but as with S(A), this is optional (363Ya). By and large the basic properties of L∞ are derived very naturally from those of S. The spaces L∞ (A), for general Boolean algebras A, are not in fact particularly important; they have too few properties not shared by all the spaces C(X) for compact Hausdorff X. The point at which it becomes helpful to interpret C(X) as L∞ (A) is when C(X) is Dedekind σ-complete. The spaces X for which this is true are difficult to picture, and alternative representations of L∞ along the lines of 363H-363I can be easier on the imagination. For Dedekind σ-complete A, there is an alternative description of members of L∞ (A) in terms of objects ‘[[u > α]]’ (363Xh); I will return to this idea in the next section. For the moment I remark only that it gives an alternative approach to 363M not necessarily depending on the representation of L∞ as a quotient L∞ /V nor on an analysis of a Stone space. I used a version of such an argument in the proof of 363M which I gave in Fremlin 74a, 43D. I spend so much time on 363M not only because Dedekind completeness is one of the basic properties of any lattice, but because it offers an abstract expression of one of the central results of Chapter 24. In 243H I showed that L∞ (µ) is always Dedekind σ-complete, and that it is Dedekind complete if µ is localizable. We can now relate this to the results of 321H and 322Be: the measure algebra of any measure is Dedekind σ-complete, and the measure algebra of a localizable measure is Dedekind complete. The ideas of the proof of 363M can of course be rearranged in various ways. One uses 353Yb: for completely regular spaces X, C(X) is Dedekind complete iff X is extremally disconnected; while for compact Hausdorff spaces, X is extremally disconnected iff it is the Stone space of a Dedekind complete algebra. With the right modification of the concept ‘extremally disconnected’ (314Yf), the same approach works for Dedekind σ-completeness.

314

Function spaces

363 Notes

363R is the ‘Nachbin-Kelley theorem’; it is commonly phrased ‘a normed space U has the Hahn-Banach extension property iff it is isomorphic, as normed space, to C(X) for some compact extremally disconnected Hausdorff space X’, but the expression in terms of L∞ spaces seems natural in the present context. The implication in one direction (Part A of the proof) calls for nothing but a check through one of the standard proofs of the Hahn-Banach theorem to make sure that the argument applies in the generalized form. Part B of the proof has ideas in it; I have tried to set it out in a way suggesting that if you can remember the construction of the set X the rest is just a matter of a little ingenuity. One way of trying to understand the multiple structures of L∞ spaces is by looking at the corresponding automorphisms. We observe, for instance, that an operator T from L∞ (A) to itself is a Banach algebra automorphism iff it is a Banach lattice automorphism preserving the standard order unit iff it corresponds to an automorphism of the algebra A (363Xb). Of course there are Banach space automorphisms of L∞ which do not respect the order or multiplicative structure; but they have to be closely related to algebra isomorphisms (363Ye). I devote a couple of exercises (363Xe, 363Yg) to indications of how the ideas here are relevant to the Lifting Theorem. If you found the formulae of the proof of 341G obscure it may help to work through the parallel argument. A lecture by W.A.J.Luxemburg on the equivalence between (i) and (iv) in 363S was one of the turning points in my mathematical apprenticeship. I introduce it here, even though the real importance of the Banach-Ulam problem lies in the metamathematical ideas it has nourished, because these formulations provide a focus for questions which arise naturally in this volume and which otherwise might prove distracting. The next group of significant ideas in this context will appear in §438.

364 L0 My next objective is to develop an abstract construction corresponding to the L0 (µ) spaces of §241. These generalized L0 spaces will form the basis of the work of the rest of this chapter and also the next; partly because their own properties are remarkable, but even more because they form a framework for the study of Archimedean Riesz spaces in general (see §368). There seem to be significant new difficulties, and I take the space to describe an approach which can be made essentially independent of the route through Stone spaces used in the last three sections. I embark directly on a definition in the new language (364A), and relate it to the constructions of §241 (364C-364E, 364J) and §§361-363 (364K). The ideas of Chapter 27 can also be expressed in this language; I make a start on developing the machinery for this in 364G, with the formula ‘[[u ∈ E]]’, ‘the region in which u belongs to E’, and some exercises (364Xd-364Xf). Following through the questions addressed in §363, I discuss Dedekind completeness in L0 (364M-364O), properties of its multiplication (364P), the expression of the original algebra in terms of L0 (364Q), the action of Boolean homomorphisms on L0 (364R) and product spaces (364S). In 364T-364W I describe representations of the L0 space of a regular open algebra. 364A Definition Let A be a Dedekind σ-complete Boolean algebra. I will write L0 (A) for the set of all functions α 7→ [[u > α]] : R → A such that (α) [[u > α]] = supβ>α [[u > β]] in A for every α ∈ R, (β) inf α∈R [[u > α]] = 0, (γ) supα∈R [[u > α]] = 1. 364B Remarks (a) My reasons for using the notation ‘[[u > α]]’ rather than ‘u(α)’ will I hope become clear in the next few paragraphs. For the moment, if you think of A as a σ-algebra of sets and of L0 (A) as the family of A-measurable real-valued functions, then [[u > α]] corresponds to the set {x : u(x) > α} (364Ja). (b) Some readers will recognise the formula ‘[[. . . ]]’ as belonging to the language of forcing, so that [[u > α]] could be read as ‘the Boolean value of the proposition “u > α”’. But a vocalisation closer to my intention might be ‘the region where u > α’.

L0

364C

315

(c) Note that condition (α) of 364A automatically ensures that [[u > α]] ⊆ [[u > α0 ]] whenever α0 ≤ α in R. (d) In fact it will sometimes be convenient to note that the conditions of 364A can be replaced by (α0 ) [[u > α]] = supq∈Q,q>α [[u > q]] in A for every α ∈ R, (β 0 ) inf n∈N [[u > n]] = 0, (γ 0 ) supn∈N [[u > −n]] = 1; the point being that we need look only at suprema and infima of countable subsets of A. (e) In order to make sense of this construction we need to match it with an alternative route to the same object, based on σ-algebras and σ-ideals of sets, as follows. 364C Proposition Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of Σ. (a) Write L0 for the space of all Σ-measurable functions from X to R. Then L0 , with its linear structure, ordering and multiplication inherited from RX , is a Dedekind σ-complete f -algebra with multiplicative identity. (b) Set W = {f : f ∈ L0 , {x : f (x) 6= 0} ∈ I}. Then (i) W is a sequentially order-closed solid linear subspace and ideal of L0 ; (ii) the quotient space L0 /W, with its inherited linear, order and multiplicative structures, is a Dedekind σ-complete Riesz space and an f -algebra with a multiplicative identity; (iii) for f , g ∈ L0 , f • ≤ g • in L0 /W iff {x : f (x) > g(x)} ∈ I, and f • = g • in L0 /W iff {x : f (x) 6= g(x)} ∈ I. proof (Compare 241A-241H.) (a) The point is just that L0 is a sequentially order-closed Riesz subspace and subalgebra of RX . The facts we need to know – that constant functions belong to L0 , that f + g, αf , f × g, supn∈N fn belong to L0 whenever f , g, fn do and {fn : n ∈ N} is bounded above – are all covered by 121E-121F. Its multiplicative identity is of course the constant function χX. (b)(i) The necessary verifications are all elementary. (ii) Because W is a solid linear subspace of the Riesz space L0 , the quotient inherits a Riesz space structure (352Jb); because W is an ideal of the ring (L0 , +, ×), L0 /W inherits a multiplication; it is a commutative algebra because L0 is; and has a multiplicative identity e = χX • because χX is the identity of L0 . To check that L0 /W is an f -algebra it is enough to observe that, for any non-negative f , g , h ∈ L0 , f • × g • = (f × g)• ≥ 0, and if f • ∧ g • = 0 then {x : f (x) > 0} ∩ {x : g(x) > 0} ∈ I, so that {x : f (x)h(x) > 0} ∩ {x : g(x) > 0} ∈ I and (f • × h• ) ∧ g • = (h• × f • ) ∧ g • = 0. Finally, L0 /W is Dedekind σ-complete, by 353J(a-iii). (iii) For f , g ∈ L0 , f • ≤ g • ⇐⇒ (f − g)+ ∈ W ⇐⇒ {x : f (x) > g(x)} = {x : (f − g)+ (x) 6= 0} ∈ I (using the fact that the canonical map from L0 to L0 /W is a Riesz homomorphism, so that ((f − g)+ )• = (f • − g • )+ ). Similarly f • = g • ⇐⇒ f − g ∈ W ⇐⇒ {x : f (x) 6= g(x)} = {x : (f − g)(x) 6= 0} ∈ I.

316

Function spaces

364D

364D Theorem Let X be a set and Σ a σ-algebra of subsets of X. Let A be a Dedekind σ-complete Boolean algebra and π : Σ → A a surjective Boolean homomorphism, with kernel a σ-ideal I; define L0 and W as in 364C, so that U = L0 /W is a Dedekind σ-complete f -algebra with multiplicative identity. (a) We have a canonical bijection T : U → L0 = L0 (A) defined by the formula [[T f • > α]] = π{x : f (x) > α} for every f ∈ L0 , α ∈ R. (b)(i) For any u, v ∈ U , [[T (u + v) > α]] = supq∈Q [[T u > q]] ∩ [[T v > α − q]] for every α ∈ R. (ii) For any u ∈ U , γ > 0, [[T (γu) > α]] = [[T u >

α γ ]]

for every α ∈ R. (iii) For any u, v ∈ U , u ≤ v ⇐⇒ [[T u > α]] ⊆ [[T v > α]] for every α ∈ R. (iv) For any u, v ∈ U + , [[T (u × v) > α]] = supq∈Q,q>0 [[T u > q]] ∩ [[T v >

α q ]]

for every α ≥ 0. (v) Writing e = (χX)• for the multiplicative identity of U , we have [[T e > α]] = 1 if α < 1, 0 if α ≥ 1. proof (a)(i) Given f ∈ L0 , set ζf (α) = π{x : f (x) > α} for α ∈ R. Then it is easy to see that ζf satisfies the conditions (α)0 -(γ)0 of 364Bd, because π is sequentially order-continuous (313Qb). Moreover, if f • = g • in U , then ζf (α) 4 ζg (α) = π({x : f (x) > α}4{x : g(x) > α}) = 0 for every α ∈ R, because {x : f (x) > α}4{x : g(x) > α} ⊆ {x : f (x) 6= g(x)} ∈ I, and ζf = ζg . So we have a well-defined member T u of L0 defined by the given formula, for any u ∈ U . (ii) Next, given w ∈ L0 , there is a u ∈ L0 /W such that T u = w. P P For each q ∈ Q, choose Fq ∈ Σ such that πFq = [[w > q]] in A. Note that if q 0 ≥ q then π(Fq0 \ Fq ) = [[u > q 0 ]] \ [[u > q]] = 0, so Fq0 \ Fq ∈ I. Set H=

S q∈Q

Fq \

T

S n∈N

q∈Q,q≥n

Fq ∈ Σ,

and for x ∈ X set f (x) = sup{q : q ∈ Q, x ∈ Fq } if x ∈ H, = 0 otherwise. (H is chosen just to make the formula here give a finite value for every x.) We have

πH = sup [[w > q]] \ inf

sup

n∈N q∈Q,q≥n

q∈Q

[[w > q]]

= 1A \ inf [[w > n]] = 1A \ 0A = 1A , n∈N

so X \ H ∈ I. Now, for any α ∈ R,

L0

364D

[

{x : f (x) > α} =

317

Fq ∪ (X \ H) if α < 0

q∈Q,q>α

[

=

Fq \ (X \ H) if α ≥ 0,

q∈Q,q>α

and in either case belongs to Σ; so that f ∈ L0 and f • is defined in L0 . Next, for any α ∈ R, [[T f • > α]] = π{x : f (x) > α} = π(

[

Fq )

q∈Q,q>α

=

sup

[[w > q]] = [[w > α]],

q∈Q,q>α

and T f • = w. Q Q (iii) Thus T is surjective. To see that it is injective, observe that if f , g ∈ L0 , then T f • = T g • =⇒ [[T f • > α]] = [[T g • > α]] for every α ∈ R =⇒ π{x : f (x) > α} = π{x : g(x) > α} for every α ∈ R =⇒ {x : f (x) > α}4{x : g(x) > α} ∈ I for every α ∈ R [ =⇒ {x : f (x) 6= g(x)} = ({x : f (x) > q}4{x : g(x) > q}) ∈ I q∈Q

=⇒ f = g . •

•

So we have the claimed bijection. (b)(i) Let f , g ∈ L0 be such that u = f • , v = g • , so that u + v = (f + g)• . For any α ∈ R,

[[T (u + v) > α]] = π{x : f (x) + g(x) > α} [ = π( {x : f (x) > q} ∩ {x : g(x) > α − q}) q∈Q

= sup π{x : f (x) > q} ∩ π{x : g(x) > α − q} q∈Q

(because π is a sequentially order-continuous Boolean homomorphism) = sup [[T u > q]] ∩ [[T v > α − q]]. q∈Q

(ii) Let f ∈ L0 be such that f • = u, so that (γf )• = γu. For any α ∈ R, α γ

[[T (γu) > α]] = π{x : γf (x) > α} = π{x : f (x) > } = [[T u >

α γ ]].

(iii) Let f , g ∈ L0 be such that f • = u and g • = v. Then

u ≤ v ⇐⇒ {x : f (x) > g(x)} ∈ I (see 364C(b-iii)) ⇐⇒

[

{x : f (x) > q ≥ g(x)} ∈ I

q∈Q

⇐⇒ {x : f (x) > α} \ {x : g(x) > α} ∈ I for every α ∈ R ⇐⇒ π{x : f (x) > α} \ π{x : g(x) > α} = 0 for every α ⇐⇒ [[T u > α]] ⊆ [[T v > α]] for every α.

318

Function spaces

364D

(iv) Now suppose that u, v ≥ 0, so that they can be expressed as f • , g • where f , g ≥ 0 in L0 (351J), and u × v = (f × g)• . If α ≥ 0, then [[T (u × v) > α]] = π(

[

α q

{x : f (x) > q} ∩ {x : g(x) > })

q∈Q,q>0

=

α q

sup π{x : f (x) > q} ∩ π{x : g(x) > } q∈Q,q>0

=

sup [[T u > q]] ∩ [[T v > q∈Q,q>0

α q ]].

(v) This is trivial, because [[T (χX)• > α]] = π{x : (χX)(x) > α} = πX = 1 if α < 1, = π∅ = 0 if α ≥ 1. 364E Theorem Let A be a Dedekind σ-complete Boolean algebra. Then L0 = L0 (A) has the structure of a Dedekind σ-complete f -algebra with multiplicative identity e, defined by saying [[u + v > α]] = supq∈Q [[u > q]] ∩ [[v > α − q]], whenever u, v ∈ L0 and α ∈ R, [[γu > α]] = [[u >

α γ ]]

whenever u ∈ L0 , γ ∈ ]0, ∞[ and α ∈ R, u ≤ v ⇐⇒ [[u > α]] ⊆ [[v > α]] for every α ∈ R, [[u × v > α]] = supq∈Q,q>0 [[u > q]] ∩ [[v >

α q ]]

whenever u, v ≥ 0 in L0 and α ≥ 0, [[e > α]] = 1 if α < 1, 0 if α ≥ 1. proof (a) By the Loomis-Sikorski theorem (314M), we can find a set Z (the Stone space of A), a σ-algebra Σ of subsets of Z (the algebra generated by the open-and-closed sets and the ideal M of meager sets) and a surjective sequentially order-continuous Boolean homomorphism π : Σ → A (corresponding to the identification between A and the quotient Σ/M). Consequently, defining L0 and W as in 364C, we have a bijection between the Dedekind σ-complete f -algebra L0 /W and L0 (364Da). Of course this endows L0 itself with the structure of a Dedekind σ-complete f -algebra; and 364Db tells us that the description of the algebraic operations above is consistent with this structure. (b) In fact the f -algebra structure is completely defined by the description offered. For while scalar multiplication is not described for γ ≤ 0, the assertion that L0 is a Riesz space implies that 0u = 0 and that γu = (−γ)(−u) for γ < 0; so if we have formulae to describe u + v and γu for γ > 0, this suffices to define the linear structure of L0 . Note that we have an element 0 in L0 defined by setting [[0 > α]] = 0 if α ≥ 0, 1 if α < 0, and the formula for u + v shows us that [[0 + u > α]] = supq∈Q [[0 > q]] ∩ [[u > α − q]] = supq∈Q,q α − q]] = [[u > α]] for every α, so that 0 is the zero of L0 . As for multiplication, if L0 is to be an f -algebra we must have [[u × v > α]] ⊇ [[0 > α]] = 1 whenever u, v ∈ (L0 )+ and α < 0, because u × v ≥ 0. So the formula offered is sufficient to determine u × v for non-negative u and v; and for others we know that

L0

364I

319

u × v = (u+ × v + ) − (u+ × v − ) − (u− × v + ) + (u− × v − ), so the whole of the multiplication of L0 is defined. 364F The rest of this section will be devoted to understanding the structure just established. I start with a pair of elementary facts. Lemma Let A be a Dedekind σ-complete Boolean algebra. (a) If u, v ∈ L0 = L0 (A) and α, β ∈ R, [[u + v > α + β]] ⊆ [[u > α]] ∪ [[v > β]]. 0

(b) If u, v ≥ 0 in L and α, β ≥ 0 in R, [[u × v > αβ]] ⊆ [[u > α]] ∪ [[v > β]]. proof (a) For any q ∈ Q, either q ≥ α and [[u > q]] ⊆ [[u > α]], or q ≤ α and [[v > α + β − q]] ⊆ [[v > β]]; thus in all cases [[u > q]] ∩ [[v > α + β − q]] ⊆ [[u > α]] ∪ [[v > β]]; taking the supremum over q, we have the result. (b) The same idea works, replacing α + β − q by αβ/q for q > 0. 364G Yet another description of L0 is sometimes appropriate, and leads naturally to an important construction (364I). Proposition Let A be a Dedekind σ-complete Boolean algebra. Then there is a bijection between L0 = L0 (A) and the set Φ of sequentially order-continuous Boolean homomorphisms from the algebra B of Borel subsets of R to A, defined by saying that u ∈ L0 corresponds to φ ∈ Φ iff [[u > α]] = φ(]α, ∞[) for every α ∈ R. proof (a) If φ ∈ Φ, then the map α 7→ φ(]α, ∞[) satisfies the conditions of 364Bd, so corresponds to an element uφ of L0 . (b) If φ, ψ ∈ Φ and uφ = uψ , then φ = ψ. P P Set A = {E : E ∈ B, φ(E) = ψ(E)}. Then A is a σ-subalgebra of B, because φ and ψ are both sequentially order-continuous Boolean homomorphisms, and contains ]α, ∞[ for every α ∈ R. Now A contains ]−∞, α] for every α, and therefore includes B (121J). But this means that φ = ψ. Q Q (c) Thus φ 7→ uφ is injective. But it is also surjective. P P As in 364E, take a set Z, a σ-algebra Σ of subsets of Z and a surjective sequentially order-continuous Boolean homomorphism π : Σ → A; let T : L0 /W → L0 be the bijection described in 364D. If u ∈ L0 , there is an f ∈ L0 such that T f • = u. Now consider φE = πf −1 [E] for E ∈ B. f −1 [E] always belongs to Σ (121Ef), so φE is always well-defined; E 7→ f −1 [E] and π are sequentially order-continuous, so φ also is; and φ(]α, ∞[) = π{z : f (z) > α} = [[u > α]] for every α, so u = uφ . Q Q Thus we have the declared bijection. 364H Definition In the context of 364G, I will write [[u ∈ E]], ‘the region where u takes values in E’, for φ(E), where φ : B → A is the homomorphism corresponding to u ∈ L0 . Thus [[u > α]] = [[u ∈ ]α, ∞[ ]]. In the same spirit I write [[u ≥ α]] for [[u ∈ [α, ∞[ ]] = inf β β]], [[u 6= 0]] = [[|u| > 0]] = [[u > 0]] ∪ [[u < 0]] and so on, so that (for instance) [[u = α]] = [[u ∈ {α}]] = [[u ≥ α]] \ [[u > α]] for u ∈ L0 , α ∈ R. 364I Proposition Let A be a Dedekind σ-complete Boolean algebra, E ⊆ R a Borel set, and h : E → R a Borel measurable function. Then whenever u ∈ L0 = L0 (A) is such that [[u ∈ E]] = 1, there is an element ¯ ¯ h(u) of L0 defined by saying that [[h(u) ∈ F ]] = [[u ∈ h−1 [F ]]] for every Borel set F ⊆ R. proof All we have to observe is that F → 7 [[u ∈ h−1 [F ]]] is a sequentially order-continuous Boolean homomorphism. (The condition ‘[[u ∈ E]] = 1’ ensures that [[u ∈ h−1 [R]]] = 1.)

320

Function spaces

364J

364J Examples Perhaps I should spell out the most important contexts in which we apply these ideas, even though they have in effect already been mentioned. (a) Let X be a set and Σ a σ-algebra of subsets of X. Then we may identify L0 (Σ) with the space L0 of Σ-measurable real-valued functions on X. (This is the case A = Σ of 364D.) For f ∈ L0 , [[f ∈ E]] (364H) ¯ ) (364I) is just the is just f −1 [E], for any Borel set E ⊆ R; and if h is a Borel measurable function, h(f 0 composition hf , for any f ∈ L . (b) Now suppose that I is a σ-ideal of Σ and that A = Σ/I. Then, as in 364D, we identify L0 (A) with ¯ • ) = (hf )• , for any Borel set E and any Borel a quotient L0 /W. For f ∈ L0 , [[f • ∈ E]] = f −1 [E]• , and h(f measurable function h : R → R. (c) In particular, if (X, Σ, µ) is a measure space with measure algebra A, then L0 (A) becomes identified with L0 (µ) as defined in §241. The same remarks as in 363I apply here; the space L0 (µ) of §241 is larger than the space L0 of the present section. But for every f ∈ L0 (µ) there is a g ∈ L0 such that g = f a.e. (241Bk), so that L0 (µ) can be identified with L0 /N, where N is the set of functions in L0 which are zero almost everywhere (241Yh). 364K Embedding S and L∞ in L0 : Proposition Let A be a Dedekind σ-complete Boolean algebra. (a) We have a canonical embedding of L∞ = L∞ (A) as an order-dense solid linear subspace of L0 = L0 (A); it is the solid linear subspace generated by the multiplicative identity e of L0 . Consequently S = S(A) is also embedded as an order-dense Riesz subspace and subalgebra of L0 . (b) This embedding respects the linear, lattice and multiplicative structures of L∞ and S. (c) For a ∈ A, χa, when regarded as a member of L0 , can be described by the formula [[χa > α]] = 1 if α < 0, = a if 0 ≤ α < 1, = 0 if 1 ≤ α. The function χ : A → L0 is additive, injective, order-continuous and a lattice homomorphism. (d) For every u ∈ (L0 )+ there is a non-decreasing sequence hun in∈N in S such that u0 ≥ 0 and supn∈N un = u. proof Let Z, Σ, L0 , W and π be as in the proof of 364E. I defined L∞ to be the space C(Z) of continuous realvalued functions on Z (363A); but because A is Dedekind σ-complete, there is an alternative representation as L∞ /W ∩ L∞ , where L∞ is the space of bounded Σ-measurable functions from Z to R (363Hb). Put like this, we clearly have an embedding of L∞ ∼ = L0 /W; and this embedding represents = L∞ /W ∩ L∞ in L0 ∼ ∞ 0 ∞ L as a Riesz subspace and subalgebra of L , because L is a Riesz subspace and subalgebra of L0 . L∞ becomes the solid linear subspace of L0 generated by (χZ)• = e, because L∞ is the solid linear subspace of L0 generated by χZ. To see that L∞ is order-dense in L0 , we have only to note that f = supn∈N f ∧ nχZ in L0 for every f ∈ L0 , and therefore (because the map f 7→ f • is sequentially order-continuous) u = supn∈N u∧ne in L0 for every u ∈ L0 . To identify χa, we have the formula χ(πE) = (χE)• , as in 363Hc; but this means that, if a = πE, [[χa > α]] = π{z : χE(z) > α} = πZ = 1 if α < 0, = πE = a if 0 ≤ α < 1, = π∅ = 0 if α ≥ 1, using the formula in 364Da. Evidently χ is injective. Because S is an order-dense Riesz subspace and subalgebra of L∞ (363C), the same embedding represents it as an order-dense Riesz subspace and subalgebra of L0 . (For ‘order-dense’, use 352Nc.) Because χ : A → L∞ is additive, order-continuous and a lattice homomorphism (363D), and the embedding map L∞ ⊆ L0 also is, χ : A → L0 has the same properties. Finally, if u ≥ 0 in L0 , we can represent it as f • where f ≥ 0 in L0 . For n ∈ N set

L0

364M

321

fn (z) = 2−n k if 2−n k ≤ f (z) < 2−n (k + 1) where 0 ≤ k < 4n , = 0 if f (z) ≥ 2n ; then hfn• in∈N is a non-decreasing sequence in S + with supremum u. 364L Corollary Let (A, µ ¯) be a measure algebra. Then S(Af ) can be embedded as a Riesz subspace 0 of L (A), which is order-dense iff (A, µ ¯) is semi-finite. proof (Recall that Af is the ring {a : µ ¯a < ∞}.) The embedding Af ⊆ A is an injective ring homomorphism, f so induces an embedding of S(A ) as a Riesz subspace of S(A), by 361J. Now S(Af ) is order-dense Pn in S(A) iff (A, µ ¯) is semi-finite. P P (i) If (A, µ ¯) is semi-finite and v > 0 in S(A), then v is expressible as j=0 βj χbj where βj ≥ 0 for each j and some βj χbj is non-zero; now there is a non-zero a ∈ Af such that a ⊆ bj , so that 0 < βj χa ∈ S(Af ) and βj χa ≤ v. As v is arbitrary, S(Af ) is quasi-order-dense, therefor order-dense (353A). (ii) If S(Af ) is order-dense in S(A) and b ∈ A \ {0}, there is a u > 0 in S(Af ) such that u ≤ χb; now there are α > 0, a ∈ Af \ {0} such that αχa ≤ u, in which case a ⊆ b. Q Q Now because S(Af ) ⊆ S(A) and S(A) is order-dense in L0 (A), we must have S(Af ) is order-dense in L0 (A) ⇐⇒ S(Af ) is order-dense in S(A) ⇐⇒ (A, µ ¯) is semi-finite. 364M Suprema and infima in L0 We know that any L0 (A) is a Dedekind σ-complete partially ordered set. There is a useful description of suprema for this ordering, as follows. Proposition Let A be a Dedekind σ-complete Boolean algebra and A a subset of L0 = L0 (A). (a) A is bounded above in L0 iff there is a sequence hcn in∈N in A, with infimum 0, such that [[u > n]] ⊆ cn for every u ∈ A. (b) If A is non-empty, then A has a supremum in L0 iff cα = supu∈A [[u > α]] is defined in A for every α ∈ R and inf n∈N cn = 0; and in this case cα = [[sup A > α]] for every α. (c) If A is non-empty and bounded above, then A has a supremum in L0 iff supu∈A [[u > α]] is defined in A for every α ∈ R. proof (a)(i) If A has an upper bound u0 , set cn = [[u0 > n]] for each n; then hcn in∈N satisfies the conditions. (ii) If hcn in∈N satisfies the conditions, set φ(α) = 1 if α < 0, = inf ci if n ∈ N, α ∈ [n, n + 1[ . i≤n

Then it is easy to check that φ satisfies the conditions of 364A, since inf n∈N cn = 0. So there is a u0 ∈ L0 such that φ(α) = [[u0 > α]] for each α. Now, given u ∈ A and α ∈ R, [[u > α]] ⊆ 1 = [[u0 > α]] if α < 0, ⊆

inf [[u > i]] ⊆ inf ci = [[u0 > α]] if n ∈ N, α ∈ [n, n + 1[ .

i≤n

i≤n

Thus u0 is an upper bound for A in L0 . (b)(i) Suppose that cα = supu∈A [[u > α]] is defined in A for every α, and that inf n∈N cn = 0. Then, for any α, supq∈Q,q>α cq = supu∈A,q∈Q,q>α [[u > q]] = supu∈A [[u > q]] = cα . Also, we are supposing that A contains some u0 , so that supn∈N c−n ⊇ supn∈N [[u0 > −n]] = 1. Accordingly there is a u∗ ∈ L0 such that [[u∗ > α]] = cα for every α ∈ R. But now, for v ∈ L0 ,

322

Function spaces

364M

v is an upper bound for A ⇐⇒ [[u > α]] ⊆ [[v > α]] for every u ∈ A, α ∈ R ⇐⇒ [[u∗ > α]] ⊆ [[v > α]] for every α ⇐⇒ u∗ ≤ v, so that u∗ = sup A in L0 . (ii) Now suppose that u∗ = sup A is defined in L0 . Of course [[u∗ > α]] must be an upper bound for {[[u > α]] : u ∈ A} for every α. ?? Suppose we have an α for which it is not the least upper bound, that is, there is a c ⊂ [[u∗ > α]] which is an upper bound for {[[u > α]] : u ∈ A}. Define φ : R → A by setting φ(β) = c ∩ [[u∗ > β]] if β ≥ α, = [[u∗ > β]] if β < α. It is easy to see that φ satisfies the conditions of 364A (we need the distributive law 313Ba to check that φ(β) = supγ>β φ(γ) if β ≥ α), so corresponds to a member v of L0 . But we now find that v is an upper bound for A (because if u ∈ A, β ≥ α then [[u > β]] ⊆ [[u > α]] ∩ [[u∗ > β]] ⊆ c ∩ [[u∗ > β]] = [[v > β]],) that v ≤ u∗ and that v 6= u∗ (because [[v > α]] = c 6= [[u∗ > α]]); but this is impossible, because u∗ is supposed to be the supremum of A. X X Thus if u∗ = sup A is defined in L0 , then supu∈A [[u > α]] is defined in A for every α ∈ R. Also, of course, inf n∈N supu∈A [[u > n]] = inf n∈N [[u∗ > n]] = 0. (c) This is now easy. If A has a supremum, then surely it satisfies the condition, by (b). If A satisfies the condition, then we have a family hcα iα∈R as required in (b); but also, by (a) or otherwise, there is a sequence hc0n in∈N such that cn ⊆ c0n for every n and inf n∈N c0n = 0, so inf n∈N cn is also 0, and both conditions in (b) are satisfied, so A has a supremum. 364N We do not have such a simple formula for general infima (though see 364Xl), but the following facts are useful. Proposition Let A be a Dedekind σ-complete Boolean algebra. (a) If u, v ∈ L0 = L0 (A), then [[u ∧ v > α]] = [[u > α]] ∩ [[v > α]] for every α ∈ R. (b) If A is a non-empty subset of (L0 )+ , then inf A = 0 in L0 iff inf u∈A [[u > α]] = 0 in A for every α > 0. proof (a) Take Z, L0 and π as in the proof of 364E. Express u as f • , v as g • where f , g ∈ L0 , so that u ∧ v = (f ∧ g)• , because the canonical map from L0 to L0 is a Riesz homomorphism (351J). Then [[u ∧ v > α]] = π{z : min(f (z), g(z)) > α} = π({z : f (z) > α} ∩ {z : g(z) > α}) = π{z : f (z) > α} ∩ π{z : g(z) > α} = [[u > α]] ∩ [[v > α]] for every α. (b)(i) If inf u∈A [[u > α]] = 0 for every α > 0, and v is any lower bound for A, then [[v > α]] must be 0 for every α > 0, so that [[v > 0]] = 0; now since [[0 > α]] = 0 for α ≥ 0, 1 for α < 0, v ≤ 0. As v is arbitrary, inf A = 0. (ii) If α > 0 is such that inf u∈A [[u > α]] is undefined, or not equal to 0, let c ∈ A be such that β 0 6= c ⊆ [[u > α]] for every u ∈ A, and consider v = αχc. Then [[v > β]] = [[χc > α ]] is 1 if β < 0, c if 0 ≤ β < α and 0 if β ≥ α. If u ∈ A then [[u > β]] is 1 if β < 0 (since u ≥ 0), at least [[u > α]] ⊇ c if 0 ≤ β < α, and always includes 0; so that v ≤ u. As u is arbitrary, inf A is either undefined in L0 or not 0. 364O

Now we have a reward for our labour, in that the following basic theorem is easy.

Theorem For a Dedekind σ-complete Boolean algebra A, L0 = L0 (A) is Dedekind complete iff A is. proof The description of suprema in 364Mc makes it obvious that if A is Dedekind complete, so that supu∈A [[u > α]] is always defined, then L0 must be Dedekind complete. On the other hand, if L0 is Dedekind complete, then so is L∞ (A) (by 364K and 353J(b-i)), so that A is also Dedekind complete, by 363Mb.

L0

364R

323

364P The multiplication of L0 I have already observed that L0 is always an f -algebra with identity; in particular (because L0 is surely Archimedean) the map u 7→ u × v is order-continuous for every v ≥ 0 (353Oa), and multiplication is commutative (353Ob, or otherwise). The multiplicative identity is χ1 (364E, 364Kc). By 353Pb, or otherwise, u × v = 0 iff |u| ∧ |v| = 0. There is one special feature of multiplication in L0 which I can mention here. Proposition Let A be a Dedekind σ-complete Boolean algebra. Then an element u of L0 has a multiplicative inverse in L0 iff |u| is a weak order unit in L0 iff [[|u| > 0]] = 1. proof If u is invertible, then |u| is a weak order unit, by 353Pc or otherwise. In this case, setting c = 1 \ [[|u| > 0]], we see that [[|u| ∧ χc > 0]] = [[|u| > 0]] ∩ c = 0 (364Na), so that |u| ∧ χc ≤ 0 and χc = 0, that is, c = 0; so [[|u| > 0]] must be 1. To complete the circuit, suppose that [[|u| > 0]] = 1. Let Z, Σ, L0 , π, M be as in the proof of 364E, and S : L0 → L0 the canonical map, so that [[Sh > α]] = π{z : h(z) > α} for every h ∈ L0 , α ∈ R. Express u as Sf where f ∈ L0 . Then π{z : |f (z)| > 0} = [[Sf > 0]] = 1, so {z : f (z) = 0} ∈ M. Set g(z) =

1 f (z)

if f (z) 6= 0,

g(z) = 0 if f (z) = 0.

Then {z : f (z)g(z) 6= 1} ∈ M so u × Sg = S(f × g) = S(χZ) = χ1 and u is invertible. Remark The repeated phrase ‘by 353x or otherwise’ reflects the fact that the abstract methods there can all be replaced in this case by simple direct arguments based on the construction in 364C-364E. 364Q Recovering the algebra: Proposition Let A be a Dedekind σ-complete Boolean algebra. For a ∈ A write Va for the band of L0 = L0 (A) generated by χa. Then a 7→ Va is a Boolean isomorphism between A and the algebra of projection bands of L0 (A). proof I copy from the argument for 363J, itself based on 361K. If a ∈ A, w ∈ L0 then w × χa ∈ Va . P P If v ∈ Va⊥ then |χa| ∧ |v| = 0, so χa × v = 0, so (w × χa) × v = 0, so |w × χa| ∧ |v| = 0; thus w × χa ∈ Va⊥⊥ , which is equal to Va because L0 is Archimedean (353B). Q Q Now, if a ∈ A, u ∈ Va and v ∈ V1\a , then 0 |u| ∧ |v| = 0 because χa ∧ χ(1 \ a) = 0; and if w ∈ L (A) then w = (w × χa) + (w × χ(1 \ a)) ∈ V1 + V1\a . So Va and V1\a are complementary projection bands in L0 . Next, if U ⊆ L0 is a projection band, then χ1 is expressible as u + v = u ∨ v where u ∈ U , v ∈ U ⊥ . Setting a = [[u > 12 ]], a0 = [[v > 21 ]] we must have a ∪ a0 = 1, a ∩ a0 = 0 (using 364M and 364Na), so that a0 = 1 \ a; also 21 χa ≤ u, so that χa ∈ U , and similarly χ(1 \ a) ∈ U ⊥ . In this case Va ⊆ U and V1\a ⊆ U ⊥ , so U must be Va precisely. Thus a 7→ Va is surjective. Finally, just as in 361K, a ⊆ b ⇐⇒ Va ⊆ Vb , so we have a Boolean isomorphism. 364R

I come at last to the result corresponding to 361J and 363F.

Theorem Let A and B be Boolean algebras, and π : A → B a sequentially order-continuous Boolean homomorphism. (a) We have a multiplicative sequentially order-continuous Riesz homomorphism Tπ : L0 (A) → L0 (B) defined by the formula [[Tπ u > α]] = π[[u > α]] for every α ∈ R, u ∈ L0 (A). (b) Defining χa ∈ L0 (A) as in 364K, Tπ (χa) = χ(πa) in L0 (B) for every a ∈ A. If we regard L∞ (A) and ∞ L (B) as embedded in L0 (A) and L0 (B) respectively, then Tπ , as defined here, agrees on L∞ (A) with Tπ as defined in 363F. (c) Tπ is order-continuous iff π is order-continuous, injective iff π is injective, surjective iff π is surjective.

324

Function spaces

364R

¯ π = Tπ h ¯ for (d) [[Tπ u ∈ E]] = π[[u ∈ E]] for every u ∈ L0 (A) and every Borel set E ⊆ R; consequently hT 0 ¯ every Borel measurable h : R → R, writing h indifferently for the associated maps from L (A) to itself and from L0 (B) to itself (364I). (e) If C is another Dedekind σ-complete Boolean algebra and θ : B → C another sequentially ordercontinuous Boolean homomorphism then Tθπ = Tθ Tπ : L0 (A) → L0 (C). proof I write T for Tπ . (a)(i) To see that T u is well-defined in L0 (B) for every u ∈ L0 (A), all we need to do is to check that the map α 7→ π[[u > α]] : R → B satisfies the conditions of 364Bd, and this is easy, because π preserves all countable suprema and infima. (ii) To see that T is linear and order-preserving and multiplicative, we can use the formulae of 364E. For instance, if u, v ∈ L0 (A), then [[T u + T v > α]] = sup [[T u > q]] ∩ [[T v > α − q]] = sup π[[u > q]] ∩ π[[v > α − q]] q∈Q

q∈Q

= π(sup [[u > q]] ∩ [[v > α − q]]) = π[[u + v > α]] = [[T (u + v) > α]] q∈Q

for every α ∈ R, so that T u + T v = T (u + v). In the same way, T (γu) = γT u whenever γ > 0, T u ≤ T v whenever u ≤ v, T u × T v = T (u × v) whenever u, v ≥ 0, so that, using the distributive laws, T is linear and multiplicative. To see that T is a sequentially order-continuous Riesz homomorphism, suppose that A ⊆ L0 (A) is a countable non-empty set with a supremum u∗ ∈ L0 (A); then T [A] is a non-empty subset of L0 (B) with an upper bound T u∗ , and

sup [[T u > α]] = sup π[[u > α]] = π(sup [[u > α]]) = π[[u∗ > α]] u∈A

u∈A

u∈A

(using 364M) = [[T u∗ > α]] for every α ∈ R. So using 364M again, T u∗ = supu∈A T u. Now this is true, in particular, for doubleton sets A, so that T is a Riesz homomorphism; and also for non-decreasing sequences, so that T is sequentially order-continuous. (b) The identification of T (χa) with χ(πa) is another almost trivial verification. It follows that T agrees with the map of 363F on S(A), if we think of S(A) as a subspace of L0 (A). Next, if u ∈ L∞ (A) ⊆ L0 (A), and γ = kuk∞ , then |u| ≤ γχ1A , so that |T u| ≤ γχ1B , and T u ∈ L∞ (B), with kT uk∞ ≤ γ. Thus T ¹L∞ (A) has norm at most 1. As it agrees with the map of 363F on S(A), which is k k∞ -dense in L∞ (A) (363C), and both are continuous, they must agree on the whole of L∞ (A). α) Suppose that π is order-continuous, and that A ⊆ L0 (A) is a non-empty set with a supremum (c)(i)(α 0 u ∈ L (A). Then for any α ∈ R, ∗

[[T u∗ > α]] = π[[u∗ > α]] = π(sup [[u > α]]) u∈A

(by 364M) = sup π[[u > α]] u∈A

(because π is order-continuous)

L0

364S

325

= sup [[T u > α]]. u∈A ∗

As α is arbitrary, T u = sup T [A], by 364M again. As A is arbitrary, T is order-continuous (351Ga). β ) Now suppose that T is order-continuous and that A ⊆ A is a non-empty set with supremum c (β in A. Then χc = supa∈A χa (364Kc) so χ(πc) = T (χc) = supa∈A T (χa) = supa∈A χ(πa). But now πc = [[χ(πc) > 0]] = supa∈A [[χ(πa) > 0]] = supa∈A πa. As A is arbitrary, π is order-continuous. α) If π is injective and u, v are distinct elements of L0 (A), then there must be some α such that (ii)(α [[u > α]] 6= [[v > α]], in which case [[T u > α]] 6= [[T v > α]] and T u 6= T v. β ) Now suppose that T is injective. It is easy to see that χ : A → L0 (A) is injective, so that (β T χ : A → L0 (B) is injective; but this is the same as χπ (by (b)), so π must also be injective. α) Suppose that π is surjective. Let Σ be a σ-algebra of sets such that there is a sequentially (iii)(α order-continuous Boolean surjection φ : Σ → A. Then πφ : Σ → B is surjective. So given w ∈ L0 (B), there is an f ∈ L0 (Σ) such that [[w > α]] = πφ{x : f (x) > α} for every α ∈ R (364D). But, also by 364D, there is a u ∈ L0 (A) such that [[u > α]] = φ{x : f (x) > α} for every α. And now of course T u = w. As w is arbitrary, T is surjective. β ) If T is surjective, and b ∈ B, there must be some u ∈ L0 (A) such that T u = χb. Now set (β a = [[u > 0]] and see that πa = [[χb > 0]] = b. As b is arbitrary, π is surjective. (d) The map E 7→ π[[u ∈ E]] is a sequentially order-continuous Boolean homomorphism, equal to [[T u ∈ E]] when E is of the form ]α, ∞[; so by 364G the two are equal for all Borel sets E. If h : R → R is a Borel measurable function, u ∈ L0 (A) and E ⊆ R is a Borel set, then ¯ u) ∈ E]] = [[T u ∈ h−1 [E]]] = π[[u ∈ h−1 [E]]] [[h(T ¯ ¯ = π[[h(u) ∈ E]] = [[T (h(u)) ∈ E]]. ¯ = hT ¯ . As E and u are arbitrary, T h (e) This is immediate from the definitions. 364S Products: Proposition Let hAi ii∈I be a family of Dedekind σ-complete Boolean algebras, with simple product A. If πi : A → Ai is the coordinate map for each Ti : L0 (A) → L0 (Ai ) the Q i, and 0 0 corresponding homomorphism, then u 7→ T u = hTi uii∈I : L (A) → i∈I L (Ai ) is a multiplicative Riesz Q space isomorphism, so L0 (A) may be identified with the f -algebra product i∈I L0 (Ai ) (352Wc). proof Because each πi is a surjective order-continuous Boolean homomorphism, 364R assures us that there are corresponding surjective multiplicative Riesz Q homomorphisms Ti . So all we need to check is that the multiplicative Riesz homomorphism T : L0 (A) → i∈I L0 (Ai ) is a bijection. If u, v ∈ L0 (A) are distinct, there must be some α ∈ R such that [[T u > α]] 6= [[T v > α]]. In this case there must be an i ∈ I such that πi [[T u > α]] 6= πi [[T v > α]], that is, [[Ti u > α]] 6= [[Ti v > α]]. So Ti u 6= Ti v and T u 6= T v. As u, v are arbitrary,QT is injective. If w = hwi ii∈I is any member of i∈I L0 (Ai ), then for α ∈ R set φ(α) = h[[wi > α]]ii∈I ∈ A. It is easy to check that φ satisfies the conditions of 364A, because, for instance, supβ>α πi φ(β) = supβ>α [[wi > β]] = [[wi > α]] = πi φ(α) for every i, so that supβ>α φ(β) = φ(α), for every α ∈ R; and the other two conditions are also satisfied because they are satisfied coordinate-by-coordinate. So there is a u ∈ L0 (A) such that φ(α) = [[u > α]] for every α, that is, πi [[u > α]] = [[wi > α]] for all α, i, that is, Ti u = wi for every i, that is, T u = w. As w is arbitrary, T is surjective and we are done.

326

Function spaces

*364T

*364T Regular open algebras I noted in 314P that for every topological space X there is a corresponding Dedekind complete Boolean algebra G of regular open sets. We have an identification of L0 (G) as a space of equivalence classes of functions, different in kind from the representations above, as follows. This is hard work (especially if we do it in full generality), but instructive. I start with a temporary definition. Definition Let (X, T) be a topological space, f : X → R a function. For x ∈ X write ω(f, x) = inf G∈T,x∈G supy,z∈G |f (y) − f (z)| (allowing ∞). *364U Theorem Let X be any topological space, and G its regular open algebra. Let U be the set of functions f : X → R such that {x : ω(f, x) < ²} is dense in X for every ² > 0. Then U is a Riesz subspace of RX , closed under multiplication, and we have a surjective multiplicative Riesz homomorphism T : U → L0 (G) defined by writing [[T f > α]] = supβ>α int {x : f (x) > β}, the supremum being taken in G, for every α ∈ R, f ∈ U . The kernel of T is the set W of functions f : X → R such that int{x : |f (x)| ≤ ²} is dense for every ² > 0, so L0 (G) can be identified, as f -algebra, with the quotient space U/W . α) The first thing to observe is that for any f ∈ RX , ² > 0 the set proof (a)(i)(α {x : ω(f, x) < ²} =

[ {G : G ⊆ X is open and non-empty and sup |f (y) − f (z)| < ²} y,z∈G

is open. β ) Next, it is easy to see that (β ω(f + g, x) ≤ ω(f, x) + ω(g, x), ω(γf, x) = |γ|ω(f, x), ω(|f |, x) ≤ ω(f, x), X

for all f , g ∈ R

and γ ∈ R.

(γγ ) Thirdly, it is useful to know that if f ∈ U and G ⊆ X is a non-empty open set, then there is a non-empty open set G0 ⊆ G on which f is bounded. P P Take any x0 ∈ G such that ω(f, x0 ) < 1; then there is a non-empty open set G0 containing x0 such that |f (y) − f (z)| < 1 for all y, z ∈ G0 , and we may suppose that G0 ⊆ G. But now |f (x)| ≤ 1 + |f (x0 )| for every x ∈ G0 . Q Q (ii) So if f , g ∈ U and γ ∈ R then 1 2

1 2

{x : ω(f + g, x) < ²} ⊇ {x : ω(f, x) < ²} ∩ {x : ω(g, x) < ²} is the intersection of two dense open sets and is therefore dense, while {x : ω(γf, x) < ²} ⊇ {x : ω(f, x)
0 and let G0 be a non-empty open subset of X. By the last remark in (i) above, there is a non-empty open set G1 ⊆ G0 such that |f | ∨ |g| is bounded on G1 ; say max(|f (x)|, |g(x)|) ≤ γ for every x ∈ G1 .

L0

*364U

Set δ = min(1,

² ) 2(1+γ)

327

> 0. Then there is an x ∈ G1 such that ω(f, x) < δ and ω(g, x) < δ. Let H, H 0

be open sets containing x such that |f (y) − f (z)| < δ for all y, z ∈ H and |g(y) − g(z)| < δ for all y, z ∈ H 0 . Consider G = G1 ∩ H ∩ H 0 . This is an open set containing x, and if y, z ∈ G then |f (y)g(y) − f (z)g(z)| ≤ |f (y) − f (z)||g(y) − g(z)| + |f (y) − f (z)||g(z)| + |f (z)||g(y) − g(z)| 2

≤ δ + δγ + γδ. Accordingly ω(f × g, x) ≤ δ(1 + 2γ) < ², while x ∈ G0 . As G0 is arbitrary, {x : ω(f × g, x) < ²} is dense; as ² is arbitrary, f × g ∈ U . Q Q Thus U is a subalgebra of RX . (b) Now, for f ∈ U , consider the map φf : R → G defined by setting φf (α) = supβ>α int {x : f (x) > β} for every α ∈ R. Then φf satisfies the conditions of 364A. P P (See 314P for the calculation of suprema and infima in G.) (i) If α ∈ R then φf (α) = sup int {x : f (x) > β} = sup int {x : f (x) > γ} β>α

γ>β>α

= sup sup int {x : f (x) > γ} = sup φf (α). β>α γ>β

β>α

(ii) If G0 ⊆ X is a non-empty open set, then there is a non-empty open set G1 ⊆ G0 such that f is bounded on G1 ; say |f (x)| < γ for every x ∈ G1 . If β > γ then G1 does not meet {x : f (x) > β}, so G1 ∩ int {x : f (x) > γ} = ∅; as β is arbitrary, G1 ∩ φf (γ) = ∅ and G0 6⊆ inf α∈R φf (α). On the other hand, G1 ⊆ {x : f (x) > −γ}, so G1 ⊆ int {x : f (x) > −γ} ⊆ φf (−γ) and G0 ∩ supα∈R φf (α) 6= ∅. As G0 is arbitrary, inf α∈R φf (α) = ∅ and supα∈R φf (α) = X. Q Q (c) Thus we have a map T : U → L0 = L0 (G) defined by setting [[T f > α]] = φf (α) for every α ∈ R, f ∈ U. It is worth noting that {x : f (x) > α + ω(f, x)} ⊆ [[T f > α]] ⊆ {x : f (x) + ω(f, x) ≥ α} for every f ∈ U , α ∈ R. P P (i) If f (x) > α + ω(f, x), set δ = 12 (f (x) − α − ω(f, x)) > 0. Then there is an open set G containing x such that |f (y) − f (z)| < ω(f, x) + δ for every y, z ∈ G, so that f (y) > α + δ for every y ∈ G, and x ∈ int{y : f (y) > α + δ} ⊆ [[T f > α]]. (ii) If f (x) + ω(f, x) < α, set δ = 12 (α − f (x) − ω(f, x)) > 0; then there is an open neighbourhood G of x such that |f (y) − f (z)| < ω(f, x) + δ for every y, z ∈ G, so that f (y) < α for every y ∈ G. Accordingly G / [[T f > α]]. Q Q does not meet {y : f (y) > β} nor {y : f (y) > β} for any β > α, G ∩ [[T f > α]] = ∅ and x ∈ (d) T is additive. P P Let f , g ∈ U and α < β ∈ R. Set δ = 15 (β − α) > 0, H = {x : ω(f, x) < δ, ω(g, x) < δ}; then H is the intersection of two dense open sets, so is itself dense and open. (i) If x ∈ H ∩ [[T (f + g) > β]], then (f + g)(x) + ω(f + g, x) ≥ β; but ω(f + g, x) ≤ 2δ (see (a-i-β) above), so f (x) + g(x) ≥ β − 2δ > α + 2δ and there is a q ∈ Q such that f (x) > q + δ ≥ q + ω(f, x),

g(x) > α − q + δ ≥ α − q + ω(g, x).

Accordingly x ∈ [[T f > q]] ∩ [[T g > α − q]] ⊆ [[T f + T g > α]].

328

Function spaces

*364U

Thus H ∩ [[T (f + g) > β]] ⊆ [[T f + T g > α]]. Because H is dense, [[T (f + g) > β]] ⊆ [[T f + T g > α]]. (ii) If x ∈ H, then

x∈

[

([[T f > q]] ∩ [[T g > β − q]])

q∈Q

=⇒ ∃ q ∈ Q, f (x) + ω(f, x) ≥ q, g(x) + ω(g, x) ≥ β − q =⇒ f (x) + g(x) + 2δ ≥ β =⇒ (f + g)(x) ≥ α + 3δ > α + ω(f + g, x) =⇒ x ∈ [[T (f + g) > α]]. Thus

S H ∩ q∈Q ([[T f > q]] ∩ [[T g > β − q]]) ⊆ [[T (f + g) > α]]. S Because H is dense and q∈Q ([[T f > q]] ∩ [[T g > β − q]]) is open, [

[[T f + T g > β]] = int

[[T f > q]] ∩ [[T g > β − q]]

q∈Q

⊆ int [[T (f + g) > α]] = [[T (f + g) > α]]. (iii) Now let β ↓ α; we have [[T (f + g) > α]] = sup [[T (f + g) > β]] ⊆ [[T f + T g > α]] β>α

= sup [[T f + T g > β]] ⊆ [[T (f + g) > α]], β>α

so [[T (f + g) > α]] = [[T f + T g > α]]. As α is arbitrary, T (f + g) = T f + T g; as f and g are arbitrary, T is additive. Q Q (e) It is now easy to see that T is linear. P P If γ > 0, f ∈ U , α ∈ R then β γ

[[T (γf ) > α]] = sup int {x : γf (x) > β} = sup int {x : f (x) > } β>α

β>α

α γ

= sup int {x : f (x) > β} = [[T f > ]] = [[γT f > α]]. β>α/γ

As α is arbitrary, T (γf ) = γT f ; because we already know that T is additive, this is enough to show that T is linear. Q Q (f ) In fact T is a Riesz homomorphism. P P If f ∈ U , α ≥ 0 then [[T (f + ) > α]] = sup int {x : f + (x) > β} = sup int {x : f (x) > β} β>α

β>α

= [[T f > α]] = [[(T f )+ > α]]. If α < 0 then [[T (f + ) > α]] = sup int {x : f + (x) > β} β>α

= X = [[(T f )+ > α]]. Q Q (g) Of course the constant function χX belongs to U , and is its multiplicative identity; and T (χX) is the multiplicative identity of L0 , because

L0

*364U

329

[[T (χX) > α]] = sup int {x : (χX)(x) > β} β>α

= X if α < 1, ∅ if α ≥ 1. By 353Pd, or otherwise, T is multiplicative. (h) The kernel of T is W . P P (i) For f ∈ U , T f = 0 =⇒ [[T |f | > 0]] = [[|T f | > 0]] = ∅ =⇒ {x : |f (x)| > ω(|f |, x)} = ∅ =⇒ int{x : |f (x)| ≤ ²} ⊇ {x : ω(|f |, x) < ²} is dense, for every ² > 0 =⇒ f ∈ W. (ii) If f ∈ W , then, first, 1 3

{x : ω(f, x) < ²} ⊇ int{x : |f (x)| ≤ ²} is dense for every ² > 0, so f ∈ U ; and next, for any β > 0, {x : |f (x)| > β} does not meet the dense open set int{x : |f (x)| ≤ β}, so [[|T f | > 0]] = [[T |f | > 0]] = supβ>0 int {x : |f (x)| > β} = ∅ and T f = 0. Q Q (i) Finally, T is surjective. P P Take any u ∈ L0 . Define f˜ : X → [−∞, ∞] by setting f˜(x) = sup{α : x ∈ [[u > α]]} for each x, counting inf ∅ as −∞. Then S {x : f˜(x) > α} = β>α [[u > β]] is open, for every α ∈ R. The set {x : f˜(x) = ∞} =

T α∈R

[[u > α]]

is nowhere dense, because inf α∈R [[u > α]] = ∅ in G; while

S {x : f˜(x) = −∞} = X \ α∈R [[u > α]]

is also nowhere dense, because supα∈R [[u > α]] = X in G. Accordingly E = int{x : f˜(x) ∈ R} is dense. Set f (x) = f˜(x) for x ∈ E, 0 for x ∈ X \ E. Let ² > 0. If G ⊆ X is a non-empty open set, there is an α ∈ R such that G 6⊆ [[u > α]], so G1 = G \ [[u > α]] 6= ∅, and f˜(x) ≤ α for every x ∈ G1 . Set α0 = supx∈G1 f˜(x) ≤ α < ∞. Because E meets G1 , α0 > −∞. Then G2 = G1 ∩ [[u > α0 − 21 ²]] is a non-empty open subset of G and α0 − 21 ² ≤ f˜(x) ≤ α0 for every x ∈ G2 . Accordingly |f (y) − f (z)| ≤ 12 ² for all y, z ∈ G2 , and ω(f, x) < ² for all x ∈ G2 . As G is arbitrary, {x : ω(f, x) < ²} is dense; as ² is arbitrary, f ∈ U . Take α < β in R, and set δ = 21 (β − α). Then H = E ∩ {x : ω(f, x) < δ} is a dense open set, and H ∩ [[T f > β]] ⊆ H ∩ {x : f (x) + ω(f, x) ≥ β} ⊆ E ∩ {x : f (x) > α} ⊆ {x : f˜(x) > α} ⊆ [[u > α]]. As H is dense, [[T f > β]] ⊆ [[u > α]]. In the other direction H ∩ [[u > β]] ⊆ H ∩ {x : f˜(x) ≥ β} = H ∩ {x : f (x) ≥ β} ⊆ {x : f (x) > α + ω(f, x)} ⊆ [[T f > α]], so [[u > β]] ⊆ [[T f > α]]. Just as in (d) above, this is enough to show that T f = u. As u is arbitrary, T is surjective. Q Q This completes the proof.

330

Function spaces

*364V

*364V Compact spaces Suppose now that X is a compact Hausdorff topological space. In this case the space U of 364U is just the space of functions f : X → R such that {x : f is continuous at x} is dense in X. P P It is easy to see that T {x : f is continuous at x} = {x : ω(f, x) = 0} = n∈N Hn where Hn = {x : ω(f, x) < 2−n T } for each n. Each Hn is an open set (see part (a-i-α) of the proof of 364U), so by Baire’s theorem (3A3G) n∈N Hn is dense iff every Hn is dense, that is, iff f ∈ U . Q Q Now W becomes {f : f ∈ U, {x : f (x) = 0} is dense}. P P (i) If f ∈ W , then T |f | = 0, so (by the formula in (c) of the proof of 364U) |f (x)| ≤ ω(|f |, x) for every x. But {x : ω(f, x) = 0} is dense, because f ∈ U , so {x : f (x) = 0} is also dense. (ii) If f ∈ U and {x : f (x) = 0} is dense, then ω(f, x) ≥ inf x∈G is open supy∈G |f (y) − f (x)| ≥ |f (x)| for every x ∈ X. So for any ² > 0, int{x : |f (x)| ≤ ²} ⊇ {x : ω(f, x) < ²} is dense, and f ∈ W . Q Q In the case of extremally disconnected spaces, we can go farther. *364W Theorem Let X be a compact Hausdorff extremally disconnected space, and G its regular open algebra. Write C ∞ = C ∞ (X) for the space of continuous functions g : X → [−∞, ∞] such that {x : g(x) = ±∞} is nowhere dense. Then we have a bijection S : C ∞ → L0 = L0 (G) defined by saying that [[Sg > α]] = {x : g(x) > α} ˙ × ˙ on C ∞ defined by for every α ∈ R. Addition and multiplication in L0 correspond to the operations +, ∞ ˙ g ×h ˙ are the unique elements of C agreeing with g + h, g × h on {x : g(x), h(x) are both saying that g +h, finite}. Scalar multiplication in L0 corresponds to the operation (γg)(x) = γg(x) for x ∈ X, g ∈ C ∞ , γ ∈ R on C ∞ (counting 0 · ∞ as 0), while the ordering of L0 corresponds to the relation g ≤ h ⇐⇒ g(x) ≤ h(x) for every x ∈ X. proof (a) For g ∈ C ∞ , set Hg = {x : g(x) ∈ R}, so that Hg is a dense open set, and define Rg : X → R by setting (Rg)(x) = g(x) if x ∈ Hg , 0 if x ∈ X \ Hg . Then Rg is continuous at every point of Hg , so belongs to the space U of 364U-364V. Set Sg = T (Rg), where T : U → L0 is the map of 364U. Then [[Sg > α]] = {x : g(x) > α} for every α ∈ R. P P (i) ω(g, x) = 0 for every x ∈ Hg , so, if β > α, Hg ∩ [[Sg > β]] ⊆ {x : x ∈ Hg , (Rg)(x) ≥ β} ⊆ {x : g(x) ≥ β} by the formula in part (c) of the proof of 364U. As [[Sg > β]] is open and Hg is dense, [[Sg > β]] ⊆ Hg ∩ [[Sg > β]] ⊆ {x : g(x) ≥ β} ⊆ {x : g(x) > α}. Now [[Sg > α]] = supβ>α [[Sg > β]] = int

S β>α

[[Sg > β]] ⊆ {x : g(x) > α}.

(ii) In the other direction, Hg ∩ {x : g(x) > α} ⊆ [[Sg > α]], by the other half of the formula in the proof of 364U. Again because {x : g(x) > α} is open and Hg is dense, {x : g(x) > α} ⊆ [[Sg > α]] = [[Sg > α]] because X is extremally disconnected (see 314S). Q Q (b) Thus we have a function S defined by the formula offered. Now if g, h ∈ C ∞ and g ≤ h, we surely have {x : g(x) > α} ⊆ {x : h(x) > α} for every α, so [[Sg > α]] ⊆ [[Sh > α]] for every α and Sg ≤ Sh. On the other hand, if g 6≤ h then Sg 6≤ Sh. P P Take x0 such that g(x0 ) > h(x0 ), and α ∈ R such that g(x0 ) > α > h(x0 ); set H = {x : g(x) > α > h(x)}; this is a non-empty open set and H ⊆ [[Sg > α]]. On the other hand, H ∩ {x : h(x) > α} = ∅ so H ∩ [[Sh > α]] = ∅. Thus [[Sg > α]] 6⊆ [[Sh > α]] and Sg 6≤ Sh. Q Q In particular, S is injective. (c) S is surjective. P P If u ∈ L0 , set

L0

364Xe

331

g(x) = sup{α : x ∈ [[u > α]]} ∈ [−∞, ∞] for every x ∈ X, taking sup ∅ = −∞. Then, for any α ∈ R, {x : g(x) > α} = other hand, S {x : g(x) < α} = β β]]}

S β>α

[[u > α]] is open. On the

is also open, because all the sets [[u > β]] are open-and-closed. So g : X → [−∞, ∞] is continuous. Also S {x : g(x) > −∞} = α∈R [[u > α]], S {x : g(x) < ∞} = α∈R X \ [[u > α]] are dense, so g ∈ C ∞ . Now, for any α ∈ R, [[Sg > α]] = {x : g(x) > α} = = int

[

[

[[u > β]]

β>α

[[u > β]] = sup [[u > β]] = [[u > α]].

β>α

β>α

So Sg = u. As u is arbitrary, S is surjective. Q Q (d) Accordingly S is a bijection. I have already checked (in part (b)) that it is an isomorphism of the order structures. For the algebraic operations, observe that if g, h ∈ C ∞ then there are f1 , f2 ∈ C ∞ such that Sg + Sh = Sf1 and Sg × Sh = Sf2 , that is, T (Rg + Rh) = T Rg + T Rh = T Rf1 ,

T (Rg × Rh) = T Rg × T Rh = T Rf2 .

But this means that T (Rg + Rh − Rf1 ) = T ((Rg × Rh) − Rf2 ) = 0, so that Rg + Rh − Rf1 , (Rg × Rh) − Rf2 belong to W and are zero on dense sets (364V). Since we know also that the set G = {x : g(x), h(x) are both finite} is a dense open set, while g, h, f1 and f2 are all continuous, we must have f1 (x) = g(x) + h(x), f2 (x) = g(x)h(x) for every x ∈ G. And of course this uniquely specifies f1 and f2 as members of C ∞ . ˙ × ˙ as described, rendering S additive and multiplicative. As for scalar Thus we do have operations +, multiplication, it is easy to check that R(γg) = γRg (at least, unless γ = 0, which is trivial), so that S(γg) = γSg for every g ∈ C ∞ . 364X Basic exercises (a) Let A be a Dedekind σ-complete Boolean algebra. For u, v ∈ L0 = L0 (A) set [[u < v]] = [[v > u]] = [[v − u > 0]], [[u ≤ v]] = [[v ≥ u]] = 1 \ [[v < u]], [[u = v]] = [[u ≤ v]] ∩ [[v ≤ u]]. (i) Show that ([[u < v]], [[u = v]], [[u > v]]) is always a partition of unity in A. (ii) Show that for any u, u0 , v, v 0 ∈ L0 , [[u ≤ u0 ]] ∩ [[v ≤ v 0 ]] ⊆ [[u + v ≤ u0 + v 0 ]] and [[u = u0 ]] ∩ [[v = v 0 ]] ⊆ [[u × v = u0 × v 0 ]]. (b) Let A be a Dedekind σ-complete Boolean algebra. (i) Show that if u, v ∈ L0 = L0 (A) and α, β ∈ R then [[u + v ≥ α + β]] ⊆ [[u ≥ α]] ∪ [[v ≥ β]]. (ii) Show that if u, v ∈ (L0 )+ and α, β ≥ 0 then [[u × v ≥ αβ]] ⊆ [[u ≥ α]] ∪ [[v ≥ β]]. (c) Let A be a Dedekind σ-complete Boolean algebra and u ∈ L0 (A). Show that {[[u ∈ E]] : E ⊆ R is Borel} is the σ-subalgebra of A generated by {[[u > α]] : α ∈ R}. >(d) Let (A, µ ¯) be a probability algebra. Show that for any u ∈ L0 (A) there is a unique Radon probability measure ν on R (the distribution of u) such that νE = µ ¯[[u ∈ E]] for every Borel set E ⊆ R. (Hint: 271B.) (e) Let (A, µ ¯) be a probability algebra, and hui ii∈I any family in L0 (A); for each i ∈ I let Bi be the closed subalgebra of AQgenerated by {[[ui > α]] : α ∈ R}. Show that the following are equiveridical: (i) µ ¯(inf i∈J [[ui > αi ]]) = i∈J µ ¯[[ui > αi ]] whenever J ⊆ I is finite and αi ∈ R for each i ∈ J (ii) hBi ii∈I is independent in the sense of 325L. (In this case we may call hui ii∈I (stochastically) independent.)

332

Function spaces

364Xf

(f ) Let (A, µ ¯) be a probability algebra and u, v two stochastically independent members of L0 (A). Show that the distribution of their sum is the convolution of their distributions. (Hint: 272S.) > (g) Let A be a Dedekind σ-complete Boolean algebra and g, h : R → R Borel measurable functions. (i) ¯ = gh, ¯ where g¯, h ¯ : L0 → L0 are defined as in 364I. (ii) Show that g + h(u) = g¯(u) + h(u), ¯ Show that g¯h ¯ g × h(u) = g¯(u) × h(u) for every u ∈ L0 = L0 (A). (iii) Show that if hhn in∈N is a sequence of Borel ¯ n (u) = h(u) ¯ measurable functions on R and supn∈N hn = h, then supn∈N h for every u ∈ L0 . (iv) Show that ¯ ¯ if h is non-decreasing and continuous on the left, then h(sup A) = sup h[A] whenever A ⊆ L0 is a non-empty set with a supremum in L0 . (h) Let A be a Dedekind σ-complete Boolean algebra. (i) Show that S(A) can be identified (α) with the set of those u ∈ L0 = L0 (A) such that {[[u > α]] : α ∈ R} is finite (β) with the set of those u ∈ L0 such that [[u ∈ I]] = 1 for some finite I ⊆ R. (ii) Show that L∞ (A) can be identified with the set of those u ∈ L0 such that [[u ∈ [−α, α]]] = 1 for some α ≥ 0, and that kuk∞ is the smallest such α. (i) Show that if A is a Dedekind σ-complete Boolean algebra, and u ∈ L0 (A), then for any α ∈ R [[u > α]] = inf β>α sup{a : a ∈ A, u × χa ≥ βχa} (compare 363Xh). (j) Let A be a Dedekind σ-complete Boolean algebra and u ≥ 0 in L0 = L0 (A). Show that u = supq∈Q qχ[[u > q]] in L0 . (k) (i) Let A be a Dedekind σ-complete Boolean algebra and A ⊆ L0 (A) a non-empty countable set with supremum w. Show that [[w ∈ E]] ⊆ supu∈A [[u ∈ E]] for every Borel set E ⊆ R. (ii) Let (A, µ ¯) be a localizable measure algebra and A ⊆ L0 (A) a non-empty set with supremum w. Show that [[w ∈ E]] ⊆ supu∈A [[u ∈ E]] for every Borel set E ⊆ R. (l) Let A be a Dedekind σ-complete Boolean algebra and A ⊆ L0 = L0 (A) a non-empty set which is bounded below in L0 . Suppose that φ0 (α) = inf u∈A [[u > α]] is defined in A for every α ∈ R. Show that v = inf A is defined in L0 , and that [[v > α]] = supβ>α φ0 (β) for every α ∈ R. (m) Let (X, Σ, µ) be a measure space and f : X → [0, ∞[ a function; set A = {g • : g ∈ L0 (µ), g ≤ f a.e.}. (i) Show that if (X, Σ, µ) either is localizable or has the measurable envelope property (213Xl), then sup A is defined in L0 (µ). (ii) Show that if (X, Σ, µ) is complete and locally determined and w = sup A is defined in L0 (µ), then w ∈ A. (n) Let A be a Dedekind σ-complete Boolean algebra. Show that if u, v ∈ L0 = L0 (A) then the following are equiveridical: (α) [[|v| > 0]] ⊆ [[|u| > 0]] (β) v belongs to the band of L0 generated by u (γ) there is a w ∈ L0 such that u × w = v. (o) Let A be a Dedekind σ-complete Boolean algebra and a ∈ A; let Aa be the principal ideal of A generated by a. Show that L0 (Aa ) can be identified, as f -algebra, with the band of L0 (A) generated by χa. (p) Let A and B be Dedekind σ-complete Boolean algebras, and π : A → B a sequentially ordercontinuous Boolean homomorphism. Let T : L0 (A) → L0 (B) be the corresponding Riesz homomorphism (364R). Show that (i) the kernel of T is the sequentially order-closed solid linear subspace of L0 (A) generated by {χa : a ∈ A, πa = 0} (ii) the set of values of T is the sequentially order-closed Riesz subspace of L0 (B) generated by {χ(πa) : a ∈ A}. (q) Let A and B be Dedekind σ-complete Boolean algebras, and π : A → B a sequentially ordercontinuous Boolean homomorphism, with T : L0 (A) → L0 (B) the associated operator. Suppose that h is a ¯ u) = T h(u) ¯ Borel measurable real-valued function defined on a Borel subset of R. Show that h(T whenever ¯ u ∈ L0 (A) and h(u) is defined in the sense of 364I. >(r) Let X and Y be sets, Σ, T σ-algebras of subsets of X, Y respectively, and I, J σ-ideals of Σ, T. Set A = Σ/I, B = T/J . Suppose that φ : X → Y is a function such that φ−1 [F ] ∈ Σ for every F ∈ T,

364Yk

L0

333

φ−1 [F ] ∈ I for every F ∈ J . Show that there is a sequentially order-continuous Boolean homomorphism π : B → A defined by saying that πF • = φ−1 [F ]• for every F ∈ T. Let T : L0 (B) → L0 (A) be the Riesz homomorphism corresponding to π. Show that if we identify L0 (B) with L0T /WJ and L0 (A) with L0Σ /WI in the manner of 364D, then T (g • ) = (gφ)• for every g ∈ L0T . (s) Use the ideas of part (d) of the proof of 364U to show that the operator T there is multiplicative, without appealing to 353P. 364Y Further exercises (a) Show directly, without using the Stone representation, that if A is any Dedekind σ-complete Boolean algebra then the formulae of 364E define a group operation + on L0 (A), and generally an f -algebra structure. (b) Let A be a Dedekind σ-complete Boolean algebra. Show that A is ccc iff L0 (A) has the countable sup property. (c) Let A be a Dedekind σ-complete Boolean algebra and u1 , . . . , un members of L0 (A). Write Bn for the algebra of Borel sets in Rn . (i) Show that there is a unique sequentially order-continuous Boolean homomorphism E 7→ [[(u1 , . . . , un ) ∈ E]] : Bn → A such that [[(u1 , . . . , un ) ∈ E]] = inf i≤n [[ui > αi ]] when Q E = i≤n ]αi , ∞[. (ii) Show that for every sequentially order-continuous Boolean homomorphism φ : Bn → A there are u1 , . . . , un ∈ L0 (A) such that φE = [[(u1 , . . . , un ) ∈ E]] for every E ∈ Bn . (d) Let A be a Dedekind σ-complete Boolean algebra, n ≥ 1 and h : Rn → R a Borel measurable ¯ : L0 (A)n → L0 (A) defined by saying that function. Show that we have a corresponding function h −1 ¯ [[h(u1 , . . . , un ) ∈ E]] = [[(u1 , . . . , un ) ∈ h [E]]] for every Borel set E ⊆ R. (e) Suppose that h1 (x, y) = x + y, h2 (x, y) = xy, h3 (x, y) = max(x, y) for all x, y ∈ R. Show that, in the ¯ 1 (u, v) = u + v, h ¯ 2 (u, v) = u × v, h ¯ 3 (u, v) = u ∨ v for all u, v ∈ L0 . language of 364Yd, h (f ) Let A and B be Dedekind σ-complete Boolean algebras, and T : L0 (A) → L0 (B) a Riesz homomorphism such that T e = e0 , where e, e0 are the multiplicative identities of L0 (A), L0 (B) respectively. Show that there is a unique sequentially order-continuous Boolean homomorphism π : A → B such that T = Tπ in the sense of 364R. (Hint: use 353Pd. Compare 375A below.) (g) Suppose, in 364U, that X = Q. (i) Show that there is an f ∈ W such that f (q) > 0 for every q ∈ Q. (ii) Show that there is a u ∈ L0 such that no f ∈ U representing u can be continuous at any point of Q. (h) Let X and Y be topological spaces and φ : X → Y a continuous function such that φ−1 [M ] is nowhere dense in X for every nowhere dense subset M of Y . (Cf. 313R.) (i) Show that we have an order-continuous Boolean homomorphism π from the regular open algebra GY of Y to the regular open algebra GX of X defined by the formula πG = int φ−1 [G] for every G ∈ GY . (ii) Show that if UX , UY are the function spaces of 364U then gφ ∈ UX for every g ∈ UY . (iii) Show that if TX : UX → L0 (GX ), TY : UY → L0 (GY ) are the canonical surjections, and T : L0 (GY ) → L0 (GX ) is the homomorphism corresponding to π, then T (TY g) = TX (gφ) for every g ∈ UY . (iv) Rewrite these ideas for the special case in which X is a dense subset of Y and φ is the identity map, showing that in this case π and T are isomorphisms. (i) Let X be a Baire space, G its algebra of regular open sets, M its ideal of meager sets, and Bb the Baire b property σ-algebra {G4A : G ⊆ X is open, A ∈ M}, so that G can be identified with B/M (314Yd). (i) Repeat the arguments of 364V in this context. (ii) Show that the space U of 364U-364V is a subspace of b and that W = U ∩ W where W = {f : f ∈ RX , {x : f (x) 6= 0} ∈ M}, so that the representations of L0 (B), 0 L (G) as U/W , L0 /W are consistent. (j) Work through the arguments of 364U and 364Yi for the case of compact Hausdorff X, seeking simplifications based on 364V. (k) Let X be an extremally disconnected compact Hausdorff space with regular open algebra G. Let U0 be the space of real-valued functions f : X → R such that int{x : f is continuous at x} is dense. Show that U0 is a Riesz subspace of the space U of 364U, and that every member of L0 (G) is represented by a member of U0 .

334

Function spaces

364Yl

(l) Let X be a Baire space. Let Q be the set of all continuous real-valued functions defined on subsets of X, and Q∗ the set of all members of Q which are maximal in the sense that there is no member of Q properly extending them. (i) Show that the domain of any member of Q∗ is a dense Gδ set. (ii) Show that ˙ f ×g, ˙ γ.f we can define addition and multiplication and scalar multiplication on Q∗ by saying that f +g, are to be the unique members of Q∗ extending the partially-defined functions f + g, f × g, γf , and that these definitions render Q∗ an f -algebra if we say that f ≤ g iff f (x) ≤ g(x) for every x ∈ dom f ∩ dom g. (iii) Show that every member of Q∗ has an extension to a member of U , as defined in 364U, and that these extensions define an isomorphism between Q∗ and L0 (G), where G is the regular open algebra of X. (iv) Show that if X is compact, Hausdorff and extremally disconnected, then every member of Q∗ has a unique extension to a member of C ∞ (X), as defined in 364W. (m) Let X be an extremally disconnected Hausdorff space, and Z any compact Hausdorff space. Show that if D ⊆ X is dense and f : D → Z is continuous, there is a continuous g : X → Z extending f . (n) Let A be a Dedekind σ-complete Boolean algebra. Write L0C = {u + iv : u, v ∈ L0 } for the complexification of L0 = L0 (A) as defined in 354Yk. (i) Writing B(C) for the Borel σ-algebra of C, show that there is a one-to-one correspondence between sequentially order-continuous Boolean homomorphisms φ : B(C) → A and members w = u + iv of L0C defined by saying that [[u > α]] ∩ [[v > β]] = φ{z : Re z > α, Im z > β}. (ii) Show that if Σ is a σ-algebra of subsets of a set X and π : Σ → A is a surjective sequentially ordercontinuous Boolean homomorphism with kernel I, then we can identify L0C with L0C /W, where L0C is the set of Σ-measurable functions from X to C, and W is the set of those f ∈ L0C such that {x : f (x) 6= 0} ∈ I. 364 Notes and comments This has been a long section, and so far all we have is a supposedly thorough grasp of the construction of L0 spaces; discussion of their properties still lies ahead. The difficulties seem to stem from a variety of causes. First, L0 spaces have a rich structure, being linear ordered spaces with multiplications; consequently all the main theorems have to check rather a lot of different aspects. Second, unlike L∞ spaces, they are not accessible by means of the theory of normed spaces, so I must expect to do more of the work here rather than in an appendix. But this is in fact a crucial difference, because it affects the proof of the central theorem 364E. The point is that a given algebra A will be expressible in the form Σ/I for a variety of algebras Σ of sets. Consequently any definition of L0 (A) as a quotient L0 (Σ)/W must include a check that the structure produced is independent of the particular pair Σ, I chosen. The same question arises with S(A) and L∞ (A). But in the case of S, I was able to use a general theory of additive functions on A (see the proof of 361L), while in the case of L∞ I could quote the result for S and a little theory of normed spaces (see the proof of 363H). The theorems of §368 will show, among other things, that a similar approach (describing L0 as a special kind of extension of S or L∞ ) can be made to work in the present situation. I have chosen, however, an alternative route using a novel technique. The price is the time required to develop skill in the technique, and to relate it to the earlier approach (364D, 364E, 364K). The reward is a construction which is based directly on the algebra A, independent of any representation (364A), and methods of dealing with it which are complementary to those of the previous three sections. In particular, they can be used in the absence of the full axiom of choice (364Ya). I have deliberately chosen the notation [[u > α]] from the theory of Forcing. I do not propose to try to explain myself here, but I remark that much of the labour of this section is a necessary basis for understanding real analysis in Boolean-valued models of set theory. The idea is that just as a function f : X → R can be described in terms of the sets {x : f (x) > α}, so can an element u of L0 (A) be described in terms of the elements [[u > α]] of A where in some sense u is greater than α. This description is well adapted to discussion of the order struction of L0 (A) (see 364M-364O), but rather ill-adapted to discussion of its linear and multiplicative structures, which leads to a large part of the length of the work above. Once we have succeeded in describing the algebraic operations on L0 in terms of the values of [[u > α]], however, as in 364E, the fundamental result on the action of Boolean homomorphisms (364R) is elegant and reasonably straightforward. The concept ‘[[u > α]]’ can be dramatically generalized to the concept ‘[[(u1 , . . . , un ) ∈ E]]’, where E is a Borel subset of Rn and u1 , . . . , un ∈ L0 (A) (364H, 364Yc). This is supposed to recall the notation Pr(X ∈ E), already used in Chapter 27. If, as sometimes seems reasonable, we wish to regard a random variable as a member of L0 (µ) rather than of L0 (µ), then ‘[[u ∈ E]]’ is the appropriate translation of ‘X −1 [E]’.

L1

365B

335

The reasons why we can reach all Borel sets E here, but then have to stop, seem to lie fairly deep. We see that we have here another potential definition of L0 (A), as the set of sequentially order-continuous Boolean homomorphisms from the Borel σ-algebra of R to A. This is suitably independent of realizations of A, but makes the f -algebra structure of L0 difficult to elucidate, unless we move to a further level of abstraction in the definitions, as in 364Ye. I take the space to describe the L0 spaces of general regular open algebras in detail (364U) partly to offer a demonstration of an appropriate technique, and partly to show that we are not limited to σ-algebras of sets and their quotients. This really is a new representation; for instance, it does not meld in any straightforward way with the constructions of 364G-364I. Of course the most important examples are compact Hausdorff spaces, for which alternative methods are available (364V-364W, 364Yk, 364Yi, 364Yl); from the point of view of applications, indeed, it is worth sorting out compact Hausdorff spaces in general (364Yj). The version in 364W is derived from Vulikh 67. But I have starred everything from 364T on, because I shall not rely on this work later for anything essential.

365 L1 Continuing my programme of developing the ideas of Chapter 24 at a deeper level of abstraction, I arrive at last at L1 . As usual, the first step is to establish a definition which can be matched both with the constructions of the previous sections and with the definition of L1 (µ) in §242 (365A-365C, 365F). Next, I give what I regard as the most characteristic internal properties of L1 spaces, including versions of the Radon-Nikod´ ym theorem (365E), before turning to abstract versions of theorems in §235 (365H, 365T) and the duality between L1 and L∞ (365I-365K). As in §§361 and 363, the construction is associated with universal mapping theorems (365L-365N) which define the Banach lattice structure of L1 . As in §§361, 363 and 364, homomorphisms between measure algebras correspond to operators between their L1 spaces; but now the duality theory gives us two types of operators (365O-365Q), of which one class can be thought of as abstract conditional expectations (365R). For localizable measure algebras, the underlying algebra can be recovered from its L1 space (365S), but the measure cannot. 365A Definition Let (A, µ ¯) be a measure algebra. For u ∈ L0 (A), write kuk1 =

R∞ 0

µ ¯[[|u| > α]] dα,

the integral being with respect to Lebesgue measure on R, and allowing ∞ as a value of the integral. (Because the integrand is monotonic, it is certainly measurable.) Set L1µ¯ = L1 (A, µ ¯) = {u : u ∈ L0 (A), kuk1 < ∞}. It is convenient to note at once that if u ∈ L1 (A, µ ¯), then µ[[|u| > α]] must be finite for almost every α > 0, and therefore for every α > 0, since it is a non-increasing function of α; so that [[u > α]] also belongs to the Boolean ring Af = {a : µ ¯a < ∞} for every α > 0. 365B Theorem Let (X, Σ, µ) be a measure space with measure algebra (A, µ ¯). Then the canonical isomorphism between L0 (µ) and L0 (A), defined in 364D and 364J, matches L1 (µ) ⊆ L0 (µ), defined in §242, with L1µ¯ ⊆ L0 (A), and the standard norm of L1 (µ) with k k1 : L1µ¯ → [0, ∞[, as defined in 365A. proof Take any f ∈ L0 = L0 (Σ) (364C); write f • for its equivalence class in L0 (µ), and u for the corresponding element of L0 (A), so that [[|u| > α]] = {x : |f (x)| > α}• in A, for every α ∈ R, and kuk1 =

R∞ 0

µ{x : |f (x)| > α} dα.

Pn (a) If f is a non-negative simple function, it is expressible as i=0 αi χEi where E0 , . . . , En are disjoint measurable sets of finite measure and αi ≥ 0 for each i; re-enumerating if necessary, we may suppose that α0 ≥ α1 ≥ . . . ≥ αn . In this case, setting αn+1 = 0,

336

Function spaces

Z kuk1 = =

µ{x : f (x) > α} dα = n X n X

365B

j n X X ( µEi )(αj − αj+1 ) j=0 i=0 n X

µEi (αj − αj+1 ) =

i=0 j=i

Z

αi µEi =

f dµ.

i=0

(b) If f is integrable, then there is a non-decreasing sequence hfn in∈N of non-negative simple functions such that |f (x)| = supn∈N fn (x) a.e.; now µ{x : |f (x)| > α} = supn∈N µ{x : fn (x) > α} for every α ∈ R, so Z

Z Z ∞ |f | = sup fn = sup µ{x : fn (x) > α}dα n∈N n∈N 0 Z ∞ = µ{x : |f (x)| > α}dα = kuk1 . 0

1

Thus in this case u ∈ L (A, µ ¯) and kuk1 = kf • k1 . It follows that every member of L1 (µ) corresponds to a member of L1 (A, µ ¯), and that the norm of L1 (µ) 1 corresponds to k k1 on L (A, µ ¯). (c) On the other hand, if u ∈ L1 (A, µ ¯), then µ{x : |f (x)| > α} = µ ¯[[|u| > α]] is finite for every α > 0 (365A). Also, if g is any simple function with 0 ≤ g ≤ |f |,

R

g dµ =

R∞ 0

µ{x : g(x) > α}dα ≤

R∞ 0

µ{x : f (x) > α}dα = kuk1 < ∞.

So f is integrable (122J). This shows that every member of L1 (A, µ ¯) corresponds to a member of L1 (µ). 365C

Accordingly we can apply everything we know about L1 (µ) spaces to L1µ¯ spaces. For instance:

Theorem For any measure algebra (A, µ ¯), L1µ¯ = L1 (A, µ ¯) is a solid linear subspace of L0 (A), and k k1 is 1 1 a norm on Lµ¯ under which Lµ¯ is an L-space. Consequently L1µ¯ is a perfect Riesz space with an ordercontinuous norm which has the Levi property, and if hun in∈N is a non-decreasing norm-bounded sequence in L1µ¯ then it converges for k k1 to supn∈N un . proof (A, µ ¯) is isomorphic to the measure algebra of some measure space (X, Σ, µ) (321J). L1 (µ) is a solid linear subspace of L0 (µ) (242Cb), so L1µ¯ is a solid linear subspace of L0 (A). L1 (µ) is an L-space (354M), so L1µ¯ also is. The rest of the properties claimed are general features of L-spaces (354N, 354E, 356P). 365D Integration Let (A, µ ¯) be any measure algebra. (a) If u ∈ L1 = L1 (A, µ ¯), then u+ , u− belong to L1 , and we may set

R

R

u = ku+ k1 − ku− k1 =

R∞ 0

µ ¯[[u > α]] dα −

R∞ 0

µ ¯[[−u > α]] dα.

Now : L1 → R is an order-continuous positive linear functional (356P), and under the translation of 365B matches the integral on L1 (µ) as defined in 242Ab. (b) Of course kuk1 =

R

|u| ≥ |

R

u| for every u ∈ L1 .

R R (c) If u ∈ L1 , a ∈ A we may set a u = u × χa. (Compare 242Ac.) If γ > 0 and 0 6= a ⊆ [[u > γ]] then there is a δ > γ such that a0 = a ∩ [[u > δ]] 6= 0, so that

R

u= a

R∞ 0

µ ¯(a ∩ [[u > α]])dα ≥

Rγ 0

In particular, setting a = [[u > γ]], µ ¯[[u > γ]] must be finite.

µ ¯a dα +

Rδ γ

µ ¯a0 > γ µ ¯a.

L1

365E

337

R f (d) If u ∈ L1 then Af = {a : µ ¯a < ∞}, as usual. P P If u ≥ 0 R u ≥ 0 iff a u ≥ 0 for every a ∈ A , writing − then u × χa ≥ 0, a u ≥ 0 for every a ∈ A. If u 6≥ 0, then [[u > 0]] 6= 0 and there is an α > 0 such that a = [[u− > α]] 6= 0. But now µ ¯a is finite ((c) above) and

R

R

R

u × χa = − u− × χa = − µ ¯(a ∩ [[u− ≥ β]])dβ ≤ −α¯ µa < 0, R R R 1 f Q If u, v ∈ L and a u = a v for every Ra ∈ A then u = v (cf. 242Ce). If u ≥ 0 in L1 then Rso a u < 0.R Q P Of course u × χa ≤ u so a u ≤ u for every a ∈ A. On setting u = sup{ a u : a ∈ Af }. P R the other hand, R an = [[u > 2−n ]], hu × χan in∈N is a non-decreasing sequence with supremum u, so u = limn→∞ an u, while µ ¯an is finite for every n. Q Q R R (e) If u ∈ L1 , u ≥ 0 and u = 0 then u = 0 (122Rc). If u ∈ L1 , u ≥ 0 and a u = 0 then u × χa = 0, that is, a ∩ [[u > 0]] = 0. R (f )R If C ⊆ L1 is non-empty and upwards-directed and supv∈C v is finite, then sup C is defined in L1 R and sup C = supv∈C v (356P). (g) RIt will occasionally be convenient to adapt the conventions of §133 to the new context; so that I may R write u = ∞ if u ∈ L0 (A), u− ∈ L1 , u+ ∈ / L1 , while u = −∞ if u+ ∈ L1 and u− ∈ / L1 . (h) On this convention, we can restate (f) as follows: if C ⊆ (L0 )+ is non-empty R R R and upwards-directed 0 and has a supremum u in L , then u = sup v in [0, ∞]. P P For if sup v is infinite, then surely v∈C v∈C R u = ∞; while otherwise we can apply (f). Q Q 365E The Radon-Nikod´ ym theorem again (a) Let (A, µ ¯) be a semi-finite measure algebra and ν : A → R an additive functional. Then the followingRare equiveridical: (i) there is a u ∈ L1 = L1 (A, µ ¯) such that νa = a u for every a ∈ A; (ii) ν is additive and continuous for the measure-algebra topology on A; (iii) ν is completely additive. (b) Let (A, µ ¯) be any measure algebra, and ν : Af → R a function. Then the following are equiveridical: (i) ν is additive and bounded and inf a∈A |νa| = 0 whenever A ⊆ Af is downwards-directed and has infimum 0; R (ii) there is a u ∈ L1 such that νa = a u for every a ∈ Af . proof (a) The equivalence of (ii) and (iii) is 327Bd. The equivalence of (i) and (iii) is just a translation of 327D into the new context. α) Set M = supa∈Af |νa| < ∞. (b)(i)⇒(ii)(α Let D ⊆ Af be a maximal disjoint set. For each d ∈ D, write Ad for the principal ideal of A generated by d, and µ ¯d for the restriction of µ ¯ to Ad , so that (Ad , µ ¯d ) is a totally finite measure algebra. Set νd = ν¹ Ad ; R then νd : Ad → R is completely additive. By (a), there is a ud ∈ L1 (Ad , µ ¯d ) such that a ud = νd a = νa for every a ⊆ d. 0 0 + Now u+ ˜+ d ∈ L (Ad ) corresponds to a member u d of L (A) defined by saying + [[˜ u+ d > α]] = [[ud > α]] = [[ud > α]] if α ≥ 0,

= 1 if α ≤ 0. Set d+ = [[ud > 0]]. If a ∈ A, then

R

u ˜+ = a d

R∞

0 k˜ u+ d k1

µ ¯(a ∩ [[˜ u+ d > α]])dα =

R∞ 0

µ ¯(a ∩ [[u+ d > α]])dα =

R a∩d

u+ d;

1 = νd+ is finite, so that u ˜+ ¯. d ∈ Lµ P β ) For any finite I ⊆ D, set vI = d∈I u (β ˜+ d . Then

taking a = 1, we see that

R

vI = ν(supd∈I d+ ) ≤ M ;

consequently the upwards-directed set A = {vI : I ⊆ D is finite} is bounded above in L1µ¯ , and we can set R R R R P P + v = sup A in L1µ¯ . If a ∈ A, then a vI = d∈I a∩d u+ d∈D a∩d ud . d for each finite I ⊆ D, so a v =

338

Function spaces

365E

Applying the same arguments to −ν, there is a w ∈ L1µ¯ such that

R

w= a

for every a ∈ A. Try u = v − w; then

R

u= a

R

P

d∈D

u+ − a∩d d

R

R

P

d∈D

u− = a∩d d

u− d

a∩d

R

P d∈D

u = a∩d d

P d∈D

ν(a ∩ d)

for every a ∈ A. (γγ ) Now take any a ∈ Af . For J ⊆ D set aJ = supd∈J a ∩ d. Let ² > 0. Then there is a finite I ⊆ D such that |

R

u − νaJ | = | a

P

ν(a ∩ d) − d∈D

P

d∈J

ν(a ∩ d)| ≤ ²

whenever I ⊆ J ⊆ D and J is finite. But now consider A = {a \ aJ : I ⊆ J ⊆ D, J is finite}. Then inf A = 0, so there is a finite J such that I ⊆ J ⊆ D and |νa − νaJ | = |ν(a \ aJ )| ≤ ². Consequently |νa − As ² is arbitrary, νa =

R a

R a

u| ≤ |νa − νaJ | + |

R a

u − νaJ | ≤ 2².

u. As a is arbitrary, (ii) is proved.

R (ii)⇒(i) From where Rwe now are, this is nearly trivial. Thinking of νa as u × χa, ν is surely additive and bounded. Also |νa| ≤ |u| × χa. If A ⊆ Af is non-empty, downwards-directed and has infimum 0, the same is true of {|u| × χa : a ∈ A}, because a 7→ |u| × χa is order-continuous, so inf a∈A |νa| ≤ inf a∈A 365F

R

|u| × χa = inf a∈A k|u| × χak1 = 0.

It will be useful later to have spelt out the following elementary facts.

Lemma Let (A, µ ¯) be a measure algebra. Write S f for the intersection S(A) ∩ L1 (A, µ ¯). Then S f is a norm1 1 dense and order-dense Riesz subspace of L = L (A, µ ¯), and can be identified with S(Af ). The function f f 1 χ : A → S ⊆ L is an injective order-continuous additive lattice homomorphism. If u ≥ 0 in L1 , there is a non-decreasing sequence hun in∈N in (S f )+ such that u = supn∈N un = limn→∞ un . f 0 proof (a) As in 364L, Pn If u ∈ S, Pn we can think of S(A ) as a Riesz subspace of S = S(A), embedded in L (A). it is expressible as i=0 αi χai where a0 , . . . , an ∈ A are disjoint and no αi is zero. Now |u| = i=0 |αi |χai , so u ∈ L1 iff µ ¯ai < ∞ for every i, that is, iff u ∈ S(Af ); thus S f = S(Af ). Now suppose that u ≥ 0 in L1 . By 364Kd, there is a non-decreasing sequence hun in∈N in S(A)+ such that u0 ≥ 0 and u = supn∈N un in L0 . Because L1 is a solid linear subspace of L0 , every un belongs to L1 and therefore to S f . By 365C, hun in∈N is norm-convergent to u. This shows also that S f is order-dense in L1 . The map χ : Af → S f is an injective order-continuous additive lattice homomorphism; because S f is regularly embedded in L1 , χ has the same properties when regarded as a map into L1 . For general u ∈ L1 , there are sequences in S f converging to u+ and to u− , so that their difference is a sequence in S f converging to u, and u belongs to the closure of S f ; thus S f is norm-dense in L1 .

Remark Of course S f here corresponds to the space of (equivalence classes of) simple functions. 365G Semi-finite algebras: Lemma Let (A, µ ¯) be a measure algebra. (a) (A, µ ¯) is semi-finite iff L1 = L1 (A, µ ¯) is order-dense Rin L0 = L0 (A). R (b) In this case, writing S f = S(A) ∩ L1 (as in 365F), u = sup{ v : v ∈ S f , 0 ≤ v ≤ u} in [0, ∞] for every u ∈ (L0 )+ . proof (a) If (A, µ ¯) is semi-finite then S f is order-dense in L0 (364L), so L1 must also be. If L1 is order-dense 0 in L , then so is S f , by 365F and 352Nc, so (A, µ ¯) is semi-finite, by the other half of 364L.

L1

365J

339

f (b) Set C = {v : v ∈ SR f , 0 ≤ v ≤ u}. R Then C is an upwards-directed set with supremum u, because S 0 is order-dense in L . So u = supv∈C v by 365Dh.

365H Measurable transformations We have a generalization of the ideas of §235 in this abstract context. Theorem Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a sequentially order-continuous Boolean homomorphism. Let T : L0 (A) → L0 (B) be the sequentially order-continuous Riesz homomorphism associated with π (364R). R (a) Suppose that w ≥ R0 in L0 (B) is such that πa w d¯ ν=µ ¯aRwhenever a ∈ A and µ ¯a < ∞. Then for any u ∈ L1 (A, µ ¯) and a ∈ A, πa T u × w d¯ ν is defined and equal to u d¯ µ . R R R (b) Suppose that w0 ≥ 0 in L0 (A) is such that a w0 d¯ µ = ν¯(πa) for every a ∈ A. Then T u d¯ ν = u×w0 d¯ µ whenever u ∈ L0 (A) and either integral is defined in [−∞, ∞]. Remark I am using the convention of 365Dg concerning ‘∞’ as the value of an integral, and the notation R ‘ u d¯ µ’ is supposed to indicate that I am considering the integral in L1 (A, µ ¯). P n proof (a) If u ∈ S f =P L1 (A, µ ¯) ∩ S(A) then u is expressible as i=0 αi χai where a0 , . . . , an have finite n measure, so that T u = i=0 αi χ(πai ) and

R

T u × w d¯ ν=

Pn

i=0

αi

R

πai

w=

Pn

i=0

αi µ ¯ ai =

R

u d¯ µ.

If u ≥ 0 in L1 (A, µ ¯) there is a non-decreasing sequence hun in∈N in S f with supremum u, so that T u = supn∈N T un and w × T u = supn∈N w × T un in L0 (B), and

R

T u × w = supn∈N

R

T un × w = supn∈N

R

un =

R

u.

1

(365C tells us that in this context T u × w ∈ L (B, ν¯).) Finally, for general u ∈ L1 (A, µ ¯),

R

Tu × w =

R

T u+ × w −

R

T u− × w =

R

u+ −

R

u− =

R

u.

(b) The argument follows the same lines: start with u = χa for a ∈ A, then with u ∈ S(A), then with u ∈ L0 (A)+ and conclude with general u ∈ L0 (A). The point is that T is a Riesz homomorphism, so that at the last step Z Z Z Z Z + − + T u = (T u) − (T u) = T (u ) − T (u− ) Z Z Z Z Z = u+ × w0 − u− × w0 = (u × w0 )+ − (u × w0 )− = u × w0 whenever either side is defined in [−∞, ∞]. 365I The duality between L1 and L∞ Let (A, µ ¯) be a measure algebra. If we identify L∞ = L∞ (A) with the solid linear subspace of L0 = L0 (A) generated by e = χ1A (364K), then we have a bilinear map (u, v) 7→ u × v : L1 × L∞ → L1 , because |u × v| ≤ kvk∞ |u| and L1 = L1 (A, µ ¯) is a solid linear subspace of L0 . Note that ku × vk1 ≤ kuk1 kvk∞ , so that the bilinear map (u, v) → 7 u × v has norm at most 1 (253A). R Consequently we have a bilinear functional (u, v) 7→ u × v : L1 × L∞ → R, which also has norm at most 1, corresponding to linear operators S : L1 → (L∞ )∗ and T : L∞ → (L1 )∗ , both of norm at most 1. Because L1 and L∞ are both Banach lattices, we have (L1 )∗ = (L1 )∼ , (L∞ )∗ = (L∞ )∼ (356Dc). Because the norm of L1 is order-continuous, (L1 )∗ = (L1 )× (356Dd). 365J Theorem Let (A, µ ¯) be a measure algebra, and set L1 = L1 (A, µ ¯), L∞ = L∞ (A). Let S : L1 → (L∞ )∗ = (L∞ )∼ , T : L∞ → (L1 )∗ = (L1 )∼ = (L1 )× be the canonical maps defined by the duality between L1 and L∞ , as in 365I. Then (a) S and T are order-continuous Riesz homomorphisms, S[L1 ] ⊆ (L∞ )× , and S is norm-preserving; (b) (A, µ ¯) is semi-finite iff T is injective, and in this case T is norm-preserving, while S is a normed Riesz space isomorphism between L1 and (L∞ )× ;

340

Function spaces

365J

(c) (A, µ ¯) is localizable iff T is bijective, and in this case T is a normed Riesz space isomorphism between L∞ and (L1 )∗ = (L1 )× . proof Take a measure space (X, Σ, µ) such that (A, µ ¯) is isomorphic to its measure algebra. Then we can identify L1 with L1 (µ), as in 365B, and L∞ with L∞ (µ), as in 363I. Moreover, these identifications are based on the canonical embeddingsRof L1 and L∞ in L0 (µ), so that the duality described in 365I corresponds to the familiar duality (u, v) 7→ u × v already used in 243F. (a)(i) If u ≥ 0 in L1 and v ≥ 0 in L∞ then u × v ≥ 0 and (T v)(u) =

R

u × v ≥ 0.

1 ×

As u is arbitrary, T v ≥ 0 in (L ) ; as v is arbitrary, T is a positive linear operator. If v ∈ L∞ , set a = [[v > 0]] ∈ A. (Remember that we are identifying L0 (µ), as defined in §241, with 0 L (A), as defined in §364.) Then v + = v × χa, so for any u ≥ 0 in L1 (T v + )(u) =

R

u × v × χa = (T v)(u × χa) ≤ (T v)+ (u).

As u is arbitrary, T v + ≤ (T v)+ . On the other hand, because T is a positive linear operator, T v + ≥ T v and T v + ≥ 0, so T v + ≥ (T v)+ . Thus T v + = (T v)+ . As v is arbitrary, T is a Riesz homomorphism (352G). (ii) Exactly the same arguments show that S is a Riesz homomorphism. (iii) Given u ∈ L1 , set a = [[u > 0]]; then kSuk ≥ (Su)(χa − χ(1 \ a)) =

R a

u−

R 1\a

u=

R

|u| = kuk1 ≥ kSuk.

So S is norm-preserving. (iv) By 355Ka, S is order-continuous. (v) If A ⊆ L∞ is a non-empty downwards-directed set with infimum 0, and u ∈ (L1 )+ , then inf v∈A u × v = 0 for every u ∈ (L1 )+ , because v 7→ u × v : L0 → L0 is order-continuous. So inf v∈A (T v)(u) = inf v∈A

R

u × v = inf v∈A ku × vk1 = 0

and the only possible non-negative lower bound for T [A] in (L1 )× is 0. As A is arbitrary, T is ordercontinuous. (vi) The ideas of (v) show also that S[L1 ] ⊆ (L∞ )× . P P If u ∈ (L1 )+ and A ⊆ L∞ is non-empty, downwards-directed and has infimum 0, then inf v∈A (Su)(v) = inf v∈A

R

u × v = 0.

As A is arbitrary, Su is order-continuous. For general u ∈ L1 , Su = Su+ − Su− belongs to (L∞ )× . Q Q (b)(i) If (A, µ ¯) is not semi-finite, let a ∈ A be such that µ ¯a = ∞ and µ ¯b = ∞ whenever 0 6= b ⊆ a. If u ∈ L1 , then [[|u| > n1 ]] has finite measure for every n ≥ 1, so must be disjoint from a; accordingly This means that

R

a ∩ [[|u| > 0]] = supn≥1 a ∩ [[|u| >

1 n ]]

= 0.

1

u × χa = 0 for every u ∈ L . Accordingly T (χa) = 0 and T is not injective.

(ii) If (A, µ ¯) is semi-finite, take any v ∈ L∞ . Then if 0 ≤ δ < kvk∞ , a = [[|v| > δ]] 6= 0. Let b ⊆ a be such that 0 < µ ¯b < ∞. Then χb ∈ L1 , and kT vk = k|T v|k = kT |v|k ≥ (T |v|)(χb)/kχbk1 ≥ δ because |v| × χb ≥ δχb, so (T |v|)(χb) ≥ δ µ ¯b = δkχbk1 . As δ is arbitrary, kT vk ≥ kvk∞ . But we already know that kT vk ≤ kvk∞ , so the two are equal. As v is arbitrary, T is norm-preserving (and, in particular, is injective). (iii) Still supposing that (A, µ ¯) is semi-finite, S[L1 ] = (L∞ )× . P P Take any h ∈ (L∞ )× . For a ∈ A, set • νa = h(χa ). By 363K, ν : A → R is completely additive. By 365Ea, there is a u ∈ L1 such that

L1

365M

(Su)(χa) =

R

u × χa =

341

R a

u = νa = h(χa)

for every a ∈ A. Because Su and h are both linear functionals on L∞ , they must agree on S(A); because they are continuous and S(A) is dense in L∞ (363C), Su = h. As h is arbitrary, S is surjective. Q Q (c) Using (b), we know that if either T is bijective or (A, µ ¯) is localizable, then (A, µ ¯) is semi-finite, so that (X, Σ, µ) is also semi-finite (322Bd). Given this, 243G tells us that T is bijective iff (X, Σ, µ) is localizable, and in this case T is norm-preserving; but of course (X, Σ, µ) is localizable iff (A, µ ¯) is (322Be). 365K Corollary If (A, µ ¯) is a localizable measure algebra, L∞ (A) is perfect. proof By 365J(b)-(c), we can identify L∞ with (L1 )× ∼ = (L∞ )×× . 365L Theorem Let (A, µ ¯) be a measure algebra and U a Banach space. Let ν : Af → U be a function. Then the following are equiveridical: (i) there is a continuous linear operator T : L1 → U , where L1 = L1 (A, µ ¯), such that νa = T (χa) for every a ∈ Af ; (ii)(α) ν is additive (β) there is an M ≥ 0 such that kνak ≤ M µ ¯a for every a ∈ Af . Moreover, in this case, T is unique and kT k is the smallest number M satisfying the condition in (ii-β). proof (a)(i)⇒(ii) If T : L1 → U is a continuous linear operator, then χa ∈ L1 for every a ∈ Af , so ν = T χ is a function from Af to U . If a, b ∈ Af and a ∩ b = 0, then χ(a ∪ b) = χa + χb in L0 = L0 (A) and therefore in L1 , so ν(a ∪ b) = T χ(a ∪ b) = T (χa + χb) = T (χa) + T (χb) = νa + νb. If a ∈ Af then kχak1 = µ ¯a (using the formula in 365A, or otherwise), so kνak = kT (χa)k ≤ kT kkχak1 = kT k¯ µa. (b)(ii)⇒(i) Now suppose that ν : Af → U is additive and that kνak ≤ M µ ¯a for every a ∈ Af . Let S f = L1 ∩ S(A), as in 365F. Then there is a linear operator T0 : S f → U such that T0 (χa) = νa for every a Af (361F). Next, kT0 uk ≤ M kuk1 for every u ∈ S f . P P If u ∈ S f ∼ = S(Af ), then u is expressible as P∈ m f β χb where b , . . . , b ∈ A are disjoint (361Eb). So j 0 m j=0 j Pm Pm µbj = M kuk1 . Q Q kT0 uk = k j=0 βj νbj k ≤ M j=0 |βj |¯ There is therefore a continuous linear operator T : L1 → U , extending T0 , and with kT k ≤ kT0 k ≤ M (2A4I). Of course we still have ν = T χ. (c) The argument in (b) shows that T0 = T ¹S f and T are uniquely defined from ν. We have also seen that if T , ν correspond to each other then kνak ≤ kT k¯ µa for every a ∈ Af , kT k ≤ M whenever kνak ≤ M µ ¯a for every a ∈ Af , so that kT k = min{M : M ≥ 0, kνak ≤ M µ ¯a for every a ∈ Af }. 365M Corollary Let (X, Σ, µ) be a measure space and U any Banach space. Set Σf = {E : E ∈ Σ, µE < ∞}. Let ν : Σf → U be a function. Then the following are equiveridical: (i) there is a continuous linear operator T : L1 → U , where L1 = L1 (µ), such that νE = T (χE)• for every E ∈ Σf ; (ii)(α) ν(E ∪ F ) = νE + νF whenever E, F ∈ Σf and E ∩ F = 0 (β) there is an M ≥ 0 auch that kνEk ≤ M µE for every E ∈ Σf . Moreover, in this case, T is unique and kT k is the smallest number M satisfying the condition in (ii-β). proof This is a direct translation of 365L. The only point to note is that if ν satisfies the conditions of (ii), and E, F ∈ Σf are such that E • = F • in the measure algebra (A, µ ¯) of (X, Σ, µ), then µ(E \ F ) = µ(F \ E) = 0, so that ν(E \ F ) = ν(F \ E) = 0 (using condition (ii-β)) and

342

Function spaces

365M

νE = ν(E ∩ F ) + ν(E \ F ) = ν(E ∩ F ) + ν(F \ E) = νF . This means that we have a function ν¯ : Af → U , where Af = {a : a ∈ A, µ ¯a < ∞} = {E • : E ∈ Σf }, defined by setting ν¯E • = νE for every E ∈ Σf . Of course we now have ν¯(a ∪ b) = ν¯a + ν¯b whenever a, b ∈ Af and a ∩ b = 0 (since we can express them as a = E • , b = F • with E ∩ F = ∅), and k¯ ν ak ≤ M µ ¯a for every a ∈ Af . Thus we have a one-to-one correspondence between functions ν : Σf → U satisfying the conditions (ii) here, and functions ν¯ : Af → U satisfying the conditions (ii) of 365L. The rest of the argument is covered by the identification between L1 (µ) and L1 (A, µ ¯) in 365B. 365N Theorem Let (A, µ ¯) be a measure algebra, U a Banach lattice, and T : L1 → U a bounded linear 1 1 operator, where L = L (A, µ ¯). Let ν : Af → U be the corresponding additive function, as in 365L. (a) T is a positive linear operator iff νa ≥ 0 in U for every a ∈ Af ; in this case, T is order-continuous. (b) If U is Dedekind complete and T ∈ L∼ (L1 ; U ), then |T | : L1 → U corresponds to |ν| : Af → U , where Pn |ν|(a) = sup{ i=0 |νai | : a0 , . . . , an ⊆ a are disjoint} for every a ∈ Af . (c) T is a Riesz homomorphism iff ν is a lattice homomorphism. proof As in 365F, let S f be L1 ∩ S(A), identified with S(Af ). (a)(i) If T is a positive linear operator and a ∈ Af , then χa ≥ 0 in L1 , so νa = T (χa) ≥ 0 in U . 1 (ii) Now suppose that νa ≥ 0 in U for every a ∈ Af , and let u ≥ there is a Pn0 in L , ² > 0 in R. Then f v ∈ S such that 0 ≤ v ≤ u and ku − vk1 ≤ ² (365F). Express v as i=0 αi χai where ai ∈ Af , αi ≥ 0 for each i. Now

kT u − T vk ≤ kT kku − vk1 ≤ ²kT k. On the other hand, Tv =

Pn i=0

αi νai ∈ U + .

As U + is norm-closed in U , and ² is arbitrary, T u ∈ U + . As u is arbitrary, T is a positive linear operator. (iii) By 355Ka, T is order-continuous. (b) The point is that |T ¹S f | = |T |¹S f . P P (i) Because the embedding S f ⊆ L1 is positive, the map f P 7→ P ¹S is a positive linear operator from L∼ (L1 ; U ) to L∼ (S f ; U ) (see 355Bd). So |T ¹S f | ≤ |T |¹S f . (ii) There is a positive linear operator T1 : L1 → U extending |T ¹S f |, by 365M and (a) above, and now T1 ¹S f dominates both T ¹S f and −T ¹S f ; since (S f )+ is dense in (L1 )+ , T1 ≥ T and T1 ≥ −T , so that T1 ≥ |T | and |T ¹S f | = T1 ¹S f ≥ |T |¹S f . Q Q Now 361H tells us that |T |(χa) = |T ¹S f |(χa) = |ν|a for every a ∈ Af . (c)(i) If T is a lattice homomorphism, then so is ν = T χ, because χ : Af → S f is a lattice homomorphism. (ii) Now suppose that χ is a lattice homomorphism. In this case T ¹S f is a Riesz homomorphism (361Gc), that is, |T v| = T |v| for every v ∈ S f . Because S f is dense in L1 and the map u 7→ |u| is continuous both in L1 and in U (354Bb), |T u| = T |u| for every u ∈ L1 , and T is a Riesz homomorphism. 365O Theorem Let (A, µ ¯) and (B, ν¯) be measure algebras. Let π : Af → Bf be a measure-preserving ring homomorphism. (a) There is a unique order-continuous norm-preserving Riesz homomorphism Tπ : L1µ¯ → L1ν¯ such that Tπ (χa) = χ(πa) whenever a ∈ Af . We have Tπ (u × χa) = Tπ u × χ(πa) whenever a ∈ Af and u ∈ L1µ¯ . R R R R (b) Tπ u = u, πa Tπ u = a u for every u ∈ L1ν¯ , a ∈ Af .

L1

365O

343

(c) [[Tπ u > α]] = π[[u > α]] for every u ∈ L1 (A, µ), α > 0. (d) Tπ is surjective iff π is. ¯ is another measure algebra and θ : Bf → C another measure-preserving Boolean homomor(e) If (C, λ) ¯ phism, then Tθπ = Tθ Tπ : L1µ¯ → L1 (C, λ). proof Throughout the proof I will write T for Tπ and S f for S(A) ∩ L1µ¯ ∼ = S(Af ) (see 365F). (a)(i) We have a map ψ : Af → L1ν¯ defined by writing ψa = χ(πa) for a ∈ Af . Because χπ(a ∪ b) = χ(πa ∪ πb) = χπa + χπb,

kχ(πa)k1 = ν¯(πa) = µ ¯a

f

whenever a, b ∈ A and a ∩ b = 0, we get a (unique) corresponding bounded linear operator T : L1µ¯ → L1ν¯ such that T χ = χπ on Af (365L). Because π : Af → Bf and χ : Bf → L1ν¯ are lattice homomorphisms, so is ψ, and T is a Riesz homomorphism (365Nc). Pn Pn (ii) If u ∈ S f , express u as i=0 αi χai where a0 , . . . , an are disjoint in Af . Then T u = i=0 αi χ(πai ) and πa0 , . . . , πan are disjoint in Bf , so Pn Pn kT uk1 = i=0 |αi |¯ ν (πai ) = i=0 |αi |¯ µai = kuk1 . Because S f is dense in L1µ¯ and u 7→ kuk1 is continuous (in both L1µ¯ and L1ν¯ ), kT uk1 = kuk1 for every u ∈ L1µ¯ , that is, T is norm-preserving. As noted in 365Na, T is order-continuous. (iii) If a, b ∈ Af then T (χa × χb) = T (χ(a ∩ b)) = χπ(a ∩ b) = χ(πa ∩ πb) = χπa × χπb = χπa × T (χb). Because T is linear and × is bilinear, T (χa × u) = χπa × T u for every u ∈ S f . Because the maps u 7→ u × χa : L1µ¯ → L1µ¯ , T : L1µ¯ → L1ν¯ and v 7→ v × χπa : L1ν¯ → L1ν¯ are all continuous, T u × χπa = T (u × χa) for every u ∈ L1µ¯ . (iv) T is unique because the formula T (χa) = χπa defines T on the norm-dense and order-dense subspace S f . (b) Because T is positive,

R

For a ∈ Af ,

R

T u = kT u+ k1 − kT u− k1 = ku+ k1 − ku− k1 =

Tu = πa

(c) If u ∈ S f , express it as

R

Pn i=0

T u × χπa =

R

T (u × χa) =

R

R

u × χa =

u.

R a

u.

αi χai where a0 , . . . , an are disjoint; then

π[[u > α]] = π(supi∈I ai ) = supi∈I πai = [[T u > α]] where I = {i : i ≤ n, αi > α}. For u ∈ (L1 )+ , take a sequence hun in∈N in S f with supremum u; then supn∈N T un = T u, so π[[u > α]] = π(sup [[un > α]]) n∈N

(364Mb; [[u > α]] ∈ Af by 365A) = sup π[[un > α]] n∈N

(because π is order-continuous, see 361Ad) = sup [[T un > α]] = [[T u > α]] n∈N

because T is order-continuous. For general u ∈ L1 , π[[u > α]] = π[[u+ > α]] = [[T (u+ ) > α]] = [[(T u)+ > α]] = [[T u > α]] because T is a Riesz homomorphism.

344

Function spaces

365O

(d)(i) Suppose that T is surjective and that b ∈ Bf . Then there is a u ∈ L1µ¯ such that T u = χb. Now b = [[T u > 21 ]] = π[[u > 12 ]] ∈ π[Af ]; as b is arbitrary, π is surjective. (ii) Suppose now that π is surjective. Then T [L1µ¯ ] is a linear subspace of L1ν¯ containing χb for every b ∈ Bf , so includes S(Bf ). If v ∈ (L1ν¯ )+ there is a sequence hvn in∈N in S(Bf )+ with supremum v. For each n, choose un such that T un = vn . Setting u0n = supi≤n ui , we get a non-decreasing sequence hu0n in∈N such that vn ≤ T u0n ≤ v for every n ∈ N. So supn∈N ku0n k1 = supn∈N kT u0n k1 ≤ kvk1 < ∞ and u = supn∈N u0n is defined in L1µ¯ , with T u = supn∈N T u0n = v. Thus (L1ν¯ )+ ⊆ T [L1µ¯ ]; consequently L1ν¯ ⊆ T [L1µ¯ ] and T is surjective. (e) This is an immediate consequence of the ‘uniqueness’ assertion in (i), because for any a ∈ Af Tθ Tπ (χa) = Tθ χ(πa) = χ(θπa), so that Tθ Tπ : L1µ¯ → L1λ¯ is a bounded linear operator taking the right values at elements χa, and must therefore be equal to Tθπ . 365P Theorem Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : Af → B an order-continuous ring homomorphism. R R (a) There is a unique positive linear operator Pπ : L1ν¯ → L1µ¯ such that a Pπ v = πa v for every v ∈ L1ν¯ , a ∈ Af . (b) Pπ is order-continuous and norm-continuous, and kPπ k ≤ 1. (c) If a ∈ Af , v ∈ L1ν¯ then Pπ (v × χπa) = Pπ v × χa. (d) If π[Af ] is order-dense in B then Pπ is a norm-preserving Riesz homomorphism; in particular, Pπ is injective. (e) If (B, ν¯) is semi-finite and π is injective, then Pπ is surjective, and there is for every u ∈ L1µ¯ a v ∈ L1ν¯ such that Pπ v = u and kvk1 = kuk1 . ¯ is another measure algebra and θ : B → C an order(f) Suppose again that (B, ν¯) is semi-finite. If (C, λ) ¯ → L1 , where I write θ0 for the restriction continuous Boolean homomorphism, then Pθπ = Pπ Pθ0 : L1 (C, λ) µ ¯ of θ to Bf . proof I write P for Pπ .

R (a)-(b) For v ∈ L1ν¯ , a ∈ Af set νv (a) = πa v. Then νv : Af → R is additive, bounded (by kvk1 ) and if A ⊆ Af is non-empty, downwards-directed and has infimum 0, then inf a∈A |νv (a)| ≤ inf a∈A

R

R

|v| × χπa = 0

because a 7→ |v| × χπa is a composition of order-continuous functions, therefore order-continuous. So R R 365Eb tells us that there is a P v ∈ L1µ¯ such that a P v = νv (a) = πa v for every a ∈ Af . By 365Dd, this formula defines P v uniquely. Consequently P must be linear (since P v1 + P v2 , αP v will always have the properties defining P (v1R + v2 ), PR (αv)). If v ≥ 0 in L1ν¯ , then a P v = πa v ≥ 0 for every a ∈ Af , so P v ≥ 0 (365Dd); thus P is positive. It must therefore be norm-continuous and order-continuous (355C, 355Ka). Again supposing that v ≥ 0, we have kP vk1 = (using 365Dd). For general v ∈

R

P v = supa∈Af

R

P v = supa∈Af a

R

πa

L1ν¯ , kP vk1 = k|P v|k1 ≤ kP |v|k1 ≤ kvk1 .

(c) For any c ∈ Af ,

v ≤ kvk1

L1

365P

R c

P v × χa =

R c∩a

Pv =

R π(c∩a)

345

v=

R

v × χπa =

πc

R c

P (v × χπa).

(d) Now suppose that π[Af ] is order-dense. Take any v, v 0 ∈ L1ν¯ such that v ∧ v 0 = 0. ?? Suppose, if possible, that u = P v ∧ P v 0 > 0. Take α > 0 such that a = [[u > α]] is non-zero. Since

R

πa

v=

R

Pv ≥

a

R

a

u > 0,

b = πa ∩ [[v > 0]] 6= 0. Let c ∈ Af be such that 0 6= πc ⊆ b; then π(a ∩ c) = πc 6= 0, so a ∩ c 6= 0, and 0
0]] and v ∧ v 0 = 0 so πc v 0 = 0. X X So P v ∧ P v 0 = 0. As v, v 0 are arbitrary, P is a Riesz homomorphism (352G). Next, if v ≥ 0 in L1ν¯ ,

R

P v = supa∈Af

R

R

P v = supa∈Af a

v= πa

R

v

because π[Af ] is upwards-directed and has supremum 1 in B. So, for general v ∈ L1ν¯ , kP vk1 =

R

|P v| =

R

P |v| =

R

|v| = kvk1 ,

and P is norm-preserving. (e) Next suppose that (B, ν¯) is semi-finite and that π is injective. R R (i) If u > 0 in L1µ¯ , there is a v > 0 in L1ν¯ such that P v ≤ u and P v ≥ v. P P Let δ > 0 be such that a = [[u > δ]] 6= 0. Then πa 6= 0. BecauseR (B, ν¯) is semi-finite, there is a non-zero b ∈ Bf such that b ⊆ πa. Set u1 = P (χb). Then u1 ≥ 0, and also a u1 = ν¯b > 0,

R

1\a

u1 = supc∈Af

R

c\a

R

u1 = supc∈Af

πc\πa

χb = 0.

So u1 × χ(1 \ a) = 0 and 0 6= [[u1 > 0]] ⊆ a. Let γ > 0 be such that [[u1 > γ]] 6= [[u1 > 0]], and set a1 = a \ [[u1 > γ]], v = δγ −1 χ(b ∩ πa1 ). Then P v = δγ −1 P (χb × χ(πa1 )) = δγ −1 P (χb) × χa1 = δγ −1 u1 × χa1 ≤ δχa ≤ u because [[u1 × χa1 > γ]] ⊆ [[u1 > γ]] ∩ a1 = 0 so u1 × χa1 ≤ γχ[[u1 > 0]] ≤ γχa. Also P v > 0 because a1 ∩ [[u1 > 0]] 6= 0, so v 6= 0; and

R

Pv ≥

R

a1

Pv =

R πa1

v=

R

v. Q Q

R R (ii) Now take any u ≥ 0 in L1µ¯ , and set B = {v : v ∈ L1ν¯ , v ≥ 0, P v ≤ u, v ≤ P v}. Let C ⊆ B be a maximal upwards-directed set (applying Zorn’s Lemma to the family P of all upwards-directed subsets of B). We have supv∈C so v0 = sup C is defined in

L1ν¯

R

v ≤ supv∈C

R

P v ≤ kuk1 ,

(365Df). Because P is order-continuous, P v0 = sup P [C] ≤ u, and

R

P v0 = supv∈C

R

P v ≤ supv∈C

R

v=

R

v0 .

if possible, that P v0 6= u. In this case, by (α), there is a v1 > 0 such that P v1 ≤ u − P v0 , R?? Suppose, R v1 ≤ P v1 . In this case, v0 + v1 ∈ B, so C 0 = C ∪ {v0 + v1 } is an upwards-directed subset of B strictly larger than C, which is impossible. X X Thus P v0 = u; also kv0 k1 =

R

v0 ≤

R

P v0 = kP v0 k1 .

(iii) Now take any u ∈ L1µ¯ . By (ii), there are non-negative v1 , v2 ∈ L1ν¯ such that P v1 = u+ , P v2 = u− , kv1 k1 ≤ ku+ k1 , kv2 k1 ≤ ku− k1 . Setting v = v1 − v2 , we have P v = u. Also we must have

346

Function spaces

365P

kvk1 ≤ kv1 k1 + kv2 k1 ≤ ku+ k1 + ku− k1 = kuk1 ≤ kvk1 , so kvk1 = kuk1 , as required. (f ) As usual, this is a consequence of the uniqueness of P . RHowever (because I do not assume that R π[Af ] ⊆ Bf ) there is an extra refinement: we need to know that b Pθ0 w = θb w for every b ∈ B, w ∈ L1λ¯ . P P Because θ is order-continuous and (B, ν¯) is semi-finite, θb = sup{θb0 : b0 ∈ Bf , b0 ⊆ b}, so if w ≥ 0 then

R

w = supb0 ∈Bf ,b0 ⊆ b θb

R

θb0

w = supb0 ∈Bf ,b0 ⊆ b

R

b0

Pθ0 w =

R

b

Pθ0 w.

Expressing w as w+ − w− , we see that the same is true for every w ∈ L1ν¯ . Q Q Now we can say that P Pθ0 is a positive linear operator from L1λ¯ to L1µ¯ such that

R

a

R

P Pθ 0 w =

πa

Pθ0 w =

R

θπa

w=

R

a

Pθπ w

whenever a ∈ Af , w ∈ L1λ¯ . 365Q Proposition Let (A, µ ¯) and (B, µ ¯) be measure algebras and π : Af → Bf a measure-preserving ring homomorphism. (a) In the language of 365O-365P above, Pπ Tπ is the identity operator on L1 (A, µ ¯). (b) If π is surjective (so that it is an isomorphism between Af and Bf ) then Pπ = Tπ−1 = Tπ−1 , Tπ = Pπ−1 = Pπ−1 . proof (a) If u ∈ L1µ¯ , a ∈ Af then

R

P T u= a π π

R

T u= πa π

R a

u.

So u = Pπ Tπ u, by 365Dd. (b) From 365Od, we know that Tπ is surjective, while Pπ Tπ is the identity, so that Pπ = Tπ−1 , Tπ = Pπ−1 . As for Tπ−1 , 365Oe tells us that Tπ−1 = Tπ−1 ; so Pπ−1 = Tπ−1 −1 = Tπ . 365R Conditional expectations It is a nearly universal rule that any investigation of L1 spaces must include a look at conditional expectations. In the present context, they take the following form. (a) Let (A, µ ¯) be a probability algebra and B a closed subalgebra; write ν¯ for the restriction µ ¯¹ B. The identity map from B to A induces operators T : L1ν¯ → L1µ¯ and P : L1µ¯ → L1ν¯ . If we take L0 (A) to be defined as the set of functions from R to A described in 364A, then L0 (B) becomes a subset of L0 (A) in the literal sense, and T is actually the identity operator associated with the subset L1ν¯ ⊆ L1µ¯ ; L1ν¯ is a norm-closed and order-closed Riesz subspace of L1µ¯ . P is a positive linear operator, while P T is the identity, so P is a projection from L1µ¯ onto L1ν¯ . P is defined by the familiar formula

R

Pu = b

R

b

u for every u ∈ L1µ¯ , b ∈ B,

so is the conditional expectation operator in the sense of 242J. (b) Just as in 233I-233J and 242K, we have a fundamental inequality concerning convex functions: if R R ¯ ¯ h : R → R is a convex function and u ∈ L1µ¯ , then h( u) ≤ h(u); and if h(u) ∈ L1µ¯ (364I), then ¯ u) ≤ P (h(u)). ¯ h(P P P I repeat the proof of 233I-233J. For each q ∈ Q, take βq ∈ R such that h(t) ≥ hq (t) = ¯ ¯ q (u) for h(q) + βq (t − q) for every t ∈ R, so that h(t) = supq∈Q hq (t) for every t ∈ R, and h(u) = supq∈Q h 0 0 every u ∈ L = L (A). (This is because S ¯ [[h(u) > α]] = [[u ∈ h−1 [ ]α, ∞[ ]]] = [[u ∈ q∈Q h−1 q [ ]α, ∞[ ]]] −1 ¯ q (u) > α]] = sup [[u ∈ h [ ]α, ∞[ ]]] = sup [[h q∈Q

q

q∈Q

¯ q (u) = h(q)e + βq (u − qe) for every u ∈ L0 , so that for every α ∈ R.) But setting e = χ1, we see that h

R

R

R

hq (u) = h(q) + βq ( u − q) = hq ( u),

L1

365T

because

R

347

¯ q (u)) = h(q)e + βq (P u − qe) = h ¯ q (P u) P (h e = 1 and P e = e. Taking the supremum over q, we get

R

R

h( u) = supq∈Q hq ( u) = supq∈Q

R

¯ q (u) ≤ h

R

¯ h(u),

¯ and if h(u) ∈ L1µ¯ then ¯ ¯ ¯ ¯ u) = sup Q h(P q∈Q hq (P u) = supq∈Q P (hq (u)) ≤ P (h(u)). Q Of course the result in this form can also be deduced from 233I-233J if we represent (A, µ ¯) as the measure algebra of a probability space (X, Σ, µ) and set T = {E : E ∈ Σ, E • ∈ B}. (c) I note here a fact which is occasionally useful. If u ∈ L1µ¯ is non-negative, then [[P u > 0]] = upr([[u > 0]], C), the upper envelope of [[u > 0]] in C as defined in 314V. P P We have only to observe that, for c ∈ C, c ∩ [[P u > 0]] = 0 ⇐⇒ χc × P u = 0 ⇐⇒

R

c

P u = 0 ⇐⇒

R

c

u = 0 ⇐⇒ c ∩ [[u > 0]] = 0.

Taking complements, c ⊇ [[P u > 0]] iff c ⊇ [[u > 0]]. Q Q 365S Recovering the algebra: Proposition (a) Let (A, µ ¯) be a localizable measure algebra. Then A is isomorphic to the band algebra of L1 = L1 (A, µ ¯). (b) Let A be a Dedekind σ-complete Boolean algebra, and µ ¯, ν¯ two measures on A such that (A, µ ¯) and (A, ν¯) are both semi-finite measure algebras. Then L1 (A, µ ¯) is isomorphic, as Banach lattice, to L1 (A, ν¯). proof (a) Because (A, µ ¯) is semi-finite, L1 is order-dense in L0 = L0 (A) (365G). Consequently, L1 and L0 have isomorphic band algebras (353D). But the band algebra of L0 is just its algebra of projection bands (because A and therefore L0 are Dedekind complete, see 364O and 353I), which is isomorphic to A (364Q). (b) Let π : A → A be the identity map. Regarding π as an order-continuous Boolean homomorphism ¯a < ∞} to (A, ν¯), we have an associated positive linear operator P = Pπ : L1ν¯ → L1µ¯ ; from Afµ¯ = {a : µ similarly, we have Q = Pπ−1 : L1µ¯ → L1ν¯ , and both P and Q have norm at most 1 (365Pb). Now 365Pf assures us that QP is the identity operator on L1ν¯ and P Q is the identity operator on L1µ¯ . So P and Q are the two halves of a Banach lattice isomorphism between L1µ¯ and L1ν¯ . 365T Having opened the question of varying measures on a single Boolean algebra, this seems an appropriate moment for a general description of how they interact. Proposition Let A be a Dedekind complete Boolean algebra, and µ ¯ : A → [0, ∞], ν¯ : A → [0, ∞] two functions such that (A, µ ¯) and (A, ν¯) are both semi-finite (therefore localizable) measure algebras. R (a) There is a unique u ∈ L0 = L0 (A) such that (if we allow ∞ as a value of the integral) a u d¯ µ = ν¯a for every a ∈ A. R R (b) For every v ∈ L1ν¯ , v d¯ ν = u × v d¯ µ. R (c) u is strictly positive (i.e., [[u > 0]] = 1) and, writing u1 for the multiplicative inverse of u, a u1 d¯ ν=µ ¯a for every a ∈ A. proof (a) Because (A, ν¯) is semi-finite, there is a partition of unity D ⊆ A such that ν¯d < ∞ for every d ∈ D. For each Rd ∈ D, the functional a 7→ ν¯(a ∩ d) : A →R R is completely additive, so there is a 1 u µ = ν¯(a ∩ d) for every a ∈ A. Because a ud d¯ µ ≥ 0 for every a, ud ≥ 0. Because R d ∈ Lµ¯ such that a ud d¯ 0 u = 0, [[ud > 0]] ⊆ d. Now u = supd∈D ud is defined in L . P P (See 368K below.) For n ∈ N, 1\d d set cn = supd∈D [[ud > n]]. If d, d0 ∈ D are distinct, then d ∩ [[ud0 > n]] = 0, so d ∩ cn = [[ud > n]]. Set c = inf n∈N cn . If d ∈ D, then d ∩ c = inf n∈N d ∩ cn = inf n∈N [[ud > n]] = 0. But c ⊆ c0 ⊆ sup D, so c = 0. By 364Ma, {ud : d ∈ D} is bounded above in L0 , so has a supremum, because L0 is Dedekind complete, by 364O. Q P Q For finite I ⊆ D set u ˜I = d∈I ud = supd∈I ud (because ud ∧ uc = 0 for distinct c, d ∈ D). Then u = sup{˜ uI : I ⊆ D, I is finite}. So, for any a ∈ A,

348

Function spaces

Z

365T

Z u d¯ µ=

sup I⊆D is finite

a

(365Dh) =

u ˜I d¯ µ a

XZ

sup I⊆D is finite

d∈I

ud d¯ µ=

a

sup I⊆D is finite

X

ν¯(a ∩ d) = ν¯a.

d∈I

Note that if a ∈ A is non-zero, then ν¯a > 0, so a ∩ [[u > 0]] 6= 0; consequently [[u > 0]] = 1. To see that u is unique, observe that if u0 has the same property then for any d ∈ D

R

a 0

(u × χd)d¯ µ = ν¯(a ∩ d) =

R

a

(u0 × χd)

for every a ∈ A, so that u × χd = u × χd; because sup D = 1 in A, u must be equal to u0 . (b) Use 365Hb, with π and T the identity maps. (c) In the same way, there is a w ∈ L0 such that observe that applying 365Hb we get

R

R

R a

w × χa × u d¯ µ=

w d¯ ν = µ ¯a for every a ∈ A. To relate u and w,

R

w × χa d¯ ν

for every a ∈ A, that is, a w × u d¯ µ=µ ¯a for every a. But from this we see that w × u × χb = χb at least when µ ¯b < ∞, so that w × u = χ1 is the multiplicative identity of L0 , and w = u1 . 365X Basic exercises (a) Let (A, µ ¯) be a measure algebra, and u ∈ L1µ¯ . Show that

R

u=

R∞ 0

µ ¯[[u > α]] dα −

R0

−∞

µ ¯(1 \ [[u > α]]) dα.

R > (b) Let (A, µ ¯) be any measure algebra, and u ∈ L1µ¯ . (i) Show that kuk1 ≤ 2 supa∈Af | a u|. (Hint: R R 246F.) (ii) Show that for any ² > 0 there is an a ∈ Af such that | u − b u| ≤ ² whenever a ⊆ b ∈ A. + > (c) Let U be an L-space. If hun in∈N is any R norm-bounded sequence in RU , show that lim inf n→∞ un = supn∈N inf m≥n um is defined in U , and that lim inf n→∞ un ≤ lim inf n→∞ un .

(d) Let U be an L-space. Let F be a filter on U + such that R{u : u ≥ 0, kuk ≤ k} belongs to F for some R k ∈ N. Show that u0 = supF ∈F inf F is defined in U , and that u0 ≤ supF ∈F inf u∈F u. iff

(e) Let (A, µ ¯) be a measure algebra and A ⊆ L1µ¯ a non-empty set. Show that A is bounded above in L1µ¯

Pn R

sup{

ui : a0 , . . . , an is a partition of unity in A, u0 , . . . , un ∈ A} R is finite, and that in this case the supremum is Rsup A. (Hint: P givenRu0 , . . . , un ∈ A, set bij = [[ui ≥ uj ]], n bi = supj6=i bij , ai = bi \ supj 0 there is a δ > 0 such that |νa| ≤ ² whenever µ ¯a ≤ δ; (iii) for every ² > 0, c ∈ Af there is a δ > 0 such that νa ≤ ² whenever a ⊆ c and µ ¯a ≤ δ; (iv) for every ² > 0 there are c ∈ Af , δ > 0 such that |νa| ≤ ² whenever f a ∈ A and µ ¯(a ∩ c) ≤ δ; (v) limn→∞ νan = 0 whenever han in∈N is a non-increasing sequence in Af with infimum 0. (g) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a sequentially order-continuous Boolean homomorphism. Let T : L0 (A) → L0R(B) be the Riesz homomorphism associated with π (364R). Suppose that w ≥ 0 inRL0 (B) is such that πa w d¯ ν = µ ¯a whenever a ∈ A. Show that for any u ∈ L0 (A, µ ¯), R T u × w d¯ ν = u d¯ µ whenever either is defined in [−∞, ∞].

365 Notes

L1

349

> (h) Let (A, µ ¯) be a measure algebra and a ∈ A; write Aa for the principal ideal it generates. Show that if π is the identity embedding of Af ∩ Aa into Af , then Tπ , as defined in 365O, identifies L1 (Aa , µ ¯¹ Aa ) with a band in L1 (A, µ ¯). > (i) Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with measure algebras (A, µ ¯) and (B, ν¯). Let φ : X → Y be an inverse-measure-preserving function and π : B → A the corresponding measure-preserving homomorphism (324M). Show that Tπ : L1ν¯ → L1µ¯ (365O) corresponds to the map g • 7→ (gφ)• : L1 (ν) → L1 (µ) of 242Xf. (j) Let (A, µ ¯) and (B, ν¯) be measure algebras. Let π : Af → Bf be a ring homomorphism such that, for some γ > 0, ν¯(πa) ≤ γ µ ¯a for every a ∈ Af . (i) Show that there is a unique order-continuous Riesz 1 1 homomorphism T : Lµ¯ → Lν¯ such that T (χa) = χ(πa) whenever a ∈ Af , and that kT k ≤ γ. (ii) Show that [[T u > α]] = π[[u > α]] for every u ∈ L1 (A, µ), α > 0. (iii) Show that T is surjective iff π is, injective iff π is. (iv) Show that T is norm-preserving iff ν¯(πa) = µ ¯a for every a ∈ Af . (k) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a measure-preserving Boolean homomorphism. Let T : L1µ¯ → L1ν¯ and P : L1ν¯ → L1µ¯ be the operators corresponding to π¹ Af , as described in 365O-365P, and T˜ : L∞ (A) → L∞ (B) the operator corresponding to π, as described in 363F. (i) Show that T (u × v) = T u × T˜v for every u ∈ L1µ¯ , v ∈ L∞ (A). (ii) Show that if π is order-continuous, then R R P v × u = v × T˜u for every u ∈ L∞ (A), v ∈ L1ν¯ . > (l) Let (X, Σ, µ) be a probability space, with measure algebra (A, µ ¯), and let T be a σ-subalgebra of Σ. Set ν = µ¹ T, B = {F • : F ∈ T} ⊆ A, ν¯ = µ ¯¹ B, so that (B, ν¯) is a measure algebra. Let π : B → A be the identity homomorphism. Show that Tπ : L1ν¯ → L1µ¯ (365O) corresponds to the canonical embedding of L1 (ν) in L1 (µ) described in 242Jb, while Pπ : L1µ¯ → L1ν¯ (365P) corresponds to the conditional expectation operator described in 242Jd. b µ (m) Let (A, µ ¯) be a semi-finite measure algebra, and (A, ˆ) its localization (322P). Show that the natural b b µ embedding of A in A induces a Banach lattice isomorphism between L1µ¯ and L1 (A, ˆ), so that the band 1 b of A. algebra of Lµ¯ can be identified with the Dedekind completion A (n) Let A be a Dedekind σ-complete Boolean algebra and µ ¯, ν¯ two functions such that (A, µ ¯), (A, ν¯) are measure algebras. Show that L1µ¯ ⊆ L1ν¯ (as subsets of L0 (A)) iff there is a γ > 0 such that ν¯a ≤ γ µ ¯a for every a ∈ A. (Hint: show that the identity operator from L1µ¯ to L1ν¯ is bounded.) (o) Let (A, µ ¯) be a measure algebra, I∞ the ideal of ‘purely infinite’ elements of A, µ ¯sf the measure on B = A/I∞ (322Xa). Let π : A → B be the canonical map. Show that Tπ , as defined in 365O, is a Banach lattice isomorphism between L1 (A, µ ¯) and L1 (B, µ ¯sf ). (p) Let (X, Σ, µ) be a a semi-finite measure space. Show that L1 (µ) is separable iff µ is σ-finite and has countable Maharam type. 365Y Further exercises (a) Let (A, µ ¯) be a semi-finite measure algebra, not {0}. Show that the topological density of L1 (A, µ ¯) (331Yf) is max(ω, τ (A), c(A)), where τ (A), c(A) are the Maharam type and cellularity of A. 365 Notes and comments You should not suppose that L1 spaces appear in the second half of this chapter because they are of secondary importance. Indeed I regard them as the most important of all function spaces. I have delayed the discussion of them for so long because it is here that for the first time we need measure algebras in an essential way. The actual definition of L1µ¯ which I give is designed for speed rather than illumination; I seek only a formula, visibly independent of any particular representation of (A, µ ¯) as the measure algebra of a measure space, for which I can prove 365B. R ∞I use a relatively primitive R argument in 365B, not appealing to Fubini’s theorem; of course the formula 0 µ{x : |f (x)| > α}dα = |f (x)|dx can also be regarded as a reversal of

350

Function spaces

365 Notes

order of integration in a double integral (252N, 252O). 365C-365D and 365Ea are now elementary. In 365Eb I take a page to describe a form of the Radon-Nikod´ ym theorem which is applicable to arbitrary measure algebras, at the cost of dealing with functionals on the ring Af rather than on the whole algebra A. This is less for the sake of applications than to emphasize one of the central properties of L1 : it depends only on Af and µ ¯¹ Af . For alternative versions of the condition 365Eb(i) see 365Xf. The convergence theorems (B.Levi’s theorem, Fatou’s lemma and Lebesgue’s dominated convergence theorem) are so central to the theory of integrable functions that it is natural to look for versions in the language here. Corresponding to B.Levi’s theorem is the Levi property of a norm in an L-space; note how the abstract formulation makes it natural to speak of general upwards-directed families rather than of nondecreasing sequences, though the sequential form is so often used that I have spelt it out (365C). In the same way, the integral becomes order-continuous rather than just sequentially order-continuous (365Da). Corresponding to Fatou’s lemma we have 365Xc-365Xd. For abstract versions of Lebesgue’s theorem I will wait until §367. In 365H I have deliberately followed the hypotheses of 235A and 235T. Of course 365H can be deduced from these if we use the Stone representations of (A, µ ¯) and (B, ν¯), so that π can be represented by a function between the Stone spaces (312P). But 365H is essentially simpler, because the technical problems concerning measurability which took up so much of §235 have been swept under the carpet. In the same way, 365Xg corresponds to 235E. Here we have a fair example of the way in which the abstract expression in terms of measure algebras can be tidier than the expression in terms of measure spaces. But in my view this is because here, at least, some of the mathematics has been left out. 365J is a re-run of 243G, but with the additional refinement that I examine the action of L1 on L∞ (the operator S) as well as the action of L∞ on L1 (the operator T ). Note that (b-iii) and (c) of the proof of 365J depend on the Radon-Nikod´ ym theorem; these parts lie a step deeper than the rest. 365L-365N correspond closely to 361F-361H and 363E. Theorems 365O-365Q lie at the centre of my picture of L1 spaces, and are supposed to show their dual nature. Starting from a semi-finite measure algebra (A, µ ¯) we have two essentially different routes to the L1 -space: we can either build it up from characteristic functions of elements of finite measure, so that it is naturally embedded in L0 (A), or we can think of it as the order-continuous dual of L∞ (A). The first is a ‘covariant’ construction (signalled by the formula Tθπ = Tθ Tπ in 365Oe) and the second is ‘contravariant’ (so that Pθπ = Pπ Pθ0 in 365Pf). The first construction is the natural one if we are seeking to copy the ideas of §242, but the second arises inevitably if we follow the ordinary paths of functional analysis and study dual spaces whenever they appear. The link between them is the Radon-Nikod´ ym theorem. I have deliberately written out 365O and 365P with different hypotheses on the homomorphism π in the hope of showing that the two routes to L1 really are different, and can be expected to tell us different things about it. I use the letter P in 365P in order to echo the language of 242J; in the most important context, in which A is actually a subalgebra of B and π is the identity map, P is a kind of conditional expectation operator (365R). I note that in the proof of 365Pe I have returned to first principles, using some of the ideas of the Radon-Nikod´ ym theorem (232E), but a different approach to the exhaustion step (converting ‘for every u > 0 there is a v > 0 such that P v ≤ u’ into ‘P is surjective’). I chose the somewhat cruder method in 232E (part (c) of the proof) in order to use the weakest possible form of the axiom of choice. In the present context such scruples seem absurd. I used the words ‘covariant’ and ‘contravariant’ above; of course this distinction depends on the side of the mirror on which we are standing; if our measure-preserving homomorphism is derived (contravariantly) from an inverse-measure-preserving transformation, then the T ’s become contravariant (365Xi). An important component of this work, for me, is the fact that not all measure-preserving homomorphisms between measure algebras can be represented by inverse-measure-preserving functions (343Jb, 343M). I have already remarked (in the notes to §244) that the properties of L1 (µ) are not much affected by peculiarities in a measure space (X, Σ, µ), because (unlike L0 or L∞ ) they really depend only on Af , the ring of elements of finite measure in the measure algebra. (See 365O-365Q and 365Xm-365Xn.) Note that while the algebra A is uniquely determined (given that (A, µ ¯) is localizable, 365Sa), the measure µ ¯ is not; if A is any algebra carrying two non-isomorphic semi-finite measures, the corresponding L1 spaces are still isomorphic (365Sb). For instance, the L1 -spaces of Lebesgue measure µ on R, and the subspace measure µ[0,1] on [0, 1], are isomorphic, though their measure algebras are not. I make no attempt here to add to the results in §§246, 247, 354 and 356 concerning uniform integrability

Lp

366D

351

and weak compactness. Once we have left measure spaces behind, these ideas belong to the theory of Banach lattices, and there is little to relate them to the questions dealt with in this section. But see 373Xj and 373Xn below.

366 Lp In this section I apply the methods of this chapter to Lp spaces, where 1 < p < ∞. The constructions proceed without surprises up to 366E, translating the ideas of §244 by the methods used in §365. Turning to the action of Boolean homomorphisms on Lp spaces, I introduce a space M 0 , which can be regarded as the part of L0 that can be determined from the ring Af of elements of A of finite measure (366F), and which includes Lp whenever 1 ≤ p < ∞. Now a measure-preserving ring homomorphism from Af to Bf acts on the M 0 spaces in a way which includes injective Riesz homomorphisms from Lp (A, µ ¯) to Lp (B, ν¯) p p and surjective positive linear operators from L (B, ν¯) to L (A, µ ¯) (366H). The latter may be regarded as conditional expectation operators (366J). The case p = 2 (366K-366L) is of course by far the most important. 366A Definition Let (A, µ ¯) be a measure algebra and suppose that 1 < p < ∞. For u ∈ L0 (A), define 0 |u| ∈ L (A) by setting p

[[|u|p > α]] = [[|u| > α1/p ]] if α ≥ 0 = 1 if α < 0. ¯ (In the language of 364I, |u|p = h(u), where h(t) = |t|p for t ∈ R.) Set ¯) = {u : u ∈ L0 (A), |u|p ∈ L1 (A, µ ¯)}, Lpµ¯ = Lp (A, µ and for u ∈ Lpµ¯ set

R

1/p

kukp = ( |u|p )1/p = k|u|p k1 . 366B Theorem Let (X, Σ, µ) be a measure space, and (A, µ ¯) its measure algebra. Then the canonical isomorphism between L0 (µ) and L0 (A) (364Jc) makes Lp (µ), as defined in §244, correspond to Lp (A, µ ¯). proof What we really have to check is that if w ∈ L0 (µ) corresponds to u ∈ L0 (A), then |w|p , as defined in 244A, corresponds to |u|p as defined in 366A. But this was noted in 364Jb. Now, because the isomorphism between L0 (µ) and L0 (A) matches L1 (µ) with L1 (A, µ ¯) (365B), we can be sure that |w|p ∈ L1 (µ) iff |u|p ∈ L1 (A, µ ¯), and that in this case ¢1/p ¡R p ¢1/p ¡R kwkp = |w|p = |u| = kukp , as required. 366C Corollary For any measure algebra (A, µ ¯) and p ∈ ]1, ∞[, Lp = Lp (A, µ ¯) is a solid linear subspace 0 of L (A). It is a Dedekind complete Banach lattice under its norm k k . Setting q = p/(p − 1), (Lp )∗ is p R q f identified with L (A, µ ¯) by the duality (u, v) 7→ u × v. Writing A for the ring {a : a ∈ A, µ ¯a < ∞}, S(Af ) p is norm-dense in L . proof Because we can find a measure space (X, Σ, µ) such that (A, µ ¯) is isomorphic to the measure algebra of µ (321J), this is just a digest of the results in 244B, 244E, 244F, 244G, 244H and 244K. (Of course S(Af ) corresponds to the space S of equivalence classes of simple functions in 244Ha, just as in 365F.) 366D

I can add a little more, corresponding to 365C and 365J.

Theorem Let (A, µ ¯) be a measure algebra, and p ∈ ]1, ∞[. (a) The norm k kp on Lp = Lp (A, µ ¯) is order-continuous. (b) Lp has the Levi property.

352

Function spaces

366D

(c) Setting q = p/(p − 1), the canonical identification of Lq = Lq (A, µ ¯) with (Lp )∗ is a Riesz space q p ∼ p × isomorphism between L and (L ) = (L ) . (d) Lp is a perfect Riesz space. proof (a) Suppose that A ⊆ Lp is non-empty, downwards-directed and has infimum 0. For u, v ≥ 0 in Lp , u ≤ v ⇒ up ≤ v p (by the definition in 366A, or otherwise), so B = {up : u ∈ A} is downwards-directed. If 1/p v0 = inf B in L1 = L1 (A, µ ¯), then v0 (defined by the formula in 366A, or otherwise) is less than or equal to every member of A, so must be 0, and v0 = 0. Accordingly inf B = 0 in L1 . Because k k1 is order-continuous (365C), 1/p

inf u∈A kukp = inf u∈A kup k1

= (inf v∈B kvk1 )1/p = 0.

As A is arbitrary, k kp is order-continuous. (b) Now suppose that A ⊆ (Lp )+ is non-empty, upwards-directed and norm-bounded. Then B = {up : 1/p u ∈ A} is non-empty, upwards-directed and norm-bounded in L1 . So v0 = sup B is defined in L1 , and v0 p is an upper bound for A in L . (c) By 356Dd, (Lp )∗ = (Lp )∼ = (Lp )× . The extra information we need is that the identification of Lq with (LRp )∗ is an order-isomorphism. P P (α) If w ∈ (Lq )+ and u ∈ (Lp )+ then u × w ≥ 0 in L1 , so (T w)(u) = u × w ≥ 0, writing T : Lq → (Lp )∗ for the canonical bijection. As u is arbitrary, T w ≥ 0. As w is arbitrary, T is a positive linear operator. (β) If w ∈ Lq and T w ≥ 0, consider u = (w− )q/p . Then u ≥ 0 in Lp and w+ × u = 0 (because [[w+ > 0]] ∩ [[u > 0]] = [[w+ > 0]] ∩ [[w− > 0]] = 0), so 0 ≤ (T w)(u) =

R

w×u=−

R

R

w− × u = − (w− )p ≤ 0,

R and (w− )p = 0. But as (w− )p ≥ 0 in L1 , this means that (w− )p and w− must be 0, that is, w ≥ 0. As w is arbitrary, T −1 is positive and T is an order-isomorphism. Q Q (d) This is an immediate consequence of (c), since p = q/(q − 1), so that Lp can be identified with (L ) = (Lq )× . From 356M we see that it is also a consequence of (a) and (b). q ∗

366E Proposition Let (A, µ ¯) be a semi-finite measure algebra, and p ∈ [1, ∞]. Set q = p/(p − 1) if 1 < p < ∞, and q = ∞ if p = 1, q = 1 if p = ∞. Then Lq (A, µ ¯) = {u : u ∈ L0 (A), u × v ∈ L1 (A, µ ¯) for every v ∈ Lp (A, µ ¯)}. proof (a) We already know that if u ∈ Lp and v ∈ Lq then u × v ∈ L1 ; this is elementary if p ∈ {1, ∞} and otherwise is covered by 366C. (b) So suppose that u ∈ L0 \ Lp . If p = 1 then of course χ1 ∈ L∞ = Lq and u × χ1 ∈ / L1 . If p > 1 set A = {w : w ∈ S(Af ), 0 ≤ w ≤ |u|}. Because µ ¯ is semi-finite, S(Af ) is order-dense in L0 (364L), and |u| = sup A. Because the norm on Lp has the Levi property (366Db, 363Ba) and A is not bounded above in Lp , supw∈A kwkp = ∞. n q R For each n n∈ N choose wn ∈ A with kwn kp > 4 . Then there is a vn ∈ LR such that kvn kq = 1 and wn × vn ≥ 4 . P P (α) If p < ∞ this is covered by 366C, since kwn kp = sup{ wn × v : kvkq ≤ 1}. (β) If p = ∞ then [[wn > 4Rn ]] 6= 0; because µ ¯ is semi-finite, there is a b ⊆ [[wn > 4n ]] such that 0 < µ ¯b < ∞, and 1 1 n k µ¯b χbk1 = 1, while wn × µ¯b χb ≥ 4 . Q Q P∞ Because Lq is complete (363Ba, 366C), v = n=0 2−n |vn | is defined in Lq . But now

R

|u| × v ≥ 2−n

R

w n × v n ≥ 2n

for every n, so u × v ∈ / L1 . Remark This result is characteristic of perfect subspaces of L0 ; see 369C and 369J. 366F The next step is to look at the action of Boolean homomorphisms, as in 365O. It will be convenient to be able to deal with all Lp spaces at once by introducing names for a pair of spaces which include all of them.

Lp

366H

353

Definition Let (A, µ ¯) be a measure algebra. Write Mµ¯0 = M 0 (A, µ ¯) = {u : u ∈ L0 (A), µ ¯[[|u| > α]] < ∞ for every α > 0}, Mµ¯1,0 = M 1,0 (A, µ ¯) = {u : u ∈ Mµ¯0 , u × χa ∈ L1 (A, µ ¯) whenever µ ¯a < ∞}. 366G Lemma Let (A, µ ¯) be any measure algebra. Write M 0 = M 0 (A, µ ¯), etc. 0 1,0 (a) M and M are Dedekind complete solid linear subspaces of L0 which include Lp for every p ∈ [1, ∞[; moreover, M 0 is closed under multiplication. (b) If u ∈ M 0 and u ≥ 0, there is a non-decreasing sequence hun in∈N in S(Af ) such that u = supn∈N un . + (c) M 1,0 = {u : u ∈ L0 , R(|u| − ²χ1) ∈ L1 for every ² > 0} = L1 + (L∞ ∩ RM 0 ). R R 1,0 ¯a < ∞, ¯a < ∞, then u ≤ v; so if a u = a v whenever µ (d) If u, v ∈ M and a u ≤ a v whenever µ u = v. proof (a) If u, v ∈ M 0 and γ ∈ R, then for any α > 0 [[|u + v| > α]] ⊆ [[|u| > 12 α]] ∪ [[|v| > 12 α]], [[|γu| > α]] ⊆ [[|u| > [[|u × v| > α]] ⊆ [[|u| >

√

α 1+|γ| ]],

α ]] ∪ [[|v| >

√ α ]]

(364F) are of finite measure. So u + v, γu and u × v belong to M 0 . Thus M 0 is a linear subspace of L0 closed under multiplication. If u ∈ M 0 , |v| ≤ |u| and α > 0, then [[|v| > α]] ⊆ [[|u| > α]] is of finite measure; thus v ∈ M 0 and M 0 is a solid linear subspace of L0 . It follows that M 1,0 also is. If u ∈ Lp = Lp (A, µ ¯), where p < ∞, and α > 0, then [[|u| > α]] = [[|u|p > αp ]] is of finite measure, so u ∈ M 0 ; moreover, if µ ¯a < ∞, then χa ∈ Lq , where q = p/(p − 1), so u × χa ∈ L1 ; thus u ∈ M 1,0 . To see that M 0 is Dedekind complete, observe that if A ⊆ (M 0 )+ is non-empty and bounded above by u0 ∈ M 0 , and α > 0, then {[[u > α]] : u ∈ A} is bounded above by [[u0 > α]] ∈ Af , so has a supremum in A (321C). Accordingly sup A is defined in L0 (364Mc) and belongs to M 0 . Finally, M 1,0 , being a solid linear subspace of M 0 , must also be Dedekind complete. (b) If u ≥ 0 in M 0 , then there is a non-decreasing sequence hun in∈N in S = S(A) such that u = supn∈N un and u0 ≥ 0 (364Kd). But now every un belongs to S ∩ M 0 = S(Af ), just as in 365F. (c)(i) If u ∈ M 1,0 and ² > 0, then a = [[|u| > ²]] ∈ Af , so u × χa ∈ L1 = L1 (A, µ ¯); but (|u| − ²χ1)+ ≤ + 1 |u| × χa, so (|u| − ²χ1) ∈ L . (ii) Suppose that u ∈ L0 and (|u| − ²χ1)+ ∈ L1 for every ² > 0. Then, given ² > 0, v = (|u| − 21 ²χ1)+ ∈ L , and µ ¯[[v > 12 ²]] < ∞; but [[|u| > ²]] ⊆ [[v > 12 ²]], so also has finite measure. Thus u ∈ M 0 . Next, if a ∈ Af , then |u × χa| ≤ χa + (|u| − χ1)+ ∈ L1 , so u ∈ M 1,0 . 1

(iii) Of course L1 and L∞ ∩ M 0 are included in M 1,0 , so their linear sum also is. On the other hand, if u ∈ M 1,0 , then u = (u+ − χ1)+ − (u− − χ1)+ + (u+ ∧ χ1) − (u− ∧ χ1) ∈ L1 + (L∞ ∩ M 0 ). R R 0 0 (d) Take to Mµ¯1,0 , µ ¯a < ∞ and u ≤ a u0 , a R R α >0 0 and set a = [[u − u > α]]. Because both u and u belong R that is, a u − u ≤ 0; so a must be 0 (365Dc). As α is arbitrary, u − u0 ≤ 0 and u ≤ u0 . If a u = a u0 for every a ∈ Af , then u0 ≤ u so u = u0 . 366H Theorem Let (A, µ ¯) and (B, ν¯) be measure algebras. For p ∈ [1, ∞], write Lpµ¯ , Lpν¯ for Lp (A, µ ¯), 0 L (B, ν¯); similarly, write Mµ¯ for M 0 (A, µ ¯), etc. Let π : Af → Bf be a measure-preserving ring homomorphism. (a)(i) We have a unique order-continuous Riesz homomorphism T = Tπ : Mµ¯0 → Mν¯0 such that T (χa) = χ(πa) for every a ∈ Af . (ii) [[T u > α]] = π[[u > α]] for every u ∈ Mµ¯0 and α > 0. (iii) T is injective and multiplicative. p

354

Function spaces

366H

(iv)R For p ∈ [1, ∞] and u ∈ Mµ¯0 , T u ∈ Lpν¯ iff u ∈ Lpµ¯ , and in this case kT ukp = kukp . In particular, T u = u whenever u ∈ L1µ¯ . (v) For u ∈ Mµ¯0 , T u ∈ Mν¯1,0 iff u ∈ Mµ¯1,0 . (b)(i) We have a unique order-continuous positive linear operator P = Pπ : Mν¯1,0 → Mµ¯1,0 such that R R P v = πa v whenever v ∈ Mν¯1,0 and a ∈ Af . a (ii) If u ∈ Mν¯0 , v ∈ Mν¯1,0 and v × T u ∈ Mν¯1,0 , then P (v × T u) = u × P v. 0 ∞ (iii) If q ∈ [1, ∞[ and v ∈ Lqν¯ , then P v ∈ Lqµ¯ and kP vkq ≤ kvkq ; if v ∈ L∞ ν ¯ ∩ Mν ¯ , then P v ∈ Lµ ¯ and kP vk∞ ≤ kvk∞ . (iv) P T u = u for every u ∈ Mµ¯1,0 ; in particular, P [Lpν¯ ] = Lpµ¯ for every p ∈ [1, ∞[. (c) If (C, λ) is another measure algebra and θ : Bf → Cf another measure-preserving ring homomorphism, then Tθπ = Tθ Tπ : Mµ¯0 → Mλ¯0 and Pθπ = Pπ Pθ : Mλ¯1,0 → Mµ¯1,0 . (d) Now suppose that π[Af ] = Bf , so that π is a measure-preserving isomorphism between the rings Af and Bf . Then (i) T is a Riesz space isomorphism between Mµ¯0 and Mν¯0 , and its inverse is Tπ−1 . (ii) P is a Riesz space isomorphism between Mν¯1,0 and Mν¯1,0 , and its inverse is Pπ−1 . (iii) The restriction of T to Mµ¯1,0 is P −1 = Pπ−1 ; the restriction of T −1 = Tπ−1 to Mν¯1,0 is P . (iv) For any p ∈ [1, ∞[, T ¹Lpµ¯ = Pπ−1 ¹Lpµ¯ and P ¹Lpν¯ = Tπ−1 ¹Lpν¯ are the two halves of a Banach lattice isomorphism between Lpµ¯ and Lpν¯ .

R

proof (a)(i) By 361J, π induces a multiplicative Riesz homomorphism T0 : S(Af ) → S(Bf ) which is orderf continuous ) and α > 0,Pthen [[T0 u > α]] = π[[u > α]]. P P Express Pn because π is (361Ad, 361Je). If u ∈ S(A n f u as i=0 αi χai where a0 , . . . , an are disjoint in A ; then T0 u = i=0 αi χ(πai ), so [[T0 u > α]] = sup{πai : i ≤ n, αi > α} = π(sup{ai : i ≤ n, αi > α}) = π[[u > α]]. Q Q Now if u0 ≥ 0 in Mµ¯0 , sup{T0 u : u ∈ S(Af ), 0 ≤ u ≤ u0 } is defined in Mν¯0 . P P Set A = {u : u ∈ S(Af ), 0 ≤ u ≤ u0 }. Because u0 = sup A (366Gb), supu∈A [[T u > α]] = supu∈A π[[u > α]] = π(supu∈A [[u > α]]) = π[[u0 > α]] is defined and belongs to Bf for any α > 0. Also inf n≥1 supu∈A [[T u > n]] = π(inf n≥1 [[u0 > n]]) = 0. By 364Mb, v0 = sup T0 [A] is defined in L0 (B), and [[v0 > α]] = π[[u0 > α]] ∈ Bf for every α > 0, so v0 ∈ Mν¯0 , as required. Q Q Consequently T0 has a unique extension to an order-continuous Riesz homomorphism T : Mµ¯0 → Mν¯0 (355F). (ii) If u0 ∈ Mµ¯0 and α > 0, then [[T u0 > α]] = [[T u+ 0 > α]] (because T is a Riesz homomorphism) =

sup

[[T u > α]]

u∈S(Af ),0≤u≤u+ 0

(because T is order-continuous and S(Af ) is order-dense in Mµ¯0 ) = π[[u0 > α]] by the argument used in (i). (iii) I have already remarked, at the beginning of the proof of (i), that T (u × u0 ) = T u × T u0 for u, u ∈ S(Af ). Because both T and × are order-continuous and S(Af ) is order-dense in Mµ¯0 , 0

T (u0 × u1 ) = sup{T (u × u0 ) : u, u0 ∈ S(Af ), 0 ≤ u ≤ u0 , 0 ≤ u0 ≤ u1 } = sup T u × T u0 = T u0 × T u1 u,u0

Lp

366H

355

whenever u0 , u1 ≥ 0 in Mµ¯0 . Because T is linear and × is bilinear, it follows that T is multiplicative on Mµ¯0 . To see that it is injective, observe that if u 6= 0 in Mµ¯0 then there is some α > 0 such that a = [[|u| > α]] 6= 0, so that 0 < χπa ≤ T |u| = |T u| and T u 6= 0. α) Suppose that p ∈ [1, ∞[ and that u ∈ Lpµ¯ . Then for any α > 0, (iv)(α [[|T u|p > α]] = [[T |u| > α1/p ]] = π[[|u| > α1/p ]] = π[[|u|p > α]]. So k|T u|p k1 =

R∞ 0

R∞

ν¯[[|T u|p > α]] dα =

0

µ ¯[[|u|p > α]] dα = k|u|p k1 < ∞

and T u ∈ Lpν¯ , with kT ukp = kukp . β ) As for the case p = ∞, if u ∈ L∞ (β µ ¯ and γ = kuk∞ > 0 then [[|u| > γ]] = 0, so [[|T u| > γ]] = π[[|u| > γ]] = 0. This shows that kT uk∞ ≤ γ. On the other hand, if 0 < α < γ then a = [[|u| > α]] 6= 0, and χa ≤ |u| so χ(πa) ≤ |T u|; as πa 6= 0 (because ν¯(πa) = µ ¯a > 0), kT uk∞ > α. This shows that kT uk∞ = kuk∞ , at least when u 6= 0; but the case u = 0 is trivial. (γγ ) Now take any p ∈ [1, ∞], and suppose that u ∈ Mµ¯0 and that T u ∈ Lpν¯ . Let hun in∈N be a non-decreasing sequence in S(Af ) with supremum |u| and u0 ≥ 0 (366Gb). Then T un ≤ T u so kun kp = kT un kp ≤ kT ukp . But this means that hun in∈N is bounded above in Lpµ¯ (366Db), so that |u| and u belong to Lpµ¯ . (δδ ) If u ∈ L1µ¯ , then

R

T u = k(T u)+ k1 − k(T u)− k1 = kT u+ k1 − kT u− k1 = ku+ k1 − ku− k1 =

R

u.

P Set a = [[|u| > ²]] ∈ Af . Then (v) If u ∈ Mµ¯1,0 and ² > 0, then T (|u| ∧ ²χ1A ) = |T u| ∧ ²χ1B . P |u| ∧ ²χ1A = ²χa + |u| − |u| × χa and [[|T u| > ²]] = πa. So T (|u| ∧ ²χ1A ) = T (²χa) + T |u| − T (|u| × χa) = ²χ(πa) + |T u| − |T u| × χ(πa) = |T u| ∧ ²χ1B . Q Q Consequently T (|u| − ²χ1A )+ = T (|u| − |u| ∧ ²χ1A ) = (|T u| − ²χ1B )+ . But this means that (|u| − ²χ1A )+ ∈ L1µ¯ iff (|T u| − ²χ1B )+ ∈ L1ν¯ . Since this is true for every ² > 0, 366Gc tells us that u ∈ Mµ¯1,0 iff T u ∈ Mν¯1,0 . α) By 365Pa, we have an order-continuous positive linear operator P0 : L1ν¯ → L1µ¯ such that (b)(i)(α R P v = πa v for every v ∈ L1ν¯ and a ∈ Af . a 0

R

β ) We now find that if v0 ≥ 0 in Mν¯1,0 and B = {v : v ∈ L1ν¯ , 0 ≤ v ≤ v0 }, then P0 [B] has a (β P Because B is upwards-directed and P0 is order-preserving, supremum in L0 (A) which belongs to Mµ¯1,0 . P P0 [B] is upwards-directed. If α > 0 and v ∈ B and a = [[P0 v > α]], then α 2

v ≤ (v0 − α2 χ1B )+ + χ1B , so Z

Z

α¯ µa ≤

P0 v = Z

=

a

Z

α 2

πa

α 2

α 2

(v0 − χ1B )+ + ν¯(πa)

v≤

α 2

(v0 − χ1B )+ + µ ¯a

and µ ¯[[P0 v > α]] ≤

2 α

R

(v0 − α2 χ1B )+ .

Thus {[[P0 v > α]] : v ∈ B} is an upwards-directed set in Af with measures bounded above in R, and cα = supv∈B [[P0 v > α]] is defined in Af . Also

356

Function spaces

inf n≥1 cn ≤ inf n≥1

2 n

R

366H

(v0 − n2 χ1B )+ = 0.

So, by 364Mb, P0 [B] has a supremum u0 ∈ L0 (A), and [[u0 > α]] = cα ∈ Af for every α > 0, so u0 ∈ Mµ¯0 . If c ∈ Af , then

R

c

u0 = supv∈B

R

c

P0 v = supv∈B

R

πc

v≤

R

πc

v0 < ∞,

so u0 ∈ Mµ¯1,0 . Q Q (γγ ) Now 355F tells us that P0 has a unique extension to an order-continuous positive linear operator P : Mν¯1,0 → Mµ¯1,0 . If v0 ≥ 0 in Mν¯1,0 and a ∈ Af , then, as remarked above,

R

R

R

P v0 = sup{ a P0 v : v ∈ L1ν¯ , 0 ≤ v ≤ v0 } = sup{ πa v : v ∈ L1ν¯ , 0 ≤ v ≤ v0 } = a R R because P is linear, a P v = πa v for every v ∈ Mν¯1,0 , a ∈ Af .

R

πa

v0 ;

(δδ ) By 366Gd, P is uniquely defined by the formula

R

R

Pv = a

πa

v whenever v ∈ Mν¯1,0 , a ∈ Af .

(ii) Because Mµ¯0 is closed under multiplication, u × P v certainly belongs to Mµ¯0 . α) Suppose (α that u, v ≥ 0. Fix c ∈ Af for the moment. Suppose that u0 ∈ S(Af ). Then we can Pn 0 express u as i=0 αi χai where ai ∈ Af for every i ≤ n. Accordingly

R

u0 × P v = c

Pn

α i=0 i

R

c∩ai

Pv =

Pn

α i=0 i

R

v × χ(πai ) × χ(πc) =

R

πc

v × T u0 .

Next, we can find a non-decreasing sequence hun in∈N in S(Af )+ with supremum u, and Z Z Z sup un × P v = sup v × T un = sup v × T un n∈N c n∈N πc n∈N Z Zπc = v × sup T un = v × T u, n∈N

πc

R

πc

using the order-continuity of T , and ×. But this means that u × P v = supn∈N un × P v is integrable over R R c and that c u × P v = πc v × T u. As c is arbitrary, u × P v = P (v × T u) ∈ Mµ¯1,0 . β ) For general u, v, (β v + × T u+ + v + × T u− + v − × T u+ + v − × T u− = |v| × T |u| = |v × T u| ∈ Mν¯1,0 (because T is a Riesz homomorphism), so we may apply (α) to each of the four products; combining them, we get P (v × T u) = u × P v, as required. (iii) Because P is a positive operator, we surely have |P v| ≤ P |v|, so it will be enough to show that kP vkq ≤ kvkq for v ≥ 0 in Lqν¯ . R R α) I take the case q = 1 first. (α In this case, for any a ∈ Af , we have a P v = πa v ≤ kvk1 . In R particular, setting an = [[P v > 2−n ]], an P v ≤ kvk1 . But P v = supn∈N P v × χan , so kP vk1 = supn∈N

then

R

an

P v ≤ kvk1 .

+ β ) Next, suppose that q = ∞, so that v ∈ (L∞ (β ? If γ > 0 and a = [[P v > γ]] 6= 0, ν ¯ ) ; say kvk∞ = γ. ?

γµ ¯a
γ]] = 0 and P v ∈ L∞ µ ¯ , with kP vk∞ ≤ kvk∞ , at least when kvk∞ > 0; but the case kvk∞ = 0 is trivial. (γγ ) I come at last to the ‘general’ case q ∈ ]1, ∞[, v ∈ Lqν¯ . In this case set p = q/(q − 1). If u ∈ Lpµ¯ then T u ∈ Lpν¯ so T u × v ∈ L1ν¯ and

Lp

366J

357

Z |

u × P v| ≤ ku × P vk1 = kP (T u × v)k1

(by (ii)) ≤ kT u × vk1 (by (α) just above)

Z =

|T u| × |v| ≤ kT ukp kvkq = kukp kvkq R by (a-iii) of this theorem. But this means that u 7→ u × P v is a bounded linear functional on Lpµ¯R, and R q p is therefore represented by some w ∈ Lµ¯ with kwkq ≤ kvkq . If a ∈ Af then χa ∈ Lµ¯ , so a w = a P v; accordingly P v is actually equal to w (by 366Gd) and kP vkq = kwkq ≤ kvkq , as claimed. (iv) If u ∈ Mµ¯1,0 and a ∈ Af , we must have

R

R

R

R

R

R

P T u = πa T u = T (χa) × T u = T (χa × u) = χa × u = a u, a R R using (a-iv) to see that χa × u is defined and equal to T (χa × u). As a is arbitrary, u ∈ Mµ¯1,0 and P T u = u. (c) As usual, in view of the uniqueness of Tθπ and Pθπ , all we have to check is that Tθ T (χa) = Tθ χ(πa) = χ(θπa) = Tθπ (χa),

R

P Pθ w = a

R

P w= πa θ

R

w= θπa

R a

Pθπ w

whenever a ∈ Af , w ∈ Mλ¯1,0 . (d)(i) By (c), Tπ−1 T = Tπ−1 π must be the identity operator on Mµ¯0 ; similarly, T Tπ−1 is the identity operator on Mν¯0 . Because T and Tπ−1 are Riesz homomorphisms, they must be the two halves of a Riesz space isomorphism. (ii) In the same way, P and Pπ−1 must be the two halves of an ordered linear space isomorphism between Mµ¯1,0 and Mν¯1,0 , and are therefore both Riesz homomorphisms. (iii) By (b-iv), P T u = u for every u ∈ Mµ¯1,0 , so T ¹Mµ¯1,0 must be P −1 . Similarly P = Pπ−1 −1 is the restriction of T −1 = Tπ−1 to Mν¯1,0 . (iv) Because T −1 [Lpν¯ ] = Lpµ¯ (by (a-iv)), and T is a bijection between Mµ¯0 and Mν¯0 , T ¹Lpµ¯ must be a Riesz space isomorphism between Lpµ¯ and Lpν¯ ; (a-iv) also tells us that it is norm-preserving. Now its inverse is P ¹Lpν¯ , by (iii) here. 366I Corollary Let (A, µ ¯) be a measure algebra, and B a σ-subalgebra of A. Then, for any p ∈ [1, ∞[, Lp (B, µ ¯¹ B) can be identified, as Banach lattice, with the closed linear subspace of Lp (A, µ ¯) generated by {χb : b ∈ B, µ ¯b < ∞}. proof The identity map b 7→ b : B → A induces an injective Riesz homomorphism T : L0 (B) → L0 (A) (364R) such that T u ∈ LpA = Lp (A, µ ¯) and kT ukp = kukp whenever p ∈ [1, ∞[ and u ∈ LpB = Lp (B, µ ¯¹ B) f (366H(a-iv)). Because S(B ), the linear span of {χb : b ∈ B, µ ¯b < ∞}, is dense in LpB (366C), the image of LpB in LpA must be the closure of the image of S(Bf ) in LpA , that is, the closed linear span of {χb : b ∈ Bf } interpreted as a subset of LpA . 366J Corollary If (A, µ ¯) is a probability algebra, B is a closed subalgebra of A, and P : L1 (A, µ ¯) → L1 (B, µ ¯¹ B) is the conditional expectation operator (365R), then kP ukp ≤ kukp whenever p ∈ [1, ∞] and u ∈ Lp (A, µ ¯). proof Because (A, µ ¯) is totally finite, M 0 (A, µ ¯) = L1 (A, µ ¯), so that the operator P of 366Hb can be identified with the conditional expectation operator of 365R. Now 366H(b-iii) gives the result. Remark Of course this is also covered by 244M.

358

Function spaces

366K

366K Corollary Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : Af → Bf a measure-preserving ring homomorphism. Let T : L2µ¯ → L2ν¯ and P : L2ν¯ → L2µ¯ be the corresponding operators, as in 366H. Then T P : L2ν¯ → L2ν¯ is an orthogonal projection, its range T P [L2ν¯ ] being isomorphic, as Banach lattice, to L2µ¯ . The kernel of T P is just

R

{v : v ∈ L2ν¯ ,

πa

v = 0 for every a ∈ Af }.

proof Most of this is simply because T is a norm-preserving Riesz homomorphism (so that T [L2µ¯ ] is isomorphic to L2µ¯ ), P T is the identity on L2µ¯ (so that (T P )2 = T P ) and kP k ≤ 1 (so that kT P k ≤ 1). These are enough to ensure that T P is a projection of norm at most 1, that is, an orthogonal projection. Also Z P v = 0 for every a ∈ Af T P v = 0 ⇐⇒ P v = 0 ⇐⇒ a Z ⇐⇒ v = 0 for every a ∈ Af . πa

366L Corollary Let (A, µ ¯) be a measure algebra, and π : Af → Af a measure-preserving ring automorphism. Then there is a corresponding Banach lattice isomorphism T of L2 = L2 (A, µ ¯) defined by writing T (χa) = χ(πa) for every a ∈ Af . Its inverse is defined by the formula

R

T −1 u = a

R

πa

u for every u ∈ L2 , a ∈ Af .

proof In the language of 366H, T = Tπ and T −1 = P . 366X Basic exercises (a) Let (A, µ ¯) be a measure algebra and p ∈ ]1, ∞[. Show that, for u ∈ L0 (A), R ∞ p−1 p u ∈ L (A, µ ¯) iff γ = p 0 α µ ¯[[|u| > α]] dα is finite, and that in this case kukp = γ 1/p . (Cf. 263Xa.) > (b) Let (A, µ ¯) be a localizable measure algebra and p ∈ [1, ∞]. Show that the band algebra of Lp (A, µ ¯) is isomorphic to A. (Cf. 365S.) (c) Let (A, µ ¯) be a measure algebra and p ∈ ]1, ∞[. Show that Lp (A, µ ¯) is separable iff L1 (A, µ ¯) is. (d) Let (A, µ ¯) be a measure algebra. (i) Show that L∞ (A) ∩ M 0 (A, µ ¯) and L∞ (A) ∩ M 1,0 (A, µ ¯), as defined ∞,0 in 366G, are equal. (ii) Call this intersection M (A, µ ¯). Show that it is a norm-closed solid linear subspace of L∞ (A), therefore a Banach lattice in its own right. b µ (e) Let (A, µ ¯) be a semi-finite measure algebra and (A, ˆ) its localization (322P). Show that the natural b b µ embedding of A in A induces a Banach lattice isomorphism between Lp (A, µ ¯) and Lp (A, ˆ) for every p ∈ b [1, ∞[, so that the band algebra of Lp (A, µ ¯) can be identified with A. b µ (f ) Let (A, µ ¯) be a semi-finite measure algebra which is not localizable (cf. 211Ya, 216D), and (A, ˆ) its b be the identity embedding, so that π is an order-continuous measure-preserving localization. Let π : A → A ∞ b RBooleanR homomorphism. Show that if we set v = χb where b ∈ A \ A, then there is no u ∈ L (A) such that u = v whenever µ ¯ a < ∞. a πa (g) In 366H, show that [[T u ∈ E]] = π[[u ∈ E]] whenever u ∈ Mµ¯0 and E ⊆ R is a Borel set such that 0 ∈ / E. > (h) Let (A, µ ¯) be a measure algebra and let G be the group of all measure-preserving ring automorphisms of Af . Let H be the group of all Banach lattice automorphisms of L2 (A, µ ¯). Show that the map π 7→ T of 366L is an injective group homomorphism from G to H, so that G is represented as a subgroup of H. (i) Let h(Ai , µ ¯i )ii∈I be any family of measure algebras, with simple product (A, µ ¯) (322K). Show that for any p ∈ [1, ∞[, Lp (A, µ ¯) can be identified, as normed Riesz space, with the solid linear subspace ¡P ¢ p 1/p < ∞} {u : kuk = i∈I ku(i)k Q of i∈I Lp (Ai , µ ¯i ).

366 Notes

Lp

359

(j) Let A be a Dedekind σ-complete Boolean algebra and µ ¯, ν¯ two functionals rendering A a semi-finite measure algebra. Show that for any p ∈ [1, ∞[, Lp (A, µ ¯) and Lp (A, ν¯) are isomorphic as normed Riesz spaces. (Hint: use 366Xe to reduce to the case in which A is Dedekind complete. Take w ∈ L0 (A) such that R w d¯ µ = ν¯a for every a ∈ A (365T). Set T u = w1/p × u for u ∈ Lp (A, µ ¯).) a (k) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras, and p ∈ [1, ∞[. Show that the following are equiveridical: (i) Lpµ¯ and Lpν¯ are isomorphic as Banach lattices; (ii) Lpµ¯ and Lpν¯ are isomorphic as Riesz spaces; (iii) A and B have isomorphic Dedekind completions. 366Y Further exercises (a) Let (A, µ ¯) be a measure algebra and suppose that 0 < p < 1. Write R Lp = Lp (A, µ ¯) for {u : u ∈ L0 (A), |u|p ∈ L1 (A, µ ¯)}, and for u ∈ Lp set τ (u) = |u|p . (i) Show that τ defines a Hausdorff linear space topology on Lp (see 2A5B). (ii) Show that if A ⊆ Lp is non-empty, downwardsdirected and has infimum 0 then inf u∈A τ (u) = 0. (iii) Show that if A ⊆ Lp is non-empty, upwards-directed and bounded in the linear topological space sense (see 245Yf) then A is bounded above. (iv) Show that (Lp )∼ = (Lp )× is just the set of continuous linear functionals from Lp to R, and is {0} iff A is atomless. (b) Let (A, µ ¯) be a measure algebra. Show that M 0 (A, µ ¯) has the countable sup property. (c) Let (A, µ ¯) be a measure algebra and define M ∞,0 (A, µ ¯) as in 366Xd. Show that (M ∞,0 (A, µ ¯))× can 1 be identified with L (A, µ ¯). (d) In 366H, show that if T˜ : M 0 (A, µ ¯) → M 0 (B, ν¯) is any positive linear operator such that T˜(χa) = χ(πa) for every a ∈ Af , then T˜ is order-continuous, so that it is equal to Tπ . (e) Let (A, µ ¯) be a measure algebra. (i) Show that there is a natural one-to-one correspondence between M 1,0 (A, µ ¯) and the set of additive functionals ν : Af → R such that ν 0 there are δ, M > 0 such that |νa| ≤ ² whenever µa ≤ δ and |νa| ≤ ²µa whenever µa ≥ M . (ii) Use this description of M 1,0 to prove 366H(b-i). (f ) In 366H, show that the following are equiveridical: (α) π[Af ] = Bf ; (β) T = Tπ is surjective; (γ) P = Pπ is injective; (δ) P is a Riesz homomorphism; (²) there is some q ∈ [1, ∞] such that kP vkq = kvkq for every v ∈ Lqν¯ ; (ζ) T P v = v for every v ∈ Mν¯1,0 . (g) Let (A, µ ¯) and (B, ν¯) be measure algebras, and suppose that π : Af → Bf is a measure-preserving ring homomorphism, as in 366H; let T : M 0 (A, µ ¯) → M 0 (B, ν¯) be the associated linear operator. Show that p 0 if 0 < p < 1 (as in 366Ya) then L (A, µ ¯) ⊆ M (A, µ ¯) and T −1 [Lp (B, ν¯)] = Lp (A, µ ¯). (h) Let (A, µ ¯) be a totally finite measure algebra. (i) For each Boolean automorphism π : A → A, let 0 0 T : L (A) → L (A) be the associated Riesz space isomorphism (364R), and let wπ ∈ L1 (A)+ be such that π R √ −1 w = µ(π a) for every a ∈ A (365Ea). Set Qπ u = Tπ u × wπ for u ∈ L0 (A). Show that kQπ uk2 = kuk2 a π for every u ∈ L2 (A). (ii) Show that if π, φ : A → A are Boolean automorphisms then Qπφ = Qπ Qφ . 366 Notes and comments The Lp spaces, for 1 ≤ p ≤ ∞, constitute the most important family of leading examples for the theory of Banach lattices, and it is not to be wondered at that their properties reflect a wide variety of general results. Thus 366Dd and 366E can both be regarded as special cases of theorems about perfect Riesz spaces (356M and 369D). In a different direction, the concept of ‘Orlicz space’ (369Xd below) generalizes the Lp spaces if they are regarded as normed subspaces of L0 invariant under measure-preserving automorphisms of the underlying algebra. Yet another generalization looks at the (non-locally-convex) spaces Lp for 0 < p < 1 (366Ya). In 366H and its associated results I try to emphasize the way in which measure-preserving homomorphisms of the underlying algebras induce both ‘direct’ and ‘dual’ operators on Lp spaces. We have already seen the phenomenon in 365P. I express this in a slightly different form in 366H, noting that we really do need the homomorphisms to be measure-preserving, for the dual operators as well as the direct operators, so we no longer have the shift in the hypotheses which appears between 365O and 365P. Of course all these refinements in the hypotheses are irrelevant to the principal applications of the results, and they make

360

Function spaces

366 Notes

substantial demands on the reader; but I believe that the demands are actually demands to expand one’s imagination, to encompass the different ways in which the spaces depend on the underlying measure algebras. In the context of 366H, L∞ is set apart from the other Lp spaces, because L∞ (A) is not in general determined by the ideal Af , and the hypotheses of 366H do not look outside Af . 366H(a-iv) and 366H(b-iii) reach only the space M ∞,0 as defined in 366Xd. To deal with L∞ we need slightly stronger hypotheses. If we are given a measure-preserving Boolean homomorphism from A to B, rather than from Af to Bf , then of course the direct operator T has a natural version acting on L∞ (A) and indeed on M 1,∞ (A, µ ¯), as in 363F and 369Xm. If we know that (A, µ ¯) is localizable, then A can be recovered from Af , and the dual operator P acts on L∞ (B), as in 369Xm. But in general we can’t expect this to work (366Xf). Of course 366H can be applied to many other spaces; for reasons which will appear in §§371 and 374, the archetypes are not really Lp spaces at all, but the spaces M 1,0 (366F) and M 1,∞ (369N). I include 366L and 366Yh as pointers to one of the important applications of these ideas: the investigation of properties of a measure-preserving homomorphism in terms of its action on Lp spaces. The case p = 2 is the most useful because the group of unitary operators (that is, the normed space automorphisms) of L2 has been studied intensively.

367 Convergence in measure Continuing through the ideas of Chapter 24, I come to ‘convergence in measure’. The basic results of §245 all translate easily into the new language (367M-367N, 367Q). The associated concept of (sequential) order-convergence can also be expressed in abstract terms (367A), and I take the trouble to do this in the context of general lattices (367A-367B), since the concept can be applied in many ways (367C-367F, 367L, 367Xa-367Xm). In the particular case of L0 spaces, which are the first aim of this section, the idea is most naturally expressed by 367G. It enables us to express some of the fundamental theorems from Volumes 1 and 2 in the language of this chapter (367J-367K). In 367O and 367P I give two of the most characteristic properties of the topology of convergence in measure on L0 ; it is one of the fundamental types of topological Riesz space. Another striking fact is the way it is determined by the Riesz space structure (367T). In 367U I set out a theorem which is the basis of many remarkable applications of the concept; for the sake of a result in §369 I give one such application (367V). 367A Order*-convergence As I have remarked before, the function spaces of measure theory have three interdependent structures: they are linear spaces, they have a variety of interesting topologies, and they are ordered spaces. Ordinary elementary functional analysis studies interactions between topologies and linear structures, in the theory of normed spaces and, more generally, of linear topological spaces. Chapter 35 in this volume looked at interactions between linear and order structures. It is natural to seek to complete the triangle with a theory of topological ordered spaces. The relative obscurity of any such theory is in part due to the difficulty of finding convincing definitions; that is, isolating concepts which lead to in elegant and useful general theorems. Among the many rival ideas, however, I believe it is possible to identify one which is particularly important in the context of measure theory. In its natural home in the theory of L0 spaces, this notion of ‘order*-convergence’ has a very straightforward expression (367G). But, suitably interpreted, the same idea can be applied in other contexts, some of which will be very useful to us, and I therefore begin with a definition which is applicable in any lattice. Definition Let P be a lattice and hpn in∈N a sequence in P , p an element of P . I will say that hpn in∈N order*-converges to p, or that p is the order*-limit of hpn in∈N , if p = inf{q : ∃ n ∈ N, q ≥ (p0 ∨ pi ) ∧ p00 ∀ i ≥ n} = sup{q : ∃ n ∈ N, q ≤ p0 ∨ (pi ∧ p00 ) ∀ i ≥ n} whenever p0 ≤ p ≤ p00 in P .

367B

Convergence in measure

361

367B Lemma Let P be a lattice. (a) A sequence in P can order*-converge to at most one point. (b) A constant sequence order*-converges to its constant value. (c) Any subsequence of an order*-convergent sequence is order*-convergent, with the same limit. (d) If hpn in∈N and hp0n in∈N both order*-converge to p, and pn ≤ qn ≤ p0n for every n, then hqn in∈N order*-converges to p. (e) If hpn in∈N is an order-bounded sequence in P , then it order*-converges to p ∈ P iff p = inf{q : ∃ n ∈ N, q ≥ pi ∀ i ≥ n} = sup{q : ∃ n ∈ N, q ≥ pi ∀ i ≥ n}. (f) If P is a Dedekind σ-complete lattice (314A) and hpn in∈N is an order-bounded sequence in P , then it order*-converges to p ∈ P iff p = supn∈N inf i≥n pi = inf n∈N supi≥n pi . proof (a) Suppose that hpn in∈N is order*-convergent to both p and p˜. Set p0 = p ∧ p˜, p00 = p ∨ p˜; then p = inf{q : ∃ n ∈ N, q ≥ (p0 ∨ pi ) ∧ p00 ∀ i ≥ n} = p˜. (b) is trivial. (c) Suppose that hpn in∈N is order*-convergent to p, and that hp0n in∈N is a subsequence of hpn in∈N . Take p , p00 such that p0 ≤ p ≤ p00 , and set 0

B = {q : ∃ n ∈ N, q ≤ p0 ∨ (pi ∧ p00 ) ∀ i ≥ n}, B 0 = {q : ∃ n ∈ N, q ≤ p0 ∨ (p0i ∧ p00 ) ∀ i ≥ n}, C = {q : ∃ n ∈ N, q ≥ (p0 ∨ pi ) ∧ p00 ∀ i ≥ n}, C 0 = {q : ∃ n ∈ N, q ≥ (p0 ∨ p0i ) ∧ p00 ∀ i ≥ n}. If q ∈ B 0 and q 0 ∈ C, then for all sufficiently large i q ≤ p0 ∨ (p0i ∧ p00 ) ≤ (p0 ∨ p0i ) ∧ p00 ≤ q 0 . As p = inf C, we must have q ≤ p; thus p is an upper bound for B 0 . On the other hand, {p0i : i ≥ n} ⊆ {pi : i ≥ n} for every n, so B ⊆ B 0 and p must be the least upper bound of B 0 , since p = sup B. Similarly, p = inf C 0 . As p0 and p00 are arbitrary, hp0n in∈N order*-converges to p. (d) Take p0 , p00 such that p0 ≤ p ≤ p00 , and set B = {q : ∃ n ∈ N, q ≤ p0 ∨ (pi ∧ p00 ) ∀ i ≥ n}, B 0 = {q : ∃ n ∈ N, q ≤ p0 ∨ (qi ∧ p00 ) ∀ i ≥ n}, C = {q : ∃ n ∈ N, q ≥ (p0 ∨ p0i ) ∧ p00 ∀ i ≥ n}, C 0 = {q : ∃ n ∈ N, q ≥ (p0 ∨ qi ) ∧ p00 ∀ i ≥ n}. If q ∈ B 0 and q 0 ∈ C, then for all sufficiently large i q ≤ p0 ∨ (qi ∧ p00 ) ≤ (p0 ∨ p0i ) ∧ p00 ≤ q 0 . As p = inf C, we must have q ≤ p; thus p is an upper bound for B 0 . On the other hand, p0 ∨ (pi ∧ p00 ) ≤ p0 ∨ (qi ∧ p00 ) for every i, so B ⊆ B 0 and p = sup B 0 . Similarly, p = inf C 0 . As p0 and p00 are arbitrary, hqn in∈N order*-converges to p. (e) Set B = {q : ∃ n ∈ N, q ≤ pi ∀ i ≥ n}, C = {q : ∃ n ∈ N, q ≥ pi ∀ i ≥ n}.

362

Function spaces

367B

(i) Suppose that hpn in∈N order*-converges to p. Let p0 , p00 be such that p0 ≤ pn ≤ p00 for every n ∈ N and p0 ≤ p ≤ p00 . Then B = {q : ∃ n ∈ N, q ≤ p0 ∨ (pi ∧ p00 ) ∀ i ≥ n}, so sup B = p. Similarly, inf C = p, so the condition is satisfied. (ii) Suppose that sup B = inf C = p. Take any p0 , p00 such that p0 ≤ p ≤ p00 and set B 0 = {q : ∃ n ∈ N, q ≤ p0 ∨ (pi ∧ p00 ) ∀ i ≥ n}, C 0 = {q : ∃ n ∈ N, q ≥ (p0 ∨ pi ) ∧ p00 ∀ i ≥ n}. If q ∈ B 0 and q 0 ∈ C, then for all large enough i q ≤ p0 ∨ (pi ∧ p00 ) ≤ p0 ∨ q 0 = q 0 because p ≤ q 0 . As inf C = p, p is an upper bound for B 0 . On the other hand, if q ∈ B, then q ≤ p, so q ≤ p0 ∨ (pi ∧ p00 ) whenever q ≤ pi , which is so for all sufficiently large i, and q ∈ B 0 . Thus B 0 ⊇ B and p must be the supremum of B 0 . Similarly, p = inf C 0 ; as p0 and p00 are arbitrary, hpn in∈N order*-converges to p. (f ) This follows at once from (e). Setting B = {q : ∃ n ∈ N, q ≤ pi ∀ i ≥ n}, B 0 = {inf i≥n pi : i ∈ N}, then B 0 ⊆ B and for every q ∈ B there is a q 0 ∈ B 0 such that q ≤ q 0 ; so sup B = sup B 0 if either is defined. Similarly, inf{q : ∃ n ∈ N, q ≥ pi ∀ i ≥ n} = inf n∈N supi≥n pi if either is defined. 367C Proposition Let U be a Riesz space and hun in∈N , hvn in∈N two sequences in U order*-converging to u, v respectively. (a) If w ∈ U , hun + win∈N order*-converges to u + w, and αun order*-converges to αu for every α ∈ R. (b) hun ∨ vn in∈N order*-converges to u ∨ v. (c) If hwn in∈N is any sequence in U , then it order*-converges to w ∈ U iff h|wn − w|in∈N order*-converges to 0. (d) hun + vn in∈N order*-converges to u + v. (e) If U is Archimedean, and hαn in∈N is a sequence in R converging to α ∈ R, then hαn un in∈N order*converges to αu. (f) Again suppose that U is Archimedean. Then a sequence hwn in∈N in U + is not order*-convergent to 0 iff there is a w ˜ > 0 such that w ˜ = supi≥n w ˜ ∧ wi for every n ∈ N. proof (a)(i) hun + win∈N order*-converges to u + w because the ordering of U is translation-invariant; the map w0 7→ w0 + w is an order-isomorphism. α) If α > 0, then the map w0 7→ αw0 is an order-isomorphism, so hαun in∈N order*-converges to (ii)(α αu. β ) If α = 0 then hαun in∈N order*-converges to αu = 0 by 367Bb. (β (γγ ) If w0 ≤ −u ≤ w00 then −w00 ≤ u ≤ w0 so u = inf{w : ∃ n ∈ N, w ≥ ((−w00 ) ∨ ui ) ∧ (−w0 ) ∀ i ≥ n} = sup{w : ∃ n ∈ N, w ≤ (−w00 ) ∨ (ui ∧ (−w0 )) ∀ i ≥ n}. Turning these formulae upside down, −u = sup{w : ∃ n ∈ N, w ≤ (w00 ∧ (−ui )) ∨ w0 ∀ i ≥ n} = inf{w : ∃ n ∈ N, w ≥ w00 ∧ ((−ui ) ∨ w0 ) ∀ i ≥ n}.

367C

Convergence in measure

363

As w0 and w00 are arbitrary, h−un in∈N order*-converges to −u. (δδ ) Putting (α) and (γ) together, hαun in∈N order*-converges to αu for every α < 0. (b) Suppose that w0 ≤ u ∨ v ≤ w00 . Set B = {w : ∃ n ∈ N, w ≤ w0 ∨ ((ui ∨ vi ) ∧ w00 ) ∀ i ≥ n}, C = {w : ∃ n ∈ N, w ≥ (w0 ∨ (ui ∨ vi )) ∧ w00 ∀ i ≥ n}, B1 = {w : ∃ n ∈ N, w ≤ (w0 ∧ u) ∨ (ui ∧ w00 ) ∀ i ≥ n}, B2 = {w : ∃ n ∈ N, w ≤ (w0 ∧ v) ∨ (vi ∧ w00 ) ∀ i ≥ n}, C1 = {w : ∃ n ∈ N, w ≥ ((w0 ∧ u) ∨ ui ) ∧ w00 ∀ i ≥ n}, C2 = {w : ∃ n ∈ N, w ≥ ((w0 ∧ v) ∨ vi ) ∧ w00 ∀ i ≥ n}, If w1 ∈ B1 and w2 ∈ B2 then w1 ∨ w2 ∈ B. P P There is an n ∈ N such that w1 ≤ (w0 ∧ u) ∨ (ui ∧ w00 ) for every i ≥ n, while w2 ≤ (w0 ∧ v) ∨ (vi ∧ w00 ) for every i ≥ n. So w1 ∨ w2 ≤ (w0 ∧ u) ∨ (w0 ∧ v) ∨ (ui ∧ w00 ) ∨ (vi ∧ w00 ) = (w0 ∧ (u ∨ v)) ∨ ((ui ∨ vi ) ∧ w00 ) (by the distributive law 352Ec) = w0 ∨ ((ui ∨ vi ) ∧ w00 ) for every i ≥ n, and w1 ∨ w2 ∈ B. Q Q Similarly, if w1 ∈ C1 and w2 ∈ C2 then w1 ∨w2 ∈ C. P P There is an n ∈ N such that w1 ≥ ((w0 ∧u)∨ui )∧w00 , 0 00 w2 ≥ ((w ∧ v) ∨ vi ) ∧ w for every i ≥ n. So w1 ∨ w2 ≥ (((w0 ∧ u) ∨ ui ) ∧ w00 ) ∨ (((w0 ∧ v) ∨ vi ) ∧ w00 ) = ((w0 ∧ u) ∨ ui ∨ (w0 ∧ v) ∨ vi ) ∧ w00 = ((w0 ∧ (u ∨ v)) ∨ (ui ∨ vi )) ∧ w00 = (w0 ∨ (ui ∨ vi )) ∧ w00 for every i ≥ n, so w1 ∨ w2 ∈ C. Q Q At the same time, of course, w ≤ w ˜ whenever w ∈ B, w ˜ ∈ C, since there is some i ∈ N such that w ≤ w0 ∨ ((ui ∨ vi ) ∧ w00 ) ≤ (w0 ∨ (ui ∨ vi )) ∧ w00 ≤ w. ˜ Since sup{w1 ∨ w2 : w1 ∈ B1 , w2 ∈ B2 } = (sup B1 ) ∨ (sup B2 ) = u ∨ v, inf{w1 ∨ w2 : w1 ∈ C1 , w2 ∈ C2 } = (inf C1 ) ∨ (inf C2 ) = u ∨ v (using the generalized distributive laws in 352E), we must have sup B = inf C = u ∨ v. As w0 and w00 are arbitrary, hun ∨ vn in∈N is order*-convergent to u ∨ v. (c) The hard parts are over. (i) If hwn in∈N order*-converges to w, then hwn − win∈N , hw − wn in∈N and h|wn − w|in∈N = h(wn − w) ∨ (w − wn )in∈N all order*-converge to 0, putting (a) and (b) together. (ii) If h|wn − w|in∈N order*-converges to 0, then so do h−|wn − w|in∈N and hwn − win∈N , by (a) and 367Bd; so hwn in∈N order*-converges to 0, by (a) again. (d) h|un − u|in∈N and h|vn − v|in∈N order*-converge to 0, by (c), so h2(|un − u| ∨ |vn − v|)in∈N also order*-converges to 0, by (b) and (a). But 0 ≤ |(un + vn ) − (u + v)| ≤ |un − u| + |vn − v| ≤ 2(|un − u| ∨ |vn − v|) for every n, so h|(un + vn ) − (u + v)|in∈N order*-converges to 0, by 367Bb and 367Bd, and hun + vn in∈N order*-converges to u + v.

364

Function spaces

367C

(e) Set βn = supi≥n |αi −α| for each n. Then hβn in∈N → 0, so inf n∈N βn |u| = 0, because U is Archimedean. Consequently hβn |u|in∈N order*-converges to 0, by 367Be. But we also have β0 |un − u| order*-converging to 0, by (c) and (a), so hβ0 |un −u|+βn |u|in∈N order*-converges to 0, by (d). As |αn un −αu| ≤ β0 |un −u|+βn |u| for every n, hαn un in∈N order*-converges to αu, as required. (f )(i) Suppose that hwn in∈N is not order*-convergent to 0. Then there are w0 , w00 such that w0 ≤ 0 ≤ w00 and either B = {w : ∃ n ∈ N, w ≤ w0 ∨ (wi ∧ w00 ) ∀ i ≥ n} does not have supremum 0, or C = {w : ∃ n ∈ N, w ≥ (w0 ∨ wi ) ∧ w00 ∀ i ≥ n} does not have infimum 0. Now 0 ∈ B, because every wi ≥ 0, and every member of B is a lower bound for C; so 0 cannot be the greatest lower bound of C. Let w ˜ > 0 be a lower bound for C. Let n ∈ N, and set Cn = {w : w ≥ (w0 ∨ wi ) ∧ w00 ∀ i ≥ n} = {w : w ≥ wi ∧ w00 ∀ i ≥ n}. Because U is Archimedean, we know that inf(Cn − An ) = 0, where An = {wi ∧ w00 : i ≥ n} (353F). Now w ˜ is a lower bound for Cn , so inf (w ˜ − wi )+ ≤ inf{(w − wi )+ : w ∈ C, i ≥ n}

i≥n

≤ inf{(w − (wi ∧ w00 ))+ : w ∈ C, i ≥ n} = inf{w − (wi ∧ w00 ) : w ∈ C, i ≥ n} = inf(Cn − An ) = 0. As this is true for every n ∈ N, w ˜ has the property declared. (ii) If w ˜ > 0 is such that w ˜ = supi≥n w ˜ ∧ wi for every n ∈ N, then {w : ∃ n ∈ N, w ≥ (0 ∨ wi ) ∧ w ˜ ∀ i ≥ n} cannot have infimum 0, and hwn in∈N is not order*-convergent to 0. 367D

As an example of the use of this concept in a relatively abstract setting, I offer the following.

Proposition Let U be a Banach lattice and hun in∈N a sequence in U which is norm-convergent to u ∈ U . Then hun in∈N has a subsequence which is order-bounded and order*-convergent to u. So if hun in∈N itself is order*-convergent, its order*-limit is u. proof Let hu0n in∈N be a subsequence of hun in∈N such that ku0n − uk ≤ 2−n for each n ∈ N. Then vn = supi≥n |u0i − u| is defined in U , and kvn k ≤ 2−n+1 , for each n (354C). Because inf n∈N kvn k = 0, inf n∈N vn must be 0, while u − vn ≤ u0i ≤ u + vn whenever i ≥ n; so hu0n in∈N order*-converges to u, by 367Be. Now if hun in∈N has an order*-limit, this must be u, by 367Ba and 367Bc. 367E Proposition Let U be a Riesz space with an order-continuous norm. Then any order-bounded order*-convergent sequence is norm-convergent. proof Suppose that hun in∈N is order*-convergent to u. Then h|un − u|in∈N is order*-convergent to 0 (367Cc), so C = {v : ∃ n ∈ N, v ≥ |ui − u| ∀ i ≥ n} has infimum 0 (367Be). Because U is a lattice, C is downwards-directed, so inf v∈C kvk = 0. But inf v∈C kvk ≥ inf n∈N supi≥n kui − uk, so limn→∞ kun − uk = 0, that is, hun in∈N is norm-convergent to u.

367H

Convergence in measure

365

367F One of the fundamental obstacles to the development of any satisfying general theory of ordered topological spaces is the erratic nature of the relations between subspace topologies of order topologies and order topologies on subspaces. The particular virtue of order*-convergence in the context of function spaces is that it is relatively robust when transferred to the subspaces we are interested in. Proposition Let U be an Archimedean Riesz space and V a regularly embedded Riesz subspace. (For instance, V might be either solid or order-dense.) If hvn in∈N is a sequence in V and v ∈ V , then hvn in∈N order*-converges to v when regarded as a sequence in V , iff it order*-converges to v when regarded as a sequence in U . proof (a) Since, in either V or U , hvn in∈N order*-converges to v iff h|vn − v|in∈N order*-converges to 0 (367Cc), it is enough to consider the case vn ≥ 0, v = 0. (b) If hvn in∈N is not order*-convergent to 0 in U , then, by 367Cf, there is a u > 0 in U such that u = supi≥n u ∧ vi for every n ∈ N (the supremum being taken in U , of course). In particular, there is a k ∈ N such that u ∧ vk > 0. Now consider the set C = {w : w ∈ V, ∃ n ∈ N, w ≥ (0 ∨ vi ) ∧ vk ∀ i ≥ n}. Then for any w ∈ C, u ∧ vk = supi≥n u ∧ vi ∧ vk ≤ w, using the generalized distributive law in U , so 0 is not the greatest lower bound of C in U . But as the embedding of V in U is order-continuous, 0 is not the greatest lower bound of C in V , and hvn in∈N cannot be order*-convergent to 0 in V . (c) Now suppose that hvn in∈N is not order*-convergent to 0 in V . Because V also is Archimedean (351Rc), there is a w > 0 in V such that w = supi≥n w ∧ vi for every n ∈ N, the suprema being taken in V . Again because V is regularly embedded in U , we have the same suprema in U , so, by 367Cf in the other direction, hvn in∈N is not order*-convergent to 0 in U . 367G I now spell out the connexion between the definition above and the concepts introduced in 245C. Proposition Let X be a set, Σ a σ-algebra of subsets of X, A a Boolean algebra and π : Σ → A a sequentially order-continuous surjective Boolean homomorphism; let I be its kernel. Write L0 for the space of Σ-measurable real-valued functions on X, and let T : L0 → L0 = L0 (A) be the canonical Riesz homomorphism (364D, 364R). Then for any hfn in∈N , f in L0 , hT fn in∈N order*-converges to T f in L0 iff X \ {x : f (x) = limn→∞ fn (x)} ∈ I. proof Set H = {x : limn→∞ fn (x) exists = f (x)}; of course H ∈ Σ. Set gn (x) = |fn (x) − f (x)| for n ∈ N, x ∈ X. (a) If X \ H ∈ I, set hn (x) = supi≥n gi (x) for x ∈ H and hn (x) = 0 for x ∈ X \ H. Then hhn in∈N is a non-increasing sequence with infimum 0 in L0 , so inf n∈N T hn = 0 in L0 , because T is sequentially order-continuous (364Ra). But as \H ∈ I, T hn ≥ T gi = |T fi − T f | whenever i ≥ n, so h|T fn − T f |in∈N order*-converges to 0, by 367Be or 367Bf, and hT fn in∈N order*-converges to T f , by 367Cc. (b) Now suppose that hT fn in∈N order*-converges to T f . Set gn0 (x) = min(1, gn (x)) for n ∈ N, x ∈ X; then hT gn0 in∈N = he ∧ |T fn − T f |in∈N order*-converges to 0, where e = T (χX). By 367Bf, inf n∈N supi≥n T gi0 = 0 in L0 . But T is a sequentially order-continuous Riesz homomorphism, so T (inf n∈N supi≥n gi0 ) = 0, that is, X \ H = {x : inf n∈N supi≥n gi0 > 0} belongs to I. 367H Corollary Let A be a Dedekind σ-complete Boolean algebra. (a) Any order*-convergent sequence in L0 = L0 (A) is order-bounded. (b) If hun in∈N is a sequence in L0 , then it is order*-convergent to u ∈ L0 iff u = inf n∈N supi≥n ui = supn∈N inf i≥n ui .

366

Function spaces

367H

proof (a) We can express A as a quotient Σ/I of a σ-algebra of sets, in which case L0 can be identified with the canonical image of L0 = L0 (Σ) (364D). If hun in∈N is an order*-convergent sequence in L0 , then it is expressible as hT fn in∈N , where T : L0 → L0 is the canonical map, and 367G tells us that hfn (x)in∈N converges for every x ∈ H, where X \ H ∈ I. If we set h(x) = supn∈N |fn (x)| for x ∈ H, 0 for x ∈ X \ H, then we see that |un | ≤ T h for every n ∈ N, so that hun in∈N is order-bounded in L0 . (b) This now follows from 367Be, because L0 is Dedekind σ-complete. 367I Proposition Suppose that E ⊆ R is a Borel set and h : E → R is a continuous function. Let A be a Dedekind σ-complete Boolean algebra and set QE = {u : u ∈ L0 , [[u ∈ E]] = 1}, where L0 = L0 (A). ¯ : QE → L0 be the function defined by h (364I). Then hh(u ¯ n )in∈N order*-converges to h(u) ¯ Let h whenever hun in∈N is a sequence in QE order*-converging to u ∈ QE . proof This is an easy consequence of 367G. We can represent A as Σ/I where Σ is a σ-algebra of subsets of some set X and I is a σ-ideal of Σ (314M); let T : L0 → L0 (A) be the corresponding homomorphism (364D, 367G). Now we can find Σ-measurable functions hfn in∈N , f such that T fn = un , T f = u, as in 367G; and the hypothesis [[un ∈ E]] = 1, [[u ∈ E]] = 1 means just that, adjusting fn and f on a member of I if necessary, we can suppose that fn (x), f (x) ∈ E for every x ∈ X. (I am passing over the trivial case ¯ n ) = T (hfn ), h(u) ¯ E = ∅, X ∈ I, A = {0}.) Accordingly h(u = T (hf ), and (because h is continuous) {x : h(f (x)) 6= limn→∞ h(fn (x))} ⊆ {x : f (x) 6= limn→∞ fn (x)} ∈ I, ¯ ¯ so hh(un )in∈N order*-converges to h(u). 367J Dominated convergence We now have a suitable language in which to express an abstract version of Lebesgue’s Dominated Convergence Theorem. Theorem Let (A, µ ¯) be a measure algebra. If hun in∈N is a sequence in L1 = L1 (A, µ ¯) which R is order-bounded R and order*-convergent in L1 , then hun in∈N is norm-convergent to u in L1 ; in particular, u = limn→∞ un . R proof The norm Rof L1 is order-continuous (365C), so hun in∈N is norm-convergent to u, by 367E. As is R norm-continuous, u = limn→∞ un . 367K The Martingale Theorem In the same way, we can re-write theorems from §275 in this language. Theorem Let (A, µ ¯) be a probability algebra, and hBn in∈N a non-decreasing sequence of closed subalgebras 0 of A. For each n ∈ N let Pn : L1 = L1µ¯ → L1 ∩ L S (Bn ) be the conditional expectation operator (365R); let B be the closed subalgebra of A generated by n∈N Bn , and P the conditional expectation operator onto L1 ∩ L0 (B). (a) If hun in∈N is a norm-bounded sequence in L1 such that Pn (un+1 ) = un for every n ∈ N, then hun in∈N is order*-convergent in L1 . (b) If u ∈ L1 then hPn uin∈N is order*-convergent to P u. proof If we represent (A, µ ¯) as the measure algebra of a probability space, these become mere translations of 275G and 275I. (Note that this argument relies on the description of order*-convergence in L0 in terms of a.e. convergence of functions, as in 367G; so that we need to know that order*-convergence in L1 matches order*-convergence in L0 , which is what 367F is for.) 367L Some of the most important applications of these ideas concern spaces of continuous functions. I do not think that this is the time to go very far along this road, but one particular fact will be useful in §376. Proposition Let X be a locally compact Hausdorff space, and hun in∈N a sequence in C(X), the space of continuous real-valued functions on X. Then hun in∈N order*-converges to 0 in C(X) iff {x : x ∈ X, lim supn→∞ |un (x)| > 0} is meager. In particular, hun in∈N order*-converges to 0 if limn→∞ un (x) = 0 for every x.

367N

Convergence in measure

367

proof (a) The following elementary fact is worth noting: if A ⊆ C(X)+ is non-empty and inf A = 0 in S C(X), then G = u∈A {x : u(x) < ²} is dense for every ² > 0. P P?? If not, take x0 ∈ X \ G. Because X is completely regular (3A3Bb), there is a continuous function w : X → [0, 1] such that w(x0 ) = 1 and w(x) = 0 for every x ∈ G. But in this case 0 < ²w ≤ u for every u ∈ A, which is impossible. X XQ Q (b) Suppose that hun in∈N order*-converges to 0. Set vn = |un | ∧ χX, so that hvn in∈N order*-converges to 0 (using 367C, as usual). Set B = {v : v ∈ C(X), ∃ n ∈ N, vi ≤ v ∀ i ≥ n}, S so that inf B = 0 in C(X) (367Be).SFor each k ∈ N, set Gk = v∈B {x : v(x) < 2−k }; then Gk is dense, by (a), and of course is open. So H = k∈N X \ Gk is a countable union of nowhere dense sets and is meager. But this means that {x : lim sup |un (x)| > 0} = {x : lim sup vn (x) > 0} n→∞

n→∞

⊆ {x : inf v(x) > 0} ⊆ H v∈B

is meager. (c) Now suppose that hun in∈N does not order*-converge to 0. By 367Cf, there is a w > 0 in C(X) such that w = supi≥n w ∧ |ui | for every n ∈ N; that is, inf i≥n (w − |ui |)+ = 0 for every n. Set Gn = {x : inf i≥n (w − |ui |)+ (x) < 2−n } = {x : supi≥n |ui (x)| > w(x) − 2−n } for each n. Then H=

T n∈N

Gn = {x : lim supn→∞ un (x) ≥ w(x)}

is the intersection of a sequence of dense open sets, and its complement is meager. Let G be the non-empty open set {x : w(x) > 0}. Then G is not meager, by Baire’s theorem (3A3Ha); so G ∩ H cannot be meager. But {x : lim supn→∞ |un (x)| > 0} includes G ∩ H, so is also not meager. Remark Unless the topology of X is discrete, C(X) is not regularly embedded in RX , and we expect to find sequences in C(X) which order*-converge to 0 in C(X) but not in RX . But the proposition tells us that if we have a sequence in C(X) which order*-converges in RX to a member of C(X), then it order*-converges in C(X). 367M Everything above concerns a particular notion of sequential convergence. There is inevitably a suggestion that there ought to be a topological interpretation of this convergence (see 367Yc, 367Yl), but I have taken care to avoid spelling one out. I come now to something which really is a topology, and is as closely involved with order-convergence as any. Convergence in measure Let (A, µ ¯) be a measure algebra. For a ∈ Af = {a : µ ¯a < ∞} and u ∈ L0 = R 0 L (A) set τa (u) = |u| ∧ χa, τa² (u) = µ ¯(a ∩ [[|u| > ²]]). Then the topology of convergence in measure on L0 is defined either as the topology generated by the pseudometrics (u, v) 7→ τa (u − v) or by saying that G ⊆ L0 is open iff for every u ∈ G there are a ∈ Af , ² > 0 such that v ∈ G whenever τa² (u − v) ≤ ². Remark The sentences above include a number of assertions which need proving. But at this point, rather than write out any of the relevant arguments, I refer you to §245. Since we know that L0 (A) can be identified with L0 (µ) for a suitable measure space (X, Σ, µ) (321J, 364Jc), everything we know about general spaces L0 (µ) can be applied directly to L0 (A) for measure algebras (A, µ ¯); and that is what I will do for the next few paragraphs. So far, all I have done is to write τa in place of the τ¯F of 245A, and call on the remarks in 245Bb and 245F. 367N Theorem (a) For any measure algebra (A, µ ¯), the topology T of convergence in measure on L0 = L0 (A) is a linear space topology, and any order*-convergent sequence in L0 is T-convergent to the same limit. (b) (A, µ ¯) is semi-finite iff T is Hausdorff. (c) (A, µ ¯) is localizable iff T is Hausdorff and L0 is complete under the uniformity corresponding to T. (d) (A, µ ¯) is σ-finite iff T is metrizable.

368

Function spaces

367N

proof 245D, 245Cb, 245E. Of course we need 322B to assure us that the phrases ‘semi-finite’, ‘localizable’, ‘σ-finite’ here correspond to the same phrases used in §245, and 367G to identify order*-convergence in L0 with the order-convergence studied in §245. 367O Proposition Let (A, µ ¯) be a measure algebra and give L0 = L0 (A) its topology of convergence in 0 measure. If A ⊆ L is a non-empty, downwards-directed set with infimum 0, then for every neighbourhood G of 0 in L0 there is a u ∈ A such that v ∈ G whenever |v| ≤ u. R proof Let a ∈ Af , ² > 0 be such that u ∈ G whenever |u|∧ χa ≤ ² (see 245Bb). Since {u ∧ χa :Ru ∈ A} is a downwards-directed set in L1 = L1 (A, µ ¯) with infimum 0 in L1 , there must be a u ∈ A such that u∧χa ≤ ² (365Da). But now [−u, u] ⊆ G, as required. 367P Theorem Let U be a Banach lattice and (A, µ ¯) a measure algebra. Give L0 = L0 (A) its topology 0 0 of convergence in measure. If T : U → L = L (A) is a positive linear operator, then it is continuous. proof Take any open set G ⊆ L0 . ?? Suppose, if possible, that T −1 [G] is not open. Then we can find u, hun in∈N ∈ U such that T u ∈ G and kun − uk ≤ 2−n , T un ∈ / GP for every n. Set H = G −P T u; then H is an ∞ ∞ open set containing 0 but not T (un − u), for any n ∈ N. Since n=0 kun − uk < ∞, v = n=0 n|un − u| is 1 defined in U , and |T (un − u)| ≤ n T v for every n ≥ 1. But by 367N (or otherwise) we know that there is some n such that w ∈ H whenever |w| ≤ n1 T v, so that T (un − u) ∈ H for some n, which is impossible. X X 367Q Proposition Let (A, µ ¯) be a σ-finite measure algebra. (a) A sequence hun in∈N in L0 = L0 (A) converges in measure to u ∈ L0 iff every subsequence of hun in∈N has a sub-subsequence which order*-converges to u. (b) A set F ⊆ L0 is closed for the topology of convergence in measure iff u ∈ F whenever there is a sequence hun in∈N in F order*-converging to u ∈ L0 . proof 245K, 245L. 367R

It will be useful later to be able to quote the following straightforward facts.

Proposition Let (A, µ ¯) be a measure algebra. Give A its measure-algebra topology (323A) and L0 = L0 (A) the topology of convergence in measure. Then the map χ : A → L0 is a homeomorphism between A and its image in L0 . proof Of course χ is injective (364Kc). The measure-algebra topology of A is defined by the Rpseudometrics ρa (b, c) = µ ¯(a ∩ (b4c)), while the topology of L0 is defined by the pseudometrics τa (u, v) = |u − v| ∧ χa, in both cases taking a to run over elements of A of finite measure; as τa (χb, χc) is always equal to ρa (b, c), we have the result. 367S Proposition Let E ⊆ R be a Borel set, and h : E → R a continuous function. Let (A, µ ¯) be a ¯ : QE → L0 = L0 (A) the associated function, where QE = {u : u ∈ L0 , [[u ∈ E]] = 1} measure algebra, and h ¯ is continuous for the topology of convergence in measure. (364I). Then h proof (Compare 245Dd.) Express (A, µ ¯) as the measure algebra of a measure space (X, Σ, µ), and write f • for the element of L0 corresponding to any f ∈ L0 . Take any u ∈ QE , any a ∈ A such that µ ¯a < ∞, and any ² > 0. Express u as f • where f : X → R is a measurable function, and a as F • where F ∈ Σ. Then f (x) ∈ E a.e.(x). For each n ∈ N, write En for {t : t ∈ E, |h(s) − h(t)| ≤ 13 ² whenever s ∈ E and |s − t| ≤ 2−n }. Then hEn in∈N is a non-decreasing sequence of Borel sets with union E, so there is an n such that µ{x : x ∈ F , f (x) ∈ / En } ≤ 31 ². R Now suppose that v ∈ QE is such that |v − u| ∧ χa ≤ 13 ²/2n . Express v as g • where g : X → R is a measurable function. Then g(x) ∈ E for almost every x, and

R F

1 3

min(1, |g(x) − f (x)|)µ(dx) ≤ ²/2n ,

367T

Convergence in measure

369

so µ{x : x ∈ F, |f (x) − g(x)| > 2−n } ≤ 13 ², and 1 ²} 3 ⊆ {x : x ∈ F, f (x) ∈ / En } ∪ {x : g(x) ∈ / E}

{x : x ∈ F, |h(g(x)) − h(f (x))| >

∪ {x : x ∈ F, |f (x) − g(x)| > 2−n } has measure at most 23 ². But this means that

R

¯ ¯ |h(v) − h(u)| ∧ χa =

R F

min(1, |hg(x) − hf (x)|)µ(dx) ≤ ².

¯ is continuous. As u, a and ² are arbitrary, h 367T Intrinsic description of convergence in measure It is a remarkable fact that the topology of convergence in measure, not only on L0 but on its order-dense Riesz subspaces, can be described in terms of the Riesz space structure alone, without referring at all to the underlying measure algebra or to integration. (Compare 324H.) There is more than one way of doing this. As far as I know, none is outstandingly convincing; I present a formulation which seems to me to exhibit some, at least, of the essence of the phenomenon. Proposition Let (A, µ ¯) be a semi-finite measure algebra, and U an order-dense Riesz subspace of L0 = 0 L (A). Suppose that A ⊆ U and u∗ ∈ U . Then u∗ belongs to the closure of A for the topology of convergence in measure iff there is an order-dense Riesz subspace V of U such that for every v ∈ V + there is a non-empty downwards-directed B ⊆ U , with infimum 0, such that for every w ∈ B there is a u ∈ A such that |u − u∗ | ∧ v ≤ w. proof (a) Suppose first that u∗ ∈ A. Take V to be U ∩ L1 (A, µ ¯); then V is an order-dense Riesz subspace of L0 , by 352Nc, and is therefore order-dense in U . (This is where I use the hypothesis that (A, µ ¯) is semi-finite, so that L1 is order-dense in L0 , by 365Ga.) Take any v ∈ V + . For each n ∈ N, set an = [[v > 2−n ]] ∈ Af . Because u∗ ∈ A, there is a un ∈ A such that µ ¯bn ≤ 2−n , where bn = an ∩ [[|un − u∗ | > 2−n ]] = [[|un − u∗ | ∧ v > 2−n ]]. Set cn = supi≥n bi ; then µ ¯cn ≤ 2−n+1 for each n, so inf n∈N cn = 0 and inf n∈N wn = 0 in L0 , where wn = v × χcn + 2−n χ1. Also |un − u∗ | ∧ v ≤ wn for each n. The wn need not belong to U , so we cannot set B = {wn : n ∈ N}. But if instead we write B = {w : w ∈ U, w ≥ v ∧ wn for some n ∈ N}, then B is non-empty and downwards-directed (because hwn in∈N is non-increasing); and inf B = v − sup{v − w : w ∈ B} = v − sup{w : w ∈ U, w ≤ (v − wn )+ for some n ∈ N} = v − sup(v − wn )+ n∈N

(because U is order-dense in L0 ) = 0. Since for every w ∈ B there is an n such that v ∧ |un − u∗ | ≤ v ∧ wn ≤ w, B witnesses that the condition is satisfied. f (b) Now suppose that the condition is satisfied. Fix a ∈ A R , ² > 0. Because V is order-dense in U and 0 therefore in L , there is a v ∈ V such that 0 ≤ v ≤ χa and v ≥ µ ¯a − ². Let B be a downwards-directed

370

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set, with infimumR 0, such that for every w ∈ B there is a u ∈ A with v ∧ |u − u∗ | ≤ w. Then there is a w ∈ B such that w ∧ v ≤ ². Now there is a u ∈ A such that |u − u∗ | ∧ v ≤ w, so that

R

|u − u∗ | ∧ χa ≤ ² +

R

|u − u∗ | ∧ v ≤ ² +

R

w ∧ v ≤ 2².

∗

As a and ² are arbitrary, u ∈ A. *367U Theorem Let (A, µ ¯) be a semi-finite measure algebra; write L1 for L1 (A, µ ¯). Let P : (L1 )∗∗ → L1 be the linear operator corresponding to the band projection from (L1 )∗∗ = (L1 )×∼ onto (L1 )×× and the canonical isomorphism between L1 and (L1 )×× . For A ⊆ L1 write A∗ for the weak* closure of the image of A in (L1 )∗∗ . Then for every A ⊆ L1 P [A∗ ] ⊆ Γ(A), where Γ(A) is the convex hull of A and Γ(A) is the closure of Γ(A) in L0 = L0 (A) for the topology of convergence in measure. proof (a) The statement of the theorem includes a number of assertions: that (L1 )∗ = (L1 )× ; that (L1 )∗∗ = ((L1 )∗ )∼ ; that the natural embedding of L1 into (L1 )∗∗ = (L1 )×∼ identifies L1 with (L1 )×× ; and that (L1 )×× is a band in (L1 )×∼ . For proofs of these see 365C, 356B, 356D, 356N and 356P. Now for the new argument. First, observe that the statement of the theorem involves the measure algebra (A, µ ¯) and the space L0 only in the definition of ‘convergence in measure’; everything else depends only on the Banach lattice structure of L1 . And since we are concerned only with the question of whether members of P [A∗ ], which is surely a subset of L1 , belong to Γ(A), 367T shows that this also can be answered in terms of the Riesz space structure of L1 . What this means is that we can suppose that (A, µ ¯) is localizable. P P b Let (A, µ ˜) be the localization of (A, µ ¯) (322P). The natural expression of A as an order-dense subalgebra of b identifies Af = {a : a ∈ A, µ b f (322O), so that L1 (A, µ b µ A ¯a < ∞} with A ¯) becomes identified with L1 (A, ˜), 1 1 b b µ by 365Od. Thus we can think of L as L (A, µ ˜), and (A, ˜) is localizable. Q Q (b) Take φ ∈ A∗ and set u0 = P φ; I have to show that u0 ∈ Γ(A). Write R for the canonical map from L to (L1 )∗∗ , so that φ belongs to the weak* closure of R[A]. Consider first the case u0 = 0. Take any c ∈ Af and ² > 0. We know that (L1 )∗ = (L1 )∼ = (L1 )× can be identified with L∞ = L∞ (A) (365Jc), so that φ ∈ (L∞ )∗ = (L∞ )∼ must be in the band orthogonal to (L∞ )× . Now we can identify (L∞ )∼ with the Riesz space M of bounded additive functionals on A, and if we do so then (L∞ )× corresponds to the space Mτ of completely additive functionals (363K). Writing Pτ : M → Mτ for the band projection, we must have Pτ (ν) = 0, where ν ∈ M is defined by setting νa = φ(χa) for each a ∈ A; consequently Pτ (|ν|) = 0 and there is an upwards-directed family C ⊆ A, with supremum 1, such that |ν|(a) = 0 for every a ∈ C (362D). Since µ ¯c = supa∈C µ ¯(a ∩ c), there is an a ∈ C such that µ ¯(c \ a) ≤ ². Consider the map Q : L1 → L1 defined by setting Qw = w × χa for every w ∈ L1 . Then its adjoint 0 Q : L∞ → L∞ (3A5Ed) can be defined by the same formula: Q0 v = v × χa for every v ∈ L∞ . Since |φ| ∈ (L∞ )∼ corresponds to |ν| ∈ M , we have 1

|φ(Q0 v)| ≤ kvk∞ |φ|(χa) = kvk∞ |ν|(a) = 0 for every v ∈ L∞ , and Q00 φ = 0, where Q00 : (L∞ )∗ → (L∞ )∗ is the adjoint of Q0 . Since Q00 is continuous for the weak* topology on (L∞ )∗ , 0 ∈ Q00 [R[A]], where Q00 R[A] is the closure for the weak* topology of (L∞ )∗ . But of course Q00 R = RQ, while the weak* topology of (L∞ )∗ corresponds, on the image R[L1 ] of L1 , to the weak topology of L1 ; so that 0 belongs to the closure of Q[A] for the weak topology of L1 . Because Q is linear, Q[Γ(A)] is convex. Since 0 belongs to the closure of Q[Γ(A)] for the weak topology of L1 , it belongs to the closure of Q[Γ(A)] for the norm topology (3A5Ee). So 0 belongs to the closure of Q[Γ(A)] for the norm topology, and there is a w ∈ Γ(A) such that kw × χak1 ≤ ²2 . But this means that µ ¯(a ∩ [[|w| ≥ ²]]) ≤ ² and µ ¯(c ∩ [[|w| ≥ ²]]) ≤ 2². Since c and ² are arbitrary, 0 ∈ Γ(A). (c) This deals with the case u0 = 0. Now the general case follows at once if we set B = A − u0 and observe that φ − Ru0 ∈ B ∗ , so 0 = P (φ − Ru0 ) ∈ Γ(B) = Γ(A) − u0 = Γ(A) − u0 .

367Xf

Convergence in measure

371

Remark This is a version of a theorem from Bukhvalov 95. *367V Corollary Let (A, µ ¯) be a localizable measure algebra. Let C be a family of convex subsets of L0 = L0 (A), all closed for the topology of convergence in measure, with the finite intersection property, and R suppose that for every non-zero a ∈ A there are a non-zero b ⊆ a and a C ∈ C such that sup |u| < ∞. u∈C b T Then C = 6 ∅. proof Because C has the finite intersection property, there is an ultrafilter F on L0 including C. Set I = {a : a ∈ A, inf F ∈F supu∈F

R

a

|u| < ∞};

because F is a filter, I is an ideal in A, and the condition on C tells us that I is order-dense. For each a ∈ I, define Qa : L0 → L0 by setting Qa u = u × χa. Then there is an F ∈ F such that Qa [F ] is a norm-bounded set in L1 , so φa = limu→F RQa u is defined in (L∞ )∗ for the weak* topology on (L∞ )∗ , writing R for the canonical map from L1 to (L∞ )∗ ∼ = (L1 )∗∗ . If P : (L∞ )∗ → L1 is the map corresponding to the band ∞ ∼ ∞ × ˜ projection P from (L ) onto (L ) , as in 367U, and C ∈ C, then 367U tells us that P (φa ) must belong to the closure of the convex set Qa [C] for the topology of convergence in measure. Moreover, if a ⊆ b ∈ I, so that Qa = Qa Qb , then P (φa ) = Qa P (φb ). P P Qa ¹L1 is a band projection on L1 , so its adjoint Q0a is a 1 ∼ 00 ∞ ∼ band projection on L = (L ) (356C) and Qa is a band projection on (L∞ )∗ ∼ = (L∞ )∼ . This means that 00 00 ˜ Qa will commute with P (352Sb). But also Qa is continuous for the weak* topology of (L∞ )∗ , so Q00a (φb ) = limu→F Q00a RQb u = limu→F RQa Qb u = φa , and Q P (φa ) = R−1 P˜ (φa ) = R−1 P˜ Q00a (φb ) = R−1 Q00a P˜ (φb ) = Qa R−1 P˜ (φb ) = Qa P (φb ). Q Q 0 ∼ What this means is that if we take a partition D of unity included in I (313K), so that L =

L0 (Ad ) (315F(iii), 364S), and define w ∈ L by saying that w × χd = P (φd ) for every d ∈ D, then we shall have w × χa = P (φa ) for every a ∈ I. But now, given a ∈ Af and ² > 0 and C ∈ C, there is a b ∈ I such that µ ¯(a \ b) ≤ ²; w × χb ∈ Qb [C], so there is a u ∈ C such that µ ¯(b ∩ [[|w − u| ≥ ²]]) ≤T²; and µ ¯ (a ∩ [[|w − u| ≥ ²]]) ≤ 2². As a, ² are arbitrary and C is closed, w ∈ C; as C is arbitrary, w ∈ C and T C 6= ∅. 0

d∈D

367X Basic exercises > (a) Let P be a lattice. (i) Show that if p ∈ P and hpn in∈N is a non-decreasing sequence in P , then hpn in∈N is order*-convergent to p iff p = supn∈N pn . (ii) Suppose that hpn in∈N a sequence in P order*-converging to p ∈ P . Show that p = supn∈N p ∧ pn = inf n∈N p ∨ pn . (iii) Let hpn in∈N , hqn in∈N be two sequences in P which are order*-convergent to p, q respectively. Show that if pn ≤ qn for every n then p ≤ q. (iv) Let hpn in∈N be a sequence in P . Show that hpn in∈N order*-converges to p ∈ P iff hpn ∨ pin∈N and hpn ∧ pin∈N order*-converge to p. (b) Let P and Q be lattices, and f : P → Q an order-preserving function. Suppose that hpn in∈N is an order-bounded sequence which order*-converges to p in P . Show that hf (pn )in∈N order*-converges to f (p) in Q if either f is order-continuous or P is Dedekind σ-complete and f is sequentially order-continuous. (c) Let P be either a Boolean algebra or a Riesz space. Suppose that hpn in∈N is a sequence in P such that hp2n in∈N and hp2n+1 in∈N are both order*-convergent to p ∈ P . Show that hpn in∈N is order*-convergent to p. (Hint: 313B, 352E.) >(d) Let A be a Boolean algebra and han in∈N , hbn in∈N two sequences in A order*-converging to a, b respectively. Show that han ∪ bn in∈N , han ∩ bn in∈N , han \ bn in∈N , han 4 bn in∈N order*-converge to a ∪ b, a ∩ b, a \ b and a 4 b respectively. (e) Let A be a Boolean algebra and han in∈N a sequence in A. Show that han in∈N does not order*-converge to 0 iff there is a non-zero a ∈ A such that a = supi≥n a ∧ ai for every n ∈ N. >(f ) (i) Let U be a Riesz space and hun in∈N an order*-convergent sequence in U + with limit u. Show that h(u) ≤ lim inf n→∞ h(un ) for every h ∈ (U × )+ . (ii) Let U be a Riesz space and hun in∈N an order-bounded order*-convergent sequence in U with limit u. Show that h(u) = limn→∞ h(un ) for every h ∈ U × . (Compare 356Xd.)

372

Function spaces

367Xg

> (g) Let U be a Riesz space with a Fatou norm k k. (i) Show that if hun in∈N is an order*-convergent sequence in U with limit u, then kuk ≤ lim inf n→∞ kun k. (Hint: h|un | ∧ |u|in∈N is order*-convergent to |u|.) (ii) Show P∞ that if hun in∈N is a norm-convergent sequence in U it has an order*-convergent subsequence. (Hint: if n=0 kun k < ∞ then hun in∈N order*-converges to 0.) (h) Let U and V be Archimedean Riesz spaces and T : U → V an order-continuous Riesz homomorphism. Show that if hun in∈N is a sequence in U which order*-converges to u ∈ U , then hT un in∈N order*-converges to T u in V . (i) Let A be a Boolean algebra and B an order-closed subalgebra. Show that if hbn in∈N is a sequence in B and b ∈ B, then hbn in∈N order*-converges to b in B iff it order*-converges to b in A. (j) Let A be a Dedekind σ-complete Boolean algebra and hun in∈N , hvn in∈N two sequences in L0 (A) which are order*-convergent to u, v respectively. Show that hun × vn in∈N order*-converges to u × v. Show that if −1 −1 u, un all have multiplicative inverses u−1 , u−1 . n then hun in∈N order*-converges to u (k) Let A be a Dedekind σ-complete Boolean algebra and I a σ-ideal of A. Show that for any han in∈N , a ∈ A, ha•n in∈N order*-converges to a• in A/I iff inf n∈N supm≥n am 4a ∈ I. > (l) Let A be a Dedekind σ-complete Boolean algebra, and hhn in∈N a sequence of Borel measurable ¯ n (u)in∈N functions from R to itself such that h(t) = limn→∞ hn (t) is defined for every t ∈ R. Show that hh 0 0 0 0 ¯ ¯ ¯ order*-converges to h(u) for every u ∈ L = L (A), where hn , h : L → L are defined as in 364I. (m) Let (A, µ ¯) be a measure algebra, and hun in∈N a sequence in L1 = L1 (A, µ ¯) which is order*-convergent 1 to u ∈ L . Show that hun in∈N is norm-convergent to u iff {un : n ∈ N} is uniformly integrable iff kuk1 = limn→∞ kun k1 . (Hint: 245H, 246J.) (n) Let U be an L-space and hun in∈N a norm-bounded sequence in U . Show that there are a v ∈ U and Pn 1 a subsequence hvn in∈N of hun in∈N such that h n+1 i=0 wi in∈N order*-converges to v for every subsequence hwn in∈N of hvn in∈N . (Hint: 276H.) > (o) Let (A, µ ¯) be a measure algebra and give L0 = L0 (A) its topology of convergence in measure. Show that u 7→ |u|, (u, v) 7→ u ∨ v, (u, v) 7→ u × v are continuous. (p) Let (A, µ ¯) be a measure algebra and p ∈ [1, ∞[. For v ∈ (Lp )+ = Lp (A, µ ¯)+ define ρv : L0 ×L0 → [0, ∞[ by setting ρv (u1 , u2 ) = k|u1 − u2 | ∧ vkp for all u1 , u2 ∈ U . Show that each ρv is a pseudometric and that the topology on L0 = L0 (A) defined by {ρv : v ∈ (Lp )+ } is the topology of convergence in measure. (q) Let (A, µ ¯) be a σ-finite measure algebra. Suppose we have a double sequence huij i(i,j)∈N×N in L0 = L (A) such that huij ij∈N order*-converges to ui in L0 for each i, while hui ii∈N order*-converges to u. Show that there is a strictly increasing sequence hn(i)ii∈N such that hui,n(i) ii∈N order*-converges to u. 0

(r) Let (X, Σ, µ) be a semi-finite measure space. Show that L0 (µ) is separable for the topology of convergence in measure iff µ is σ-finite and has countable Maharam type. (Cf. 365Xp.) (s) Let (A, µ ¯) be a measure algebra. (i) Show that if han in∈N is order*-convergent to a ∈ A, then han in∈N → a for the measure-algebra topology. (ii) Show that if (A, µ ¯) is σ-finite, then (α) a sequence converges to a for the topology of A iff every subsequence has a sub-subsequence which is order*-convergent to to a (β) a set F ⊆ A is closed for the topology of A iff a ∈ F whenever there is a sequence han in∈N in F which is order*-convergent to a ∈ A. (t) Let (A, µ ¯) be a measure algebra which is not σ-finite. Show that there is a set A ⊆ L0 (A) such that the limit of any order*-convergent sequence in A belongs to A, but A is not closed for the topology of convergence in measure. (u) Let U be a Banach lattice with an order-continuous norm. (i) Show that a sequence hun in∈N is norm-convergent to u ∈ U iff every subsequence has a sub-subsequence which is order-bounded and order*convergent to u. (ii) Show that a set F ⊆ U is closed for the norm topology iff u ∈ F whenever there is an order-bounded sequence hun in∈N in F order*-converging to u ∈ U .

367Yj

Convergence in measure

373

(v) Let (A, µ ¯) be a probability algebra. For u ∈ L0 = L0 (A) let νu be the distribution of u (364Xd). Show that u 7→ νu is continuous when L0 is given the topology of convergence in measure and the space of probability distributions on R is given the vague topology (274Ld). (w) Let (A, µ ¯) be a probability algebra and hun in∈N a stochastically independent sequence in L0 , all with the Cauchy distribution νC,1 with centre 0 and scale parameter 1 (285Xm). For each n let Cn be the convex hull of {ui : i ≥ n}, and Cn its closure for the topology of convergence in measure. Show that every u ∈ C0 has distribution νC,1 . (Hint: consider first u ∈ C0 .) Show that C0 is bounded for the topology of T convergence in measure. Show that n∈N Cn = ∅. (x) If U is a linear space and C ⊆ U is a convex set, a function f : C → R is convex if f (αx + (1 − α)y) ≤ αf (x) + (1 − α)f (y) whenever x, y ∈ C and α ∈ [0, 1]. Let (A, µ ¯) be a localizable measure algebra and C ⊆ L1 (A, µ ¯) a non-empty convex norm-bounded set which is closed in L0 (A) for the topology of convergence in measure. Show that any convex function f : C → R which is lower semi-continuous for the topology of convergence in measure is bounded below and attains its infimum. 367Y Further exercises (a) Give an example of an Archimedean Riesz space U and an order-bounded sequence hun in∈N in U which is order*-convergent to 0, but such that there is no non-increasing sequence hvn in∈N , with infimum 0, such that un ≤ vn for every n ∈ N. (b) Let P be a distributive lattice. Show that if hpn in∈N is a sequence in P order*-converging to p ∈ P , then hpn ∨ qin∈N , hpn ∧ qin∈N order*-converge to p ∨ q, p ∧ q respectively for any q ∈ P . (c) Let P be any lattice. (i) Show that there is a topology on P for which a set A ⊆ P is closed iff p ∈ A whenever there is a sequence in A which is order*-convergent to p. Show that any closed set for this topology is sequentially order-closed. (ii) Now let Q be another lattice, with the topology defined in the same way, and f : P → Q an order-preserving function. Show that if f is topologically continuous it is sequentially order-continuous. (d) Let us say that a lattice P is (2, ∞)-distributive if (α) whenever A, B ⊆ P are non-empty sets with infima p, q respectively, then inf{a ∨ b : a ∈ A, b ∈ B} = p ∨ q (β) whenever A, B ⊆ P are non-empty sets with suprema p, q respectively, then sup{a ∧ b : a ∈ A, b ∈ B} = p ∧ q. Show that, in this case, if hpn in∈N order*-converges to p and hqn in∈N order*-converges to q, hpn ∨ qn in∈N order*-converges to p ∨ q. (e) Give an example of a lattice P with two sequences hpn in∈N , hqn in∈N , both order*-convergent to p, such that hpn ∨ qn in∈N is not order*-convergent to p. (f ) (i) Give an example of a Riesz space U with an order-dense Riesz subspace V of U and a sequence hvn in∈N in V such that hvn in∈N order*-converges to 0 in V but does not order*-converge in U . (ii) Give an example of a Riesz space U with an order-dense Riesz subspace V of U and a sequence hvn in∈N in V , order-bounded in V , such that hvn in∈N order*-converges to 0 in U but does not order*-converge in V . (g) Let U be an Archimedean f -algebra. Show that if hun in∈N , hvn in∈N are sequences in U order*converging to u, v respectively, then hun × vn in∈N order*-converges to u × v. (h) Let A be a Dedekind σ-complete Boolean algebra and r ≥ 1. Let E ⊆ Rr be a Borel set and write QE = {(u1 , . . . , ur ) : [[(u1 , . . . , ur ) ∈ E]] = 1} ⊆ L0 (A)r (364Yc). Let h : E → R be a continuous function ¯ : QE → L0 the corresponding map (364Yd). Show that if hun in∈N is a sequence in QE which is and h ¯ n )in∈N is order*-convergent to h(u). ¯ order*-convergent to u ∈ QE (in the lattice (L0 )r ), then hh(u (i) Let X be a completely regular Baire space (definition: 314Yd), and hun in∈N a sequence in C(X). Show that hun in∈N order*-converges to 0 in C(X) iff {x : lim supn→∞ |un (x)| > 0} is meager in X. (j) (i) Give an example of a sequence hun in∈N in C([0, 1]) such that limn→∞ un (x) = 0 for every x ∈ [0, 1], but {un : n ∈ N} is not order-bounded in C([0, 1]). (ii) Give an example of an order-bounded sequence hun in∈N in C(Q) such that limn→∞ un (q) = 0 for every q ∈ Q, but supi≥n ui = χQ in C(Q) for every n ∈ N. (iii) Give an example of a sequence hun in∈N in C([0, 1]) such that hun in∈N order*-converges to 0 in C([0, 1]), but limn→∞ un (q) > 0 for every q ∈ Q ∩ [0, 1].

374

Function spaces

367Yk

(k) Write out an alternative proof of 367K/367Yi based on the fact that, for a Baire space X, C(X) can be identified with an order-dense Riesz subspace of a quotient of the space of Σ-measurable functions, where Σ is the algebra of subsets of X with the Baire property, as in 364Yi. (l) Let A be a ccc weakly (σ, ∞)-distributive Boolean algebra. Show that there is a topology on A such that the closure of any A ⊆ A is precisely the set of limits of order*-convergent sequences in A. (m) Give an example of a set X and a double sequence humn im,n∈N in RX such that limn→∞ umn (x) = um (x) exists for every m ∈ N and x ∈ X, limm→∞ um (x) = 0 for every x ∈ X, but there is no sequence hvk ik∈N in {umn : m, n ∈ N} such that limk→∞ vk (x) = 0 for every x. (n) Let U be any Banach lattice with an order-continuous norm. For v ∈ U + define ρv : U × U → [0, ∞[ by setting ρv (u1 , u2 ) = k|u1 − u2 | ∧ vk for all u1 , u2 ∈ U . Show that every ρv is a pseudometric on U , and that {ρv : v ∈ U + } defines a Hausdorff linear space topology on U . (o) Let U be any Riesz space. For h ∈ (Uc∼ )+ (356Ab), v ∈ U + define ρvh : U × U → [0, ∞[ by setting ρvh (u1 , u2 ) = h(|u1 − u2 | ∧ v) for all u1 , u2 ∈ U . Show that each ρvh is a pseudometric on U , and that {ρvh : h ∈ (Uc∼ )+ , v ∈ U + } defines a linear space topology on U . (p) Let (A, µ ¯) be a σ-finite measure algebra. Show that the function (α, u) 7→ [[u > α]] : R × L0 → A is Borel measurable when L0 = L0 (A) is given the topology of convergence in measure and A is given its measure-algebra topology. (Hint: if a ∈ A, γ ≥ 0 then {(α, u) : µ ¯(a ∩ [[u > α]]) > γ} is open.) (q) Let G be the regular open algebra of R. Show that there is no Hausdorff topology T on L0 (G) such that hun in∈N is T-convergent to u whenever hun in∈N is order*-convergent to u. (Hint: Let H be any T-open set containing 0. Enumerate Q as hqn in∈N . Find inductively a non-decreasing sequence hGn in∈N in G such that χGn ∈ H, qn ∈ Gn for every n. Conclude that χR ∈ H.) (r) Give an example of a Banach lattice with a norm which is not order-continuous, but in which every order-bounded order*-convergent sequence is norm-convergent. (s) Let A be a Dedekind σ-complete Boolean algebra and r ≥ 1. Let E ⊆ Rr be a Borel set and write QE = {(u1 , . . . , ur ) : [[(u1 , . . . , ur ) ∈ E]] = 1} ⊆ L0 (A)r (364Yc). Let h : E → R be a continuous function ¯ : QE → L0 the corresponding map (364Yd). Show that if h ¯ is continuous if L0 is given its topology and h 0 r of convergence in measure and (L ) the product topology. (t) Show that 367U is true for all measure algebras, whether semi-finite or not. 367 Notes and comments I have given a very general definition of ‘order*-convergence’. The general theory of convergence structures on ordered spaces is complex and full of traps for the unwary. I have tried to lay out a safe path to the results which are important in the context of this book. But the propositions here are necessarily full of little conditions (e.g., the requirement that U should be Archimedean, in 367F) whose significance may not be immediately obvious. In particular, the definition is very much better adapted to distributive lattices than to others (367Yb, 367Yd, 367Ye). It is useful in the study of Riesz spaces and Boolean algebras largely because these satisfy strong distributive laws (313B, 352E). The special feature which distinguishes the definition here from other definitions of order-convergence is the fact that it can be applied to sequences which are not order-bounded. For order-bounded sequences there are useful simplifications (367Be-f), but the Martingale Theorem (357J), for instance, if we want to express it in terms of its natural home in the Riesz space L1 , refers to sequences which are hardly ever order-bounded. The * in the phrase ‘order*-convergent’ is supposed to be a warning that it may not represent exactly the concept you expect. I think nearly any author using the phrase ‘order-convergent’ would accept sequences fulfilling the conditions of 367Bf; but beyond this no standard definitions have taken root. The fact that order*-convergent sequences in an L0 space are order-bounded (367H) is actually one of the characteristic properties of L0 . Related ideas will be important in the next section (368A, 368M). It is one of the outstanding characteristics of measure algebras in this context that they provide nontrivial linear space topologies on their L0 spaces, related in striking ways to the order structure. Not all L0

Embedding Riesz spaces in L0

368B

375

spaces have such topologies (367Yq). It is not known whether a topology corresponding to ‘convergence in measure’ can be defined on L0 (A) for any A which is not a measure algebra; this is the ‘control measure problem’, which I will discuss in §393 (see 393L). 367T shows that the topology of convergence in measure on L0 (A) is (at least for semi-finite measure algebras) determined by the Riesz space structure of L0 ; and that indeed the same is true of its order-dense Riesz subspaces. This fact is important for a full understanding of the representation theorems in §369 below. If a Riesz space U can be embedded as an order-dense subspace of any such L0 , then there is already a ‘topology of convergence in measure’ on U , independent of the embedding. It is therefore not surprising that there should be alternative descriptions of the topology of convergence in measure on the important subspaces of L0 (367Xp, 367Yn). For σ-finite measure algebras, the topology of convergence in measure is easily described in terms of order-convergence (367Q). For other measure algebras, the formula fails (367Xt). 367Yq shows that trying to apply the same ideas to Riesz spaces in general gives rise to some very curious phenomena. 367V enables us to prove results which would ordinarily be associated with some form of compactness. Of course compactness is indeed involved, as the proof through 367U makes clear; but it is weak* compactness in (L1 )∗∗ , rather than in the space immediately to hand. I hardly mention ‘uniform integrability’ in this section, not because it is uninteresting, but because I have nothing to add at this point to 246J and the exercises in §246. But I do include translations of Lebesgue’s Dominated Convergence Theorem (367I) and the Martingale Theorem (367J) to show how these can be expressed in the language of this chapter.

368 Embedding Riesz spaces in L0 In this section I turn to the representation of Archimedean Riesz spaces as function spaces. Any Archimedean Riesz space U can be represented as an order-dense subspace of L0 (A), where A is its band algebra (368E). Consequently we get representations of Archimedean Riesz spaces as quotients of subspaces of RX (368F) and as subspaces of C ∞ (X) (368G), and a notion of ‘Dedekind completion’ (368I-368J). Closely associated with these is the fact that we have a very general extension theorem for order-continuous Riesz homomorphisms into L0 spaces (368B). I give a characterization of L0 spaces in terms of lateral completeness (368M, 368Yd), and I discuss weakly (σ, ∞)-distributive Riesz spaces (368N-368S). 368A Lemma Let A be a Dedekind σ-complete Boolean algebra, and A ⊆ (L0 )+ a set with no upper bound in L0 , where L0 = L0 (A). If either A is countable or A is Dedekind complete, there is a v > 0 in L0 such that nv = supu∈A u ∧ nv for every n ∈ N. proof The hypothesis ‘A is countable or A is Dedekind complete’ ensures that cα = supu∈A [[u > α]] is defined for each α. By 364Ma, c = inf n∈N cn = inf α∈R cα is non-zero. Now for any n ≥ 1, α ∈ R [[supu∈A (u ∧ nχc) > α]] = supu∈A [[u > α]] ∩ [[χc >

α n ]]

= [[χc >

α n ]],

because if 0 ≤ α ≤ k ∈ N then supu∈A [[u > α]] ⊇ ck ⊇ c ⊇ [[χc >

α n ]],

while if α < 0 then (because A is a non-empty subset of (L0 )+ ) supu∈A [[u > α]] = 1 = [[χc >

α n ]].

So supu∈A u ∧ nχc = nχc for every n ≥ 1, and we can take v = χc. (The case n = 0 is of course trivial.) 368B Theorem Let A be a Dedekind complete Boolean algebra, U an Archimedean Riesz space, V an order-dense Riesz subspace of U and T : V → L0 = L0 (A) an order-continuous Riesz homomorphism. Then T has a unique extension to an order-continuous Riesz homomorphism T˜ : U → L0 . proof (a) The key to the proof is the following: if u ≥ 0 in U , then {T v : v ∈ V, 0 ≤ v ≤ u} is bounded above in L0 . P P?? Suppose, if possible, otherwise. Then by 368A there is a w > 0 in L0 such that

376

Function spaces

368B

nw = supv∈A nw ∧ T v for every n ∈ N, where A = {v : v ∈ V, 0 ≤ v ≤ u}. In particular, there is a v0 ∈ A such that w0 = w ∧ T v0 > 0. Because U is Archimedean, inf k≥1 k1 u = 0, so v0 = supk≥1 (v0 − k1 u)+ . Because V is order-dense in U , v0 = sup B where 1 k

B = {v : v ∈ V, 0 ≤ v ≤ (v0 − u)+ for some k ≥ 1}. Because T is order-continuous, T v0 = sup T [B] in L0 , and there is a v1 ∈ B such that w1 = w0 ∧ T v1 > 0. Let k ≥ 1 be such that v1 ≤ (v0 − k1 u)+ . Then for any m ∈ N, mv1 ∧ u ≤ (mv1 ∧ kv0 ) + (mv1 ∧ (u − kv0 )+ ) (352Fa) 1 k

≤ kv0 + (m + k)(v1 ∧ ( u − v0 )+ ) = kv0 . So for any v ∈ A, m ∈ N, mw1 ∧ T v = mw1 ∧ mT v1 ∧ T v ≤ T (mv1 ∧ v) ≤ T (mv1 ∧ u) ≤ T (kv0 ) = kT v0 . But this means that, for m ∈ N, mw1 = mw1 ∧ mw = supv∈A mw1 ∧ (mw ∧ T v) = supv∈A mw1 ∧ T v ≤ kT v0 , which is impossible because L0 is Archimedean and w1 > 0. X XQ Q (b) Because L0 is Dedekind complete, sup{T v : v ∈ V, 0 ≤ v ≤ u} is defined in L0 for every u ∈ U . By 355F, T has a unique extension to an order-continuous Riesz homomorphism from U to L0 . 368C Corollary Let A and B be Dedekind complete Boolean algebras and U , V order-dense Riesz subspaces of L0 (A), L0 (B) respectively. Then any Riesz space isomorphism between U and V extends uniquely to a Riesz space isomorphism between L0 (A) and L0 (B); and in this case A and B must be isomorphic as Boolean algebras. proof If T : U → V is a Riesz space isomorphism, then 368B tells us that we have (unique) order-continuous Riesz homomorphisms S : L0 (A) → L0 (B) and S 0 : L0 (B) → L0 (A) extending T , T −1 respectively. Now S 0 S : L0 (A) → L0 (A) is an order-continuous Riesz homomorphism agreeing with the identity on U , so must be the identity on L0 (A); similarly SS 0 is the identity on L0 (B), and S is a Riesz space isomorphism. To see that A and B are isomorphic, recall that by 364Q they can be identified with the algebras of projection bands of L0 (A) and L0 (B), which must be isomorphic. 368D Corollary Suppose that A is a Dedekind σ-complete Boolean algebra, and that U is an orderdense Riesz subspace of L0 (A) which is isomorphic, as Riesz space, to L0 (B) for some Dedekind complete Boolean algebra B. Then U = L0 (A) and A is isomorphic to B (so, in particular, is Dedekind complete). proof The identity mapping U → U is surely an order-continuous Riesz homomorphism, so by 368B extends to an order-continuous Riesz homomorphism T˜ : L0 (A) → U . Now T˜ must be injective, because if u 6= 0 in L0 (A) there is a u0 ∈ U such that 0 < u0 ≤ |u|, so that 0 < u0 ≤ |T˜u|. So we must have U = L0 (A) and T˜ the identity map. By 368C, or otherwise, A ∼ = B. 368E Theorem Let U be any Archimedean Riesz space, and A its band algebra (353B). Then U can be embedded as an order-dense Riesz subspace of L0 (A). proof (a) If U = {0} then A = {0}, L0 = L0 (A) = {0} and the result is trivial; I shall therefore suppose henceforth that U is non-trivial. Note that by 352Q A is Dedekind complete. Let C ⊆ U + \ {0} be a maximal disjoint set (in the sense of 352C); to obtain such a set apply Zorn’s lemma to the family of all disjoint subsets of U + \ {0}. Now I can write down the formula for the embedding T : U → L0 immediately, though there will be a good deal of work to do in justification: for u ∈ U and α ∈ R, [[T u > α]] will be the band in U generated by

Embedding Riesz spaces in L0

368E

377

{e ∧ (u − αe)+ : e ∈ C}. (For once, I allow myself to use the formula [[. . . ]] without checking immediately that it represents a member of L0 ; all I claim for the moment is that [[T u > α]] is a member of A determined by u and α.) (b) Before getting down to the main argument, I make some remarks which will be useful later. (i) If u > 0 in U , then there is some e ∈ C such that u ∧ e > 0, since otherwise we ought to have added u to C. Thus C ⊥ = {0}. (ii) If u ∈ U and e ∈ C and α ∈ R, then v = e ∧ (αe − u)+ belongs to [[T u > α]]⊥ . P P If e0 ∈ C, then 0 either e 6= e so v ∧ e0 ∧ (u − αe0 )+ ≤ e ∧ e0 = 0, or e0 = e and v ∧ e0 ∧ (u − αe0 )+ ≤ (αe − u)+ ∧ (u − αe)+ = 0. Accordingly [[T u > α]] is included in the band {v}⊥ and v ∈ [[T u > α]]⊥ . Q Q (c) Now I must confirm that the formula given for [[T u > α]] is consistent with the conditions laid down in 364A. P P Take u ∈ U . (i) If α ≤ β then 0 ≤ e ∧ (u − βe)+ ≤ e ∧ (u − αe)+ ∈ [[T u > α]] so e ∧ (u − βe)+ ∈ [[T u > α]] for every e ∈ C and [[T u > β]] ⊆ [[T u > α]]. (ii) Given α ∈ R, set W = supβ>α [[T u > β]] in A, that is, the band in U generated by {e ∧ (u − βe)+ : e ∈ C, β > α}. Then for each e ∈ C, supβ>α e ∧ (u − βe)+ = e ∧ (u − inf β>α e)+ = e ∧ (u − αe)+ using the general distributive laws in U (352E), the translation-invariance of the order (351D) and the fact that U is Archimedean (to see that αe = inf β>α βe). So e ∧ (u − αe)+ ∈ W ; as e is arbitrary, [[T u > α]] ⊆ W and [[T u > α]] = W . (iii) Now set W = inf n∈N [[T u > n]]. For any e ∈ C, n ∈ N we have e ∧ (ne − u)+ ∈ [[T u > n]]⊥ ⊆ W ⊥ , so that e ∧ (e − n1 u+ )+ ≤ e ∧ (e − n1 u)+ ∈ W ⊥ for every n ≥ 1 and e = supn≥1 e ∧ (e − n1 u+ )+ ∈ W ⊥ . Thus C ⊆ W ⊥ and W ⊆ C ⊥ = {0}. So we have inf n∈N [[T u > n]] = 0. (iv) Finally, set W = supn∈N [[T u > −n]]. Then e ∧ (e − n1 u− )+ ≤ e ∧ (e + n1 u)+ ≤ e ∧ (u + ne)+ ∈ W for every n ≥ 1 and e ∈ C, so e = supn≥1 e ∧ (e − n1 u− )+ ∈ W for every e ∈ C and W ⊥ = {0}, W = U . Thus all three conditions of 364A are satisfied. Q Q (d) Thus we have a well-defined map T : U → L0 . I show next that T (u + v) = T u + T v for all u, v ∈ U . P P I rely on the formulae in 364E and 364Fa, and on partitions of unity in A, constructed as follows. Fix n ≥ 1 for the moment. Then we know that supi∈Z [[T u > So setting

i n ]]

= 1,

inf i∈Z [[T u >

i n ]]

= 0.

378

Function spaces

Vi = [[T u >

i n ]] \ [[T u

i+1 n ]]

>

368E

= [[T u >

i n ]]

∩ [[T u >

i+1 ⊥ n ]] ,

hVi ii∈Z is a partition of unity in A. Similarly, hWi ii∈Z is a partition of unity, where i n ]]

Wi = [[T v >

i+1 ⊥ n ]] .

∩ [[T v >

Now, for any i, j, k ∈ Z such that i + j ≥ k, Vi ∩ Wj ⊆ [[T u >

∩ [[T v >

j n ]]

⊆ [[T u + T v >

[[T u + T v >

k n ]]

⊇ supi+j≥k Vi ∩ Wj .

i n ]]

i+j n ]]

⊆ [[T u + T v >

k n ]];

thus

On the other hand, if q ∈ Q and k ∈ Z, there is an i ∈ Z such that [[T u > q]] ∩ [[T v >

k+1 n

− q]] ⊆ [[T u >

i n ]]

∩ [[T v >

i n

≤q


k+1 n ]]

⊆ supi+j≥k Vi ∩ Wj ⊆ [[T u + T v >

k n ]].

Also, if 0 < w ∈ Vi ∩ Wj , e ∈ C then w ∧ e ∧ (u −

i+1 + e) n

= w ∧ e ∧ (v −

j+1 + e) n

= 0,

so that w ∧ e ∧ (u + v −

i+j+2 + e) n

=0

because (u + v −

i+j+2 + e) n

by 352Fc. But this means that Vi ∩ Wj ∩ [[T (u + v) > [[T (u + v) >

i+1 + e) n

≤ (u −

k+1 n ]]

i+j+2 n ]]

+ (v −

j+1 + e) n

= {0}. Turning this round,

∩ supi+j≤k−1 Vi ∩ Wj = 0,

and because supi,j∈Z Vi ∩ Wj = U in A, [[T (u + v) >

k+1 n ]]

⊆ supi+j≥k Vi ∩ Wj .

Finally, if i + j ≥ k and 0 < w ∈ Vi ∩ Vj , then there is an e ∈ C such that w1 = w ∧ e ∧ (u − ni e)+ > 0; there is an e0 ∈ C such that w2 = w1 ∧ e0 ∧ (v − nj e0 )+ > 0; of course e = e0 , and i n

j n

i+j + e) n

0 < w2 ≤ e ∧ (u − e)+ ∧ (v − )+ ≤ e ∧ (u + v − ∈ [[T (u + v) > using 352Fc. This shows that w 6∈ [[T (u + v) >

k ⊥ n ]] ;

i+j ]] n

k n

⊆ [[T (u + v) > ]]

as w is arbitrary, Vi ∩ Wj ⊆ [[T (u + v) >

supi+j≥k Vi ∩ Wj ⊆ [[T (u + v) >

k n ]].

Putting all these four together, we see that [[T (u + v) >

k+1 n ]]

⊆ supi+j≥k Vi ∩ Wj ⊆ [[T u + T v >

k n ]],

[[T u + T v >

k+1 n ]]

⊆ supi+j≥k Vi ∩ Wj ⊆ [[T (u + v) >

k n ]]

for all n ≥ 1, k ∈ Z. But this means that we must have [[T u + T v > β]] ⊆ [[T (u + v) > α]], whenever α < β. Consequently

[[T (u + v) > β]] ⊆ [[T u + T v > α]]

k n ]];

so we get

Embedding Riesz spaces in L0

368F

379

[[T u + T v > α]] = sup [[T u + T v > β]] ⊆ [[T (u + v) > α]] β>α

= sup [[T (u + v) > β]] ⊆ [[T u + T v > α]] β>α

and [[T u + T v > α]] = [[T (u + v) > α]] for every α, that is, T (u + v) = T u + T v. Q Q (e) The hardest part is over. If u ∈ U , γ > 0 and α ∈ R, then for any e ∈ C 1 γ

α γ

1 γ

min(1, )(e ∧ (γu − αe)+ ) ≤ e ∧ (u − e)+ ≤ max(1, )(e ∧ (γu − αe)+ ), so [[T (γu) > α]] = [[T u >

α γ ]]

= [[γT u > α]];

as α is arbitrary, γT u = T (γu); as γ and u are arbitrary, T is linear. (We need only check linearity for γ > 0 because we know from the additivity of T that T (−u) = −T u for every u.) (f ) To see that T is a Riesz homomorphism, take any u ∈ U and α ∈ R and consider the band [[T u > α]] ∪ [[−T u > α]] = [[|T u| > α]] (by 364Mb). This is the band generated by {e ∧ (u − αe)+ : e ∈ C} ∪ {e ∧ (−u − αe)+ : e ∈ C}. But this must also be the band generated by {(e ∧ (u − αe)+ ) ∨ (e ∧ (−u − αe)+ ) : e ∈ C} = {e ∧ (|u| − αe)+ : e ∈ C}, which is [[T |u| > α]]. Thus [[|T u| > α]] = [[T |u| > α]] for every α and |T u| = T |u|. As u is arbitrary, T is a Riesz homomorphism. (g) To see that T is injective, take any non-zero u ∈ U . Then there must be some e ∈ C such that |u| ∧ e 6= 0, and some α > 0 such that |u| ∧ e 6≤ αe, so that e ∧ (|u| − αe)+ 6= 0 and [[T |u| > α]] 6= {0} and T |u| 6= 0 and T u 6= 0. Thus T embeds U as a Riesz subspace of L0 . (h) Finally, I must check that T [U ] is order-dense in L0 . P P Let p > 0 in L0 . Then there is some α > 0 such that V = [[p > α]] 6= 0. Take u > 0 in V . Let e ∈ C be such that u ∧ e > 0. Then v = u ∧ αe > 0. Now e ∧ (v − αe)+ = 0; but also e0 ∧ v = 0 for every e0 ∈ C distinct from e, so that [[T v > α]] = {0}. Also v ∈ V , so e0 ∧ (v − βe0 )+ ∈ V whenever e0 ∈ C and β ≥ 0, and [[T v > β]] ⊆ V for every β ≥ 0. Accordingly we have [[T v > β]] = {0} ⊆ [[p > β]] if β ≥ α, ⊆ V ⊆ [[p > β]] if 0 ≤ β < α, = U = [[p > β]] if β < 0, and T v ≤ p. Also T v > 0, by (g). As p is arbitrary, T [U ] is order-dense in L0 . Q Q 368F Corollary A Riesz space U is Archimedean iff it is isomorphic T to a Riesz subspace of some reduced power RX |F, where X is a set and F is a filter on X such that n∈N Fn ∈ F whenever hFn in∈N is a sequence in F. proof (a) If U is an Archimedean Riesz space, then by 368E there is a space of the form L0 = L0 (A) such that U can be embedded into L0 . As in the proof of 364E, L0 is isomorphic to some space of the form L0 (Σ)/W, where Σ is a σ-algebra of subsets of some set X and W = {f : f ∈ L0 , {x : f (x) 6= T 0} ∈ I}, I being a σ-ideal of Σ. But now F = {A : A ∪ E = X for some E ∈ I} is a filter on X such that n∈N Fn ∈ F for every sequence hFn in∈N in F. (I am passing over the trivial case X ∈ I, since then U must be {0}.) And L0 /W is (isomorphic to) the image of L0 in RX |F, since W = {f : f ∈ L0 , {x : f (x) = 0} ∈ F}. Thus U is isomorphic to a Riesz subspace of RX |F. (b) On the other hand, if F is a filter on X closed under countable intersections, then W = {f : f ∈ RX , {x : f (x) = 0} ∈ F} is a sequentially order-closed solid linear subspace of the Dedekind σ-complete Riesz space RX , so that RX |F = RX /W is Dedekind σ-complete (353J(a-iii)) and all its Riesz subspaces must be Archimedean (353H, 351Rc).

380

Function spaces

368G

368G Corollary Every Archimedean Riesz space U is isomorphic to an order-dense Riesz subspace of some space C ∞ (X), where X is an extremally disconnected compact Hausdorff space. proof Let Z be the Stone space of the band algebra A of U . Because A is Dedekind complete (352Q), Z is extremally disconnected and A can be identified with the regular open algebra G of Z (314S). By 364W, L0 (G) can be identified with C ∞ (Z). So the embedding of U as an order-dense Riesz subspace of L0 (A) (368E) can be regarded as an embedding of U as an order-dense Riesz subspace of C ∞ (Z). 368H Corollary Any Dedekind complete Riesz space U is isomorphic to an order-dense solid linear subspace of L0 (A) for some Dedekind complete Boolean algebra A. proof Embed U in L0 = L0 (A) as in 368E; because U is order-dense in L0 and (in itself) Dedekind complete, it is solid (353K). 368I Corollary Let U be an Archimedean Riesz space. Then U can be embedded as an order-dense Riesz subspace of a Dedekind complete Riesz space V in such a way that the solid linear subspace of V generated by U is V itself, and this can be done in essentially only one way. If W is any other Dedekind complete Riesz space and T : U → W is an order-continuous positive linear operator, there is a unique positive linear operator T˜ : V → W extending T . proof By 368E, we may suppose that U is actually an order-dense Riesz subspace of L0 (A), where A is a Dedekind complete Boolean algebra. In this case, we can take V to be the solid linear subspace generated by U , that is, {v : |v| ≤ u for some u ∈ U }; being a solid linear subspace of the Dedekind complete Riesz space L0 (A), V is Dedekind complete, and of course U is order-dense in V . If W is any other Dedekind complete Riesz space and T : U → W is an order-continuous positive linear operator, then for any v ∈ V + there is a u0 ∈ U such that v ≤ u0 , so that T u0 is an upper bound for {T u : u ∈ U, 0 ≤ u ≤ v}; as W is Dedekind complete, supu∈U,0≤u≤v T u is defined in W . By 355F, T has a unique extension to an order-continuous positive linear operator from V to W . In particular, if V1 is another Dedekind complete Riesz space in which U can be embedded as an orderdense Riesz subspace, this embedding of U extends to an embedding of V ; since V is Dedekind complete, its copy in V1 must be a solid linear subspace, so if V1 is the solid linear subspace of itself generated by U , we get an identification between V and V1 , uniquely determined by the embeddings of U in V and V1 . 368J Definition If U is an Archimedean Riesz space, a Dedekind completion of U is a Dedekind complete Riesz space V together with an embedding of U in V as an order-dense Riesz subspace of V such that the solid linear subspace of V generated by U is V itself. 368I tells us that every Archimedean Riesz space U has an essentially unique Dedekind completion, so that we may speak of ‘the’ Dedekind completion of U . 368K

This is a convenient point at which to give a characterization of the Riesz spaces L0 (A).

Lemma Let A be a Dedekind σ-complete Boolean algebra. Suppose that A ⊆ L0 (A)+ is disjoint. If either A is countable or A is Dedekind complete, A is bounded above in L0 (A). proof If A = ∅, this is trivial; suppose that A is not empty. For n ∈ N, set an = supu∈A [[u > n]]; this is always defined; set a = inf n∈N an . Now a = 0. P P?? Otherwise, there must be a u ∈ A such that a0 = a ∩ [[u > 0]] 6= 0, since a ⊆ a0 . But now, for any n, and any v ∈ A \ {u}, a0 ∩ [[v > n]] ⊆ [[u > 0]] ∩ [[v > 0]] = 0, so that a0 ⊆ [[u > n]]. As n is arbitrary, inf n∈N [[u > n]] 6= 0, which is impossible. X XQ Q By 364Ma, A is bounded above. 368L Definition A Riesz space U is called laterally complete or universally complete if A is bounded above whenever A ⊆ U + is disjoint.

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381

368M Theorem Let U be an Archimedean Riesz space. Then the following are equiveridical: (i) there is a Dedekind complete Boolean algebra A such that U is isomorphic to L0 (A); (ii) U is Dedekind σ-complete and laterally complete; (iii) whenever V is an Archimedean Riesz space, V0 is an order-dense Riesz subspace of V and T0 : V0 → U is an order-continuous Riesz homomorphism, there is a positive linear operator T : V → U extending T0 . proof (a)(i)⇒(ii) and (i)⇒(iii) are covered by 368K and 368B. (b)(ii)⇒(i) Assume (ii). By 368E, we may suppose that U is actually an order-dense Riesz subspace of L0 = L0 (A) for a Dedekind complete Boolean algebra A. α) If u ∈ U + and a ∈ A then u × χa ∈ U . P (α P Set A = {v : v ∈ U, 0 ≤ v ≤ χa}, and let C ⊆ A be a maximal disjoint set; then w = sup C is defined in U , and is also the supremum in L0 . Set b = [[w > 0]]. As w ≤ χa, b ⊆ a. ?? If b 6= a, then χ(a \ b) > 0, and there is a v 0 ∈ U such that 0 < v 0 ≤ χ(a \ b); but now v 0 ∈ A and v 0 ∧ w = 0, so v 0 ∧ v = 0 for every v ∈ C, and we ought to have added v 0 to C. X X Thus [[w > 0]] = a. Now consider u0 = supn∈N u ∧ nw; as U is Dedekind σ-complete, u0 ∈ U . Since [[u0 > 0]] ⊆ a, u0 ≤ u × χa. On the other hand, u × χ[[w >

1 n ]]

× χ[[u ≤ n]] ≤ u ∧ n2 w ≤ u0

for every n ≥ 1, so, taking the supremum over n, u × χa ≤ u0 . Accordingly u × χa = u0 ∈ U , as required. Q Q β ) If w ≥ 0 in L0 , there is a u ∈ U such that 21 w ≤ u ≤ w. P P Set (β A = {u : u ∈ U, 0 ≤ u ≤ w}, C = {a : a ∈ A, a ⊆ [[u − 12 w ≥ 0]] for some u ∈ A}. Then sup A = w, so C is order-dense in A. (If a ∈ A \ {0}, either a ∩ [[w > 0]] = 0 and a ⊆ [[0 − 12 w ≥ 0]], so a ∈ C, or there is a u ∈ U such that 0 < u ≤ w × χa. In the latter case there is some n such that 2n u ≤ w and 2n+1 u 6≤ w, and now c = a ∩ [[2n u − 21 w ≥ 0]] is a non-zero member of C included in a.) Let D ⊆ C be a partition of unity and for each d ∈ D choose ud ∈ A such that d ⊆ [[ud − 12 w ≥ 0]]. By (α), ud × χd ∈ U for every d ∈ D, so u = supd∈D ud × χd ∈ U . Now u ≤ w, but also [[u − 21 w ≥ 0]] ⊇ d for every d ∈ D, so is equal to 1, and u ≥ 21 w, as required. Q Q (γγ ) Given w ≥ 0 in L0 , we can therefore choose hun in∈N , hvn in∈N inductively such that v0 = 0 and un ∈ U ,

1 (w 2

− vn ) ≤ un ≤ w − vn ,

vn+1 = vn + un

for every n ∈ N. Now hvn in∈N is a non-decreasing sequence in U and w − vn ≤ 2−n w for every n, so w = supn∈N vn ∈ U . As w is arbitrary, (L0 )+ ⊆ U and U = L0 is of the right form. (c)(iii)⇒(i) As in (b), we may suppose that U is an order-dense Riesz subspace of L0 . But now apply condition (iii) with V = L0 , V0 = U and T0 the identity operator. There is an extension T : L0 → U . If v ≥ 0 in L0 , T v ≥ T0 = u whenever u ∈ U and u ≤ v, so T v ≥ v, since v = sup{u : u ∈ U, 0 ≤ u ≤ v} in L0 . Similarly, T (T v − v) ≥ T v − v. But as T v ∈ U , T (T v) = T v and T (T v − v) = 0, so v = T v ∈ U . As v is arbitrary, U = L0 . 368N Weakly (σ, ∞)-distributive Riesz spaces We are now ready to look at the class of Riesz spaces corresponding to the weakly (σ, ∞)-distributive Boolean algebras of §316. Definition Let U be a Riesz space. Then U is weakly (σ, ∞)-distributive if whenever hAn in∈N is a S sequence of non-empty downwards-directed subsets of U + , each with infimum 0, and n∈N An has an upper bound in U , then {u : u ∈ U , for every n ∈ N there is a v ∈ An such that v ≤ u}

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has infimum 0 in U . S Remark Because the definition looks only at sequences hAn in∈N such that n∈N An is order-bounded, we can invert it, as follows: a Riesz space U is weakly (σ, ∞)-distributive iff whenever hAn in∈N is a sequence of non-empty upwards-directed subsets of U + , all with supremum u0 , then {u : u ∈ U + , for every n ∈ N there is a v ∈ An such that u ≤ v} also has supremum u0 . 368O Lemma Let U be an Archimedean Riesz space. Then the following are equiveridical: (i) U is not weakly (σ, ∞)-distributive; (ii) there are a u > 0 in U and a sequence hAn in∈N of non-empty downwards-directed sets, all with infimum 0, such that supn∈N un = u whenever un ∈ An for every n ∈ N. proof (ii)⇒(i) is immediate from the definition of ‘weakly (σ, ∞)-distributive’. For (i)⇒(ii), suppose that U is not weakly (σ, ∞)-distributive.S Then there is a sequence hAn in∈N of non-empty downwards-directed sets, all with infimum 0, such that n∈N An is bounded above, but A = {w : w ∈ U , for every n ∈ N there is a v ∈ An such that v ≤ w} does not have infimum 0. Let u > 0 be a lower bound for A, and set A0n = {u ∧ v : v ∈ An } for each n ∈ N. Then each A0n is a non-empty downwards-directed set with infimum 0. Let hun in∈N be a sequence such that un ∈ A0n for every n. Express each un as u ∧ vn where vn ∈ An . Let B be the set of upper bounds of {vn : n ∈ N}. Then inf w∈B,n∈N w − vn = 0, because U is Archimedean (353F), while B ⊆ A, so u ≤ w for every w ∈ B. If u0 is any upper bound for {un : n ∈ N}, then u − u0 ≤ u − u ∧ vn = (u − vn )+ ≤ (w − vn )+ = w − vn for every n ∈ N, w ∈ B. So u0 ≥ u. Thus u = supn∈N un . As hun in∈N is arbitrary, u and hA0n in∈N witness that (ii) is true. 368P Proposition (a) A regularly embedded Riesz subspace of an Archimedean weakly (σ, ∞)-distributive Riesz space is weakly (σ, ∞)-distributive. (b) An Archimedean Riesz space with a weakly (σ, ∞)-distributive order-dense Riesz subspace is weakly (σ, ∞)-distributive. (c) If U is a Riesz space such that U × separates the points of U , then U is weakly (σ, ∞)-distributive; in particular, U ∼ and U × are weakly (σ, ∞)-distributive for every Riesz space U . proof (a) Suppose that U is an Archimedean Riesz space and that V ⊆ U is a regularly embedded Riesz subspace which is not weakly (σ, ∞)-distributive. Then 368O tells us that there are a v > 0 in V and a sequence hAn in∈N of non-empty downwards-directed subsets of V , all with infimum 0 in V , such that supn∈N vn = v in V whenever vn ∈ An for every n ∈ N. Because V is regularly embedded in U , inf An = 0 Q in U for every n and supn∈N vn = v in U for every sequence hvn in∈N ∈ n∈N An , so U is not weakly (σ, ∞)distributive. Turning this round, we have (a). (b) Let U be an Archimedean Riesz space which is not weakly (σ, ∞)-distributive, and V an orderdense Riesz subspace of U . By 368O again, there are a u > 0 in U and a sequence hAn in∈N of non-empty downwards-directed sets in U , all with infimum 0, such that supn∈N un = u whenever un ∈ An for every n. Let v ∈ V be such that 0 < v ≤ u. Set Bn = {w : w ∈ V , there is some u ∈ An such that v ∧ u ≤ w ≤ v} for each n ∈ N. Because An is downwards-directed, w ∧ w0 ∈ Bn for all w, w0 ∈ Bn ; v ∈ Bn , so Bn 6= ∅; and inf Bn = 0 in V . P P Setting C = {w : w ∈ V + , there is some u ∈ An such that w ≤ (v − u)+ }, then (because V is order-dense) any upper bound for C in U is also an upper bound of {(v − u)+ : u ∈ An }. But supu∈An (v − u)+ = (v − inf An )+ = v,

368Q

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so v = sup C in U and inf Bn = inf{v − w : w ∈ C} = 0 in U and in V . Q Q Now if vn ∈ Bn for every n ∈ N, we can choose un ∈ An such that v ∧ un ≤ vn ≤ v for every n, so that v = v ∧ u = v ∧ supn∈N un = supn∈N v ∧ un ≤ supn∈N vn ≤ v, and v = supn∈N vn . Thus hBn in∈N witnesses that V is not weakly (σ, ∞)-distributive. (c) Now suppose that U × separates the points of U . In this case U is surely Archimedean (356G). ?? If U is not weakly (σ, ∞)-distributive, there are a u > 0 in U and a sequence hAn in∈N of non-empty downwardsdirected sets, all with infimum 0, such that supn∈N un = u whenever un ∈ An for each n. Take f ∈ U × such that f (u) 6= 0; replacing f by |f | if necessary, we may suppose that f > 0. Set δ = f (u) > 0. For each n ∈ N, there is a un ∈ An such that f (un ) ≤ 2−n−2 δ. But in this case hsupi 0 in S and a sequence hAn in∈N of non-empty downwards-directed sets in S, all with infimum 0, such that u = supn∈N un whenever un ∈ An for every n. Let α > 0 be such that c = [[u > α]] 6= 0 (361Eg), and consider Bn = {[[v > α]] : v ∈ An } ⊆ A for each n ∈ N. Then each Bn is downwards-directed (because An is), and inf Bn = 0 in A (because if b is a lower bound of Bn , αχb ≤ v for every v ∈ An ). Because A is weakly (σ, ∞)-distributive, there must be some a ∈ A such that a 6⊇ c but there is, for every n ∈ N, a bn ∈ Bn such that a ⊇ bn . Take vn ∈ An such that bn = [[vn > α]], so that vn ≤ αχ1 ∨ kvn k∞ χbn ≤ αχ1 ∨ kuk∞ χa. Since u = supn∈N vn , u ≤ αχ1 ∨ kuk∞ χa. But in this case c = [[u > α]] ⊆ a, contradicting the choice of a. X X Thus S must be weakly (σ, ∞)-distributive if A is. (ii) Now suppose that S is weakly (σ, ∞)-distributive, and let hBn in∈N be a sequence of non-empty downwards-directed subsets of A, all with infimum 0. Set An = {χb : b ∈ Bn } for each n; then An ⊆ S is non-empty, downwards-directed and has infimum 0 in S, because χ : A → S is order-continuous (361Ef). Set A = {v : v ∈ S, for every n ∈ N there is a u ∈ An such that u ≤ v}, B = {b : b ∈ A, for every n ∈ N there is an a ∈ Bn such that a ⊆ b}. ?? If 0 is not the greatest lower bound of B, take a non-zero lower bound c. P Because S is weakly (σ, ∞)n distributive, inf A = 0, and there is a v ∈ A such that χc 6≤ v. Express v as i=0 αi χai , where hai ii≤n is disjoint, and set a = sup{ai : i ≤ n, αi ≥ 1}; then χa ≤ v, so c 6⊆ a. For each n there is a bn ∈ Bn such that χbn ≤ v. But in this case bn ⊆ a for each n ∈ N, so that a ∈ B; which means that c is not a lower bound for B. X X Thus inf B = 0 in A. As hBn in∈N is arbitrary, A is weakly (σ, ∞)-distributive.

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368Q

(iii) Thus S is weakly (σ, ∞)-distributive iff A is. But S is order-dense in L∞ = L∞ (A) (363C), therefore regularly embedded (352Ne), so 368Pa-b tell us that S is weakly (σ, ∞)-distributive iff L∞ is. (b) In the same way, because S can be regarded as an order-dense Riesz subspace of L0 = L0 (A) (364K), L is weakly (σ, ∞)-distributive iff S is, that is, iff A is. 0

368R Corollary An Archimedean Riesz space is weakly (σ, ∞)-distributive iff its band algebra is weakly (σ, ∞)-distributive. proof Let U be an Archimedean Riesz space and A its band algebra. By 368E, U is isomorphic to an order-dense Riesz subspace of L0 = L0 (A). By 368P, U is weakly (σ, ∞)-distributive iff L0 is; and by 368Qb L0 is weakly (σ, ∞)-distributive iff A is. 368S Corollary If (A, µ ¯) is a semi-finite measure algebra, any regularly embedded Riesz subspace (in particular, any solid linear subspace and any order-dense Riesz subspace) of L0 (A) is weakly (σ, ∞)distributive. proof By 322F, A is weakly (σ, ∞)-distributive; by 368Qb, L0 (A) is weakly (σ, ∞)-distributive; by 368Pa, any regularly embedded Riesz subspace is weakly (σ, ∞)-distributive. 368X Basic exercises (a) Let X be an uncountable set and Σ the countable-cocountable σ-algebra of subsets of X. Show that there is a family A ⊆ L0 = L0 (Σ) such that u ∧ v = 0 for all distinct u, v ∈ A but A has no upper bound in L0 . Show moreover that if w > 0 in L0 then there is an n ∈ N such that nw 6= supu∈A u ∧ nw. b its Dedekind completion (314U). Show that L∞ (A) b can be (b) Let A be any Boolean algebra, and A ∞ identified with the Dedekind completions of S(A) and L (A). (c) Explain how to prove 368K from 368A. (d) Show that any product of weakly (σ, ∞)-distributive Riesz spaces is weakly (σ, ∞)-distributive. (e) Let A be a Dedekind complete weakly (σ, ∞)-distributive Boolean algebra. Show that a set A ⊆ L0 = L (A) is order-bounded iff h2−n un in∈N order*-converges to 0 in L0 whenever hun in∈N is a sequence in A. (Hint: use 368A. If v > 0 and v = supu∈A v ∧ 2−n u for every n, we can find a w > 0 and a sequence hun in∈N in A such that w ≤ 2−n un for every n.) 0

(f ) Give a direct proof of 368S, using the ideas of 322F, but not relying on it or on 368Q. 368Y Further exercises (a) (i) Use 364U-364V to show that if X is any compact Hausdorff space then C(X) can be regarded as an order-dense Riesz subspace of L0 (G), where G is the regular open algebra of X. (ii) Use 353M to show that any Archimedean Riesz space with order unit can be embedded as an order-dense Riesz subspace of some L0 (G). (iii) Let U be an Archimedean Riesz space and C ⊆ U + a maximal disjoint set, as in part (a) of the proof of 368E. For e ∈ C let Ue be the solid linear subspace of U generated by e, and let V be the solidQlinear subspace of U generated by C. Show that Q Q V can be embedded as an order-dense Riesz subspace of e∈C Ue and therefore in e∈C L0 (Ge ) ∼ = L0 ( e∈C Ge ) for a suitable family of regular open algebras Ge . (iv) Now use 368B to complete a proof of 368E. (b) Let U be any Archimedean Riesz space. Let V be the family of pairs (A, B) of non-empty subsets of U such that B is the set of upper bounds of A and A is the set of lower bounds of B. Show that V can be given the structure of a Dedekind complete Riesz space defined by the formulae (A1 , B1 ) + (A2 , B2 ) = (A, B) iff A1 + A2 ⊆ A, B1 + B2 ⊆ B, α(A, B) = (αA, αB) if α > 0, (A1 , B1 ) ≤ (A2 , B2 ) iff A1 ⊆ A2 . Show that u 7→ (]−∞, u] , [u, ∞[) defines an embedding of U as an order-dense Riesz subspace of V, so that V may be identified with the Dedekind completion of U .

368 Notes

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(c) Work through the proof of 364U when X is compact, Hausdorff and extremally disconnected, and show that it is easier than the general case. Hence show that 368Yb can be used to shorten the proof of 368E sketched in 368Ya. (d) Let U be a Riesz space. Show that the following are equiveridical: (i) U is isomorphic, as Riesz space, to L0 (A) for some Dedekind σ-complete Boolean algebra A (ii) U is Dedekind σ-complete and has a weak order unit and whenever A ⊆ U + is countable and disjoint then A is bounded above in U . (e) Let U be a weakly (σ, ∞)-distributive Riesz space and V a Riesz subspace of U which is either solid or order-dense. Show that V is weakly (σ, ∞)-distributive. (f ) Show that C([0, 1]) is not weakly (σ, ∞)-distributive. (Compare 316K.) (g) Let A be a ccc weakly (σ, ∞)-distributive Boolean algebra. Suppose we have a double sequence haij i(i,j)∈N×N in A such that haij ij∈N order*-converges to ai in A for each i, while hai ii∈N order*-converges to a. Show that there is a strictly increasing sequence hn(i)ii∈N such that hai,n(i) ii∈N order*-converges to a. (h) Let U be a weakly (σ, ∞)-distributive Riesz space with the countable sup property. Suppose we have an order-bounded double sequence huij i(i,j)∈N×N in U such that huij ij∈N order*-converges to ui in U for each i, while hui ii∈N order*-converges to u. Show that there is a strictly increasing sequence hn(i)ii∈N such that hui,n(i) ii∈N order*-converges to u. (i) Let A be a ccc weakly (σ, ∞)-distributive Dedekind complete Boolean algebra. Show that there is a topology on L0 = L0 (A) such that the closure of any A ⊆ L0 is precisely the set of order*-limits of sequences in A. (j) Let U be a weakly (σ, ∞)-distributive Riesz space and f : U → R a positive linear functional; write fτ for the component of f in U × . (i) Show that for any u ∈ U + there is an upwards-directed A ⊆ [0, u], with supremum u, such that fτ (u) = supv∈A f (v). (See 356Xe, 362D.) (ii) Show that if f is strictly positive, so is fτ . (Compare 391D.) 368 Notes and comments 368A-368B are manifestations of a principle which will reappear in §375: Dedekind complete L0 spaces are in some sense ‘maximal’. If we have an order-dense subspace U of such an L0 , then any Archimedean Riesz space including U as an order-dense subspace can itself be embedded in L0 (368B). In fact this property characterizes Dedekind complete L0 spaces (368M). Moreover, any Archimedean Riesz space U can be embedded in this way (368E); by 368C, the L0 space (though not the embedding) is unique up to isomorphism. If U and V are Archimedean Riesz spaces, each embedded as an order-dense Riesz subspace of a Dedekind complete L0 space, then any order-continuous Riesz homomorphism from U to V extends uniquely to the L0 spaces (368B). If one Dedekind complete L0 space is embedded as an order-dense Riesz subspace of another, they must in fact be the same (368D). Thus we can say that every Archimedean Riesz space U can be extended to a Dedekind complete L0 space, in a way which respects order-continuous Riesz homomorphisms, and that this extension is maximal, in that U cannot be order-dense in any larger space. The proof of 368E which I give is long because I am using a bare-hands approach. Alternative methods shift the burdens. For instance, if we take the trouble to develop a direct construction of the ‘Dedekind completion’ of a Riesz space (368Yb), then we need prove the theorem only for Dedekind complete Riesz spaces. A more substantial aid is the representation theorem for Archimedean Riesz spaces with order unit (353M); I sketch an argument in 368Ya. The drawback to this approach is the proof of Theorem 364U, which seems to be quite as long as the direct proof of 368E which I give here. Of course we need 364U only for compact Hausdorff spaces, which are usefully easier than the general case (364V, 368Yc). 368G is a version of Ogasawara’s representation theorem for Archimedean Riesz spaces. Both this and 368F can be regarded as expressions of the principle that an Archimedean Riesz space is ‘nearly’ a space of functions. I have remarked before on the parallels between the theories of Boolean algebras and Archimedean Riesz spaces. The notion of ‘weak (σ, ∞)-distributivity’ is one of the more striking correspondences. (Compare, for

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368 Notes

instance, 316Xm with 368Pa.) What is really important to us, of course, is the fact that the function spaces of measure theory are mostly weakly (σ, ∞)-distributive, by 368S. Of course this is easy to prove directly (368Xf), but I think that the argument through 368Q gives a better idea of what is really happening here. Some of the features of ‘order*-convergence’, as defined in §367, are related to weak (σ, ∞)-distributivity; in 368Yi I describe a topology which can be thought of as an abstract version of the topology of convergence in measure on the L0 space of a σ-finite measure algebra (367N).

369 Banach function spaces In this section I continue the work of §368 with results which involve measure algebras. The first step is a modification of the basic representation theorem for Archimedean Riesz spaces. If U is any Archimedean Riesz space, it can be represented as a subspace of L0 = L0 (A), where A is its band algebra (368E); now if U × separates the points of U , there is a measure rendering A a localizable measure algebra (369A, 369Xa). Moreover, we get a simultaneous representation of U × as a subspace of L0 (369C-369D), the duality between them corresponding exactly to the familiar duality between Lp and Lq . Still drawing inspiration from the classical Lp spaces, we have a general theory of ‘associated Fatou norms’ (369F-369M, 369R). I include notes on the spaces M 1,∞ , M ∞,1 and M 1,0 (369N-369Q), which will be particularly useful in the next chapter. 369A Theorem Let U be a Riesz space such that U × separates the points of U . Then U can be embedded as an order-dense Riesz subspace of L0 (A) for some localizable measure algebra (A, µ ¯). proof (a) Consider the canonical map S : U → U ×× . We know that this is a Riesz homomorphism onto an order-dense Riesz subspace of U ×× (356I). Because U × separates the points of U , S is injective. Let A be the band algebra of U ×× and T : U ×× → L0 = L0 (A) an injective Riesz homomorphism onto an order-dense Riesz subspace V of L0 , as in 368E. The composition T S : U → L0 is now an injective Riesz homomorphism, so embeds U as a Riesz subspace of L0 , which is order-dense because V is order-dense in L0 and T S[U ] is order-dense in V (352Nc). Thus all that we need to find is a measure µ ¯ on A rendering it a localizable measure algebra. (b) Note that V is isomorphic, as Riesz space, to U ×× , which is Dedekind complete (356B), so V must be solid in L0 (353K). Also V × must separate the points of V (356L). Let D be the set of those d ∈ A such that the principal ideal Ad is measurable in the sense that there is some ν¯ for which (Ad , ν¯) is a totally finite measure algebra. Then D is order-dense in A. P P Take any non-zero a ∈ A. Because V is order-dense, there is a non-zero v ∈ V such that v ≤ χa. Take h ≥ 0 in V × such that h(v) > 0. Then there is a v 0 such that 0 < v 0 ≤ v and h(w) > 0 whenever 0 < w ≤ v 0 in V (356H). Let α > 0 be such that d = [[v 0 > α]] 6= 0. Then χb ≤ α1 v 0 ∈ V whenever b ∈ Ad . Set ν¯b = h(χb) ∈ [0, ∞[ for b ∈ Ad . Because the map b 7→ χb : A → L0 is additive and order-continuous, the map b P 7→ χb : Ad → V also ∞ is, and ν¯ = hχ must be additive and order-continuous; in particular, ν¯(supn∈N bn ) = n=0 ν¯bn whenever 0 hbn in∈N is a disjoint sequence in Ad . Moreover, if b ∈ Ad is non-zero, then 0 < αχb ≤ v , so ν¯b = h(χb) > 0. Thus (Ad , ν¯) is a totally finite measure algebra, and d ∈ D, while 0 6= d ⊆ a. As a is arbitrary, D is order-dense. Q Q (c) By 313K, there is a partition of unity C ⊆ D. For each c ∈ C, let ν¯c : Ac → [0, ∞[ be P a functional such that (Ac , ν¯c ) is a totally finite measure algebra. Define µ ¯ : A → [0, ∞]Pby setting µ ¯a = c∈C ν¯c (a ∩ c) for every a ∈ A. Then (A, µ ¯) is a localizable measure algebra. P P (i) µ ¯0 = c∈C ν¯0 = 0. (ii) If han in∈N is a disjoint sequence in A with supremum a, then P P P∞ µ ¯a = c∈C ν¯c (a ∩ c) = c∈C,n∈N ν¯c (an ∩ c) = n=0 µ ¯an . (iii) If a ∈ A \ {0}, then there is a c ∈ C such that a ∩ c 6= 0, so that µ ¯a ≥ ν¯c (a ∩ c) > 0. Thus (A, µ ¯) is a measure algebra. (iv) Moreover, in (iii), µ ¯(a ∩ c) = ν¯c (a ∩ c) is finite. So (A, µ ¯) is semi-finite. (v) A is Dedekind complete, being a band algebra (352Q), so (A, µ ¯) is localizable. Q Q

369C

Banach function spaces

387

369B Corollary Let U be a Banach lattice with order-continuous norm. Then U can be embedded as an order-dense solid linear subspace of L0 (A) for some localizable measure algebra (A, µ ¯). proof By 356Dd, U × = U ∗ , which separates the points of U , by the Hahn-Banach theorem (3A5Ae). So 369A tells us that U can be embedded as an order-dense Riesz subspace of an appropriate L0 (A). But also U is Dedekind complete (354Ee), so its copy in L0 (A) must be solid, as in 368H. 369C The representation in 369A is complemented by the following result, which is a kind of generalization of 365J and 366Dc. Theorem Let (A, µ ¯) be a semi-finite measure algebra, and U ⊆ L0 = L0 (A) an order-dense Riesz subspace. Set V = {v : v ∈ L0 , v × u ∈ L1 for every u ∈ U }, writing L1 for L1 (A, µ ¯) ⊆ L0 . Then V is a solid linear subspace of L0 , and we have an order-continuous injective Riesz homomorphism T : V → U × defined by setting (T v)(u) =

R

u × v for all u ∈ U , v ∈ V .

×

The image of V is order-dense in U . If (A, µ ¯) is localizable, then T is surjective, so is a Riesz space isomorphism between V and U × . proof (a)(i) Because × : L0 × L0 → L0 is bilinear and L1 is a linear subspace of L0 , V is a linear subspace of L0 . If u ∈ U , v ∈ V , w ∈ L0 and |w| ≤ |v|, then |w × u| = |w| × |u| ≤ |v| × |u| = |v × u| ∈ L1 ; as L1 is solid, w × u ∈ L1 ; as u is arbitrary, w ∈ V ; this shows that V is solid.

R (ii) By the definition of V , (T v)(u) is defined in R for all u ∈ U , v ∈ V . Because × is bilinear and is linear, T v : U → R is linear for every v ∈ V , and T is a linear functional from V to the space of linear operators from U to R. R (iii) If u ≥ 0 in U and v ≥ 0 in V , then u × v ≥ 0 in L1 and (T v)(u) = u × v ≥ 0. This shows that T is a positive linear operator from V to U ∼ . (iv) If v ≥ 0 in V and A ⊆ U is a non-empty downwards-directed set with infimum 0 in U , then inf A = 0 in L0 , because U is order-dense (352Nb). Consequently inf u∈A u × v = 0 in L0 and in L1 (364P), and (because

R

inf u∈A (T v)(u) = inf u∈A

R

u×v =0

is order-continuous). As A is arbitrary, T v is order-continuous. As v is arbitrary, T [V ] ⊆ U × .

(v) If v ∈ V and u0 ≥ 0 in U , set a = [[v > 0]]. Then v + = v×χa. SetR A = {u : u ∈ U, 0 ≤ u ≤ u0 ×χa}. Because U is order-dense in L0 , u0 × χa = sup A in L0 . Because × and are order-continuous, Z + (T v) (u0 ) ≥ sup (T v)(u) = sup v × u u∈A u∈A Z Z = v × u0 × χa = v + × u0 = (T v + )(u0 ). As u0 is arbitrary, (T v)+ ≥ T v + . But because T is a positive linear operator, we must have T v + ≥ (T v)+ , so that T v + = (T v)+ . As v is arbitrary, T is a Riesz homomorphism. (vi) Now T is injective. P P If v 6= R 0 in V , there is a u > 0 in U such that u ≤ |v|, because U is order-dense. In this case u × |v| > 0 so u × |v| > 0. Accordingly |T v| = T |v| 6= 0 and T v 6= 0. Q Q (b) Putting (a-i) to (a-vi) together, we see that T is an injective Riesz homomorphism from V to U × . All this is easy. The point of the theorem is the fact that T [V ] is order-dense in U × . P P Take h > 0 in U × . Let U1 be the solid linear subspace of L0 generated by U . Then U is an order-dense Riesz subspace of U1 , h : U → R is an order-continuous positive linear functional, and sup{h(u) : u ∈ U, 0 ≤ ˜ of h to U1 such that h ˜ ∈ U × (355F). u ≤ v} is defined in R for every v ≥ 0 in U1 ; so we have an extension h 1

388

Function spaces

369C

Set S1 = S(A) ∩ U1 ; then S1 is an order-dense Riesz subspace of U1 , because S(A) is order-dense in L0 and U1 is solid in L0 . Note that S1 is the linear span of {χc : c ∈ I}, where I = {c : c ∈ A, χc ∈ U1 }, and that I is an ideal in A. ˜ 6= 0; there must therefore be a u0 ∈ S1 such that h(u ˜ 0 ) > 0, and a d ∈ I such that Because h 6= 0, h ˜ ˜ ˜ h(χd) > 0. For a ∈ A, set νa = hχ(d ∩ a). Because ∩ , χ and h are all order-continuous, so is ν, and ν : A → R is a non-negative completely additive functional. By 365Ea, there is a v ∈ L1 (A, µ ¯) such that

R

a

˜ v = νa = hχ(d ∩ a)

R ˜ for every a ∈ A; of course v ≥ 0. We have u × v ≤ h(u) whenever u = χa for a ∈ I, and therefore for every + + 0 0 0 u ∈ (S1 ) . If u ∈ U , then A = {u : u ∈ S1 , 0 ≤ u ≤ u} is upwards-directed, sup A = u and

R

˜ 0 ) = h(u) ˜ v × u0 ≤ supu0 ∈A h(u = h(u) R is finite, so v × u = supu∈A0 v × u0 belongs to L1 (365Df) and v × u ≤ h(u). As u is arbitrary, v ∈ V and R ˜ T v ≤ h. At the same time, d v = h(χd) > 0, so T v > 0. As h is arbitrary, T [V ] is order-dense. Q Q It follows that T is order-continuous (352Nb), as can also be easily proved by the argument of (a-iv) above. supu0 ∈A

(c) Now suppose that (A, µ ¯) is localizable, that is, that A is Dedekind complete. T −1 : T [V ] → V is a Riesz space isomorphism, so certainly an order-continuous Riesz homomorphism; because V is a solid linear subspace of L0 , T −1 is still an injective order-continuous Riesz homomorphism when regarded as a map from T [V ] to L0 . Since T [V ] is order-dense in U × , T −1 has an extension to an order-continuous Riesz homomorphism Q : U × → L0 (368B). But Q[U × ] ⊆ V . P P Take h ≥ 0 in U × and u ≥ 0 in U . Then B = {g : g ∈ T [V ],R 0 ≤ g ≤ h} is upwards-directed and has supremum h. For g ∈ B, we know that u × T −1 g ∈ L1 and u × T −1 g = g(u), by the definition of T . But this means that supg∈B

R

u × T −1 g = supg∈B g(u) = h(u) < ∞.

Since {u × T −1 g : g ∈ B} is upwards-directed, it follows that u × Qh = supg∈B u × Qg = supg∈B u × T −1 g ∈ L1 by 365Df again. As u is arbitrary, Qh ∈ V . As h is arbitrary (and Q is linear), Q[U × ] ⊆ V . Q Q Also Q is injective. P P If h ∈ U × is non-zero, there is a v ∈ V such that 0 < T v ≤ |h|, so that |Qh| = Q|h| ≥ QT v = v > 0 and Qh 6= 0. Q Q Since QT is the identity on V , Q and T must be the two halves of a Riesz space isomorphism between V and U × . 369D Corollary Let U be any Riesz space such that U × separates the points of U . Then there is a localizable measure algebra (A, µ ¯) such that the pair (U, U × ) can be represented by a pair (V, W ) of order0 dense Riesz subspaces of L = L0 (A) such that W = {w : w ∈ L0 , v × w ∈ L1 for every v ∈ V }, writing L1 for L1 (A, µ ¯). In this case, U ×× becomes represented by V˜ = {v : v ∈ L0 , v × w ∈ L1 for every w ∈ W } ⊇ V . proof Put 369A and 369C together. The construction of 369A finds (A, µ ¯) and an order-dense V which is isomorphic to U , and 369C identifies W with V × . To check that W is order-dense, take any u > 0 in L0 . There is a v ∈ V such that 0 < v ≤ u. There is an h ∈ (V × )+ such that h(v) > 0, so there is a w ∈ W + such that w × v 6= 0, that is, w ∧ v 6= 0. But now w ∧ v ∈ W , because W is solid, and 0 < w ∧ v ≤ u. Remark Thus the canonical embedding of U in U ×× (356I) is represented by the embedding V ⊆ V˜ ; U , or V , is ‘perfect’ iff V = V˜ . 369E Kakutani’s Theorem (Kakutani 41) If U is any L-space, there is a localizable measure algebra (A, µ ¯) such that U is isomorphic, as Banach lattice, to L1 = L1 (A, µ ¯). R R proof U is a perfect Riesz space, and U × = U ∗ has an order unit defined by saying that u = kuk for u ≥ 0 (356P). By 369D, we can find a localizable measure algebra (A, µ ¯) and an identification of the pair (U, U × ), as dual Riesz spaces, with a pair (V, W ) of subspaces of L0 = L0 (A); and V will be {v : v×w ∈ L1 for

369H

Banach function spaces

389

every w ∈ W }. But W , like U × , must have an order unit; call it e. Because W is order-dense, [[e > 0]] must be 1 and e must have a multiplicative inverse 1e in L0 (364P). This means that V must be {v : v × e ∈ L1 }, so that v 7→ v × e is a Riesz space isomorphism between V and L1 , which gives a Riesz space isomorphism between U and L1 . Moreover, if we write k k0 for the norm on V corresponding to the norm of U , we have kuk =

R

|u| for u ∈ U ,

kvk0 =

R

|v| × e =

R

|v × e| for v ∈ V .

1

Thus the Riesz space isomorphism between U and L is norm-preserving, and U and L1 are isomorphic as Banach lattices. 369F The Lp spaces are leading examples for a general theory of normed subspaces of L0 , which I proceed to sketch in the rest of the section. Definition Let A be a Dedekind σ-complete Boolean algebra. An extended Fatou norm on L0 = L0 (A) is a function τ : L0 → [0, ∞] such that (i) τ (u + v) ≤ τ (u) + τ (v) for all u, v ∈ L0 ; (ii) τ (αu) = |α|τ (u) for all u ∈ L0 , α ∈ R (counting 0 · ∞ as 0, as usual); (iii) τ (u) ≤ τ (v) whenever |u| ≤ |v| in L0 ; (iv) supu∈A τ (u) = τ (v) whenever A ⊆ (L0 )+ is a non-empty upwards-directed set with supremum v in 0 L ; (v) τ (u) > 0 for every non-zero u ∈ L0 ; (vi) whenever u > 0 in L0 there is a v ∈ L0 such that 0 < v ≤ u and τ (v) < ∞. 369G Proposition Let (A, µ ¯) be a Dedekind σ-complete Boolean algebra and τ an extended Fatou norm on L0 = L0 (A). Then Lτ = {u : u ∈ L0 , τ (u) < ∞} is an order-dense solid linear subspace of L0 , and τ , restricted to Lτ , is a Fatou norm under which Lτ is a Banach lattice. If hun in∈N is a non-decreasing norm-bounded sequence in (Lτ )+ , then it has a supremum in Lτ ; if A is Dedekind complete, then Lτ has the Levi property. proof (a) By (i), (ii) and (iii) of 369F, Lτ is a solid linear subspace of L0 ; by (vi), it is order-dense. Hypotheses (i), (ii), (iii) and (v) show that τ is a Riesz norm on Lτ , while (iv) shows that it is a Fatou norm. (b)(i) Suppose that hun in∈N is a non-decreasing norm-bounded sequence in (Lτ )+ . Then u = supn∈N un is defined in L0 . P P?? Otherwise, there is a v > 0 in L0 such that kv = supn∈N kv ∧un for every k ∈ N (368A). By (v)-(vi) of 369F, there is a v 0 such that 0 < v 0 ≤ v and 0 < τ (v 0 ) < ∞. Now kv 0 = supn∈N kv 0 ∧ un for every k, so kτ (v 0 ) = τ (kv 0 ) = supn∈N τ (kv 0 ∧ un ) ≤ supn∈N τ (un ) for every k, using 369F(iv), and supn∈N τ (un ) = ∞, contrary to hypothesis. X XQ Q By 369F(iv) again, τ (u) = supn∈N τ (un ) < ∞, so that u ∈ Lτ and u = supn∈N un in Lτ . (ii) It follows that Lτ is complete P Let hun in∈N be a sequence in Lτ such that τ (un+1 −un ) ≤ Pn under τ . P 2 for every n ∈ N. Set vmn = i=m |ui+1 − ui | for m ≤ n; then τ (vmn ) ≤ 2−m+1 for every n, so by (i) just above vm = supn∈N vmn is defined in Lτ , and τ (vm ) ≤ 2−m+1 . Now vm = |um+1 − um | + vm+1 for each m, so hum − vm im∈N is non-decreasing and hum + vm im∈N is non-increasing, while um − vm ≤ um ≤ um + vm for every m. Accordingly u = supm∈N um − vm is defined in Lτ and |u − um | ≤ vm for every m. But this means that limm→∞ τ (u − um ) ≤ limm→∞ τ (vm ) = 0 and u = limm→∞ um in Lτ . As hun in∈N is arbitrary, Lτ is complete. Q Q −n

(c) Now suppose that A is Dedekind complete and A ⊆ (Lτ )+ is a non-empty upwards-directed normbounded set in Lτ . By the argument of (b-i) above, using the other half of 368A, sup A is defined in L0 and belongs to Lτ . As A is arbitrary, Lτ has the Levi property. 369H Associate norms: Definition Let (A, µ ¯) be a semi-finite measure algebra, and τ an extended Fatou norm on L0 = L0 (A). Define τ 0 : L0 → [0, ∞] by setting τ 0 (u) = sup{ku × vk1 : v ∈ L0 , τ (v) ≤ 1}

390

Function spaces

369H

for every u ∈ L0 ; then τ 0 is the associate of τ . (The word suggests a symmetric relationship; it is justified by the next theorem.) 369I Theorem Let (A, µ ¯) be a semi-finite measure algebra, and τ an extended Fatou norm on L0 = L (A). Then (i) its associate τ 0 is also an extended Fatou norm on L0 ; (ii) τ is the associate of τ 0 ; (iii) ku × vk1 ≤ τ (u)τ 0 (v) for all u, v ∈ L0 . 0

proof (a) Before embarking on the proof that τ 0 is an extended Fatou seminorm on L0 , I give the greater part of the argument needed to show that τ = τ 00 , where τ 00 (u) = sup{ku × wk1 : w ∈ L0 , τ 0 (w) ≤ 1} for every u ∈ L0 . (i) Set B = {u : u ∈ L1 , τ (u) ≤ 1}. Then B is a convex set in L1 = L1 (A, µ ¯) and is closed for the norm topology of L1 . P P Suppose that u 1 belongs to the closure of B in L . Then for each n ∈ N we can choose un ∈ B such that ku − un k1 ≤ 2−n . Set vmn = inf m≤i≤n |ui | for m ≤ n, and vm = inf n≥m vmn = inf n≥m |un | ≤ |u| for m ∈ N. The sequence hvm im∈N is non-decreasing, τ (vm ) ≤ τ (um ) ≤ 1 for every m, and P∞ P∞ k|u| − vm k1 ≤ supn≥m k|u| − vmn k1 ≤ i=m k|u| − |ui |k1 ≤ i=m ku − ui k1 → 0 as m → ∞. So |u| = supm∈N vm in L0 , τ (u) = τ (|u|) = supm∈N τ (vm ) ≤ 1 and u ∈ B. Q Q (ii) Now take any u0 ∈ L0 such that τ (u0 ) > 1. Then, writing Af for {a : µ ¯a < ∞}, A = {u : u ∈ S(Af ), 0 ≤ u ≤ u0 } is an upwards-directed set with supremum u0 (this is where I use the hypothesis that (A, µ ¯) is semi-finite, so that S(Af ) is order-dense in L0 ), and supu∈A τ (u) = τ (u0 ) > 1. Take u1 ∈ A such that τ (u1 ) > 1, that is, u1 ∈ / B. By the Hahn-Banach theorem (3A5Cc), there is a continuous linear functional f : L1 → R such that f (u1 ) > 1 but f (u) ≤ 1 for every u ∈ B. Because (L1 )∗ = (L1 )∼ (356Dc), |f | is defined in (L1 )∗ , and of course |f |(u1 ) ≥ f (u1 ) > 1,

|f |(u) = sup{f (v) : |v| ≤ u} ≤ 1

whenever u ∈ B and u ≥ 0. Set c = [[u1 > 0]], so that µ ¯c < ∞, and define νa = |f |(χ(a ∩ c)) 1 for every R a ∈ A. Then ν is a completely additive real-valued functional on A, so there is a w ∈ L such that νa = a w for every a ∈ A (365Ea). Because νa ≥ 0 for every a, w ≥ 0. Now R w = |f |(χa × χc) a

for every a ∈ A, so

R

w × u = |f |(u × χc) ≤ |f |(u) ≤ 1

for every u ∈ S(A)+ ∩ B. But if τ (v) ≤ 1, then Av = {u : u ∈ S(A)+ ∩ B, u ≤ |v|} is an upwards-directed set with supremum |v|, so that kw × vk1 = supu∈Av

R

w × u ≤ 1.

369K

Banach function spaces

Thus τ 0 (w) ≤ 1. On the other hand,

R

kw × u0 k1 ≥

w × u0 ≥

R

391

w × u1 = |f |(u1 ) > 1,

so τ 00 (u0 ) > 1. (iii) This shows that, for u ∈ L0 , τ 00 (u) ≤ 1 =⇒ τ (u) ≤ 1. (c) Now I return to the proof that τ 0 is an extended Fatou norm. It is easy to check that it satisfies conditions (i)-(iv) of 369F; in effect, these depend only on the fact that k k1 is an extended Fatou norm. For (v)-(vi), take v > 0 in L0 . Then there is a u such that 0 ≤ u ≤ v and 0 < τ (u) < ∞; set α = 1/τ (u). Then τ (2αu) > 1, so that τ 00 (2αu) > 1 and there is a w ∈ L0 such that τ 0 (w) ≤ 1, k2αu × wk1 > 1. But now set v1 = v ∧ |w|; then v ≥ v1 ≥ u ∧ |w| > 0, 0

while τ (v1 ) < ∞. Also v ∧ αu 6= 0 so τ 0 (v) ≥ kv1 × αuk1 > 0. As v is arbitrary, τ 0 satisfies 369F(v)-(vi). (d) Accordingly τ 00 is also an extended Fatou norm. Now in (a) I showed that τ 00 (u) ≤ 1 =⇒ τ (u) ≤ 1. It follows easily that τ (u) ≤ τ 00 (u) for every u (since otherwise there would be some α such that τ 00 (αu) = ατ 00 (u) < 1 < ατ (u) = τ (αu).) On the other hand, we surely have τ (u) ≤ 1 =⇒ ku × vk1 ≤ 1 whenever τ 0 (v) ≤ 1 =⇒ τ 00 (u) ≤ 1, so we must also have τ 00 (u) ≤ τ (u) for every u. Thus τ 00 = τ , as claimed. (e) Of course we have ku × vk1 ≤ 1 whenever τ (u) ≤ 1 and τ 0 (v) ≤ 1. It follows easily that ku × vk1 ≤ τ (u)τ 0 (v) whenever u, v ∈ L0 and both τ (u), τ 0 (v) are non-zero. But if one of them is zero, then u × v = 0, because both τ and τ 0 satisfy (v) of 369F, so the result is trivial. 369J Theorem Let (A, µ ¯) be a semi-finite measure algebra, and τ an extended Fatou norm on L0 = L (A), with associate θ. Then 0

Lθ = {v : v ∈ L0 , u × v ∈ L1 for every u ∈ Lτ }. proof (a) If v ∈ Lθ and u ∈ Lτ , then ku × vk1 is finite, by 369I(iii), so u × v ∈ L1 . (b) P If v ∈ / Lθ then for every n ∈ N there is a un such that τ (un ) ≤ 1 and kun × vk1 ≥ 2n . Set n wn = i=0 2−i |ui | for each n. Then hwn in∈N Ris a non-decreasing sequence and τ (wn ) ≤ 2 for each n, so w = supn∈N wn is defined in Lτ , by 369G; now w × |v| ≥ n + 1 for every n, so w × v ∈ / L1 . 369K Corollary Let (A, µ ¯) be a localizable measure algebra, and τ an extended Fatou norm on L0 = L0 (A), with associate θ. Then Lθ may be identified, as normed Riesz space, with (Lτ )× ⊆ (Lτ )∗ , and Lτ is a perfect Riesz space. proof Putting 369J and 369C together, we have an identification between Lθ and (Lτ )× . Now 369I tells us that τ is the associate of θ, so that we can identify Lτ with (Lθ )× , and Lτ is perfect, as in 369D. By the definition of θ, we have, for any v ∈ Lθ , θ(v) = sup ku × vk1 τ (u)≤1

=

sup τ (u)≤1,kwk∞ ≤1

Z

Z u × v × w = sup τ (u)≤1

which is the norm of the linear functional on Lτ corresponding to v.

u × v,

392

Function spaces

369L

369L Lp I remarked above that the Lp spaces are leading examples for this theory; perhaps I should spell out the details. Let (A, µ ¯) be a semi-finite measure algebra and p ∈ [1, ∞]. Extend the functional k kp , as defined in 364K, 365A and 366A, by saying that kukp = ∞ if u ∈ L0 \ Lp . (For p = 1 this is already done by the convention in 365A.) To see that k kp is now an extended Fatou norm, as defined in 369F, we note that conditions (i)-(iii) and (v) there are true just because Lp is a solid linear subspace on which k kp is a Riesz norm, (iv) is true because k kp is a Fatou norm with the Levi property (363Ba, 365C, 366D), and (vi) is true because S(Af ) is included in Lp and order-dense in L0 (364L). As usual, set q = p/(p − 1) if 1 < p < ∞, ∞ if p = 1, and 1 if p = ∞. Then k kq is the associate extended Fatou norm of k kp . P P By 365Jb and 366C, kvkq = sup{ku × vk1 : kukp ≤ 1} for every v ∈ Lq . But as Lq 0 is order-dense in L , kvkq =

sup w∈Lq ,|w|≤v

kwkq

Z

= sup{

Z q

|u| × |w| : w ∈ L , w ≤ |v|, kukp ≤ 1} = sup{

|u| × |v| : kukp ≤ 1}

for every v ∈ L0 . Q Q 369M Proposition Let (A, µ ¯) be a semi-finite measure algebra and τ an extended Fatou norm on L0 = L0 (A, µ ¯). Then (a) the embedding Lτ ⊆ L0 is continuous for the norm topology of Lτ and the topology of convergence in measure on L0 ; (b) τ : L0 → [0, ∞] is lower semi-continuous, that is, all the balls {u : τ (u) ≤ γ} are closed for the topology of convergence in measure; (c) if hun in∈N is a sequence in L0 which is order*-convergent to u ∈ L0 (definition: 367A), then τ (u) ≤ lim inf n→∞ τ (un ). proof (a) This is a special case of 367P. (b) Set Bγ = {u : τ (u) ≤ γ}. If u ∈ L0 \ Bγ , then A = {|u| × χa : a ∈ Af } is an upwards-directed set with supremum |u|, so there is an a ∈ Af such that τ (u × χa) > γ. ?? If u is in the closure of Bγ for the topology of convergence in measure, then for every k ∈ N there is a vk ∈ Bγ such that µ ¯(a ∩ [[|u − vk | > 2−k ]]) ≤ 2−k (see the formulae in 367M). Set vk0 = |u| ∧ inf i≥k |vi | for each k, and v ∗ = supk∈N vk0 . Then τ (vk0 ) ≤ τ (vk ) ≤ γ for each k, and hvk ik∈N is non-decreasing, so τ (v ∗ ) ≤ γ. But P∞

a ∩ [[|u| − v ∗ > 2−k ]] ⊆ a ∩ supi≥k [[|u − vi | > 2−k ]]

−i has measure at most for each k, so a ∩ [[|u| − v ∗ > 0]] must be 0, that is, |u| × χa ≤ v ∗ and i=k 2 τ (|u| × χa) ≤ γ; contrary to the choice of a. X X Thus u cannot belong to the closure of Bγ . As u is arbitrary, Bγ is closed.

(c) If hun in∈N order*-converges to u, it converges in measure (367Na). If γ > lim inf n→∞ τ (un ), there is a subsequence of hun in∈N in Bγ , and τ (u) ≤ γ, by (b). As γ is arbitary, τ (u) ≤ lim inf n→∞ τ (un ). 369N contexts.

I now turn to another special case which we have already had occasion to consider in other

Definition Let (A, µ ¯) be a measure algebra. Set Mµ¯∞,1 = M ∞,1 (A, µ ¯) = L1 (A, µ ¯) ∩ L∞ (A), Mµ¯1,∞ = M 1,∞ (A, µ ¯) = L1 (A, µ ¯) + L∞ (A), and

369O

Banach function spaces

393

kuk∞,1 = max(kuk1 , kuk∞ ) 0

for u ∈ L (A). Remark I hope that the notation I have chosen here will not completely overload your short-term memory. The idea is that in M p,q the symbol p is supposed to indicate the ‘local’ nature of the space, that is, the nature of u × χa where u ∈ M p,q and µ ¯a < ∞, while q indicates the nature of |u| ∧ χ1 for u ∈ M p,q . Thus 1,∞ M is the space of u such that u × χa ∈ L1 for every a ∈ Af , |u| ∧ χ1 ∈ L∞ ; in M 1,0 we demand further that |u| ∧ χ1 ∈ M 0 (366F); while in M ∞,1 we ask that |u| ∧ χ1 ∈ L1 , u × χa ∈ L∞ for every a ∈ Af . 369O Proposition Let (A, µ ¯) be a semi-finite measure algebra. (a) k k∞,1 is an extended Fatou norm on L0 = L0 (A). (b) Its associate k k1,∞ may be defined by the formulae kuk1,∞ = min{kvk1 + kwk∞ : v ∈ L1 , w ∈ L∞ , v + w = u} Z = min{α + (|u| − αχ1)+ : α ≥ 0} Z ∞ = min(1, µ ¯[[|u| > α]])dα 0 0

1

1

for every u ∈ L , writing L = L (A, µ ¯), L∞ = L∞ (A). (c) {u : u ∈ L0 , kuk1,∞ < ∞} = M 1,∞ = M 1,∞ (A, µ ¯), {u : u ∈ L0 , kuk∞,1 < ∞} = M ∞,1 = M ∞,1 (A, µ ¯). (d) Writing Af = {a : µ ¯a < ∞}, S(Af ) is norm-dense in M ∞,1 and S(A) is norm-dense in M 1,∞ . (e) For any p ∈ [1, ∞], kuk1,∞ ≤ kukp ≤ kuk∞,1 0

for every u ∈ L . Remark By writing ‘min’ rather than ‘inf’ in the formulae of part (b) I mean to assert that the infima are attained. proof (a) This is easy; all we need to know is that k k1 and k k∞ are extended Fatou norms. (b) We have four functionals on L0 to look at; let me give them names: τ1 (u) = sup{ku × vk1 : kvk1,∞ ≤ 1}, τ2 (u) = inf{ku0 k1 + ku00 k∞ : u = u0 + u00 }, τ3 (u) = inf α≥0 (α + τ4 (u) =

R∞ 0

R

(|u| − αχ1)+ ),

min(1, µ ¯[[|u| > α]])dα.

(I write ‘inf’ here to avoid the question of attainment for the moment.) Now we have the following. (i) τ1 (u) ≤ τ2 (u). P P If kvk1,∞ ≤ 1 and u = u0 + u00 , then ku × vk1 ≤ ku0 × vk1 + ku00 × vk1 ≤ ku0 k1 kvk∞ + ku00 k∞ kvk1 ≤ ku0 k1 + ku00 k∞ . Taking the supremum over v and the infinum over u0 and u00 , τ1 (u) ≤ τ2 (u). Q Q (ii) τ2 (u) ≤ τ4 (u). P P If τ4 (u) = ∞ this is trivial. Otherwise, take w such that kwk∞ ≤ 1 and u = |u| × w. Set α0 = inf{α : µ ¯[[|u| > α]] ≤ 1}, and try u0 = w × (|u| − α0 χ1)+ , Then u = u0 + u00 ,

u00 = w × (|u| ∧ α0 χ1).

394

Function spaces

Z

369O

∞

0

µ ¯[[|u0 | > α]]dα

ku k1 = 0

Z

∞

= Z0 ∞ =

µ ¯[[|u| > α + α0 ]]dα Z ∞ µ ¯[[|u| > α]]dα = min(1, µ ¯[[|u| > α]])dα,

α0

α0

ku00 k∞ ≤ α0 =

R α0 0

min(1, [[|u| > α]])dα,

so τ2 (u) ≤ ku0 k1 + ku00 k∞ ≤ τ4 (u). Q Q (iii) τ4 (u) ≤ τ3 (u). P P For any α ≥ 0, Z α Z ∞ τ4 (u) = min(1, µ ¯[[|u| > β]])dβ + min(1, µ ¯[[|u| > β]])dβ 0 α Z ∞ ≤α+ µ ¯[[|u| > α + β]]dβ 0 Z ∞ =α+ µ ¯[[|u| > α + β]]dβ 0 Z ∞ Z =α+ µ ¯[[(|u| − αχ1)+ > β]]dβ = α + (|u| − αχ1)+ . 0

Taking the infimum over α, τ4 (u) ≤ τ3 (u). Q Q (iv) τ3 (u) ≤ τ1 (u). α) It is enough to consider the case 0 < τ1 (u) < ∞, because if τ1 (u) = 0 then u = 0 and evidently P P(α τ3 (0) = 0, while if τ1 (u) = ∞ the required inequality is trivial. Furthermore, since τ3 (u) = τ3 (|u|) and τ1 (u) = τ1 (|u|), it is enough to consider the case u ≥ 0. R 1 β ) Note next that if µ (β ¯a < ∞, then k max(1,¯ ¯a)τ1 (u). µa) χak∞,1 ≤ 1, so that a u ≤ max(1, µ (γγ ) Set c = [[u > 2τ1 (u)]]. If a ⊆ c and µ ¯a < ∞, then 2τ1 (u)¯ µa ≤

R

a

u ≤ max(1, µ ¯a)τ1 (u),

so µ ¯a ≤ 21 . As (A, µ ¯) is semi-finite, it follows that µ ¯c ≤

1 2

(322Eb).

(δδ ) I may therefore write α0 = inf{α : α ≥ 0, µ ¯[[u > α]] ≤ 1}. Now [[u > α0 ]] = supα>α0 [[u > α]], so µ ¯[[u > α0 ]] = supα>α0 µ ¯[[u > α]] ≤ 1. (²²) If α ≥ α0 then (u − α0 χ1)+ ≤ (α − α0 )χ[[u > α0 ]] + (u − αχ1)+ , so α0 +

R

(u − α0 χ1)+ ≤ α0 + (α − α0 )¯ µ[[u > α0 ]] +

R

(u − αχ1)+ ≤ α +

If 0 ≤ α < α0 then, for every β ∈ [0, α0 − α[, (u − α0 χ1)+ + β[[u > α + β]] ≤ (u − αχ1)+ , while µ ¯[[u > α + β]] > 1, so

R

(u − α0 χ1)+ + β + α ≤ α +

R

(u − αχ1)+ ;

R

(u − αχ1)+ .

369P

Banach function spaces

taking the supremum over β, α0 +

R Thus α0 + (u − α0 χ1)+ = τ3 (u).

R

(u − α0 χ1)+ ≤ α +

R

395

(u − αχ1)+ .

(ζζ ) If α0 = 0, take v = χ[[u > 0]]; then kvk∞,1 = µ ¯[[u > 0]] ≤ 1 and τ3 (u) =

R

u = ku × vk1 ≤ τ1 (u).

(ηη ) If α0 > 0, set γ = µ ¯[[u > α0 ]]. Take any β ∈ [0, α0 [. Then µ ¯([[u > β]] \ [[u > α0 ]]) > 1 − γ, so there is a b ⊆ [[u > β]] \ [[u > α0 ]] such that 1 − γ < µ ¯b < ∞. Set v = χ[[u > α0 ]] + 1−γ µ ¯ b χb. Then kvk∞,1 = 1 so τ1 (u) ≥

R

u×v ≥

R

(u − α0 χ1)+ + α0 γ + β

1−γ µ ¯b µ ¯b

= τ3 (u) − (1 − γ)(α0 − β).

As β is arbitrary, τ1 (u) ≥ τ3 (u) in this case also. Q Q (v) Thus τ1 (u) = τ2 (u) = τ3 (u) = τ4 (u) for every u ∈ L0 , and I may write kuk1,∞ for their common value; as the associate of k k∞,1 , k k1,∞ is an extended Fatou norm. As for the attainment of the infima, R the argument of (iv-²) above shows that, at least when 0 < kuk1,∞ < ∞, there is an α0 such that α0 + (|u| − α0 )+ = kuk1,∞ . This omits the cases kuk1,∞ ∈ {0, ∞}; but in either of these cases we can set α0 = 0 to see that the infimum is attained for trivial reasons. For the other infimum, observe that the argument of (ii) produces u0 , u00 such that u = u0 + u00 and ku0 k1 + ku00 k∞ ≤ τ4 (u). (c) This is now obvious from the definition of k k∞,1 and the characterization of k k1,∞ in terms of k k1 and k k∞ . (d) To see that S = S(A) is norm-dense in M 1,∞ , we need only note that S is dense in L∞ and S ∩ L1 is dense in L1 ; so that given v ∈ L1 , w ∈ L∞ and ² > 0 there are v 0 , w0 ∈ S such that k(v + w) − (v 0 + w0 )k1,∞ ≤ kv − v 0 k1 + kw − w0 k∞ ≤ ². As for M ∞,1 , if u ≥ 0 in M ∞,1 and r ∈ N, set vr = supk∈N 2−r kχ[[u > 2−r k]]; then each vr ∈ S fR = S(ARf ), ku − vr k∞ ≤ 2−r , and hvr ir∈N is a non-decreasing sequence with supremum u, so that limr→∞ vr = u and limr→∞ ku−vr k∞,1 = 0. Thus (S f )+ is dense in (M ∞,1 )+ . As usual, it follows that S f = (S f )+ −(S f )+ is dense in M ∞,1 = (M ∞,1 )+ − (M ∞,1 )+ . (e)(i) If p = 1 or p = ∞ this is immediate from the definition of k k∞,1 and the characterization of k k1,∞ in (b). So suppose henceforth that 1 < p < ∞. R (ii) If kuk∞,1 ≤ 1 then kukp ≤ 1. P P Because kuk∞ ≤ 1, |u|p ≤ |u|, so that |u|p ≤ kuk1 ≤ 1 and kukp ≤ 1. Q Q On considering scalar multiples of u, we see at once that kukp ≤ kuk∞,1 for every u ∈ L0 . (ii) Now set q = p/(p − 1). Then kukp = sup{ku × vk1 : kvkq ≤ 1} (369L) ≥ sup{ku × vk1 : kvk∞,1 ≤ 1} = kuk1,∞ because k k1,∞ is the associate of k k∞,1 . This completes the proof. 369P

In preparation for some ideas in §372, I go a little farther with M 1,0 , as defined in 366F.

Proposition Let (A, µ ¯) be a measure algebra. (a) M 1,0 = M 1,0 (A, µ ¯) is a norm-closed solid linear subspace of M 1,∞ = M 1,∞ (A, µ ¯). (b) The norm k k1,∞ is order-continuous on M 1,0 . (c) S(Af ) and L1 (A, µ ¯) are norm-dense and order-dense in M 1,0 . proof (a) Of course M 1,0 , being a solid linear subspace of L0 included in M 1,∞ , is a solid linear subspace of M 1,∞ . To see that it is norm-closed, take any point u of its closure. Then for any ² > 0 there is a v ∈ M 1,0

396

Function spaces

369P

such that ku − vk1,∞ ≤ ²; now (|u − v| − ²χ1)+ ∈ L1 , so [[|u − v| > 2²]] has finite measure; also [[|v| > ²]] has finite measure, so [[|u| > 3²]] ⊆ [[|u − v| > 2²]] ∪ [[|v| > ²]] (364Fa) has finite measure. As ² is arbitrary, u ∈ M 1,0 ; as u is arbitrary, M 1,0 is closed. (b) Suppose that A ⊆ M 1,0 is non-empty and downwards-directed and has infimum 0. Let ² > 0. Set B = {(u − ²χ1)+ : u ∈ A}. Then B ⊆ L1 (by 366Gc); B is non-empty and downwards-directed and has infimum 0. Because k k1 is order-continuous (365C), inf v∈B kvk1 = 0 and there is a u ∈ A such that k(u − ²χ1)+ k1 ≤ ², so that kuk1,∞ ≤ 2². As ² is arbitrary, inf u∈A kuk1,∞ = 0; as A is arbitrary, k k1,∞ is order-continuous on M 1,0 . (c) By 366Gb, S(Af ) is order-dense in M 1,0 . Because the norm of M 1,0 is order-continuous, S(Af ) is also norm-dense (354Ef). Now S(Af ) ⊆ L1 ⊆ M 1,0 , so L1 must also be norm-dense and order-dense. 369Q Corollary Let (A, µ ¯) be a localizable measure algebra. Set M 1,∞ = M 1,∞ (A, µ ¯), etc. 1,∞ × 1,0 × (a) (M ) and (M ) can both be identified with M ∞,1 . (b) (M ∞,1 )× can be identified with M 1,∞ ; M 1,∞ and M ∞,1 are perfect Riesz spaces. proof Everything is covered by 369O and 369K except the identification of (M 1,0 )× with M ∞,1 . For this I return to 369C. Of course M 1,0 is order-dense in L0 , because it includes L1 , or otherwise. Setting V = {v : v ∈ L0 , u × v ∈ L1 for every u ∈ M 1,0 }, 369C identifies V with (M 1,0 )× . Of course M ∞,1 ⊆ V just because M 1,0 ⊆ M 1,∞ . Also V ⊆ M ∞,1 . P P Let v ∈ V . (i) ?? If v ∈ / L∞ , then an = [[|v| > 4n ]] 6= 0 for everyP n. For each n, choose ∞ non-zero bn ⊆ an such thatR µ ¯bn < ∞ (using the fact that (A, µ ¯) is semi-finite). Set u = n=0 2−n (¯ µbn )−1 χbn ; 1 1,0 n 1 then u ∈ L ⊆ M , but |v| × u ≥ 2 for every n ∈ N, so |v| × u ∈ / L , which is impossible, because v ∈ V . X X ThusRv ∈ L∞ . (ii) ?? If v ∈ / L1 , then (again because (A, µ ¯) isR semi-finite, so that |v| = supa∈Af |v| × χa) supa∈Af a |v| = ∞. For each n ∈ N choose an ∈ Af such that an |v| ≥ 4n , and set u = supn∈N 2−n χan ∈ R M 1,0 ; then u×|v| ≥ 2n for each n, so again v ∈ / V.X X Thus v ∈ L1 . (iii) Putting these together, v ∈ M ∞,1 ; ∞,1 as v is arbitrary, V ⊆ M .Q Q So M ∞,1 = V can be identified with (M 1,0 )× . 369R The detailed formulae of 369O are of course special to the norms k k1 , k k∞ , but the general phenomenon is not. Theorem Let (A, µ ¯) be a localizable measure algebra, and τ1 , τ2 two extended Fatou norms on L0 = L0 (A) 0 with associates τ1 , τ20 . Then we have an extended Fatou norm τ defined by the formula τ (u) = min{τ1 (v) + τ2 (w) : v, w ∈ L0 , v + w = u} for every u ∈ L0 , and its associate τ 0 is given by the formula τ 0 (u) = max(τ10 (u), τ20 (u)) for every u ∈ L0 . Moreover, the corresponding function spaces are Lτ = Lτ1 + Lτ2 ,

0

0

0

Lτ = Lτ1 ∩ Lτ2 .

proof (a) For the moment, define τ by setting τ (u) = inf{τ1 (v) + τ2 (w) : v + w = u} for u ∈ L0 . It is easy to check that, for u, u0 ∈ L0 and α ∈ R, τ (u + u0 ) ≤ τ (u) + τ (u0 ),

τ (αu) = |α|τ (u),

τ (u) ≤ τ (u0 ) if |u| ≤ |u0 |.

(For the last, remember that in this case u = u0 × z where kzk∞ ≤ 1.) (b) Take any non-empty, upwards-directed set A ⊆ (L0 )+ , with supremum u0 . Suppose that γ = supu∈A τ (u) < ∞. For u ∈ A, n ∈ N set Cun = {v : v ∈ L0 , 0 ≤ v ≤ u0 , τ1 (v) + τ2 (u − v)+ ≤ γ + 2−n }.

369R

Banach function spaces

397

Then (i) every Cun is non-empty (because τ (u) ≤ γ); (ii) every Cun is convex (because if v1 , v2 ∈ Cun and α ∈ [0, 1] and v = αv1 + (1 − α)v2 , then (u − v)+ = (α(u − v1 ) + (1 − α)(u − v2 ))+ ≤ α(u − v1 )+ + (1 − α)(u − v2 )+ , so τ1 (v) + τ2 (u − v)+ ≤ ατ1 (v1 ) + (1 − α)τ1 (v2 ) + ατ2 (u − v1 )+ + (1 − α)τ2 (u − v2 )+ ≤ γ + 2−n ); (iii) if u, u0 ∈ A, m, n ∈ N and u ≤ u0 , m ≤ n then Cu0 n ⊆ Cum ; (iv) every Cun is closed for the topology of convergence in measure. P P?? Suppose otherwise. Then we can find a v in the closure of Cun for the topology of convergence in measure, but such that τ1 (v) + τ2 (u − v)+ > γ + 2−n . In this case τ1 (v) = sup{τ1 (v × χa) : a ∈ Af },

τ2 (u − v)+ = sup{τ2 ((u − v)+ × χa) : a ∈ Af },

so there is an a ∈ Af such that τ1 (v × χa) + τ2 ((u − v)+ × χa) > γ + 2−n . Now there is a sequence hvk ik∈N in Cun such that µ ¯(a ∩ [[|v − vk | ≥ 2−k ]]) ≤ 2−k for every k. Setting vk0 = inf i≥k vi ,

wk = inf i≥k (u − vi )+

we have τ1 (vk0 ) + τ2 (wk ) ≤ τ1 (vk ) + τ2 (u − vk )+ ≤ γ + 2−n for each k, and hvk0 ik∈N , hwk ik∈N are non-decreasing. So setting v ∗ = supk∈N v∧vk0 , w∗ = supk∈N (u−v)+ ∧wk , we get τ1 (v ∗ ) + τ2 (w∗ ) ≤ γ + 2−n . But v ∗ ≥ v × χa and w∗ ≥ (u − v)+ × χa, so τ1 (v × χa) + τ2 ((u − v)+ × χa) ≤ γ + 2−n , contrary to the choice of a. X XQ QT Applying 367V, we find that u∈A,n∈N Cun is non-empty. If v belongs to the intersection, then τ1 (v) + τ2 (u − v)+ ≤ γ for every u ∈ A; since {(u − v)+ : u ∈ A} is an upwards-directed set with supremum (u0 − v)+ , and τ2 is an extended Fatou norm, τ1 (v) + τ2 (u0 − v)+ ≤ γ. (c) This shows both that the infimum in the definition of τ (u) is always attained (since this is trivial if τ (u) = ∞, and otherwise we consider A = {|u|}), and also that τ (sup A) = supu∈A τ (u) whenever A ⊆ (L0 )+ is a non-empty upwards-directed set with a supremum. Thus τ satisfies conditions (i)-(iv) of 369F. Condition (vi) there is trivial, since (for instance) τ (v) ≤ τ1 (v) for every v. As for 369F(v), suppose that u > 0 in L0 . Take u1 such that 0 < u1 ≤ u and τ10 (u1 ) ≤ 1, u2 such that 0 < u2 ≤ u1 and τ20 (u2 ) ≤ 1. In this case, if u2 = v + w, we must have τ1 (v) + τ2 (w) ≥ kv × u1 k1 + kw × u2 k1 ≥ ku2 × u2 k1 ; so that τ (u) ≥ ku2 × u2 k1 > 0. Thus all the conditions of 369F are satisfied, and τ is an extended Fatou norm on L0 . (d) The calculation of τ 0 is now very easy. Since surely we have τ ≤ τi for both i, we must have τ 0 ≥ τi0 for both i. On the other hand, if u, z ∈ L0 , then there are v, w such that u = v +w and τ (u) = τ1 (v)+τ2 (w), so that

398

Function spaces

369R

ku × zk1 ≤ kv × zk1 + kw × zk1 ≤ τ1 (v)τ10 (z) + τ2 (w)τ20 (z) ≤ τ (u) max(τ10 (z), τ20 (z)); as u is arbitrary, τ 0 (z) ≤ max(τ10 (z), τ20 (z)). So τ 0 = max(τ10 , τ20 ), as claimed. (e) Finally, it is obvious that 0

0

0

Lτ = {z : τ 0 (z) < ∞} = {z : τ10 (z) < ∞, τ20 (z) < ∞} = Lτ1 ∩ Lτ2 , while the fact that the infimum in the definition of τ is always attained means that Lτ ⊆ Lτ1 + Lτ2 , so that we have equality here also. 369X Basic exercises > (a) Let A be a Dedekind σ-complete Boolean algebra. Show that the following are equiveridical: (i) there is a function µ ¯ such that (A, µ ¯) is a semi-finite measure algebra; (ii) (L∞ )× ∞ ∞ separates the points of L = L (A); (iii) for every non-zero a ∈ A there is a completely additive functional ν : A → R such that νa 6= 0; (iv) there is some order-dense Riesz subspace U of L0 = L0 (A) such that U × separates the points of U ; (v) for every order-dense Riesz subspace U of L0 there is an order-dense Riesz subspace V of U such that V × separates the points of V . (b) Let us say that a function φ : R → ]−∞, ∞] is convex if φ(αx + (1 − α)y) ≥ αφ(x) + (1 − α)φ(y) for all x, y ∈ I and α ∈ [0, 1], interpreting 0 · ∞ as 0, as usual. For any convex function φ : R → ]−∞, ∞] which is not always infinite, set φ∗ (y) = supx∈R xy − φ(x) for every y ∈ I. (i) Show that φ∗ : R → ]−∞, ∞] is convex and lower semi-continuous. (Hint: 233Xh.) (ii) Show that if φ is lower semi-continuous then φ = φ∗∗ . (Hint: It is easy to check that φ∗∗ ≤ φ. For the reverse inequality, set I = {x : φ(x) < ∞}, and consider x ∈ int I, x ∈ I \ int I and x ∈ / I separately; 233Ha is useful for the first.) > (c) For the purposes of this exercise and the next, say that a Young’s function is a non-negative non-constant lower semi-continuous convex function φ : [0, ∞[ → [0, ∞] such that φ(0) = 0 and φ(x) is finite for some x > 0. (Warning! the phrase ‘Young’s function’ has other meanings.) (i) Show that in this case φ is non-decreasing and continuous on the left and φ∗ , defined by saying that φ∗ (y) = supx≥0 xy − φ(x) for every y ≥ 0, is again a Young’s function. (ii) Show that φ∗∗ = φ. Say that φ and φ∗ are complementary. (iii) Compute φ∗ in the cases (α) φ(x) = x (β) φ(x) = max(0, x − 1) (γ) φ(x) = x2 (δ) φ(x) = xp where 1 < p < ∞. > (d) Let φ, ψ = φ∗ be complementary Young’s functions in the sense of 369Xc, and (A, µ ¯) a semi-finite measure algebra. Set R R ¯ ¯ B = {u : u ∈ L0 , φ(|u|) ≤ 1}, C = {v : v ∈ L0 , ψ(|v|) ≤ 1}. (For finite-valued φ, φ¯ : (L0 )+ → L0 is given by 364I. Devise an appropriate convention for the Rcase in which φ takes the value ∞.) (i) Show that B and C are order-closed solid convex sets, and that |u × v| ≤ 2 for all u ∈ B, v ∈ C. (Hint: for ‘order-closed’, use 364Xg(iv).) (ii) Show that there is a unique extended FatouR norm τφ on L0 for which B is the unit ball. (iii) Show that if u ∈ L0 \ B there is a v ∈ C such that |u × v| > 1. (Hint: start with the case in which u ∈ S(A)+ .) (iv) Show that τψ ≤ τφ0 ≤ 2τψ , where τψ is the extended Fatou norm corresponding to ψ and τφ0 is the associate of τφ , so that τψ and τφ0 can be interpreted as equivalent norms on the same Banach space. (U and V are complementary Orlicz spaces; I will call τφ , τψ Orlicz norms.) (e) Let U be a Riesz space such that U × separates the points of U , and suppose that k k is a Fatou norm on U . (i) Show that there is a localizable measure algebra (A, µ ¯) with an extended Fatou norm τ on L0 (A) such that U can be identified, as normed Riesz space, with an order-dense Riesz subspace of Lτ . (ii) Hence, or otherwise, show that kuk = supf ∈U × , kf k≤1 |f (u)| for every u ∈ U . (iii) Show that if U is Dedekind complete and has the Levi property, then U becomes identified with Lτ itself, and in particular is a Banach lattice (cf. 354Xn). (f ) Let (A, µ ¯) be a semi-finite measure algebra, and τ an extended Fatou norm on L0 (A). Show that τ the norm of L is order-continuous iff the norm topology of Lτ agrees with the topology of convergence in measure on any order-bounded subset of Lτ .

369Xr

Banach function spaces

399

(g) Let (A, µ ¯) be a σ-finite measure algebra of countable Maharam type, and τ an extended Fatou norm on L0 (A) such that the norm of Lτ is order-continuous. Show that Lτ is separable in its norm topology. R (h) Let (A, µ ¯) be an atomless semi-finite measure algebra. Show that kuk1,∞ = max{ a |u| : a ∈ A, µ ¯a ≤ 1} for every u ∈ L0 (A). (Hint: take a ⊇ [[|u| > α0 ]] in part (b-iv) of the proof of 369O.) (i) Let (A, µ ¯) be any semi-finite measure algebra. Show that if τφ is any Orlicz norm (369Xd), then there is a γ > 0 such that kuk1,∞ ≤ γτφ (u) ≤ γ 2 kuk∞,1 for every u ∈ L0 (A), so that Mµ¯∞,1 ⊆ Lτφ ⊆ Mµ¯1,∞ . (j) Let (A, µ ¯) be a semi-finite measure algebra. Show that the subspaces Mµ¯1,∞ , Mµ¯∞,1 of L0 (A) can be expressed as a complementary pair of Orlicz spaces, and that the norm k k∞,1 can be represented as an Orlicz norm, but k k1,∞ cannot. > (k) Let (A, µ ¯) be a measure algebra and U a Banach space. (i) Suppose that ν : A → U is an additive function such that kνak ≤ min(1, µ ¯a) for every a ∈ A. Show that there is a unique bounded linear operator 1,∞ T : Mµ¯ → U such that T (χa) = νa for every a ∈ A. (ii) Suppose that ν : Af → U is an additive function such that kνak ≤ max(1, µ ¯a) for every a ∈ Af . Show that there is a unique bounded linear operator ∞,1 T : Mµ¯ → U such that T (χa) = νa for every a ∈ Af . (l) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras, and π : Af → Bf a measure-preserving ring homomorphism, as in 366H, with associated maps T : Mµ¯0 → Mν¯0 , P : Mν¯1,0 → Mµ¯1,0 . Show that kT uk∞,1 = kuk∞,1 for every u ∈ Mµ¯∞,1 , kP vk∞,1 ≤ kvk∞,1 for every v ∈ Mν¯∞,1 . (m) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a measure-preserving Boolean homomorphism. (i) Show that there is a unique Riesz homomorphism T : Mµ¯1,∞ → Mν¯1,∞ such that T (χa) = χ(πa) for every a ∈ A and kT uk1,∞ = kuk1,∞ for every u ∈ Mµ¯1,∞ . (ii) Now suppose that (A, µ ¯) is localizable and π is order-continuous. Show that there is a unique positive linear operator P : Mν¯1,∞ → Mµ¯1,∞ R R such that a P v = πa v for every a ∈ Af , v ∈ Mν¯1,∞ , and that kP vk∞ ≤ kvk∞ for every v ∈ L∞ (B), kP vk1,∞ ≤ kvk1,∞ for every v ∈ Mν¯1,∞ . (Compare 365P.) (n) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras, and φ : [0, ∞[ → [0, ∞] a Young’s function, as in 369Xd; write τφ for the corresponding Orlicz norm on either L0 (A) or L0 (B). Let π : A → B be a measure-preserving Boolean homomorphism, with associated map T : Mµ¯1,∞ → Mν¯1,∞ , as in 369Xm. (i) Show that τφ (T u) = τφ (u) for every u ∈ Mµ¯1,∞ . (ii) Show that if (A, µ ¯) is localizable, π is order-continuous and P : Mν¯1,∞ → Mµ¯1,∞ is the map of 369Xm(ii), then τφ (P v) ≤ τφ (v) for every v ∈ Mν¯1,∞ . (Hint: 365R.) > (o) Let (A, µ ¯) be any semi-finite measure algebra and τ1 , τ2 two extended Fatou norms on L0 (A). Show that u 7→ max(τ1 (u), τ2 (u)) is an extended Fatou norm. b µ (p) Let (A, µ ¯) be a semi-finite measure algebra, and (A, ˜) its localization (322P). Show that the Dedekind 1,∞ 1,∞ b completion of M (A, µ ¯) can be identified with M (A, µ ˜). (q) Let (A, µ ¯) be a localizable measure algebra. (i) Show that if B is any closed subalgebra of A such that sup{b : b ∈ B, µ ¯b < ∞} = 1 in A, we have an order-continuous positive linear operator PB : Mµ¯1,∞ → Mµ¯1,∞ ¹B R R such that b PB u = b u whenever u ∈ Mµ¯1,∞ , b ∈ B and µ ¯b < ∞. (ii) Show that if hBn in∈N is a nondecreasing sequence of closed subalgebras of A such that sup{b : b ∈ B0 , µ ¯b < ∞} = 1 in A, and B is the S closure of n∈N Bn , then hPBn uin∈N is order*-convergent to PB u for every u ∈ Mµ¯1,∞ . (Cf. 367K.) (r) Let φ1 and φ2 be Young’s functions (369Xc) and (A, µ ¯) a semi-finite measure algebra. Set φ(x) = max(φ1 (x), φ2 (x)) for x ∈ [0, ∞[. (i) Show that φ is a Young’s function. (ii) Writing τφ1 τφ2 , τφ for the corresponding extended Fatou norms on L0 (A) (369Xd), show that τφ ≥ max(τφ1 , τφ2 ) ≥ 21 τφ , so that τ ∗ τ ∗ Lτφ = Lτφ1 ∩ Lτφ2 and Lτφ∗ = L φ1 + L φ2 , writing φ∗ for the Young’s function complementary to φ. (iii) Repeat with ψ = φ1 + φ2 in place of φ.

400

Function spaces

369Y

369Y Further exercises (a) Let (A, µ ¯) be a localizable measure algebra and A ⊆ L0 = L0 (A) a countable set. Show that the solid linear subspace U of L0 generated by A is a perfect Riesz space. (Hint: 0 + 0 + Rreduce to thencase ninR which U is order-dense. If A = {un : n ∈ N}, w ∈ (L ) \ U find vn 0∈ (L ) such that vn × u ≥ 2 ≥ 4 vn × |ui | for every i ≤ n. Show that v = supn∈N vn is defined in L and corresponds to a member of U × .) (b) Let U be a Banach lattice and suppose that p ∈ [1, ∞[ is such that ku + vkp = kukp + kvkp whenever u, v ∈ U and |u| ∧ |v| = 0. Show that U is isomorphic, as Banach lattice, to Lpµ¯ for some localizable measure algebra (A, µ ¯). (Hint: start by using 354Yd to show that the norm of U is order-continuous, as in 354Yj.) (c) Let φ : [0, ∞[ → [0, ∞[ be a strictly increasing Young’s function (369Xc) such that φ(0) = 0 and supt>0 φ(2t)/φ(t) is finite. Show that the associated Orlicz norms τφ (369Xd) are always order-continuous on their function spaces. (d) Let φ : [0, ∞[ → [0, ∞] be a Young’s function, and suppose that the corresponding Orlicz norm on L0 (AL ), where (AL , µ ¯L ) is the measure algebra of Lebesgue measure on R, is order-continuous on its function space Lτφ . Show that there is an M ≥ 0 such that φ(2t) ≤ M φ(t) for every t ≥ 0. (e) Let (A, µ ¯) be a semi-finite measure algebra and φ : [0, ∞[ → [0, ∞[ a Young’s function such that the Orlicz norm τφ is order-continuous on Lτφ . Show that if F is a filter on Lτφ , then F → u ∈ Lτφ for the norm τφ iff (i) F → u for the topology of convergence in measure (ii) lim supv→F τφ (v) ≤ τφ (u). (Compare 245Xk.) (f ) Give an example of an extended Fatou norm τ on L0 (AL ), where (AL , µ ¯L ) is the measure algebra of Lebesgue measure on [0, 1], such that (i) τ gives rise to an order-continuous norm on its function space Lτ (ii) there is a sequence hun in∈N in Lτ , converging in measure to u ∈ Lτ , such that limn→∞ τ (un ) = τ (u) but hun in∈N does not converge to u for the norm on Lτ . (g) Let (A, µ ¯) be a semi-finite measure algebra, and τ an Orlicz norm on L0 (A). Show that Lτ has the Levi property, whether or not A is Dedekind complete. (h) Let (A, µ ¯) be any measure algebra. Show that (Mµ¯1,0 )× can be identified with Mµ¯∞,1 . (Hint: show that neither M 1,0 nor M ∞,1 is changed by moving first to the semi-finite version of (A, µ ¯), as described in 322Xa, and then to its localization.) (i) Give an example to show that the result of 369R may fail if (A, µ ¯) is only semi-finite, not localizable. 369 Notes and comments The representation theorems 369A-369D give a very concrete form to the notion of ‘perfect’ Riesz space: it is just one which can be expressed as a subspace of L0 (A), for some localizable measure algebra (A, µ ¯), in such a way that it is its own second dual, where the duality here is between subspaces of L0 , taking U 0 = {v : u × v ∈ L1 for every u ∈ U }. (I see that in this expression I ought somewhere to mention that both U and U 0 are assumed to be order-dense in L0 .) Indeed I believe that the original perfect spaces were the ‘vollkommene R¨aume’ of G.K¨othe, which were subspaces of RN , corresponding to the measure algebra PN with counting measure, so that U 0 or U × was {v : u × v ∈ `1 for every u ∈ U }. I have presented Kakutani’s theorem on the representation of L-spaces as a corollary of 369A and 369C. As usual in such things, this is a reversal of the historical relationship; Kakutani’s theorem was one of the results which led to the general theory. If we take the trouble to re-work the argument of 369A in this context, we find that the L-space condition ‘ku + vk = kuk + kvk whenever u, v ≥ 0’ can be relaxed to ‘ku + vk = kuk + kvk whenever u ∧ v = 0’ (369Yb). The complete list of localizable measure algebras provided by Maharam’s theorem (332B, 332J) now gives us a complete list of L-spaces. Just as perfect Riesz spaces come in dual pairs, so do some of the most important Banach lattices: those with Fatou norms and the Levi property for which the order-continuous dual separates the points. (Note that the dual of any space with a Riesz norm has these properties; see 356Da.) I leave the details of representing such spaces to you (369Xe). The machinery of 369F-369K gives a solid basis for studying such pairs.

369 Notes

Banach function spaces

401

Among the extended Fatou norms of 369F the Orlicz norms (369Xd, 369Yc-369Ye) form a significant subfamily. Because they are defined in a way which is to some extent independent of the measure algebra involved, these spaces have some of the same properties as Lp spaces in relation to measure-preserving homomorphisms (369Xm-369Xn). In §§373-374 I will elaborate on these ideas. Among the Orlicz spaces, we have a largest and a smallest; these are just M 1,∞ = L1 + L∞ and M ∞,1 = L1 ∩ L∞ (369N-369O, 369Xi, 369Xj). Of course these two are particularly important. There is an interesting phenomenon here. It is easy to see that k k∞,1 = max(k k1 , k k∞ ) is an extended Fatou norm and that the corresponding Banach lattice is L1 ∩L∞ ; and that the same ideas work for any pair of extended Fatou norms (369Xo). To check that the dual of L1 ∩ L∞ is precisely the linear sum L∞ + L1 a little more is needed, and the generalization of this fact to other extended Fatou norms (369Q) seems to go quite deep. In view of our ordinary expectation that properties of these normed function spaces should be reflected in perfect Riesz spaces in general, I mention that I believe I have found an example, dependent on the continuum hypothesis, of two perfect Riesz subspaces U , V of RN such that their linear sum U + V is not perfect.

402

Linear operators between function spaces

Chapter 37 Linear operators between function spaces As everywhere in functional analysis, the function spaces of measure theory cannot be properly understood without investigating linear operators between them. In this chapter I have collected a number of results which rely on, or illuminate, the measure-theoretic aspects of the theory. §371 is devoted to a fundamental property of linear operators on L-spaces, if considered abstractly, that is, of L1 -spaces, if considered in the language of Chapter 36, and to an introduction to the class T of operators which are norm-decreasing for both k k1 and k k∞ . This makes it possible to prove a version of (Birkhoff’s) Ergodic Theorem for operators which need not be positive (372D). In §372 I give various forms of this theorem, for linear operators between function spaces, for measure-preserving Boolean homomorphisms between measure algebras, and for inverse-measurepreserving functions between measure spaces, with an excursion into the theory of continued fractions. In §373 I make a fuller analysis of the class T , with a complete characterization of those u, v such that v = T u for some T ∈ T . Using this we can describe ‘rearrangement-invariant’ function spaces and extended Fatou norms (§374). Returning to ideas left on one side in §§364 and 368, I investigate positive linear operators defined on L0 spaces (§375). In the final section of the chapter (§376), I look at operators which can be defined in terms of kernels on product spaces.

371 The Chacon-Krengel theorem The first topic I wish to treat is a remarkable property of L-spaces: if U and V are L-spaces, then every continuous linear operator T : U → V is order-bounded, and k|T |k = kT k (371D). This generalizes in various ways to other V (371B, 371C). I apply the result to a special type of operator between M 1,0 spaces which will be conspicuous in the next section (371F-371H). 371A Lemma Let U be an L-space, V a Banach lattice and T : U → V a bounded linear operator. Take u ≥ 0 in U and set Pn Pn B = { i=0 |T ui | : u0 , . . . , un ∈ U + , i=0 ui = u} ⊆ V + . Then B is upwards-directed and supv∈B kvk ≤ kT kkuk.

Pm proof (a) Suppose that v, v 0 ∈ B. Then we have u0 , . . . , um , u00 , . . . , u0n ∈ U + such that i=0 ui = P P Pn n m 0 0 0 |T u |. Now there are v ≥ 0 in U , for i ≤ m and j ≤ n, such |T u | and v = u = u, v = ij i j i=0 j=0 j P Pm Pn Pj=0 m n 0 that ui = j=0 vij for i ≤ m and uj = i=0 vij for j ≤ n (352Fd). We have u = i=0 j=0 vij , so that Pm Pn v 00 = i=0 j=0 |T vij | ∈ B. But Pm Pm Pn Pm Pm v = i=0 |T ui | = i=0 |T ( j=0 vij )| ≤ i=0 j=0 |T vij | = v 00 , and similarly v 0 ≤ v 00 . As v and v 0 are arbitrary, B is upwards-directed. Pm Pm (b) The other part is easy. If v ∈ B is expressed as i=0 |T ui | where ui ≥ 0, i=0 ui = u then Pm Pm kvk ≤ i=0 kT ui k ≤ kT k i=0 kui k = kT kkuk because U is an L-space. 371B Theorem Let U be an L-space and V a Dedekind complete Banach lattice U with a Fatou norm. Then the Riesz space L∼ (U ; V ) = L× (U ; V ) is a closed linear subspace of the Banach space B(U ; V ) and is in itself a Banach lattice with a Fatou norm. proof (a) I start by noting that L∼ (U ; V ) = L× (U ; V ) ⊆ B(U ; V ) just because V has a Riesz norm and U is a Banach lattice with an order-continuous norm (355C, 355Kb). (b) The first new step is to check that k|T |k ≤ kT k for any T ∈ L∼ (U ; V ). P P Start with any u ∈ U + . Set Pn Pn B = { i=0 |T ui | : u0 , . . . , un ∈ U + , i=0 ui = u} ⊆ V + ,

371C

The Chacon-Krengel theorem

403

Pn Pn as in 371A. If u0 , . . . , un ≥ 0 are such that i=0 ui = u, then |T ui | ≤ |T |ui for each i, so that i=0 |T ui | ≤ P n i=0 |T |ui = |T |u; thus B is bounded above by |T |u and sup B ≤ |T |u. On the other hand, if |v| ≤ u in U , then v + + v − + (u − |v|) = u, so |T v + | + |T v − | + |T (u − |v|)| ∈ B and |T v| = |T v + + T v − | ≤ |T v + | + |T v − | ≤ sup B. As v is arbitrary, |T |u ≤ sup B and |T |u = sup B. Consequently k|T |uk ≤ k sup Bk = supw∈B kwk ≤ kT kkuk because V has a Fatou norm and B is upwards-directed. For general u ∈ U , k|T |uk ≤ k|T ||u|k ≤ kT kk|u|k = kT kkuk. This shows that k|T |k ≤ kT k. Q Q (c) Now if |S| ≤ |T | in L∼ (U ; V ), and u ∈ U , we must have kSuk ≤ k|S||u|k ≤ k|T ||u|k ≤ k|T |kk|u|k ≤ kT kkuk; as u is arbitrary, kSk ≤ kT k. This shows that the norm of L∼ (U ; V ), inherited from B(U ; V ), is a Riesz norm. (d) Suppose next that T ∈ B(U ; V ) belongs to the norm-closure of L∼ (U ; V ). For each n ∈ N choose Tn ∈ L∼ (U ; V ) such that kT − Tn k ≤ 2−n . Set Sn = |Tn+1 − Tn | ∈ L∼ (U ; V ) for each n. Then for each n, so S =

P∞ n=0

kSn k = kTn+1 − Tn k ≤ 3 · 2−n−1 Sn is defined in the Banach space B(U ; V ). But if u ∈ U + , we surely have P∞ Su = n=0 Sn u ≥ 0

in V . Moreover, if u ∈ U + and |v| ≤ u, then for any n ∈ N Pn Pn |Tn+1 v − T0 v| = | i=0 (Ti+1 − Ti )v| ≤ i=0 Si u ≤ Su, and T0 v − Su ≤ Tn+1 v ≤ T0 v + Su; letting n → ∞, we see that −|T0 |u − Su ≤ T0 v − Su ≤ T v ≤ T0 v + Su ≤ |T0 |u + Su. So |T v| ≤ |T0 |u + Su whenever |v| ≤ u. As u is arbitrary, T ∈ L∼ (U ; V ). This shows that L∼ (U ; V ) is closed in B(U ; V ) and is therefore a Banach space in its own right; putting this together with (b), we see that it is a Banach lattice. (e) Finally, the norm of L∼ (U ; V ) is a Fatou norm. P P Let A ⊆ L∼ (U ; V )+ be a non-empty, upwards∼ directed set with supremum T0 ∈ L (U ; V ). For any u ∈ U , kT0 uk = k|T0 u|k ≤ kT0 |u|k = k supT ∈A T |u|k by 355Ed. But {T |u| : T ∈ A} is upwards-directed and the norm of V is a Fatou norm, so kT0 uk ≤ supT ∈A kT |u|k ≤ supT ∈A kT kkuk. As u is arbitrary, kT0 k ≤ supT ∈A kT k. As A is arbitrary, the norm of L∼ (U ; V ) is Fatou. Q Q 371C Theorem Let U be an L-space and V a Dedekind complete Banach lattice with a Fatou norm and the Levi property. Then B(U ; V ) = L∼ (U ; V ) = L× (U ; V ) is a Dedekind complete Banach lattice with a Fatou norm and the Levi property. In particular, |T | is defined and k|T |k = kT k for every T ∈ B(U ; V ). proof (a) Let T : U → V be any bounded linear operator. Then T ∈ L∼ (U ; V ). P P Take any u ≥ 0 in U . Set Pn Pn B = { i=0 |T ui | : u0 , . . . , un ∈ U + , i=0 ui = u} ⊆ V + as in 371A. Then 371A tells us that B is upwards-directed and norm-bounded. Because V has the Levi property, B is bounded above. But just as in part (b) of the proof of 371B, any upper bound of B is also an upper bound of {T v : |v| ≤ u}. As u is arbitrary, T ∈ L∼ (U ; V ). Q Q

404

Linear operators between function spaces

371C

(b) Accordingly L∼ (U ; V ) = B(U ; V ). By 371B, this is a Banach lattice with a Fatou norm, and equal to L× (U ; V ). To see that it also has the Levi property, let A ⊆ L∼ (U ; V ) be any non-empty norm-bounded upwards-directed set. For u ∈ U + , {T u : T ∈ A} is non-empty, norm-bounded and upwards-directed in V , so is bounded above in V . By 355Ed, A is bounded above in L∼ (U ; V ). 371D Corollary Let U and V be L-spaces. Then L∼ (U ; V ) = L× (U ; V ) = B(U ; V ) is a Dedekind complete Banach lattice with a Fatou norm and the Levi property. 371E Remarks Note that both these theorems show that L∼ (U ; V ) is a Banach lattice with properties similar to those of V whenever U is an L-space. They can therefore be applied repeatedly, to give facts about L∼ (U1 ; L∼ (U2 ; V )) where U1 , U2 are L-spaces and V is a Banach lattice, for instance. I hope that this formula will recall some of those in the theory of bilinear maps and tensor products (see 253Xa-253Xb). 371F The class T (0) For the sake of applications in the next section, I introduce now a class of operators of great intrinsic interest. Definition Let (A, µ ¯), (B, ν¯) be measure algebras. Recall that M 1,0 (A, µ ¯) is the space of those u ∈ (0) 1 ∞ L (A, µ ¯) + L (A) such that µ ¯[[|u| > α]] < ∞ for every α > 0 (366F-366G, 369P). Write T (0) = Tµ¯,¯ν for the set of all linear operators T : M 1,0 (A, µ ¯) → M 1,0 (B, ν¯) such that T u ∈ L1 (B, ν¯) and kT uk1 ≤ kuk1 for 1 ∞ every u ∈ L (A, µ ¯), T u ∈ L (B) and kT uk∞ ≤ kuk∞ for every u ∈ L∞ (A) ∩ M 1,0 (A, µ ¯). 371G Proposition Let (A, µ ¯) and (B, ν¯) be measure algebras. (0) (0) ¯); M 1,0 (B, ν¯)). If T0 : L1 (A, µ ¯) → L1 (B, ν¯) is (a) T = Tµ¯,¯ν is a convex set in the unit ball of B(M 1,0 (A, µ ∞ a linear operator of norm at most 1, and T0 u ∈ L (B) and kT0 uk∞ ≤ kuk∞ for every u ∈ L1 (A, µ ¯)∩L∞ (A), (0) then T0 has a unique extension to a member of T . (b) If T ∈ T (0) then T is order-bounded and |T |, taken in L∼ (M 1,0 (A, µ ¯); M 1,0 (B, ν¯)) = L× (M 1,0 (A, µ ¯); M 1,0 (B, ν¯)), also belongs to T (0) . (c) If T ∈ T (0) then kT uk1,∞ ≤ kuk1,∞ for every u ∈ M 1,0 (A, µ ¯). (d) If T ∈ T (0) , p ∈ [1, ∞[ and w ∈ Lp (A, µ ¯) then T w ∈ Lp (B, ν¯) and kT wkp ≤ kwkp . (0) (0) ¯ is another measure algebra then ST ∈ T (0) (e) If (C, λ) ¯ ,¯ ν and every S ∈ Tν ¯. ¯ for every T ∈ Tµ ¯ ,λ µ ¯ ,λ proof I write Mµ¯1,0 , Lpν¯ for Mµ¯1,0 , Lp (B, ν¯), etc. (a)(i) If T ∈ T (0) and u ∈ Mµ¯1,0 then there are v ∈ L1µ¯ , w ∈ L∞ µ ¯ such that u = v + w and kvk1 + kwk∞ = kuk1,∞ (369Ob); so that kT uk1,∞ ≤ kT vk1 + kT wk∞ ≤ kvk1 + kwk∞ ≤ kuk1,∞ . As u is arbitrary, T is in the unit ball of B(Mµ¯1,0 ; Mν¯1,0 ). ∞ (0) (ii) Because the unit balls of B(L1µ¯ ; L1ν¯ ) and B(L∞ . µ ¯ ; Lν ¯ ) are convex, so is T

(iii) Now suppose that T0 : L1µ¯ → L1ν¯ is a linear operator of norm at most 1 such that kT0 uk∞ ≤ kuk∞ 1 for every u ∈ L1µ¯ ∩ L∞ µ ¯ . By the argument of (i), T0 is a bounded operator for the k k1,∞ norms; since Lµ ¯ is 1,0 1,0 1,0 dense in Mµ¯ (369Pc), T0 has a unique extension to a bounded linear operator T : Mµ¯ → Mν¯ . Of course kT uk1 = kT0 uk1 ≤ kuk1 for every u ∈ L1µ¯ . 1,0 Now suppose that u ∈ L∞ ¯ ; set γ = kuk∞ . Let ² > 0, and set µ ¯ ∩ Mµ v = (u+ − ²χ1)+ − (u− − ²χ1)+ ; then |v| ≤ |u| and ku − vk∞ ≤ ² and v ∈ L1µ¯ ∩ L∞ µ ¯ . Accordingly kT u − T vk1,∞ ≤ ku − vk1,∞ ≤ ²,

kT vk∞ = kT0 vk∞ ≤ kvk∞ ≤ γ.

So if we set w = (|T u − T v| − ²χ1)+ ∈ L1ν¯ , kwk1 ≤ ²; while

371G

The Chacon-Krengel theorem

405

|T u| ≤ |T v| + w + ²χ1 ≤ (γ + ²)χ1 + w, so k(|T u| − (γ + ²)χ1)+ k1 ≤ kwk1 ≤ ². As ² is arbitrary, |T u| ≤ γχ1, that is, kT uk∞ ≤ kuk∞ . As u is arbitrary, T ∈ T (0) . (b) Because Mµ¯1,0 has an order-continuous norm (369Pb), L∼ (Mµ¯1,0 ; Mν¯1,0 ) = L× (Mµ¯1,0 ; Mν¯1,0 ) (355Kb). Take any T ∈ T (0) and consider T0 = T ¹L1µ¯ : L1µ¯ → L1ν¯ . This is an operator of norm at most 1. By 371D, T0 is order-bounded, and k|T0 |k ≤ 1, where |T0 | is taken in L∼ (L1µ¯ ; L1µ¯ ) = B(L1µ¯ ; L1ν¯ ). Now if u ∈ L1µ¯ ∩ L∞ ν ¯ , ||T0 |u| ≤ |T0 ||u| = sup|u0 |≤|u| |T0 u0 | ≤ kuk∞ χ1, so k|T0 |uk∞ ≤ kuk∞ . By (a), there is a unique S ∈ T (0) extending |T0 |. Now Su+ ≥ 0 for every u ∈ L1µ¯ , so Su+ ≥ 0 for every u ∈ Mµ¯1,0 (since the function u 7→ (Su+ )+ − Su+ : Mµ¯1,0 → Mν¯1,0 is continuous and zero on the dense set L1µ¯ ), that is, S is a positive operator; also S|u| ≥ |T u| for every u ∈ L1µ¯ , so Sv ≥ S|u| ≥ |T u| whenever u, v ∈ Mµ¯1,0 and |u| ≤ v. This means that T : Mµ¯1,0 → Mν¯1,0 is order-bounded. Because Mν¯1,0 is Dedekind complete (366Ga), |T | is defined in L∼ (Mµ¯1,0 ; Mµ¯1,0 ). If v ≥ 0 in L1µ¯ , then |T |v = sup|u|≤v T u = sup|u|≤v T0 u = |T0 |v = Sv. Thus |T | agrees with S on L1µ¯ . Because Mµ¯1,0 is a Banach lattice (or otherwise), |T | is a bounded operator, therefore continuous (2A4Fc), so |T | = S ∈ T (0) , which is what we needed to know. (c) We can express u as v + w where kvk1 + kwk∞ = kuk1,∞ ; now w = u − v ∈ Mµ¯1,0 , so we can speak of T w, and kT uk1,∞ = kT v + T wk1,∞ ≤ kT vk1 + kT wk∞ ≤ kvk1 + kwk∞ = kuk1,∞ , as required. (d) (This is a modification of 244M.) (i) Suppose that T , p, w are as described, and that in addition T is positive. The function t 7→ |t|p is convex (233Xc), so we can find families hβq iq∈Q , hγq iq∈Q of real numbers such that |t|p = supq∈Q βq +γq (t−q) for every t ∈ R (233Hb). Then |u|p = supq∈Q βq χ1 + γq (u − qχ1) for every u ∈ L0 . (The easiest way to check this is perhaps to think of L0 as a quotient of a space of functions, as in 364D; it is also a consequence of 364Xg(iii).) We know that |w|p ∈ L1µ¯ , so we may speak of T (|w|p ); while w ∈ Mµ¯1,0 (366Ga), so we may speak of T w. For any q ∈ Q, 0p ≥ βq −qγq , that is, qγq −βq ≥ 0, while γq w−|w|p ≤ (qγq −βq )χ1 and k(γq w−|w|p )+ k∞ ≤ qγq − βq . Now this means that T (γq w − |w|p ) ≤ T (γq w − |w|p )+ ≤ kT (γq w − |w|p )+ k∞ χ1 ≤ k(γq w − |w|p )+ k∞ χ1 ≤ (qγq − βq )χ1. Turning this round again, βq χ1 + γq (T w − qχ1) ≤ T (|w|p ). R R Taking the supremum over q, |T w|p ≤ T (|w|p ), so that |T w|p ≤ |w|p (because kT vk1 ≤ kvk1 for every v ∈ L1 ). Thus T w ∈ Lp and kT wkp ≤ kwkp . (ii) For a general T ∈ T (0) , we have |T | ∈ T (0) , by (b), and |T w| ≤ |T ||w|, so that kT wkp ≤ k|T ||w|kp ≤ kwkp , as required. (e) This is elementary, because kST uk1 ≤ kT uk1 ≤ kuk1 , whenever u ∈ L1µ¯ , v ∈ L∞ µ ¯ ∩

Mµ¯1,0 .

kST vk∞ ≤ kT uk∞ ≤ kuk∞

406

Linear operators between function spaces (0)

371H (0)

371H Remark In the context of 366H, Tπ ¹Mµ¯1,0 ∈ Tµ¯,¯ν , while Pπ ∈ Tν¯,¯µ . Thus 366H(a-iv), 366H(b-iii) are special cases of 371Gd. 371X Basic exercises >(a) Let U be an L-space, V a Banach lattice with an order-continuous norm and T : U → V a bounded linear operator. Let B be the unit ball of U . Show that |T |[B] ⊆ T [B]. (b) Let U and V be Banach spaces. (i) Show that the space Kw (U ; V ) of weakly compact linear operators from U to V (definition: 3A5Kb) is a closed linear subspace of B(U ; V ). (ii) Show that if U is an L-space and V is a Banach lattice with an order-continuous norm, then Kw (U ; V ) is a norm-closed Riesz subspace of L∼ (U ; V ). (c) Let (A, µ ¯) be a semi-finite measure algebra and set U = L1 (A, µ ¯). Show that L∼ (U ; U ) = B(U ; U ) is a Banach lattice with a Fatou norm and the Levi property. Show that its norm is order-continuous iff A is finite. (Hint: consider operators u 7→ u × χa, where a ∈ A.) > (d) Let U be a Banach lattice, and V a Dedekind complete M -space. Show that L∼ (U ; V ) = B(U ; V ) is a Banach lattice with a Fatou norm and the Levi property. (e) Let U and V be Riesz spaces, of which V is Dedekind complete, and let T ∈ L∼ (U ; V ). Define T ∈ L∼ (V ∼ ; U ∼ ) by writing T 0 (h) = hT for h ∈ V ∼ . (i) Show that |T |0 ≥ |T 0 | in L∼ (V ∼ ; U ∼ ). (ii) Show × + × + 0 that |T |0 h = |T 0 |h for every Pn h ∈ V . (Hint: show thatPifn u ∈ U , h ∈ (V ) then (|T |h)(u) and h(|T |u) are both equal to sup{ i=0 gi (T ui ) : |gi | ≤ h, ui ≥ 0, i=0 ui = u}.) 0

> (f ) Using 371D, but nothing about uniformly integrable sets beyond the definition (354P), show that if U and V are L-spaces, A ⊆ U is uniformly integrable in U , and T : U → V is a bounded linear operator, then T [A] is uniformly integrable in V . 371Y Further exercises (a) Let U and V be Banach spaces. (i) Show that the space K(U ; V ) of compact linear operators from U to V (definition: 3A5Ka) is a closed linear subspace of B(U ; V ). (ii) Show that if U is an L-space and V is a Banach lattice with an order-continuous norm, then K(U ; V ) is a norm-closed Riesz subspace of L∼ (U ; V ). (See Krengel 63.) (b) Let (A, µ ¯) be a measure algebra, U a Banach space, and T : L1 (A, µ ¯) → U a bounded linear operator. Show that T is a compact linear operator iff {

1 T (χa) µ ¯a

: a ∈ A, 0 < µ ¯a < ∞} is relatively compact in U .

(c) Let (A, µ ¯) be a probability algebra, and set L1 = L1 (A, µ ¯). Let han in∈N be a stochastically R R independent sequence of elements of A of measure 21 , and define T : L1 → RN by setting T u(n) = u − 2 an u for each n. Show that T ∈ B(L1 ; c 0 ) \ L∼ (L1 ; c 0 ), where c 0 is the Banach lattice of sequences converging to 0. (See 272Yd.) (d) Regarding T of 371Yc as a map from L1 to `∞ , show that |T 0 | 6= |T |0 in L∼ ((`∞ )∗ , L∞ (A)). (e) (i) In `2 define ei by setting ei (i) = 1, ei (j) = 0 if j 6= i. Show that if T ∈ L∼ (`2 ; `2 ) then (|T |ei |ej ) = |(T ei |ej )| for all i, j ∈ N. (ii) Show that for each n ∈ N there is an orthogonal (2n¶× 2n )-matrix µ An An 1 An such that every coefficient of An has modulus 2−n/2 . (Hint: An+1 = √ .) (iii) Show 2 −An An that there is a linear isometry S : `2 → `2 such that |(Sei |ej )| = 2−n/2 if 2n ≤ i, j < 2n+1 . (iv) Show that S∈ / L∼ (`2 ; `2 ). 371 Notes and comments The ‘Chacon-Krengel theorem’, properly speaking (Chacon & Krengel 64), is 371D in the case in which U = L1 (µ), V = L1 (ν); of course no new ideas are required in the generalizations here, which I have copied from Fremlin 74a. Anyone with a training in functional analysis will automatically seek to investigate properties of operators T : U → V in terms of properties of their adjoints T 0 : V ∗ → U ∗ , as in 371Xe and 371Yd. When U is

372B

The ergodic theorem

407

an L-space, then U ∗ is a Dedekind complete M -space, and it is easy to see that this forces T 0 to be orderbounded, for any Banach lattice V (371Xd). But since no important L-space is reflexive, this approach cannot reach 371B-371D without a new idea of some kind. It can however be adapted to the special case in 371Gb (Dunford & Schwartz 57, VIII.6.4). In fact the results of 371B-371C are characteristic of L-spaces (Fremlin 74b). To see that they fail in the simplest cases in which U is not an L-space and V is not an M -space, see 371Yc-371Ye.

372 The ergodic theorem I come now to one of the most remarkable topics in measure theory. I cannot do it justice in the space I have allowed for it here, but I can give the basic theorem (372D-372E) and a variety of the corollaries through which it is regularly used (372F-372K), together with brief notes on one of its most famous and characteristic applications (to continued fractions, 372M-372O) and on ‘ergodic’ and ‘mixing’ transformations (372P372R). In the first half of the section (down to 372G) I express the arguments in the abstract language of measure algebras and their associated function spaces, as developed in Chapter 36; the second half, from 372H onwards, contains translations of the results into the language of measure spaces and measurable functions, the more traditional, and more readily applicable, forms. 372A Lemma Let U be a reflexive Banach space and T : U → U a bounded linear operator of norm at most 1. Then V = {u + v − T u : u, v ∈ U, T v = v} is dense in U . proof Of course V is a linear subspace of U . ?? Suppose, if possible, that it is not dense. Then there is a non-zero h ∈ U ∗ such that h(v) = 0 for every v ∈ V (3A5Ad). Take u ∈ U such that h(u) 6= 0. Set 1 Pn i un = i=0 T u n+1

0

for each n ∈ N, taking T to be the identity operator; because kT i uk ≤ kT i kkuk ≤ kT ki kuk ≤ kuk for each i, kun k ≤ kuk for every n. Note also that T i+1 u − T i u ∈ V for every i, so that h(T i+1 u − T i u) = 0; accordingly h(T i u) = h(u) for every i, and h(un ) = u for every n. Let F be any non-principal ultrafilter on N. Because U is reflexive, v = limn→F un is defined in U for the weak topology on U (3A5Gc). Now T v = v. P P For each n ∈ N, P 1 1 n i+1 T un − un = u − T i u) = (T n+1 u − u) i=0 (T n+1

n+1

2 has norm at most n+1 kuk. So hT un − un in∈N → 0 for the norm topology U and therefore for the weak topology, and surely limn→F T un − un = 0. On the other hand (because T is continuous for the weak topology, 2A5If)

T v = limn→F T un = limn→F (T un − un ) + limn→F un = 0 + v = v, where all the limits are taken for the weak topology. Q Q But this means that v ∈ V , while h(v) = limn→F h(un ) = h(u) 6= 0, contradicting the assumption that h ∈ V ◦ . X X 372B Lemma Let (A, µ ¯) be a measure algebra, and T : L1 → L1 a positive linear operator of norm at 1 1 most 1, where L = L (A, µ ¯). Take any u ∈ L1 and m ∈ N, and set a = [[u > 0]] ∪ [[u + T u > 0]] ∪ [[u + T u + T 2 u > 0]] ∪ . . . ∪ [[u + T u + . . . + T m u > 0]].

408

Then

Linear operators between function spaces

R a

372B

u ≥ 0.

proof Set u0 = u, u1 = u + T u, . . . , um = u + T u + . . . + T m u, v = supi≤m ui , so that a = [[v > 0]]. Consider u + T (v + ). We have T (v + ) ≥ T v ≥ T ui for every i ≤ m (because T is positive), so that u + T (v + ) ≥ u + T ui = ui+1 for i < m, and u + T (v + ) ≥ sup1≤i≤m ui . Also u + T (v + ) ≥ u because T (v + ) ≥ 0, so u + T (v + ) ≥ v. Accordingly

R

a

u≥

R

a

v−

R

a

T (v + ) =

R

v+ −

R

a

T (v + ) ≥ kv + k1 − kT v + k1 ≥ 0

because kT k ≤ 1. 372C Maximal Ergodic Theorem Let (A, µ ¯) be a measure algebra, and T : L1 → L1 a linear 1 1 operator, where L = L (A, µ ¯), P such that kT uk1 ≤ kuk1 for every u ∈ L1 and kT uk∞ ≤ kuk∞ for every n 1 1 ∞ u ∈ L ∩ L (A). Set An = n+1 i=0 T i for each n ∈ N. Then for any u ∈ L1 , u∗ = supn∈N An u is defined in L0 (A), and α¯ µ[[u∗ > α]] ≤ kuk1 for every α > 0. proof (a) To begin with, suppose that T is positive and that u ≥ 0 in L1 . Note that if v ∈ L1 ∩ L∞ , then kT i vk∞ ≤ kvk∞ for every i ∈ N, so kAn vk∞ ≤ kvk∞ for every n; in particular, An (χa) ≤ χ1 for every n and every a of finite measure. For m ∈ N and α > 0, set amα = [[supi≤m Ai u > α]]. Then α¯ µamα ≤ kuk1 . P P Set a = amα , w = u − αχa. Of course supi≤m Ai u belongs to L1 , so µ ¯a is finite 1 and w ∈ L . For any i ≤ m, Ai w = Ai u − αAi (χa) ≥ Ai u − αχ1, so [[Ai w > 0]] ⊇ [[Ai u > α]]. Accordingly a ⊆ b, where By 372B,

R b

b = supi≤m [[Ai w > 0]] = supi≤m [[w + T w + . . . + T i w > 0]]. w ≥ 0. But this means that α¯ µa = α

R

χa = b

R

u− b

R

w≤ b

R b

u ≤ kuk1 ,

as claimed. Q Q It follows that if we set cα = supn∈N anα , µ ¯cα ≤ α−1 kuk1 for every α > 0 and inf α>0 cα = 0. But ∗ this is exactly the criterion in 364Mb for u = supn∈N An u to be defined in L0 . And [[u∗ > α]] = cα , so α¯ µ[[u∗ > α]] ≤ kuk1 for every α > 0, as required. (b) Now consider the case of general T , u. In this case T is order-bounded and k|T |k ≤ 1, where |T | is the modulus of T in L∼ (L1 ; L1 ) = B(L1 ; L1 ) (371D). If w ∈ L1 ∩ L∞ , then ||T |w| ≤ |T ||w| = sup|w0 |≤|w| |T w0 | ≤ kwk∞ χ1,

Pn 1 i so k|T |wk∞ ≤ kwk∞ . Thus |T | also satisfies the conditions of the theorem. Setting Bn = n+1 i=0 |T | , ∼ 1 1 0 Bn ≥ An in L (L ; L ) and Bn |u| ≥ An u for every n. But by (a), v = supn∈N Bn |u| is defined in L and α¯ µ[[v > α]] ≤ k|u|k1 = kuk1 for every α > 0. Consequently u∗ = supn∈N An u is defined in L0 and u∗ ≤ v, so that α¯ µ[[u∗ > α]] ≤ kuk1 for every α > 0. 372D We are now ready for a very general form of the Ergodic Theorem. I express it in terms of the space M 1,0 from 366F and the class T (0) of operators from 371F. If these formulae are unfamiliar, you may like to glance at the statement of 372E before looking them up. The Ergodic Theorem: first form Let (A, µ ¯) be a measure algebra, and set M 1,0 = M 1,0 (A, µ ¯), T (0) = P (0) n 1 i 1,0 Tµ¯,¯µ ⊆ B(M 1,0 ; M 1,0 ) as in 371F-371G. Take any T ∈ T (0) , and set An = n+1 → M 1,0 for i=0 T : M 1,0 every n. Then for any u ∈ M , hAn uin∈N is order*-convergent (definition: 367A) and k k1,∞ -convergent to a member P u of M 1,0 . The operator P : M 1,0 → M 1,0 is a projection onto the linear subspace {u : u ∈ M 1,0 , T u = u}, and P ∈ T (0) . proof (a) It will be convenient to start with some elementary remarks. First, every An belongs to T (0) , by 371Ge and 371Ga. Next, hAn uin∈N is order-bounded in L0 = L0 (A) for any u ∈ M 1,0 ; this is because if

372D

The ergodic theorem

409

u = v + w, where v ∈ L1 = L1 (A, µ ¯) and w ∈ L∞ = L∞ (A), then hAn vin∈N and hAn (−v)in∈N are bounded above, by 372C, while hAn win∈N is norm- and order-bounded in L∞ . Accordingly I can uninhibitedly speak of P ∗ (u) = inf n∈N supi≥n Ai u and P∗ (u) = supn∈N inf i≥n Ai u for any u ∈ M 1,0 , these both being defined in L0 . (b) Write V1 for the set of those u ∈ M 1,0 such that hAn uin∈N is order*-convergent in L0 ; that is, P (u) = P∗ (u) (367Be). It is easy to see that V is a linear subspace of M 1,0 (use 367Ca and 367Cd). Also it is closed for k k1,∞ . Pn 1 i P P We know that |T |, taken in L∼ (M 1,0 ; M 1,0 ), belongs to T (0) (371Gb); set Bn = n+1 i=0 |T | for each i. Suppose that u0 ∈ V 1 . Then for any ² > 0 there is a u ∈ V1 such that ku0 − uk1,∞ ≤ ²2 . Write P u = P ∗ (u) = P∗ (u) for the order*-limit of hAn uin∈N . Express u0 − u as v + w where v ∈ L1 , w ∈ L∞ and kvk1 + kwk∞ ≤ 2²2 . Set v ∗ = supn∈N Bn |v|. Then µ ¯[[v ∗ > ²]] ≤ 2², by 372C. Next, if w∗ = supn∈N Bn |w|, we surely have ∗ 2 w ≤ 2² χ1. Now ∗

|An u0 − An u| = |An v + An w| ≤ Bn |v| + Bn |w| ≤ v ∗ + w∗ for every n ∈ N, that is, An u − v ∗ − w∗ ≤ An u0 ≤ An u + v ∗ + w∗ for every n. Because hAn uin∈N order*-converges to P u, P u − v ∗ − w∗ ≤ P∗ (u0 ) ≤ P ∗ (u0 ) ≤ P u + v ∗ + w∗ , and P ∗ (u0 ) − P∗ (u0 ) ≤ 2(v ∗ + w∗ ). On the other hand, µ ¯[[2(v ∗ + w∗ ) > 2² + 4²2 ]] ≤ µ ¯[[v ∗ > ²]] + µ ¯[[w∗ > 2²2 ]] = µ ¯[[v ∗ > ²]] ≤ 2² (using 364Fa for the first inequality). So µ ¯[[P ∗ (u0 ) − P∗ (u0 ) > 2²(1 + 2²)]] ≤ 2². Since ² is arbitrary, hAn u0 in∈N order*-converges to P ∗ (u0 ) = P∗ (u0 ), and u0 ∈ V1 . As u0 is arbitrary, V1 is closed. Q Q (c) Similarly, the set V2 of those u ∈ M 1,0 for which hAn uin∈N is norm-convergent is a linear subspace of M 1,0 , and it also is closed. P P This is a standard argument. If u0 ∈ V 2 and ² > 0, there is a u ∈ V2 such that ku0 − uk1,∞ ≤ ². There is an n ∈ N such that kAi u − Aj uk1,∞ ≤ ² for all i, j ≥ n, and now kAi u0 − Aj u0 k1,∞ ≤ 3² for all i, j ≥ n, because every Ai has norm at most 1 in B(M 1,0 ; M 1,0 ) (371Gc). As ² is arbitrary, hAi u0 in∈N is Cauchy; because M 1,0 is complete, it is convergent, and u0 ∈ V2 . As u0 is arbitrary, V2 is closed. Q Q (d) Now let V be {u+v−T u : u ∈ M 1,0 ∩L∞ , v ∈ M 1,0 , T v = v}. Then V ⊆ V1 ∩V2 . P P If u ∈ M 1,0 ∩L∞ , then for any n ∈ N An (u − T u) =

1 n+1 (u

− T n+1 u) → 0

for k k∞ , and therefore is both order*-convergent and convergent for k k1,∞ ; so u − T u ∈ V1 ∩ V2 . On the other hand, if T v = v, then of course An v = v for every n, so again v ∈ V1 ∩ V2 . Q Q (e) Consequently L2 = L2 (A, µ ¯ ) ⊆ V1 ∩ V2 . P P L2 ∩ V1 ∩ V2 is a linear subspace; but also it is closed for the 2 norm topology of L , because the identity map from L2 to M 1,0 is continuous (369Oe). We know also that T ¹L2 is an operator of norm at most 1 from L2 to itself (371Gd). Consequently W = {u + v − T u : u, v ∈ L2 , T v = v} is dense in L2 (372A). On the other hand, given u ∈ L2 and ² > 0, there is a u0 ∈ L2 ∩ L∞ such that ku−u0 k2 ≤ ² (take u0 = (u∧γχ1)∨(−γχ1) for any γ large enough), and now k(u−T u)−(u0 −T u0 )k2 ≤ 2². Thus W 0 = {u0 + v − T u0 : u0 ∈ L2 ∩ L∞ , v ∈ L2 , T v = v} is dense in L2 . But W 0 ⊆ V1 ∩ V2 , by (d) above. Thus L2 ∩ V1 ∩ V2 is dense in L2 , and is therefore the whole of L2 . Q Q (f ) L2 ⊇ S(Af ) is dense in M 1,0 , by 369Pc, so V1 = V2 = M 1,0 . This shows that hAn uin∈N is normconvergent and order*-convergent for every u ∈ M 1,0 . By 367D, the limits are the same; by 367F, hAn uin∈N is order*-convergent when regarded as a sequence in M 1,0 . Write P u for the common value of the limits.

410

Linear operators between function spaces

372D

(g) Of course we now have kP uk∞ ≤ supn∈N kAn uk∞ ≤ kuk∞ for every u ∈ L∞ ∩ M 1,0 , while kP uk1 ≤ lim inf kAn uk1 ≤ kuk1 1

for every u ∈ L , by Fatou’s Lemma. So P ∈ T (0) . If u ∈ M 1,0 and T u = u, then surely P u = u, because An u = u for every u. On the other hand, for any u ∈ M 1,0 , T P u = P u. P P Because hAn uin∈N is norm-convergent to P u, kT P u − P uk1,∞ = limn→∞ kT An u − An uk1,∞ = limn→∞

1 kT n+1 u − uk1,∞ n+1

= 0. Q Q

Thus, writing U = {u : T u = u}, P [M 1,0 ] = U and P u = u for every u ∈ U . 372E The Ergodic Theorem: second form Let (A, µ ¯) be a measure algebra, and let T : L1 → L1 , where L1 = L1 (A, µ ¯), be a linear operator ofPnorm at most 1 such that T u ∈ L∞ = L∞ (A) and kT uk∞ ≤ n 1 i 1 1 1 kuk∞ whenever u ∈ L1 ∩ L∞ . Set An = n+1 i=0 T : L → L for every n. Then for any u ∈ L , hAn uin∈N 1 1 1 is order*-convergent to an element P u of L . The operator P : L → L is a projection of norm at most 1 onto the linear subspace {u : u ∈ L1 , T u = u}. proof By 371Ga, there is an extension of T to a member T˜ of T (0) . So 372D tells us that hAn uin∈N is order*-convergent to some P u ∈ L1 for every u ∈ L1 , and P : L1 → L1 is a projection of norm at most 1, because P is the restriction of a projection P˜ ∈ T (0) . Also we still have T P u = P u for every u ∈ L1 , and P u = u whenever T u = u, so the set of values P [L1 ] of P must be exactly {u : u ∈ L1 , T u = u}. Remark In 372D and 372E I have used the phrase ‘order*-convergent’ from §367 without always being specific about the partially ordered set in which it is to be interpreted. But, as remarked in 367F, the notion is robust enough for the omission to be immaterial here. Since both M 1,0 and L1 are solid linear subspaces of L0 , a sequence in M 1,0 is order*-convergent to a member of M 1,0 (when order*-convergence is interpreted in the partially ordered set M 1,0 ) iff it is order*-convergent to the same point (when convergence is interpreted in the set L0 ); and the same applies to L1 in place of M 1,0 . 372F Corollary Let (A, µ ¯) be a measure algebra, and π : Af → Af a measure-preserving ring homof morphism, where A = {a : µ ¯a < ∞}. Let T : M 1,0 → M 1,0 be the corresponding Riesz homomorphism, Pn 1 1,0 1,0 i where M = M (A, µ ¯) (366H, in particular part (a-v)). Set An = n+1 T for n ∈ N. Then for every i=0 u ∈ M 1,0 , hAn uin∈N is order*-convergent and k k1,∞ -convergent to some v such that T v = v. proof By 366H(a-iv), T ∈ T (0) , as defined in 371F. So the result follows at once from 372D. 372G Corollary Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean 1 homomorphism. Let T : L → L1 be the corresponding Riesz homomorphism, where L1 = L1 (A, µ ¯). Set Pn 1 i An = n+1 i=0 T for n ∈ N. Then for every u ∈ L1 , hAn uin∈N is order*-convergent and k k1 -convergent. If we set P u = limn→∞ An u for each u, P is the conditional expectation operator corresponding to the closed subalgebra C = {a : πa = a} of A. proof (a) The first part is just a special case of 372F; the point is that because (A, µ ¯) is totally finite, L∞ (A) ⊆ L1 , so M 1,0 (A, µ ¯) = L1 . Also (because µ ¯1 = 1) kuk∞ ≤ kuk1 for every u ∈ L∞ , so the norm k k1,∞ is actually equal to k k1 . (b) For the last sentence, recall that C is a closed subalgebra of A (cf. 333R). By 372D or 372E, P is a projection operator onto the subspace {u : T u = u}. Now [[T u > α]] = π[[u > α]] (365Oc), so T u = u iff [[u > α]] ∈ C for every α ∈ R, that is, iff u belongs to the canonical image of L1 (C, µ ¯¹ C) in L1 (365R). To 1 identify P u further, observe that if u ∈ L , a ∈ C then

R

Tu =

R

R

Tu = a u R i R R R R R (365Ob). Consequently a T u = a u for every i ∈ N, a An u = a u for every n ∈ N, and a P u = a u (because P u is the limit of hAn uin∈N for k k1 ). But this is enough to define P u as the conditional expectation of u on C (365R). a

πa

372K

The ergodic theorem

411

372H The Ergodic Theorem is most often expressed in terms of transformations of measure spaces. In the next few corollaries I will formulate such expressions. The translation is straightforward, in view of the following. Lemma Let (X, Σ, µ) be a measure space with measure algebra (A, µ ¯). For h ∈ L0 = L0 (µ) write h• for 0 0 the corresponding member of L = L (A) (364Jc). Now let φ : X → X be an inverse-measure-preserving function, π : A → A the corresponding sequentially order-continuous measure-preserving homomorphism defined by setting πE • = (φ−1 [E])• for E ∈ Σ (324M), and T : L0 → L0 the Riesz homomorphism defined by setting T (χa) = χ(πa) for a ∈ A (364R). Then T h• = (hφ)• for any h ∈ L0 . ˜ : X → R be a Σ-measurable function which is equal to h almost everywhere. Because φ−1 [E] proof Let h ˜ a.e., and hφ ˜ is measurable, so hφ ∈ L0 . For any α ∈ R, is negligible for every negligible set E, hφ = hφ ˜ • > α]] = π[[h ˜ • > α]] = π{x : h(x) ˜ [[T h• > α]] = [[T h > α}• ˜ ˜ • > α]] = [[(hφ)• > α]]. = {x : h(φ(x)) > α}• = [[(hφ) 372I Corollary Let (X, Σ, µ) be a measure space and φ : X → X an inverse-measure-preserving function. Let f be a real-valued function which is integrable over X. Then 1 Pn i g(x) = limn→∞ i=0 f (φ (x)) n+1

is defined for almost every x ∈ X, and gφ(x) = g(x) for almost every x. proof Let (A, µ ¯) be the measure algebra of (X, Σ, µ), and π : A → A, T : L0 (A) → L0 (A) the homomorphisms correspondingPto φ, as in 372H. Set u = f • in L1 (A, µ ¯P ). Then for any i ∈ N, T i u = (f φi )• (372H), n n 1 1 i • so setting An = n+1 i=0 T , An u = gn , where gn (x) = n+1 i=0 f (φi (x)) whenever this is defined. Now we know from 372E or 372F that hAn uin∈N is order*-convergent to some v such that T v = v, so hgn in∈N must be convergent almost everywhere (367G), and taking g = limn→∞ gn where this is defined, g • = v. Accordingly (gφ)• = T v = v = g • and gφ = g a.e., as claimed. 372J

The following straightforward facts will be useful in the next corollary and elsewhere.

Lemma Let (X, Σ, µ) be a measure space with measure algebra (A, µ ¯). Let φ : X → X be an inversemeasure-preserving function and π : A → A the associated homomorphism, as in 372H. Set C = {c : c ∈ A, πc = c}, T = {E : E ∈ Σ, φ−1 [E]4E is negligible} and T0 = {E : E ∈ Σ, φ−1 [E] = E}. Then T and T0 are σ-subalgebras of Σ; T0 ⊆ T, T = {E : E ∈ Σ, E • ∈ C}, and C = {E • : E ∈ T0 }. proof It is easy to see that T and T0 are σ-subalgebras of Σ and that T0 ⊆ T = {E : E • ∈ C}. So we have only to check that if c ∈ C there is an E ∈ T0 such that E • = c. P P Start with any F ∈ Σ such that F • = c. −i −i • Now F 4φ [F ] is negligible for every i ∈ N, because (φ [F ]) = π i c = c. So if we set S T E = n∈N i≥n φ−i [F ] = {x : there is an n ∈ N such that φi (x) ∈ F for every i ≥ n}, E • = c. On the other hand, it is easy to check that E ∈ T0 . Q Q 372K Corollary Let (X, Σ, µ) be a probability space and φ : X → X an inverse-measure-preserving function. Let f be a real-valued function which is integrable over X. Then 1 Pn i g(x) = limn→∞ i=0 f (φ (x)) n+1

is defined for almost every x ∈ X; gφ = g a.e., and g is a conditional expectation of f on the σ-algebra T = {E : E ∈ Σ, φ−1 [E]4E is negligible}. If either f is Σ-measurable and defined everywhere in X or φ[E] is negligible for every negligible set E, then g is a conditional expectation of f on the σ-algebra T0 = {E : E ∈ Σ, φ−1 [E] = E}. proof (a) We know by 372I that g is defined almost everywhere and that gφ = g a.e. In the language of the proof of 372I, g • = v is the conditional expectation of u = f • on the closed subalgebra C = {a : a ∈ A, πa = a} = {F • : F ∈ T} = {F • : F ∈ T0 },

412

Linear operators between function spaces

372K

by 372G and 372J. So v must be expressible as h• where h : X → R is T0 -measurable and is a conditional expectation of f on T0 (and also on T). Since every set of measure zero belongs to T, g = h µ¹ T-a.e., and g is also a conditional expectation of f on T. (b) Suppose now that f is defined everywhere and Σ-measurable. Here I come to a technical obstruction. The definition of ‘conditional expectation’ in 233D asks for g to be µ¹ T0 -integrable, and since µ-negligible sets do not need to be µ¹ T0 -negligible we have some more checking to do, to confirm that {x : x ∈ dom g, g(x) = h(x)} is µ¹ T0 -conegligible as well as µ-conegligible. (i)T For n ∈ N, set Σn = {φ−n [E] : E ∈ Σ}; then Σn is a σ-subalgebra of Σ, including T0 . Set Σ∞ = n∈N Σn , still a σ-algebra including T0 . Now any negligible set E ∈ Σ∞ is µ¹ T0 -negligible. P P For each n ∈ N choose Fn ∈ Σ such that E = φ−n [Fn ]. Because φ is inverse-measure-preserving, every Fn is negligible, so that T S E ∗ = m∈N n∈N,j≥m φ−j [Fn ] T is negligible. Of course E = m∈N φ−m [Fm ] is included in E ∗ . Now T S T S φ−1 [E ∗ ] = m∈N n∈N,j≥m φ−j−1 [Fn ] = m≥1 n∈N,j≥m φ−j [Fn ] = E ∗ because

S n∈N,j≥1

φ−j [Fn ] ⊆

S n∈N,j≥0

φ−j [Fn ].

So E ∗ ∈ T0 and E is included in a negligible member of T0 , which is what we needed to know. Q Q P n 1 ◦ i (ii) We are assuming that f is Σ-measurable and defined everywhere, so that gn = n+1 i=0 f φ ∗ ∗ is Σ-measurable and defined everywhere. If we set g = lim supn→∞ gn , then g : X → [−∞, ∞] is Σ∞ measurable. P P For any m ∈ N, f ◦ φi is Σm -measurable for every i ≥ m, since {x : f (φi (x)) > α} = φ−m [{x : i−m f (φ (x)) > α}] for every α. Accordingly 1 Pn ◦ i g ∗ = lim supn→∞ i=m f φ n+1

∗

is Σm -measurable. As m is arbitrary, g is Σ∞ -measurable. Q Q Since h is surely Σ∞ -measurable, and h = g ∗ µ-a.e., (i) tells us that h = g ∗ µ¹ T0 -a.e. But similarly h = lim inf n→∞ gn µ¹ T0 -a.e., so we must have h = g µ¹ T0 -a.e.; and g, like h, is a conditional expectation of f on T0 . (c) Finally, suppose that φ[E] is negligible for every negligible set E. Then every µ-negligible S set isnµ¹ T0 ∗ negligible. P P If E is µ-negligible, then φ[E], φ2 [E] = φ[φ[E]], . . . are allSnegligible, so E = n∈N φ [E] is T negligible, and there is a measurable negligible set F ⊇ E ∗ . Now F∗ = m∈N n≥m φ−n [F ] is a negligible set in T0 including E, so E is µ¹ T0 -negligible. Q Q Consequently g = h µ¹ T0 -a.e., and in this case also g is a conditional expectation of f on T0 . 372L Remark Parts (b)-(c) of the proof above are dominated by the technical question of the exact definition of ‘conditional expectation of f on T0 ’, and it is natural to be impatient with such details. The kind of example I am concerned about is the following. Let C ⊆ [0, 1] be the Cantor set (134G), and φ : [0, 1] → [0, 1] a Borel measurable function such that φ[C] = [0, 1] and φ(x) = x for x ∈ [0, 1] \ C. (For instance, we could take φ agreeing with the Cantor function on C (134H).) Because C is negligible, φ is inverse-measure-preserving for Lebesgue measure µ, and if f is any Lebesgue integrable function then 1 Pn i g(x) = limn→∞ i=0 f (φ (x)) is defined and equal to f (x) for every x ∈ dom f \ C. But for x ∈ C we n+1

can, by manipulating φ, arrange for g(x) to be almost anything; and if f is undefined on C then g will also be undefined on C. On the other hand, C is not µ¹ T0 -negligible, because the only member of T0 including R C is [0, 1]. So we cannot be sure of being able to form g d(µ¹ T0 ). If instead of Lebesgue measure itself we took its restriction µB to the algebra of Borel subsets of [0, 1], then φ would still be inverse-measure-preserving for µB , but we should now have to worry about the possibility that f ¹C was non-measurable, so that g¹C came out to be non-measurable, even if everywhere defined, and g was not µB ¹ T0 -virtually measurable.

372Mc

The ergodic theorem

413

In the statement of 372K I have offered two ways of being sure that the problem does not arise: check that φ[E] is negligible whenever E is negligible (so that all negligible sets are µ¹ T0 -negligible), or check that f is defined everywhere and Σ-measurable. Even if these conditions are not immediately satisfied in a particular application, it may be possible to modify the problem so that they are. For instance, completing the measure will leave φ inverse-measure-preserving (343Ac), will not change the integrable functions but will make them all measurable (212F, 212Bc), and may enlarge T0 enough to make a difference. If our function f is measurable (because the measure is complete, or otherwise) we can extend it to a measurable function defined everywhere (121I) and the corresponding extension of g will be µ¹ T0 -integrable. Alternatively, if the difficulty seems to lie in the behaviour of φ rather than in the behaviour of f (as in the example above), it may help to modify φ on a negligible set. 372M Continued fractions A particularly delightful application of the results above is to a question which belongs as much to number theory as to analysis. It takes a bit of space to describe, but I hope you will agree with me that it is well worth knowing in itself, and that it also illuminates some of the ideas above. (a) Set X = [0, 1] \ Q. For x ∈ X, set φ(x) = < x1 >, the fractional part of x1 , and k1 (x) = x1 − φ(x), the integer part of x1 ; then φ(x) ∈ X for each x ∈ X, so we may define kn (x) = k1 (φn−1 (x)) for every n ≥ 1. The strictly positive integers k1 (x), k2 (x), k3 (x), . . . are the continued fraction coefficients of x. Of course kn+1 (x) = kn (φ(x)) for every n ≥ 1. Now define hpn (x)in∈N , hqn (x)in∈N inductively by setting p0 (x) = 0, q0 (x) = 1,

p1 (x) = 1,

pn (x) = pn−2 (x) + kn (x)pn−1 (x) for n ≥ 1,

q1 (x) = k1 (x),

qn (x) = qn−2 (x) + kn (x)qn−1 (x) for n ≥ 1.

The continued fraction approximations to x are the quotients pn (x)/qn (x). (I do not discuss rational x, because for my purposes here these are merely distracting. But if we set k1 (0) = ∞, φ(0) = 0 then the formulae above produce the conventional values for kn (x) for rational x ∈ [0, 1[. As for the pn and qn , use the formulae above until you get to x = pn (x)/qn (x), φn (x) = 0, kn+1 (x) = ∞, and then set pm (x) = pn (x), qm (x) = qn (x) for m ≥ n.) (b) The point is that the quotients rn (x) = pn (x)/qn (x) are, relatively speaking, good rational approximations to x. (See 372Yf.) We always have rn+1 (x) < x < rn (x) for every odd n ≥ 1 (372Xj). If x = π − 3, then the first few coefficients are k1 = 7, 1 7

r1 = ,

k2 = 15, r2 =

15 , 106

k3 = 1, r3 =

16 ; 113

the first and third of these corresponding to the classical approximations π l x = e − 2, we get k1 = 1, r1 = 1,

k2 = 2, 2 3

r2 = ,

k3 = 1, 3 4

r3 = ,

k4 = 1, 5 7

r4 = ,

k5 = 4, r5 =

23 , 32

22 , 7

πl

355 . 113

Or if we take

k6 = 1, r6 =

28 ; 39

17 86 note that the obvious approximations 24 , 120 derived from the series for e are not in fact as close as the 5 28 even terms 7 , 39 above, and involve larger numbers.

(c) Now we need a variety of miscellaneous facts about these coefficients, which I list here. (i) For any x ∈ X, n ≥ 1 we have pn−1 (x)qn (x) − pn (x)qn−1 (x) = (−1)n ,

φn (x) =

(induce on n), so x=

pn (x)+pn−1 (x)φn (x) . qn (x)+qn−1 (x)φn (x)

pn (x)−xqn (x) xqn−1 (x)−pn−1 (x)

414

Linear operators between function spaces

372Mc

(ii) Another easy induction on n shows that for any finite string m = (m1 , . . . , mn ) of strictly positive integers the set Dm = {x : x ∈ X, ki (x) = mi for 1 ≤ i ≤ n} is an interval in X on which φn is monotonic, being strictly increasing if n is even and strictly decreasing if n is odd. (For the inductive step, note just that D(m1 ,... ,mn ) = [ m11+1 , m11 ] ∩ φ−1 [D(m2 ,... ,mn ) ].) (iii) We also need to know that the intervals Dm of (ii) are small; specifically, that if m = (m1 , . . . , mn ), the length of Dm is at most 2−n+1 . P P All the coefficients pi , qi , for i ≤ n, take constant values p∗i , qi∗ on Dm , since they are determined from the coefficients ki which are constant on Dm by definition. Now every ∗ x ∈ Dm is of the form (p∗n + tp∗n−1 )/(qn∗ + tqn−1 ) for some t ∈ X (see (i) above) and therefore lies between ∗ ∗ ∗ ∗ pn−1 /qn−1 and pn /qn . But the distance between these is ∗ ∗ ¯ ¯ p∗n qn−1 −p∗n−1 qn ¯ ¯= 1 , ∗ ∗ ∗ ∗ qn qn−1

qi∗

qn qn−1

∗ qi−1

∗ + qi−2

∗ ∗ ∗ by the first formula in (i). Next, noting that ≥ for each i ≥ 2, we see that qi∗ qi−1 ≥ 2qi−1 qi−2 ∗ for i ≥ 2, and therefore that qn∗ qn−1 ≥ 2n−1 , so that the length of Dm is at most 2−n+1 . Q Q

372N Theorem Set X = [0, 1] \ Q, and define φ : X → X as in 372M. Then for every Lebesgue integrable function f on X, limn→∞

1 n+1

Pn

f (φi (x)) = i=0

1 ln 2

R 1 f (t) 0 1+t

dt

for almost every x ∈ X. proof (a) The integral just written, and the phrase ‘almost every’, refer of course to Lebesgue measure; but the first step is to introduce another measure, so I had better give a name µL to measure on R Lebesgue 1 X. Let ν be the indefinite-integral measure on X defined by saying that νE = ln12 E 1+x µL (dx) whenever 1 this is defined. The coefficient ln12 is of course chosen to make νX = 1. Because 1+x > 0 for every x ∈ X, dom ν = dom µL and ν has just the same negligible sets as µL (234D); I can therefore safely use the terms ‘measurable set’, ‘almost everywhere’ and ‘negligible’ without declaring which measure I have in mind each time. (b) Now φ is inverse-measure-preserving when regarded as a function from (X, ν) to itself. P P For each h h 1 1 1 k ≥ 1, set Ik = k+1 , k . On X ∩ Ik , φ(x) = x − k. Observe that φ¹Ik : X ∩ Ik → X is bijective and differentiable relative to its domain in the sense of §262. Consider, for any measurable E ⊆ X, Z Z 1 1 µL (dy) = |φ0 (x)|µL (dx) (y+k)(y+k+1) (φ(x)+k)(φ(x)+k+1) E Ik ∩φ−1 [E] Z x2 1 = µL (dx) = ln 2 · ν(Ik ∩ φ−1 [E]), 2 Ik ∩φ−1 [E]

x+1 x

using 263D (or more primitive results, of course). But P∞ P∞ 1 1 = k=1 − k=1 (y+k)(y+k+1)

y+k

1 y+k+1

=

1 y+1

for every y ∈ [0, 1], so νE =

1 ln 2

∞ Z X

1 µL (dy) (y+k)(y+k+1) E k=1

=

∞ X

ν(Ik ∩ φ−1 [E]) = νφ−1 [E].

k=1

As E is arbitrary, ν is inverse-measure-preserving. Q Q (c) The next thing we need to know is that if E ⊆ X and φ−1 [E] = E then E is either negligible or conegligible. P P I use the sets Dm of 372M(c-ii). (i) For any string m = (m1 , . . . , mn ) of strictly positive integers, we have

372N

The ergodic theorem

x=

415

p∗n +p∗n−1 φn (x) n ∗ +q ∗ qn n−1 φ (x)

for every x ∈ Dm , where p∗n , etc., are defined from m as in 372M(c-iii). Recall also that φn is strictly monotonic on Dm . So for any interval I ⊆ [0, 1] (open, closed or half-open) with endpoints α < β, ∗ φ−n [I] ∩ Dm will be of the form X ∩ J, where J is an interval with endpoints (p∗n + p∗n−1 α)/(qn∗ + qn−1 α), ∗ (p∗n +p∗n−1 β)/(qn∗ +qn−1 β) in some order. This means that we can estimate µL (φ−n [I]∩Dm )/µL Dm , because it is the modulus of ∗ pn +p∗ n−1 β ∗ +q ∗ qn n−1 β ∗ p∗ n +pn−1 ∗ +q ∗ qn n−1

∗ p∗ n +pn−1 α ∗ +q ∗ qn n−1 α

p∗ n ∗ qn

−

−

=

∗ ∗ ∗ (β−α)qn (qn +qn−1 ) ∗ ∗ ∗ β) (q +qn−1 α)(qn +qn−1 ∗ n

≥

∗ (β−α)qn ∗ ∗ qn +qn−1

1 2

≥ (β − α).

Now look at 1 2

A = {E : E ⊆ [0, 1] is Lebesgue measurable, µL (φ−n [E] ∩ Dm ) ≥ µL E · µL Dm }. Clearly the union of two disjoint members of A belongs to A. Because A contains every subinterval of [0, 1] it includes the algebra E of subsets of [0, 1] consisting of finite unions of intervals. Next, the union of any non-decreasing sequence in A belongs to A, and the intersection of a non-increasing sequence likewise. But this means that A must include the σ-algebra generated by E (136G), that is, the Borel σ-algebra. But also, if E ∈ A and H ⊆ [0, 1] is negligible, then 1 2

1 2

µL (φ−n [E4H] ∩ Dm ) = µL (φ−n [E] ∩ Dm ) ≥ µL E · µL Dm = µL (E4H) · µL Dm and E4H ∈ A. And this means that every Lebesgue measurable subset of [0, 1] belongs to A (134Fb). (ii) ?? Now suppose, if possible, that E is a measurable subset of X and that φ−1 [E] = E and E is neither negligible nor conegligible in X. Set γ = 21 µL E > 0. By Lebesgue’s density theorem (223B) there is some x ∈ X \ E such that limδ↓0 ψ(δ) = 0, where ψ(δ) = 1 2γ

1 µL (E 2δ

∩ [x − δ, x + δ]) for δ > 0. Take n so

−n+1

whenever 0 < δ ≤ 2 , and set mi = ki (x) for i ≤ n, so that x ∈ Dm . Taking the large that ψ(δ) < least δ such that Dm ⊆ [x − δ, x + δ], we must have δ ≤ 2−n+1 , because the length of Dm is at most 2−n+1 (372M(c-iii)), while µL Dm ≥ δ, because Dm is an interval. Accordingly µL (E ∩ Dm ) ≤ µL (E ∩ [x − δ, x + δ]) = 2δψ(δ) < γδ ≤ γµL Dm . But we also have µL (E ∩ Dm ) = µL (φ−n [E] ∩ Dm ) ≥ γµL Dm , by (i). X X This proves the result. Q Q (d) The final fact we need in preparation is that φ[E] is negligible for every negligible E ⊆ X. This is because φ is differentiable relative to its domain (see 263D(ii)). (e) Now let f be any µL -integrable function. Because consequently, using (b) above and 372K, g(x) = limn→∞

1 n+1

1 1+x

≤ 1 for every x, f is also ν-integrable (235M);

Pn i=0

f (φi (x))

is defined for almost every x ∈ X, and is a conditional expectation of f (with respect to the measure ν) on the σ-algebra T0 = {E : E is measurable, φ−1 [E] = E}. But we haveRjust seenR that T0 consists only of negligible and conegligible sets, so g must be essentially constant; since g dν = f dν, we must have 1 n→∞ n+1

lim

n X

Z f (φi (x)) =

i=0

for almost every x (using 235M to calculate

R

f dν).

f dν =

1 ln 2

Z 0

1

f (t) µL (dt) 1+t

416

Linear operators between function spaces

372O

372O Corollary For almost every x ∈ [0, 1] \ Q, 1 n

limn→∞ #({i : 1 ≤ i ≤ n, ki (x) = k}) =

1 (2 ln(k ln 2

+ 1) − ln k − ln(k + 2))

for every k ≥ 1, where k1 (x), . . . are the continued fraction coefficients of x. 1 proof In 372N, set f = χ(X ∩ [ k+1 , k1 ]). Then (for i ≥ 1) f (φi (x)) = 1 if ki (x) = k and zero otherwise. So

lim

1

n→∞ n

#({i : 1 ≤ i ≤ n, ki (x) = k}) 1 n→∞ n

= lim = =

1 ln 2

Z

n X

1 0

1 n→∞ n+1

f (φi (x)) = lim

i=1

f (t) dt 1+t

=

1 ln 2

Z

1/k 1/k+1

n X

f (φi (x))

i=0

1 dt 1+t

1 1 1 (ln(1 + ) − ln(1 + )) ln 2 k k+1

=

1 (2 ln(k ln 2

+ 1) − ln k − ln(k + 2)),

for almost every x ∈ X. 372P Mixing and ergodic transformations This seems an appropriate moment for some brief notes on two special types of measure-preserving homomorphism or inverse-measure-preserving function. Definitions (a) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. (i) π is ergodic if {a : πa = a} = {0, 1}, that is, if the fixed-point subalgebra of π is trivial. (ii) π is mixing (sometimes strongly mixing) if limn→∞ µ ¯(π n a ∩ b) = µ ¯a · µ ¯b for all a, b ∈ A. (b) Let (X, Σ, µ) be a probability space and φ : X → X an inverse-measure-preserving function. (i) φ is ergodic (also called metrically transitive, indecomposable) if every measurable set E such that φ−1 [E] = E is either negligible or conegligible. (ii) φ is mixing if limn→∞ µ(F ∩ φ−n [E]) = µE · µF for all E, F ∈ Σ. (c) Remarks (i) The reason for introducing ‘ergodic’ homomorphisms in this section is of course 372G/372K; if π in 372G, or φ in 372K, is ergodic, then the limit P u or g must be (essentially) constant, being a conditional expectation on a trivial subalgebra. (ii) In the definition (b-i) I should perhaps emphasize that we look only at measurable sets E. We certainly expect that there will generally be many sets E for which φ−1 [E] = E, since any union of orbits of φ will have this property. (iii) Part (c) of the proof of 372N was devoted to showing that the function φ there was ergodic; see also 372Xw. For another ergodic transformation see 372Xo. For examples of mixing transformations see 333P, 372Xm, 372Xn, 372Xr, 372Xu, 372Xv. 372Q

The following facts are elementary.

Proposition (a) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. (i) If π is mixing, it is ergodic. (ii) Let T : L0 = L0 (A) → L0 be the Riesz homomorphism such that T (χa) = χπa for every a ∈ A. Then the following are equiveridical: (α) π is ergodic; (β) the only u ∈ L0 such that R T u = u are the multiples Pn 1 i 1 1 of χ1; (γ) for every u ∈ L = L (A, µ ¯), h n+1 i=0 T uin∈N order*-converges to ( u)χ1. (b) Let (X, Σ, µ) be a probability space, with measure algebra (A, µ ¯). Let φ : X → X be an inversemeasure-preserving function and π : A → A the associated homomorphism such that πE • = (φ−1 [E])• for every E ∈ Σ.

372R

The ergodic theorem

417

(i) φ is mixing iff π is, and in this case it is ergodic. (ii) The following are (α) φ is ergodic; R (β) π is ergodic; (γ) for every µ-integrable realPnequiveridical: 1 i valued function f , h n+1 f (φ (x))i converges to f for almost every x ∈ X. n∈N i=0 proof (a)(i) If π is mixing and πa = a, then 0=µ ¯(a \ a) = limn→∞ µ ¯(π n a \ a) = µ ¯a · µ ¯(1 \ a), so one of µ ¯a, µ ¯(1 \ a) must be zero, and a ∈ {0, 1}. Thus π is ergodic. α)⇒(β β ) T u = u iff π[[u > α]] = [[u > α]] for every α; if π is ergodic, this means that [[u > α]] ∈ {0, 1} (ii)(α for every α, and u must be of the form γχ1, where γ = inf{α : [[u > α]] = 0}. Pn 1 i β )⇒(γγ ) If (β) is true and u ∈ L1 , then we know from 372G that h n+1 (β i=0 T uin∈N is order*convergent and k k1 -convergent to some v such that T v = v; by (β), v is of the form γχ1; and γ=

R

v = limn→∞

1 n+1

Pn R i=0

T iu =

R

u.

α) Assuming (γ), take any a ∈ A such that πa = a, and consider u = χa. Then T i u = χa for (γγ )⇒(α every i, so χa = limn→∞

1 n+1

Pn i=0

R

T i u = ( u)χ1 = µ ¯a · χ1,

and a must be either 0 or 1. (b)(i) Simply translating the definitions, we see that π is mixing iff φ is. In this case φ is ergodic, as in (a-i). α)⇒(β β ) If πa = a there is an E such that φ−1 [E] = E and E • = a, by 372L; now µ (ii)(α ¯a = µE ∈ {0, 1}, so a ∈ {0, 1}. β )⇒(γγ ) Set u = f • ∈ L1 . In the language of (a), T i u = (f φi )• for each i, by 372H, so that (β 1 Pn 1 Pn i • i ( i=0 f φ ) = n+1 i=0 T u n+1 R R R Pn 1 i is order*-convergent to ( u)χ1 = ( f )χ1, and n+1 f a.e. i=0 f φ → α) If φ−1 [E] = E then, applying (γ) to f = χE, we see that χE = µE · χX a.e., so that E is (γγ )⇒(α either negligible or conegligible. 372R

There is a useful sufficient condition for a homomorphism or function to be mixing.

Proposition T (a) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. If n∈N π n [A] = {0, 1}, then π is mixing. (b) Let (X, Σ, µ) be a probability space, and φ : X → X an inverse-measure-preserving function. Set T = {E : for every n ∈ N there is an F ∈ Σ such that E = φ−n [F ]}. If every element of T is either negligible or conegligible, φ is mixing. proof (a) Let T : L0 = L0 (A) → L0 be the Riesz homomorphism associated with π. Take any a, b ∈ A and any non-principal ultrafilter F on N. Then hT n (χa)in∈N is a bounded sequence in the reflexive space L2µ¯ = L2 (A, µ ¯), so v = limn→F T n (χa) is defined for the weak topology of L2µ¯ . Now for each n ∈ N set n Bn = π [A]. This is a closed subalgebra of A (314F(a-i)), and contains π i a for every i ≥ n. So if we identify L2 (Bn , µ ¯¹ Bn ) with the corresponding subspace of L2µ¯ (366I), it contains T i (χa) for every i ≥ n; but also it is norm-closed, therefore weakly closed (3A5Ee), so contains v. This means that [[v > α]] must belong to T Bn for every α and every n. But in this case [[v > α]] ∈ n∈N Bn = {0, 1} for every α, and v is of the form γχ1. Also γ=

R

v = limn→F

So limn→F µ ¯(π n a ∩ b) = limn→F

R

R

T n (χa) = µ ¯a.

T n (χa) × χb =

R

v × χb = γ µ ¯b = µ ¯a · µ ¯b.

418

Linear operators between function spaces

372R

But this is true of every non-principal ultrafilter F on N, so we must have limn→∞ µ ¯(π n a ∩ b) = µ ¯a · µ ¯b (3A3Lc). As a and b are arbitrary, π is mixing. T (b) The point is that if a ∈ n∈N π n [A], there is an E ∈ T such that E • = a. P P For each n ∈ N there is n • −n • an an ∈ A such that π an = a; say an = Fn where Fn ∈ Σ. Then φ [Fn ] = a. Set T T S S E = m∈N n≥m φ−n [Fn ], Ek = m≥k n≥m φ−(n−k) [Fn ] for each k; then E • = a and φ−k [Ek ] =

S

T m≥k

n≥m

φ−n [Fn ] =

T

S m∈N

n≥m

φ−n [Fn ] = E

for every Q T k, so E ∈ T. Q So n∈N An = {0, 1} and π and φ are mixing. 372X Basic exercisesP(a) Let U be any reflexive Banach space, and T : U → U an operator of norm n 1 i at most 1. Set An = n+1 i=0 T for each n ∈ N. Show that P u = limn→∞ An u is defined (as a limit for the norm topology) for every u ∈ U , and that P : U → U is a projection onto {u : T u = u}. (Hint: show that {u : P u is defined} is a closed linear subspace of U containing T u − u for every u ∈ U .) (This is a version of the mean ergodic theorem.) Pn (0) 1 i >(b) Let (A, µ ¯) be a measure algebra, and T ∈ Tµ¯,¯µ ; set An = n+1 i=0 T for n ∈ N. Take any p ∈ [1, ∞[ p p and u ∈ L = L (A, µ ¯). Show that hAn uin∈N is order*-convergent and k kp -convergent to some v ∈ Lp . (Hint: put 372Xa together with 372D.) (c) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. R R Let P : L1 → L1 be the operator defined as in 365P/366Hb, where L1 = L1 (A, µ ¯), so that a P u = πa u Pn 1 i 1 1 1 for u ∈ L1 , a ∈ A. Set An = n+1 i=0 P : L → L for each i. Show that for any u ∈ L , hAn uin∈N is order*-convergent and k k1 -convergent to the conditional expectation of u on the subalgebra {a : πa = a}. (d) Show that if f is any Lebesgue integrable function on R, and y ∈ R \ {0}, then Pn 1 limn→∞ n+1 k=0 f (x + ky) = 0 for almost every x ∈ R. (e) Let (X, Σ, µ) be a measure space and φ : X → X an inverse-measure-preserving function. Set T = {E : E ∈ Σ, µ(φ−1 [E]4E) = 0}, T0 = {E : E ∈ Σ, φ−1 [E] = E}. (i) Show that T = {E4F : E ∈ T0 , F ∈ Σ, µF = 0}. (ii) Show that a set A ⊆ X is µ¹ T0 -negligible iff φn [A] is µ-negligible for every n ∈ N. R > (f ) Let ν be a Radon probability measure on R such that |t|ν(dt) is finite (cf. 271C-271F). On X = RN let λ be the product measure obtained when each factor is given the measure ν. Define φ : X → X by setting φ(x)(n) = x(n + 1) for x ∈ X, R n ∈ N. (i) Show that φ is inverse-measure-preserving. (Hint: 254G. PnSee also 1 ◦ i 372Xu below.) (iii) Set γ = tν(dt), the expectation of the distribution ν. By considering n+1 i=0 f φ , Pn 1 where f (x) = x(0) for x ∈ X, show that limn→∞ n+1 i=0 x(i) = γ for λ-almost every x ∈ X. > (g) Use the Ergodic Theorem to prove Kolmogorov’s Strong Law of Large Numbers (273I), as follows. Given a complete probability space (Ω, Σ, µ) and an independent identically distributed sequence hFn in∈N of measurable functions from Ω to R, set X = RN and F (ω) = hFn (ω)in∈N for ω ∈ Ω. Show that if we give each copy of R the distribution of F0 then F is inverse-measure-preserving for µ and the product measure λ on X. Now use 372Xf. 1 (h) Show that the continued fraction coefficients of √ are 1, 2, 1, 2, . . . . 2

>(i) For x ∈ X = [0, 1] \ Q let k1 (x), k2 (x), . . . be its continued-fraction coefficients. Show that x 7→ hkn+1 (x) − 1in∈N is a bijection between X and NN which is a homeomorphism if X is given its usual topology (as a subset of R) and NN is given its usual product topology (each copy of N being given the discrete topology).

372Xt

The ergodic theorem

419

(j) For any irrational x ∈ [0, 1] let k1 (x), k2 (x), . . . be its continued-fraction coefficients and pn (x), qn (x) the numerators and denominators of its continued-fraction approximations, as described in 372M. Write rn (x) = pn (x)/qn (x). (i) Show that x lies between rn (x) and rn+1 (x) for every n ∈ N. (ii) Show that rn+1 (x) − rn (x) = (−1)n /qn (x)qn+1 (x) for every n ∈ N. (iii) Show that |x − rn (x)| ≤ 1/qn (x)2 kn (x) for every n ≥ 1. (iv) Hence show that for almost every γ ∈ R, the set {(p, q) : p ∈ Z, q ≥ 1, |γ − pq | ≤ ²/q 2 } is infinite for every ² > 0. (k) Let (A, µ ¯) be an atomless probability algebra. Show that the following are equiveridical: (i) A is homogeneous; (ii) there is an ergodic measure-preserving Boolean homomorphism π : A → A; (iii) there is a mixing measure-preserving automorphism π : A → A. (Hint: 333P.) (l) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. (i) Show that if n ≥ 1 then π is mixing iff π n is mixing. (ii) Show that if n ≥ 1 and π n is ergodic then π is ergodic. (iii) Show that if π is an automorphism then it is ergodic, or mixing, iff π −1 is. > (m) Consider the tent map φα (x) = α min(x, 1 − x) for x ∈ [0, 1], α ∈ [0, 2]. Show that φ2 is inversemeasure-preserving and mixing for Lebesgue measure on [0, 1]. (Hint: show that φn+1 (x) = φ2 () for 2 −n −n n ≥ 1, and hence that µ(I ∩ φ−n [J]) = µI · µJ whenever I is of the form [2 k, 2 (k + 1)] and J is an 2 interval.) (n) Consider the quadratic map ψβ (x) = βx(1 − x) for x ∈ [0, 1], β ∈ [0, 4]. Show that ψ4 p is inversemeasure-preserving and mixing for the Radon measure on [0, 1] with density function t 7→ 1/π t(1 − t). (Hint: show that the transformation t 7→ sin2 πt 2 turns it into the tent map.) Show that for almost every x, limn→∞

1 #({i n+1

: i ≤ n, ψ4i (x) ≤ α}) =

2 π

√ arcsin α

for every α ∈ [0, 1]. (o) Let (X, Σ, µ) be Lebesgue measure on [0, 1[, and fix an irrational number α ∈ [0, 1[. (i) Set φ(x) = x +1 α for every x ∈ [0, 1[, where x +1 α is whichever of x + α, x + α − 1 belongs to [0, Pn1[. Showi that φ is 1 inverse-measure-preserving. (ii) Show that if I ⊆ [0, 1[ is an interval then limn→∞ n+1 i=0 χI(φ (x)) = µI for almost every x ∈ [0, 1[. (Hint: this is Weyl’s Equidistribution Theorem (281N).) (iii) Show that φ is ergodic. (Hint: take the conditional expectation operator P of 372G, and look at P (χI • ) for intervals I.) (iv) Show that φn is ergodic for any n ∈ Z \ {0}. (v) Show that φ is not mixing. (p) Let (A, µ ¯) be a probability algebra and π : A → A a mixing measure-preserving homomorphism. Let T : L0 (A) → L0 (A) be the corresponding homomorphism. Let p, q ∈ [1, ∞] be such that p1 + 1q = 1. Show R R R that limn→∞ T n u × v = u v whenever u ∈ Lp (A, µ ¯) and v ∈ Lq (A, µ ¯). (Hint: start with u, v ∈ S(A).) (q) Let (X, Σ, µ) be a probability space and φ : X → XR a mixing inverse-measure-preserving function. R R Let p, q ∈ [1, ∞] be such that p1 + 1q = 1. Show that limn→∞ f (φn (x))g(x)dx = f g whenever f ∈ Lp (µ) and g ∈ Lq (µ). (r) Give [0, 1[ Lebesgue measure µ, and let k ≥ 2 be an integer. Define φ : [0, 1[ → [0, 1[ by setting φ(x) = , the fractional part of kx. Show that φ is inverse-measure-preserving. Show that φ is mixing. (Hint: if I = [k −n i, k −n (i + 1)[, J = [k −n j, k −n (j + 1)[ then µ(I ∩ φ−m [J]) = µI · µJ for all m ≥ n.) (s) Let (X, Σ, µ) be a probability space and φ : X → X an ergodic inverse-measure-preserving function. R Let f be a µ-virtually measurable function defined almost everywhere on X such that f dµ = ∞. Show 1 Pn i that limn→∞ i=0 f (φ (x)) = ∞ for almost every x ∈ X. (Hint: look at the corresponding limits for n+1

fm = f ∧ mχX.) (t) For irrational x ∈ [0, 1], write k1 (x), k2 (x), . . . for the continued-fraction coefficients of R x. Show that Pn limn→∞ n1 i=1 ki (x) = ∞ for almost every x. (Hint: take φ, ν as in 372N, and show that k1 dν = ∞.)

420

Linear operators between function spaces

372Xu

(u) Let (X, Σ, µ) be any probability space, and let λ be the product measure on X N . Define φ : X N → X N by setting φ(x)(n) = x(n+1). Show that φ is inverse-measure-preserving. Show that φ satisfies the conditions of 372R, so is mixing. (v) Let (X, Σ, µ) be any probability space, and let λ be the product measure on X Z . Define φ : X Z → X Z by setting φ(x)(n) = x(n + 1). Show that φ is inverse-measure-preserving. Show that φ is mixing. (Hint: show that if C, C 0 are basic cylinder sets then µ(C ∩ φ−n [C 0 ]) = µC · µC 0 for all n large enough.) Show that φ does not satisfy the conditions of 372R. (Compare 333P.) (w) In 372N, let T1 be the family {E : for every n ∈ N there is a measurable set F ⊆ X such that φ−n [F ] = E}. Show that every member of T1 is either negligible or conegligible. (Hint: the argument of part (c) of the proof of 372N still works.) Hence show that φ is mixing for the measure ν. 372Y Further exercises (a) In 372D, show that the null space of the limit operator P is precisely the closure in M 1,0 of the subspace {T u − u : u ∈ M 1,0 }. (0)

(b) Let (A, µ ¯) be a measure algebra and T ∈ Tµ¯,¯µ . Take p ∈ ]1, ∞[ and u ∈ Lp (A, µ ¯), and set u∗ = Pn 1 i supn∈N n+1 i=0 |T u|. (i) Show that for any γ > 0, µ ¯[[u∗ > γ]] ≤

2 γ

R

[[|u|>γ/2]]

|u|. p

(Hint: apply 372C to (|u| − 12 γχ1)+ .) (ii) Show that ku∗ kp ≤ 2( )1/p kukp . (Hint: show that p−1 R∞ α¯ µ[[|u| > α]] + α µ ¯[[|u| > β]]dβ; see 365A. Use 366Xa to show that ku∗ kpp ≤ 2p

R∞ 0

γ p−2

R∞

γ/2

R [[|u|>α]]

|u| =

µ ¯[[|u| > β]]dβdγ + 2p kukpp ,

and reverse the order of integration. Compare 275Yc.) (This is Wiener’s dominated ergodic theorem.) (0)

¯) such that (c) Let (A, µ ¯) be a probability algebra and T an operator in Tµ¯,¯µ . Take u ∈ L1 = L1 (A, µ 1 ¯ is the corresponding function from L0 (A) to h(|u|) ∈ L , where h(t) = tP ln t for t ≥ 1, 0 for t ≤ 1, and h n 1 i ∗ 1 itself. Set u∗ = supn∈N n+1 i=0 |T u|. Show that u ∈ L . (Hint: use the method of 372Yb to show that R∞ R ∗ ¯ µ ¯[[u > γ]]dγ ≤ 2 h(u).) 2 (d) In 372G, suppose that A is atomless. Show that there is always an a ∈ A such that µ ¯a ≤ 12 and i inf i≤n π a 6= 0 for every n, so that (except in trivial cases) hAn (χa)in∈N will not be k k∞ -convergent. (e) Show that an irrational x ∈ ]0, 1[ has an eventually periodic sequence of continued fraction coefficients iff it is a solution of a quadratic equation with integral coefficients. (f ) In the language of 372M-372O and 372Xj, show the following. (i) For any x ∈ X, n ≥ 2, qn (x)qn−1 (x) ≥ 2n−1 , pn (x)pn+1 (x) ≥ 2n−1 , so that qn+1 (x)pn (x) ≥ 2n−1 and |1 − x/rn (x)| ≤ 2−n+1 , | ln x − ln rn (x)| ≤ 2−n+2 . Also |x − rn (x)| ≥ 1/qn (x)qn+2 (x). (ii) For any x ∈ X, n ≥ 1, pn+1 (x) = qn (φ(x)) and Qn−1 Pn−1 qn (x) i=0 rn−i (φi (x)) = 1. (iii) For any x ∈ X, n ≥ 1, | ln qn (x) + i=0 ln φi (x)| ≤ 4. (iv) For almost every x ∈ X, limn→∞

1 n

ln qn (x) = −

1 ln 2

R1

ln t

0 1+t

dt =

π2 . 12 ln 2

(Hint: 225Xi, 282Xo.) (v) For almost every x ∈ X, limn→∞ n1 ln |x − rn (x)| = −π 2 /6 ln 2. (vi) For almost every x ∈ X, 11−n ≤ |x − rn (x)| ≤ 10−n and 3n ≤ qn (x) ≤ 4n for all but finitely many n. (g) In 372M, show that for any measurable setRE ⊆ X, limn→∞ µRL φ−n [E] = νE. (Hint: recall that φ is mixing for ν (372Xw). Hence show that limn→∞ φ−n [E] g dν = νE · g dν for any integrable g. Apply this to a Radon-Nikod´ ym derivative of µL with respect to ν.) (I understand that this result is due to Gauss.)

372 Notes

The ergodic theorem

421

(h) Let h(Xi , Σi , µi )ii∈I be any family of probability spaces, with product (X, Λ, λ). Suppose that for each i ∈ I we are given an inverse-measure-preserving function φi : Xi → Xi . Show that there is a corresponding inverse-measure-preserving function φ : X → X given by setting φ(x)(i) = φi (x(i)) for x ∈ X, i ∈ I. Show that if each φi is mixing so is φ. (i) Give an example of an ergodic measure-preserving automorphism φ : [0, 1[ → [0, 1[ such that φ2 is not ergodic. (Hint: set φ(x) = 21 (1 + φ0 (2x)) for x < 12 , x − 12 for x ≥ 12 . See also 387Xg.) (j) Show that there is an ergodic φ : [0, 1] → [0, 1] such that (ξ1 , ξ2 ) 7→ (φ(ξ1 ), φ(ξ2 )) : [0, 1]2 → [0, 1]2 is not ergodic. (Hint: 372Xo.) (k) Let M be an r × r matrix with integer coefficients and non-zero determinant, where r ≥ 1. Let r r r φ : [0, 1[ → [0, 1[ be the function such that φ(x) − M x ∈ Zr for every x ∈ [0, 1[ . Show that φ is inverser measure-preserving for Lebesgue measure on [0, 1[ . 372 Notes and comments I have chosen an entirely conventional route to the Ergodic Theorem here, through the Mean Ergodic Theorem (372Xa) or, rather, the fundamental lemma underlying it (372A), and the Maximal Ergodic Theorem (372B-372C). What is not to be found in every presentation is the generality here. I speak of arbitrary T ∈ T (0) , the operators which are contractions both for k k1 and for k k∞ , not requiring T to be positive, let alone correspond to a measure-preserving homomorphism. (I do not mention T (0) in the statement of 372C, but of course it is present in spirit.) The work we have done up to this point puts this extra generality within easy reach, but as the rest of the section shows, it is not needed for the principal examples. Only in 372Xc do I offer an application not associated with a measure-preserving homomorphism or an inverse-measure-preserving function. The Ergodic Theorem is an ‘almost-everywhere pointwise convergence theorem’, like the strong law(s) of large numbers and the martingale theorem(s) (§273, §275). Indeed Kolmogorov’s form of the strong law can be derived from the Ergodic Theorem (372Xg). There are some very strong family resemblances. For instance, the Maximal Ergodic Theorem corresponds to the most basic of all the martingale inequalities (275D). Consequently we have similar results, obtained by similar methods, concerning the domination of sequences starting from members of Lp (372Yb, 275Yc), though the inequalities are not identical. (Compare also 372Yc with 275Yd.) There are some tantalising reflections of these traits in results surrounding Carleson’s theorem on the pointwise convergence of square-integrable Fourier series (see §286 notes), but Carleson’s theorem seems to be much harder than the others. Other forms of the strong law (273D, 273H) do not appear to fit into quite the same pattern, but I note that here, as with the Ergodic Theorem, we begin with a study of square-integrable functions (see part (e) of the proof of 372D). After 372D, there is a contraction and concentration in the scope of the results, starting with a simple replacement of M 1,0 with L1 (372E). Of course it is almost as easy to prove 372D from 372E as the other way about; I give precedence to 372D only because M 1,0 is the space naturally associated with the class T (0) of operators to which these methods apply. Following this I turn to the special family of operators to which the rest of the section is devoted, those associated with measure-preserving homomorphisms (372F), generally on probability spaces (372G). This is the point at which we can begin to identify the limit as a conditional expectation as well as an invariant element. Next comes the translation into the language of measure spaces and inverse-measure-preserving functions, all perfectly straightforward in view of the lemmas 372H (which was an exercise in §364) and 372J. These turn 372F into 372I and 372G into the main part of 372K. In 372K-372L I find myself writing at some length about a technical problem. The root of the difficulty is in the definition of ‘conditional expectation’. Now it is generally accepted that any pure mathematician has ‘Humpty Dumpty’s privilege’: ‘When I use a word, it means just what I choose it to mean – neither more nor less’. With any privilege come duties and responsibilities; in this context, the duty to be self-consistent, and the responsibility to try to use terms in ways which will not mystify or mislead the unprepared reader. Having written down a definition of ‘conditional expectation’ in Volume 2, I must either stick to it, or go back and change it, or very carefully explain exactly what modification I wish to make here. I don’t wish to suggest that absolute consistency – in terminology or anything else – is supreme among mathematical virtues. Surely it is better to give local meanings to words, or tolerate ambiguities, than to suppress ideas

422

Linear operators between function spaces

372 Notes

which cannot be formulated effectively otherwise, and among ‘ideas’ I wish to include the analogies and resonances which a suitable language can suggest. But I do say that it is always best to be conscious of what one is doing – I go farther: one of the things which mathematics is for, is to raise our consciousness of what our thoughts really are. So I believe it is right to pause occasionally over such questions. In 372M-372O (see also 372Xj, 372Xt, 372Xw, 372Yf, 372Yg) I make an excursion into number theory. This is a remarkable example of the power of advanced measure theory to give striking results in other branches of mathematics. Everything here is derived from Billingsley 65, who goes farther than I have space for, and gives references to more. Here let me point to 372Xi; almost accidentally, the construction offers a useful formula for a homeomorphism between two of the most important spaces of descriptive set theory, which will be important to us in Volume 4. I end the section by introducing two terms, ‘ergodic’ and ‘mixing’ transformation, not because I wish to use them for any new ideas (apart from the elementary 372R, these must wait for §§385-386) but because it may help if I immediately classify some of the inverse-measure-preserving functions we have seen (372Xm372Xo, 372Xr, 372Xt-372Xv). Of course in any application of any ergodic theorem it is of great importance to be able to identify the limits promised by the theorem, and the point about an ergodic transformation is just that our averages converge to constant limits (372Q). Actually proving that a given inverse-measurepreserving function is ergodic is rarely quite trivial (see 372N, 372Xn, 372Xo), though a handful of standard techniques cover a large number of cases, and it is usually obvious when a map is not ergodic, so that if an invariant region does not leap to the eye one has a good hope of ergodicity. I take the opportunity to mention two famous functions from [0, 1] to itself, the ‘tent’ and ‘quadratic’ maps (372Xm, 372Xn). In the formulae φα , ψβ I include redundant parameters; this is because the real importance of these functions lies in the way their behaviour depends, in bewildering complexity, on these parameters. It is only at the extreme values α = 2, β = 4 that the methods of this volume can tell us anything interesting.

373 Decreasing rearrangements I take a section to discuss operators in the class T (0) of 371F-371H and §372 and two associated classes T , T × (373A). These turn out to be intimately related to the idea of ‘decreasing rearrangement’ (373C). In 373D-373F I give elementary properties of decreasing rearrangements; then in 373G-373O I show how they may be used to characterize the set {T u : T ∈ T } for a given u. TheRargument uses a natural topology on the set T (373K). I conclude with remarks on the possible values of T u × v for T ∈ T (373P-373Q) and (0) (0) identifications between Tµ¯,¯ν , Tν¯,¯µ and Tµ¯×,¯ν (373R-373T). 373A Definition Let (A, µ ¯) and (B, ν¯) be measure algebras. Recall that Mµ¯1,∞ = L1 (A, µ ¯) + L∞ (A) 0 + is the set of those u ∈ L (A) such that (|u| − αχ1) is integrable for some α, its norm k k1,∞ being defined by the formulae kuk1,∞ = min{kvk1 + kwk∞ : v ∈ L1 , w ∈ L∞ , v + w = u} Z = min{α + (|u| − αχ1)+ : α ≥ 0} (369Ob). (a) Tµ¯,¯ν will be the space of linear operators T : Mµ¯1,∞ → Mν¯1,∞ such that kT uk1 ≤ kuk1 for every u ∈ L1µ¯ and kT uk∞ ≤ kuk∞ for every u ∈ L∞ (A). (Compare the definition of T (0) in 371F.) (b) If B is Dedekind complete, so that Mµ¯1,∞ , being a solid linear subspace of the Dedekind complete space L0 (B), is Dedekind complete, Tµ¯×,¯ν will be Tµ¯,¯ν ∩ L× (Mµ¯1,∞ ; Mµ¯1,∞ ).

373B

Decreasing rearrangements

423

373B Proposition Let (A, µ ¯) and (B, ν¯) be measure algebras. (a) T = Tµ¯,¯ν is a convex set in the unit ball of B(Mµ¯1,∞ ; Mν¯1,∞ ). (0) (b) If T ∈ T then T ¹Mµ¯1,0 belongs to Tµ¯,¯ν , as defined in 371F. So if T ∈ T , p ∈ [1, ∞[ and u ∈ Lpµ¯ then p T u ∈ Lν¯ and kT ukp ≤ kukp . (c) If B is Dedekind complete and T ∈ T , then T ∈ L∼ (Mµ¯1,∞ ; Mν¯1,∞ ) and T1 ∈ T whenever T1 ∈ L∼ (Mµ¯1,∞ ; Mν¯1,∞ ) and |T1 | ≤ |T |; in particular, |T | ∈ T . (d) If π : A → B is a measure-preserving Boolean homomorphism, then we have a corresponding operator T ∈ T defined by saying that T (χa) = χ(πa) for every a ∈ A. If π is order-continuous, then so is T . ¯ is another measure algebra and T ∈ T , S ∈ T ¯ then ST ∈ T ¯ . (e) If (C, λ) ν ¯ ,λ µ ¯ ,λ proof (a) As 371G, parts (a-i) and (a-ii) of the proof. (b) If u ∈ Mµ¯1,0 and ² > 0, then u is expressible as u0 + u00 where ku00 k∞ ≤ ² and u0 ∈ L1µ¯ . (Set u00 = (u+ ∧ ²χ1) − (u− ∧ ²χ1).) So (|T u| − ²χ1)+ ≤ |T u − T u00 | ∈ L1ν¯ . As ² is arbitrary, T u ∈ Mν¯1,0 ; as u is arbitrary, T ¹Mµ¯1,0 ∈ T (0) . Now the rest is a consequence of 371Gd. (c) Because Mν¯1,∞ is a solid linear subspace of L0 (B), which is Dedekind complete because B is, L∼ (Mµ¯1,∞ ; Mν¯1,∞ ) is a Riesz space (355Ea). Take any u ≥ 0 in Mµ¯1,∞ . Let α ≥ 0 be such that (u − αχ1)+ ∈ L1µ¯ . Because T ¹L1µ¯ belongs to B(L1µ¯ ; L1ν¯ ) = L∼ (L1µ¯ ; L1ν¯ ) (371D), w0 = sup{T v : v ∈ L1µ¯ , 0 ≤ v ≤ (u − αχ1)+ } is defined in L1ν¯ . Now if v ∈ Mµ¯1,∞ and 0 ≤ v ≤ u, we must have T v = T (v − αχ1)+ + T (v ∧ αχ1) ≤ w0 + αχ1 ∈ Mν¯1,∞ . Thus {T v : 0 ≤ v ≤ u} is bounded above in Mν¯1,∞ . As u is arbitrary, T ∈ L∼ (Mµ¯1,∞ ; Mν¯1,∞ ) (355Ba). Now take T1 such that |T1 | ≤ |T | in L∼ (Mµ¯1,∞ ; Mν¯1,∞ ). 371D also tells us that k|T ¹L1µ¯ |k = kT ¹L1µ¯ k, so that kT1 uk1 = k|T1 u|k1 ≤ k|T1 ||u|k1 ≤ k|T ||u|k1 = k sup T vk1 ≤ k sup vk1 = kuk1 |v|≤|u|

for every u ∈ then

L1µ¯

|v|≤|u|

(using one of the formulae in 355Eb for the first equality). At the same time, if u ∈ L∞ (A), |T1 u| ≤ |T1 ||u| ≤ |T ||u| = sup T v |v|≤|u|

≤ sup kT vk∞ χ1 ≤ sup kvk∞ χ1 = kuk∞ χ1, |v|≤|u|

|v|≤|u|

so kT1 uk∞ ≤ kuk∞ . Thus T1 ∈ T . (d) By 365O and 363F, we have norm-preserving positive linear operators T1 : L1µ¯ → L1ν¯ and T∞ : L (A) → L∞ (B) defined by saying that T1 (χa) = χ(πa) whenever µ ¯a < ∞ and T∞ (χa) = χ(πa) for every a ∈ A. If u ∈ S(Af ) = L1µ¯ ∩ S(A) (365F), then T1 u = T∞ u, because both T1 and T∞ are linear and they agree on {χa : µ ¯a < ∞}. If u ≥ 0 in Mµ¯∞,1 = L1µ¯ ∩ L∞ (A), there is a non-decreasing sequence hun in∈N in f S(A ) such that u = supn∈N un and ∞

limn→∞ ku − un k1 = limn→∞ ku − un k∞ = 0 (see the proof of 369Od), so that T1 u = supn∈N T1 un = supn∈N T∞ un = T∞ u. Accordingly T1 and T∞ agree on L1µ¯ ∩L∞ (A). But this means that if u ∈ Mµ¯1,∞ is expressed as v+w = v 0 +w0 , where v, v 0 ∈ L1µ¯ and w, w0 ∈ L∞ (A), we shall have

424

Linear operators between function spaces

373B

T1 v 0 + T∞ w0 = T1 v + T∞ w + T1 (v 0 − v) − T∞ (w − w0 ) = T1 v + T∞ w, because v 0 − v = w − w0 ∈ Mµ¯∞,1 . Accordingly we have an operator T : Mµ¯1,∞ → Mν¯1,∞ defined by setting T (v + w) = T1 v + T∞ w whenever v ∈ L1µ¯ , w ∈ L∞ (A). This formula makes it easy to check that T is linear and positive, and it clearly belongs to T . To see that T is uniquely defined, observe that T ¹L1µ¯ and T ¹L∞ (A) are uniquely defined by the values T takes on S(Af ), S(A) respectively, because these spaces are dense for the appropriate norms. Now suppose that π is order-continuous. Then T1 and T∞ are also order-continuous (365Oa, 363Ff). If A ⊆ Mµ¯1,∞ is non-empty and downwards-directed and has infimum 0, take u0 ∈ A and γ > 0 such that (u0 − γχ1)+ ∈ L1µ¯ . Set A1 = {(u − γχ1)+ : u ∈ A, u ≤ u0 },

A∞ = {u ∧ γχ1 : u ∈ A}.

Then A1 ⊆ L1µ¯ and A∞ ⊆ L∞ (A) are both downwards-directed and have infimum 0, so inf T1 [A1 ] = inf T∞ [A∞ ] = 0 in L0 (B). But this means that inf(T1 [A1 ] + T∞ [A∞ ]) = 0 (351Dc). Now any w ∈ T1 [A1 ] + T∞ [A∞ ] is expressible as T (u − γχ1)+ + T (u0 ∧ γχ1) where u, u0 ∈ A; because A is downwardsdirected, there is a v ∈ A such that v ≤ u ∧ u0 , in which case T v ≤ w. Accordingly T [A] must also have infimum 0. As A is arbitrary, T is order-continuous. (e) is obvious, as usual. 373C Decreasing rearrangements The following concept is fundamental to any understanding of ¯) for the set of those u ∈ L0 (A) the class T . Let (A, µ ¯) be a measure algebra. Write Mµ¯0,∞ = M 0,∞ (A, µ such that µ ¯[[|u| > α]] is finite for some α ∈ R. (See 369N for the ideology of this notation.) It is easy to see that Mµ¯0,∞ is a solid linear subspace of L0 (A). Let (AL , µ ¯L ) be the measure algebra of Lebesgue measure , defined by setting u∗ = g • , where on [0, ∞[. For u ∈ Mµ¯0,∞ its decreasing rearrangement is u∗ ∈ Mµ¯0,∞ L g(t) = inf{α : α ≥ 0, µ ¯[[|u| > α]] ≤ t} for every t > 0. (This is always finite because inf α∈R µ ¯[[|u| > α]] = 0, by 364A(β) and 321F.) I will maintain this usage of the symbols AL , µ ¯L , u∗ for the rest of this section. 373D Lemma Let (A, µ ¯) be a measure algebra. 0,∞ (a) For any u ∈ Mµ¯ , its decreasing rearrangement u∗ may be defined by the formula [[u∗ > α]] = [0, µ ¯[[|u| > α]][ for every α ≥ 0, •

that is, µ ¯L [[u∗ > α]] = µ ¯[[|u| > α]] for every α ≥ 0. (b) If |u| ≤ |v| in Mµ¯0,∞ , then u∗ ≤ v ∗ ; in particular, |u|∗ = u∗ . Pn Pn • ∗ ¯ai [ . (c)(i) If u = P i=0 αi χ [0, µ i=0 αi χai , where a0 ⊇ a1 ⊇ . . . ⊇ an and αi ≥ 0 for each i, then u = n ∗ i=0 αi χai where Pn (ii) If u = P a0 , . . . , an are disjoint and |α0 | ≥ |α1 | ≥ . . . ≥ |αn |, then u = • |α |χ [β , β [ , where β = ¯ai for i ≤ n + 1. i i i+1 i i=0 j 0

Rt 0

u∗ = inf α≥0 αt +

R

(|u| − αχ1)+ .

(h) If A ⊆ (Mµ¯0,∞ )+ is non-empty and upwards-directed and has supremum u0 ∈ Mµ¯0,∞ , then u∗0 = supu∈A u∗ .

373D

Decreasing rearrangements

425

proof (a) Set g(t) = inf{α : µ ¯[[|u| > α]] ≤ t} as in 373C. If α ≥ 0, g(t) > α ⇐⇒ µ ¯[[|u| > β]] > t for some β > α ⇐⇒ µ ¯[[|u| > α]] > t (because [[|u| > α]] = supβ>α [[|u| > β]]), so [[u∗ > α]] = {t : g(t) > α}• = [0, µ ¯[[|u| > α]][ . •

Of course this formula defines u∗ . (b) This is obvious, either from the definition in 373C or from (a) just above. Pn • (c)(i) Setting v = i=0 αi χ [0, µ ¯ai [ , we have [[v > α]] = 0 if

n X

αi ≤ α

i=0 •

= [0, µ ¯aj [ if

j−1 X

αi ≤ α
α]][ . (ii) A similar argument applies. (If any aj has infinite measure, then ai is irrelevant for i > j.) (d) Fix γ > 0 for the moment, and consider A = {E : E ⊆ ]γ, ∞[ is a Borel set, µ ¯L [[u∗ ∈ E]] = µ ¯[[|u| ∈ E]]}, I = {]α, ∞[ : α ≥ γ}. Then I ⊆ A (by (a)), I ∩ J ∈ I for S all I, J ∈ I, E \ F ∈ A whenever E, F ∈ A and F ⊆ E (because u ∈ Mµ¯0 , so µ ¯[[|u| ∈ E]] < ∞), and n∈N En ∈ A whenever hEn in∈N is a non-decreasing sequence in A. So, by the Monotone Class Theorem (136B), A includes the σ-algebra of subsets of ]γ, ∞[ generated by I; but this must contain E ∩ ]γ, ∞[ for every Borel set E ⊆ R. Accordingly, for any Borel set E ⊆ ]0, ∞[, µ ¯L [[u∗ ∈ E]] = supn∈N µ ¯L [[u∗ ∈ E ∩ ]2−n , ∞[ ]] = µ ¯[[|u| ∈ E]]. (e) For any α > 0, Eα = {t : h(t) > α} is a Borel subset of ]0, ∞[. If u ∈ Mµ¯0 then, using (d) above, ∗ ¯ ∗ ) > α]] = µ ¯ ¯ µ ¯L [[h(u ¯L [[u∗ ∈ Eα ]] = µ ¯[[u ∈ Eα ]] = µ ¯[[h(u) > α]] = µ ¯L [[(h(u)) > α]]. ∗ ¯ ¯ ∗ ) are equivalence classes of non-increasing functions, they must be equal. As both (h(u)) and h(u If h is continuous on the left, then Eα = ]γ, ∞[ for some γ, so we no longer need to use (d), and the argument works for any u ∈ (Mµ¯0,∞ )+ .

(f ) Apply (e) with h(β) = max(0, β − α). (g) Express u∗ as g • , where g(s) = inf{α : µ ¯[[|u| > α]] ≤ s} for every s > 0. Because g is non-increasing, it is easy to check that, for t > 0,

Rt 0

g = tg(t) +

for every α ≥ 0; so that

R∞ 0

max(0, g(s) − g(t))ds ≤ αt +

Rt 0

u∗ = minα≥0 αt +

R

R∞ 0

max(0, g(s) − α)ds

(u∗ − αχ1)+ .

426

Linear operators between function spaces

373D

Now Z

Z

∞

(u∗ − αχ1)+ =

µ ¯L [[(u∗ − αχ1)+ > β]]dβ 0

Z

Z

∞

(|u| − αχ1)+

+

=

µ ¯[[(|u| − αχ1) > β]]dβ = 0

for every α ≥ 0, using (f) and 365A, and

Rt 0

u∗ = minα≥0 αt +

R

(|u| − αχ1)+ .

(h) µ ¯[[u0 > α]] = µ ¯(supu∈A [[u > α]]) = supu∈A µ ¯[[u > α]] for any α > 0, using 364Mb and 321D. So [[u∗0 > α]] = [0, µ ¯[[u0 > α]][ = supu∈A [0, µ ¯[[u > α]][ = [[supu∈A u∗ > α]] •

•

for every α, and u∗0 = supu∈A u∗ . R

R

u∗ × v ∗ for all u, v ∈ Mµ¯0,∞ . Pm Pn proof (a) Consider first the case u, v ≥ 0 in S(A). Then we may express u, v as i=0 αi χai , j=0 βj χbj where a0 ⊇ a1 ⊇ . . . ⊇ am , b0 ⊇ . . . ⊇ bn in A and αi , βj ≥ 0 for all i, j (361Ec). Now u∗ , v ∗ are given by Pn Pm • • ¯ bj [ ¯ai [ , v ∗ = j=0 βj χ [0, µ u∗ = i=0 αi χ [0, µ 373E Theorem Let (A, µ ¯) be a measure algebra. Then

|u × v| ≤

(373Dc). So Z u×v = =

m X n X i=0 j=0 m X n X

αi βj µ ¯(ai ∩ bj ) ≤

m X n X

αi βj min(¯ µai , µ ¯bj )

i=0 j=0

Z

αi βj µL ([0, µ ¯ai [ ∩ [0, µ ¯bj [) =

u∗ × v ∗ .

i=0 j=0

(b) For the general case, we have non-decreasing sequences hun in∈N , hvn in∈N in S(A)+ with suprema |u|, |v| respectively (364Kd), so that |u × v| = |u| × |v| = supn∈N |u| × vn = supm,n∈N um × vn = supn∈N un × vn and

R

|u × v| =

R

supn∈N un × vn = supn∈N

R

un × vn ≤ supn∈N

R

u∗n × vn∗ ≤

using 373Db. 373F Theorem Let (A, µ ¯) be a measure algebra, and u any member of Mµ¯0,∞ . (a) For any p ∈ [1, ∞], u ∈ Lpµ¯ iff u∗ ∈ Lpµ¯L , and in this case kukp = ku∗ kp . (b)(i) u ∈ Mµ¯0 iff u∗ ∈ Mµ¯0L ; (ii) u ∈ Mµ¯1,∞ iff u∗ ∈ Mµ¯1,∞ , and in this case kuk1,∞ = ku∗ k1,∞ ; L (iii) u ∈ Mµ¯1,0 iff u∗ ∈ Mµ¯1,0 ; L (iv) u ∈ Mµ¯∞,1 iff u∗ ∈ Mµ¯∞,1 , and in this case kuk∞,1 = ku∗ k∞,1 . L proof (a)(i) Consider first the case p = 1. In this case

R

|u| =

R∞ 0

µ ¯[[|u| > α]]dα =

R∞

0 ∗ p

µ ¯L [[u∗ > α]]dα =

(ii) If 1 < p < ∞, then by 373De we have (|u|p )∗ = (u ) , so that kukpp =

R

|u|p =

R

(|u|p )∗ =

R

(u∗ )p = ku∗ kpp

R

u∗ .

R

u∗ × v ∗ ,

373H

Decreasing rearrangements

427

if either kukp or ku∗ kp is finite. (iii) As for p = ∞, kuk∞ ≤ γ ⇐⇒ [[|u| > γ]] = 0 ⇐⇒ [[u∗ > γ]] = 0 ⇐⇒ ku∗ k∞ ≤ γ. (b)(i) u ∈ Mµ¯0 ⇐⇒ µ ¯[[|u| > α]] < ∞ for every α > 0 ⇐⇒ µ ¯L [[u∗ > α]] < ∞ for every α > 0 ⇐⇒ u∗ ∈ Mµ¯0L . (ii) For any α ≥ 0,

R

(|u| − αχ1)+ =

R

(u∗ − αχ1)+

as in the proof of 373Dg. So kuk1,∞ = ku∗ k1,∞ if either is finite, by the formula in 369Ob. (iii) This follows from (i) and (ii), because M 1,0 = M 0 ∩ M 1,∞ . (iv) Allowing ∞ as a value of an integral, we have Z kuk1,∞ = min{α + (|u| − αχ1)+ : α ≥ 0} Z = min{α + (u∗ − αχ1)+ : α ≥ 0} = ku∗ k1,∞ . by 369Ob; in particular, u ∈ Mµ¯1,∞ iff u∗ ∈ Mµ¯1,∞ L 373G Lemma Let (A, µ ¯) and (B, ν¯) be measure algebras. If 1,∞ either u ∈ Mµ¯ and T ∈ Tµ¯,¯ν (0)

or u ∈ Mµ¯1,0 and T ∈ Tµ¯,¯ν , Rt then 0 (T u)∗ ≤ 0 u∗ for every t ≥ 0. Rt

proof Set T1 = T ¹L1µ¯ , so that kT1 k ≤ 1 in B(L1µ¯ ; L1ν¯ ), and |T1 | is defined in B(L1µ¯ ; L1ν¯ ), also with norm at most 1. If α ≥ 0, then we can express u as u1 + u2 where |u1 | ≤ (|u| − αχ1)R+ and |u2 | ≤ αχ1. (Let w ∈ L∞ (A) be such that kwk∞ ≤ 1, u = |u| × w; set u2 = w × (|u| ∧ αχ1).) So if (|u| − αχ1)+ < ∞, |T u| ≤ |T u1 | + |T u2 | ≤ |T1 ||u1 | + αχ1 and

R

R

R

R

(|T u| − αχ1)+ ≤ |T1 ||u1 | ≤ |u1 | ≤ (|u| − αχ1)+ . Rt Rt The formula of 373Dg now tells us that 0 (T u)∗ ≤ 0 u∗ for every t. 373H Lemma Let (A, µ ¯) be a measure algebra, and θ : Af → R an additive functional, where Af = {a : µ ¯a < ∞}. (a) The following are equiveridical: 1

(α) limt↓0 supµ¯a≤t |θa| = limt→∞ supµ¯a≤t |θa| = 0, t R 1,0 1,0 (β) there is some u ∈ M = Mµ¯ such that θa = a u for every a ∈ Af , and in this case u is uniquely defined. (b) Now suppose that (A, µ ¯) is localizable. Then the following are equiveridical: (α) limt↓0 supµ¯a≤t |θa| = 0, (β) there is some u ∈ M 1,∞ and again this u is uniquely defined.

1

lim supt→∞ supµ¯a≤t |θa| < ∞, t R 1,∞ = Mµ¯ such that θa = a u for every a ∈ Af ,

f f proof (a)(i) Assume (α). R For a, c ∈ A , setf θc (a) = θ(a ∩ c). Then for each c ∈ A , there is a unique 1 uc ∈ Lµ¯ such that θc a = a uc for every a ∈ A (365Eb). Because uc is unique we must have uc = ud × χc

428

Linear operators between function spaces

373H

whenever c ⊆ d ∈ Af . Next, given α > 0, there is a t0 ≥ 0 such that |θa| ≤ α¯ µa whenever a ∈ Af and f + µ ¯a ≥ t0 ; so that µ ¯[[uc > α]] ≤ t0 for every c ∈ A , and e(α) = supc∈Af [[uc > α]] is defined in Af . Of course + 0 0 e(α) = [[ue(1) > α]] for every α ≥ 1, so inf α∈R e(α) = 0, and v1 = supc∈Af u+ c is defined in L = L (A) (364Mb). Because [[v1 > α]] = e(α) ∈ Af for each α > 0, v1 ∈ Mµ¯0 . For any a ∈ Af , + v1 × χa = supc∈Af u+ c × χa = ua ,

R R f so v1 ∈ M 1,0 and a v1 = a u+ a for every a ∈ A . R R − f Similarly, v2 = supc∈Af uc is defined in M 1,0 and a v2 = a u− a for every a ∈ A . So we can set 1,0 u = v1 − v2 ∈ M and get

R

a

u=

R

a

ua = θa

for every a ∈ Af . Thus (β) is true.

R + (ii) ¯a ≤ δ (365Ea), so R Assume (β). If ² > 0, there is a δ > 0 such that a (|u| − ²χ1) ≤ ²R whenever µ that | a u| ≤ ²(1 + µ ¯a) whenever µ ¯a ≤ δ. As ² is arbitrary, limt↓0 supµ¯a≤t | a u| = 0. Moreover, whenever R R t > 0 and µ ¯a ≤ t, 1t | a u| ≤ ² + 1t (|u| − ²χ1)+ . Thus lim supt→∞

1 t

supµ¯a≤t |

R

a

u| ≤ ².

As ² is arbitrary, θ satisfies the conditions in (α). (iii) The uniqueness of u is a consequence of 366Gd. (b) The argument for (b) uses the same ideas.

R (i) Assume (α). Again, for each c ∈ Af , we have uc ∈ L1 such that θc a = a uc for every a ∈ Af ; again, set e(α) = supc∈Af [[u+ c > α]], which is defined because A is supposed to be Dedekind complete. This time, there are t0 , γ ≥ 0 such that |θa| ≤ γ µ ¯a whenever a ∈ Af and µ ¯a ≥ t0 ; so that µ ¯[[uc > γ]] ≤ t0 for every c ∈ Af , and µ ¯e(γ) < ∞. Accordingly inf α≥γ e(α) = inf α≥γ [[u+ e(γ) > α]] = 0, 0 0 + 1 f and once more v1 = supc∈Af u+ c is defined in L = L (A). As before, v1 × χa = ua ∈ L for any a ∈ A , 1,∞ f 1,∞ − , with v2 ×χa = u− Because [[v1 > γ]] = e(γ) ∈ A , v1 ∈ M . Similarly, v2 = supc∈Af uc is defined in M a f 1,∞ for every a ∈ A . So u = v1 − v2 ∈ M , and

R

a

u=

R

a

ua = θa

for every a ∈ Af .

R + (ii) Take γ ≥ 0 such that β = (|u| R Assume (β). R − γχ1) is finite. If ² > 0, there is a δ > 0 such + that a (|u| − γχ1) ≤ ² whenever µ ¯a ≤ δ, so that | a u| ≤ ² + γ µ ¯a whenever µ ¯a ≤Rδ. As ² is arbitrary, R R limt↓0 supµ¯a≤t | a u| = 0. Moreover, whenever t > 0 and µ ¯a ≤ t, then 1t | a u| ≤ γ + 1t (|u| − ²χ1)+ . Thus R lim supt→∞ 1t supµ¯a≤t | a u| ≤ γ < ∞, R and the function a 7→ a u satisfies the conditions in (β). (iii) u is uniquely defined because u × χa must be ua , as defined in (i), for every a ∈ Af , and (A, µ ¯) is semi-finite. 373I Lemma Suppose that u, v, w ∈ Mµ¯0,∞ are all equivalence classes of non-negative non-increasing Rt Rt RL R functions. If 0 u ≤ 0 v for every t ≥ 0, then u × w ≤ v × w. P 4n proof For n ∈ N, i ≤ 4n set ani = [[w > 2−n i]]; set wn = i=1 2−n χani . Then each ani is of the form [0, t]• , so

R

u × wn =

P 4n

i=1

2−n

R

ani

u≤

P4n

i=1

2−n

R

ani

v=

Also hwn in∈N is a non-decreasing sequence with supremum w, so

R

u × w = supn∈N

R

u × wn ≤ supn∈N

R

v × wn =

R

R

v × wn .

v × w.

373L

Decreasing rearrangements

429

373J Corollary Suppose that (A, µ ¯) and (B, ν¯) are measure algebras and v ∈ Mν¯0,∞ . If (0)

either u ∈ Mµ¯1,0 and T ∈ Tµ¯,¯ν then

R

or u ∈ Mµ¯1,∞ and T ∈ Tµ¯,¯ν R |T u × v| ≤ u∗ × v ∗ .

proof Put 373E, 373G and 373I together. 373K The very weak operator topology of T Let (A, µ ¯) and (B, ν¯) be two measure algebras. For u ∈ Mµ¯1,∞ , w ∈ Mν¯∞,1 set ρuw (S, T ) = |

R

Su × w −

R

T u × w| for all S, T ∈ T = Tµ¯,¯ν .

Then ρuw is a pseudometric on T . I will call the topology generated by {ρuw : u ∈ Mµ¯1,∞ , w ∈ Mν¯∞,1 } (2A3F) the very weak operator topology on T . 373L Theorem Let (A, µ ¯) be a measure algebra and (B, ν¯) a localizable measure algebra. Then T = Tµ¯,¯ν is compact in its very weak operator topology. proof Let F be an ultrafilter on T . If u ∈ Mµ¯1,∞ , w ∈ Mν¯∞,1 then |

R

T u × w| ≤

R

u∗ × w ∗ < ∞

R ∗ for every T ∈ T (373J); u × w∗ is finite because u∗ ∈ M 1,∞ and w∗ ∈ M ∞,1 (373F). R R In particular, { T u × R w : T ∈ T } is bounded. Consequently hu (w) = limT →F T u × w is defined in R (2A3Se). Because w 7→ T u × w is additive for every T ∈ T , so is hu . Also |hu (w)| ≤

R

u∗ × w∗ ≤ ku∗ k1,∞ kw∗ k∞,1 = kuk1,∞ kwk∞,1

for every w ∈ Mν¯∞,1 . Rt |hu (χb)| ≤ 0 u∗ whenever b ∈ Bf and ν¯b ≤ t. So limt↓0 supν¯b≤t |hu (χb)| ≤ limt↓0 lim supt→∞

1 t

Rt 0

u∗ = 0,

supν¯b≤t |hu (χb)| ≤ lim supt→∞

1 t

Rt 0

u∗ < ∞.

R Of course b 7→ hu (χb) is additive, so by 373H there is a unique Su ∈ Mν¯1,∞ such that hu (χb) = b Su for R every b ∈ Bf . Since both hu and w 7→ Su × w are linear and continuous on Mν¯∞,1 , and S(Bf ) is dense in Mν¯∞,1 (369Od),

R

Su × w = hu (w) = limT →F

Mν¯∞,1 .

R

Tu × w

Mµ¯1,∞ .

for every w ∈ And this is true for every u ∈ R R For any particular w ∈ Mν¯∞,1 , all the maps u 7→ T u × w are linear, so u 7→ Su × w also is; that is, S : Mµ¯1,∞ → Mν¯1,∞ is linear. Now S ∈ T . P P (α) If u ∈ L1µ¯ and b, c ∈ Bf , then Z Z Z Z T u × (χb − χc) ≤ sup T u × (χb − χc) Su − Su = lim b

c

T →F

T ∈T

≤ sup kT uk1 kχb − χck∞ ≤ kuk1 . T ∈T

But, setting e = [[Su > 0]], we have Z Z Z |Su| = Su − Su e 1\e Z Su + = sup b∈Bf ,b ⊆ e

b

sup c∈Bf ,c ⊆ 1\e

(−Su) ≤ kuk1 .

430

Linear operators between function spaces

(β) If u ∈ L∞ (A), then |

R

Su| ≤ supT ∈T | b

R

373L

T u × χb| ≤ supT ∈T kT uk∞ ν¯b ≤ kuk∞ ν¯b

for every b ∈ Bf . So [[Su > kuk∞ ]] = [[−Su > kuk∞ ]] = 0 and kSuk∞ ≤ kuk∞ . (Note that both parts of this argument depend on knowing that (B, ν¯) is semi-finite, so that we cannot be troubled by purely infinite elements of B.) Q Q Of course we now have limT →F ρuw (T, S) = 0 for all u ∈ Mµ¯1,0 , w ∈ Mν¯∞,1 , so that S = lim F in T . As F is arbitrary, T is compact (2A3R). 373M Corollary Let (A, µ ¯) be a measure algebra and (B, ν¯) a localizable measure algebra, and u any member of Mµ¯1,∞ . Then B = {T u : T ∈ Tµ¯,¯ν } is compact in Mν¯1,∞ for the topology Ts (Mν¯1,∞ , Mν¯∞,1 ). proof The point is just that the map T 7→ T u : Tµ¯,¯ν → Mν¯1,0 is continuous for the very weak operator topology on Tµ¯,¯ν and Ts (Mν¯1,∞ , Mν¯∞,1 ). So B is a continuous image of a compact set, therefore compact (2A3Nb). 373N Corollary Let (A, µ ¯) be a measure algebra, (B, ν¯) a localizable measure algebra and u any member of Mµ¯1,∞ ; set B = {T u : T ∈ Tµ¯,¯ν }. If hvn in∈N is any non-decreasing sequence in B, then supn∈N vn is defined in Mν¯1,∞ and belongs to B. R proof By 373M, hvn in∈N must have a cluster point v ∈ B for Ts (Mν¯1,∞ , Mν¯∞,1 ). Now for any b ∈ Bf , b v R R R must be a cluster point of h n vn in∈N , because w 7→ b w is continuous for Ts (Mν¯1,∞ , MνR¯∞,1 ). But h b vnRin∈N is a non-decreasing sequence, so its only possible cluster point is its supremum; thus b v = limn→∞ b vn . Consequently v × χb must be the supremum of {vn × χb : n ∈ N} in L1 . And this is true for every b ∈ Bf ; as (B, ν¯) is semi-finite, v is the supremum of hvn in∈N in L0 (B) and in Mν¯1,∞ . 373O Theorem Let (A, µ ¯), (B, ν¯) be measure algebras and u ∈ Mµ¯1,∞ , v ∈ Mν¯1,∞ . Then the following are equiveridical: (i) there is a T ∈ Tµ¯,¯ν such that T u = v, Rt Rt (ii) 0 v ∗ ≤ 0 u∗ for every t ≥ 0. In particular, given u ∈ Mµ¯1,∞ , there are S ∈ Tµ¯,¯µL , T ∈ Tµ¯L ,¯µ such that Su = u∗ , T u∗ = u. proof (i)⇒(ii) is Lemma 373G. Accordingly I shall devote the rest of the proof to showing that (ii)⇒(i). (a) If (A, µ ¯), (B, ν¯) are measure algebras and u ∈ Mµ¯1,∞ , v ∈ Mν¯1,∞ , I will say that v 4 u if there is a T ∈ Tµ¯,¯ν such that T u = v, and that v ∼ u if v 4 u and u 4 v. (Properly speaking, I ought to write (u, µ ¯) 4 (v, ν¯), because we could in principle have two different measures on the same algebra. But I do not think any confusion is likely to arise in the argument which follows.) By 373Be, 4 is transitive and ∼ is an equivalence relation. Now we have the following facts. P There (b) If (A, µ ¯) is a measure algebra and u1 , u2 ∈ Mµ¯1,∞ are such that |u1 | ≤ |u2 |, then u1 4 u2 . P 1,∞ ∞ is a w ∈ L (A) such that u1 = w × u2 and kwk∞ ≤ 1. Set T v = w × v for for v ∈ Mµ¯ ; then T ∈ Tµ¯,¯µ and T u2 = u1 . Q Q So u ∼ |u| for every u ∈ Mµ¯1,∞ . (c) If (A, µ ¯P ) is a measure algebra and u ≥ 0 in S(A), then u 4 u∗ . P P If u = 0 this is trivial. Otherwise, n express u as i=0 αi χai where a0 , . . . , an are disjoint and non-zero and α0 > α1 . . . > αn > 0 ∈ R. If µ ¯ai = ∞ for any i, take m to be minimal subject to µ ¯am = ∞; otherwise, set m = n. Then u∗ = Pm Pj−1 • ¯ai for 1 ≤ j ≤ m + 1. i=0 αi χ [βi , βi+1 [ , where β0 = 0, βj = i=0 µ For i < m, and for i = m if µ ¯am < ∞, define hi : Mµ¯1,∞ → R by setting hi (v) =

1 µ ¯ai

R

ai

v

for every v ∈ Mµ¯1,∞ . If µ ¯am = ∞, then we need a different idea to define hm , as follows. Let I be {a : a ∈ A, µ ¯(a∩am ) < ∞}. Then I is an ideal of A not containing am , so there is a Boolean homomorphism π : A → {0, 1} such that πa = 0 for a ∈ I and πam = 1 (311D). This induces a corresponding k k∞ -continuous

373O

Decreasing rearrangements

431

linear operator h : L∞ (A) → L∞ ({0, 1}) ∼ ¯a < ∞, and accordingly = R, as in 363F. Now h(χa) = 0 whenever µ h(v) = 0 whenever v ∈ Mµ¯∞,1 , since S(Af ) is dense in Mµ¯∞,1 for k k∞,1 and therefore also for k k∞ . But this means that h has a unique extension to a linear functional hm : Mµ¯1,∞ → R such that hm (v) = 0 for every v ∈ L1µ¯ , while hm (χam ) = 1 and |h(v)| ≤ kvk∞ for every v ∈ L∞ (A). Having defined hi for every i ≤ m, define T : Mµ¯1,∞ → Mµ¯1,∞ by setting L Pm • T v = i=0 hi (v)χ [βi , βi+1 [ for every v ∈ Mµ¯1,∞ . For any i ≤ m, v ∈ L1µ¯ ,

R βi+1 βi

|T v| = |hi (v)|¯ µai ≤

R ai

|v|;

summing over i, kT vk1 ≤ kvk1 . Similarly, for any i ≤ m, v ∈ L∞ (B), |hi (v)| ≤ kvk∞ , so kT vk∞ ≤ kvk∞ . Thus T ∈ Tµ¯,¯µL . Since u∗ = T u, we conclude that u∗ 4 u, as claimed. Q Q (d) If (A, µ ¯) is a measure algebra and u ≥ 0 in Mµ¯1,∞ , then u∗ 4 u. P P Let hun in∈N be a non-decreasing sequence in S(A) with u0 ≥ 0 and supn∈N un = u. Then hu∗n in∈N is a non-decreasing sequence in Mµ¯1,∞ with L supremum u∗ , by 373Db and 373Dh. Also u∗n 4 un 4 u for every n, by (b) and (c) of this proof. By 373N, u∗ 4 u. Q Q (e) If (A, µ ¯) is a measure algebra and u ≥ 0 in S(A), then u 4 u∗ . P P The argument is very similar to that of (c). Again, the result is trivial if u = 0; suppose that u > 0 and define αi , ai , m, βi as before. This Pm ˜ amd u ˜∗ = u∗ . Set time, set a0i = ai for i < m, a0m = supm≤j≤n aj , u ˜ = i=0 αi χa0i ; then u ≤ u hi (v) =

1 βi+1 −βi

R βi+1 βi

v

if i ≤ m, βi+1 < ∞ (that is, µ ¯ai < ∞) and v ∈ Mµ¯1,∞ ; and if µ ¯am = ∞, set L hm (v) = limk→F

1 k

Rk 0

v

for some non-principal ultrafilter F on N. As before, we have Rβ |hi (v)|¯ µa0i ≤ βii+1 |v|, whenever v ∈ L1µ¯L , i ≤ m, while |hi (v)| ≤ kvk∞ whenever v ∈ L∞ (AL ) and i ≤ m. So we can define Pm , and get T ∈ Tµ¯L ,¯µ by setting T v = i=0 hi (v)χa0i for every v ∈ Mµ¯1,∞ L u4u ˜ = T u ∗ 4 u∗ . Q Q (f ) If (A, µ ¯) is a measure algebra and u ≥ 0 in Mµ¯1,∞ , then u 4 u∗ . P P This time I seek to copy the ideas of (d); there is a new obstacle to circumvent, since (A, µ ¯) might not be localizable. Set α0 = inf{α : α ≥ 0, µ ¯[[u > α]] < ∞},

e = [[u > α0 ]].

−n

Then e = supn∈N [[u > α0 + 2 ]] is a countable supremum of elements of finite measure, so the principal ideal Ae , with its induced measure µ ¯e , is σ-finite. Now let hun in∈N be a non-decreasing sequence in S(A) with u0 ≥ 0 and supn∈N un = u; set u ˜ = u × χe and u ˜n = un × χe, regarded as members of S(Ae ), for each n. In this case u ˜n 4 u ˜∗n 4 u∗ for every n. Because (Ae , µ ¯e ) is σ-finite, therefore localizable, 373N tells us that u ˜ 4 u∗ . ∗ Let S ∈ Tµ¯L ,¯µe be such that Su = u ˜. As in part (e), choose a non-principal ultrafilter F on N and set h(v) = limk→F

1 k

Rk 0

v

for v ∈ Mµ¯1,∞ . Now define T : Mµ¯1,∞ → Mµ¯1,∞ by setting L L T v = Sv + h(v)χ(1 \ e),

432

Linear operators between function spaces

373O

here regarding Sv as a member of Mµ¯1,∞ . (I am taking it to be obvious that Mµ¯1,∞ can be identified with e {w × χe : w ∈ Mµ¯1,∞ }.) Then it is easy to see that T ∈ Tµ¯L ,¯µ . Also u ≤ T u∗ , because h(u∗ ) = inf{α : µ ¯L [[u∗ > α]] < ∞} = α0 , while u × χ(1 \ e) ≤ α0 χ(1 \ e). So we get u 4 T u∗ 4 u∗ . Q Q R Rt t (g) Now suppose that u, v ≥ 0 in Mµ¯1,∞ , that 0 u∗ ≥ 0 v ∗ for every t ≥ 0, and that v is of the form L Pn ¯L ai < ∞ for each i. Then v 4 u. i=1 αi χai where α1 > . . . > αn > 0, a1 , . . . , an ∈ AL are disjoint and µ P P Induce on n. If n = 0 then v = 0 and the result is trivial. For the inductive step to n ≥ 1, if v ∗ ≤ u∗ we have v ∼ v ∗ 4 u∗ ∼ u, Rt using (b), (d) and (f) above. Otherwise, look at φ(t) = 1t 0 u∗ for t > 0. We have φ(t) ≥

1 t

Rt 0

v ∗ = α1

for t ≤ β = µ ¯a1 , while limt→∞ φ(t) < α1 , because (limt→∞ φ(t))χ1 ≤ u∗ and v ∗ ≤ α1 χ1 and v ∗ 6≤ u∗ . Becaasue φ is continuoue, there is a γ ≥ β such that φ(γ) = α1 . Define T0 ∈ Tµ¯L ,¯µL by setting T0 w = (

1 γ

Rγ 0

•

. Then T0 u∗ 4 u∗ ∼ u, and for every w ∈ Mµ¯1,∞ L T0 u∗ × χ [0, γ[ = ( •

We need to know that Z

t

0

Rt 0

T0 u∗ ≥

Rt 0

•

w)χ [0, γ[ + (w × χ [γ, ∞[ )

1 γ

Rγ 0

u∗ )χ [0, γ[ = α1 χ [0, γ[ . •

•

v ∗ for every t; this is because

Z t T0 u∗ = α1 t ≥ v ∗ whenever t ≤ γ, 0 Z γ Z t Z t Z t = T0 u∗ + T0 u∗ = u∗ ≥ v ∗ whenever t ≥ γ. 0

γ

0

0

Set u1 = T0 u∗ × χ [β, ∞[ , •

v1 = v ∗ × χ [β, ∞[ . •

Then u∗1 , v1∗ are just translations of T0 u∗ , v ∗ to the left, so that

Rt

R β+t

R β+t

R β+t

R β+t

Rt

T0 u∗ = 0 T0 u∗ − α1 β ≥ 0 v ∗ − α1 β = β v ∗ = 0 v1∗ Pn Pi • ¯aj for each j. So by the inductive for every t ≥ 0. Also v1 = j=1 µ i=2 αi χ [βi−1 , βi [ where βi = hypothesis, v1 4 u1 . Let S ∈ Tµ¯L ,¯µL be such that Su1 = v1 , and define T ∈ Tµ¯L ,¯µL by setting 0

u∗1 =

β

•

•

•

T w = w × χ [0, β[ + S(w × χ [β, ∞[ ) × χ [β, ∞[

∗ ∗ ∗ ∗ for every w ∈ Mµ¯1,∞ Q µL . Then T T0 u = v , so v ∼ v 4 u ∼ u, as required. Q L ,¯ R R t t (h) We are nearly home. If u, v ≥ 0 in Mµ¯1,∞ and 0 v ∗ ≤ 0 u∗ for every t ≥ 0, then v 4 u. P P Let L f + hvn in∈N be a non-decreasing sequence in S(AL ) with supremum v. Then vn∗ ≤ v ∗ for each n, so (g) tells us that vn 4 u for every n. By 373N, for the last time, v 4 u. Q Q

(i) Finally, suppose that (A, µ ¯) and (B, ν¯) are arbitrary measure algebras and that u ∈ Mµ¯1,∞ , v ∈ Mν¯1,∞ Rt ∗ Rt ∗ are such that 0 v ≤ 0 u for every t ≥ 0. By (b), v 4 |v|; by (f), |v| 4 |v|∗ ; by 373Db, |v|∗ = v ∗ ; by (h) of this proof, v ∗ 4 u∗ ; by (d), u∗ = |u|∗ 4 |u|; and by (b) again, |u| 4 u. 373P Theorem Let (A, µ ¯) be a measure algebra and (B,Rν¯) a semi-finite algebra. Then for R ∗ measure 1,∞ 0 ∗ any u ∈ Mµ¯ and v ∈ Mν¯ , there is a T ∈ T = Tµ¯,¯ν such that T u × v = u × v .

373P

Decreasing rearrangements

433

proof (a) It is convenient to dispose immediately of some elementary questions. R R (i) We need only find a T ∈ T such that |T u × v| ≥ u∗ × v ∗ . P P Take v0 ∈ L∞ (B) such that 1,∞ |T u × v| = v0 × T u × v and kv0 k∞ ≤ 1, and set T1 w = v0 × T w for w ∈ Mµ¯ ; then T1 ∈ T and

R

T1 u × v =

R

R

|T u × v| ≥

u∗ × v ∗ ≥

R

T1 u × v

by 373J. Q Q (ii) Consequently it will be enough to consider v ≥ 0, since of course |v|∗ = v ∗ .

R

|T u × v| =

R

|T u × |v||, while

R R (iii) It will be enough to consider u = u∗ . P P If we can find T ∈ Tµ¯L ,¯ν such that T u∗ ×v = (u∗ )∗ ×v ∗ , then we know from 373O that there is an S ∈ Tµ¯,¯µL such that Su = u∗ , so that T S ∈ T and

R

T Su × v =

R

(u∗ )∗ × v ∗ =

R

u∗ × v ∗ . Q Q

(iv) It will be enough to consider localizable (B, ν¯). P P Assuming that v ≥ 0, following (ii) above, set e = [[v > 0]] = supn∈N [[v > 2−n ]], and let ν¯e be the restriction of ν¯ to the principal ideal Be generated by e. Then if we write v˜ for the member of L0 (Be ) corresponding to v (so that [[˜ v > α]] = [[v > α]] for every ∗ ∗ α > 0), v ˜ = v . Also (B , ν ¯ ) is σ-finite, therefore localizable. Now if we can find T ∈ Tµ¯,¯νe such that e e R R T u × v˜ = u∗ × v˜∗ , then ST will belong to Tµ¯,¯ν , where S : L0 (Be ) → L0 (B) is the canonical embedding defined by the formula [[Sw > α]] = [[w > α]] if α ≥ 0, = [[w > α]] ∪ (1 \ e) if α < 0, and

R

ST u × v =

R

T u × v˜ =

R

u∗ × v˜∗ =

R

u∗ × v ∗ . Q Q

(b) So let us suppose henceforth that µ ¯ = µ ¯L , u = u∗ is the equivalence class of a non-increasing non-negative function, v ≥ 0 and (B, ν¯) is localizable. For n, i ∈ N set bni = [[v > 2−n i]],

βni = ν¯bni ,

cni = bni \ bn,i+1 ,

γni = ν¯cni = βni − βn,i+1

(because βni < ∞ if i > 0; this is really where I use the hypothesis that v ∈ M 0 ). For n ∈ N set Kn = {i : i ≥ 1, γni > 0}, Tn w =

X ¡ 1 Z βni ¢ w χcni γni βn,i+1

i∈Kn

for w ∈ Mµ¯1,∞ ; this is defined in L0 (B) because Kn is countable and hcni ii∈N is disjoint. Of course Tn : L 1,∞ 0 Mµ¯L → L (B) is linear. If w ∈ L∞ (AL ) then

¯

kTn wk∞ = supi∈Kn ¯ and if w ∈ L1µ¯L then kTn wk1 =

P

¯ 1 R βni ¯

i∈Kn γ ni

1 γni

¯

βn,i+1

¯

R βni

βn,i+1

w¯ν¯cni =

w¯ ≤ kwk∞ ,

P i∈Kn

¯R βni ¯

βn,i+1

¯

w¯ ≤ kwk1 ;

so Tn w ∈ Mν¯1,∞ for every w ∈ Mµ¯1,∞ , and Tn ∈ T . It will be helpful to observe that L

R

cni

Tn w =

R βni

βn,i+1

w

whenever i ≥ 1, since if i ∈ / Kn then both sides are 0. Note next that for every n, i ∈ N, bni = bn+1,2i ,

βni = βn+1,2i ,

cni = cn+1,2i ∪ cn+1,2i+1 ,

γni = γn+1,2i + γn+1,2i+1 ,

434

Linear operators between function spaces

so that, for i ≥ 1,

R cni

R βni

Tn u =

βn,i+1

u=

R

373P

Tn+1 u.

cni

This means that if T is any cluster R point of hTn in∈N in T for the very R weak operator topology (and such a cluster point exists, by 373L), cmi T u must be a cluster point of h cmi Tn uin∈N , and therefore equal to R T u, for every m ∈ N, i ≥ 1. cmi m Consequently, if m ∈ N, Z |T u × v| ≥

∞ Z X

|T u| × v ≥ cmi

i=0

=

∞ X i=1 ∞ X

Z 2−m i|

T u| = cmi

Z −m

2

Z −m

2

i

∞ X

Z 2−m i

Tm u cmi

i=1

Z

βmi

i

|T u| cmi

i=0

(because cmi ⊆ [[v > 2−m i]]) ≥

∞ X

u≥

u × (v ∗ − 2−m χ1)+

βm,i+1

i=0

because [βm,i+1 , βmi ]• ⊆ [[v ∗ ≤ 2−m (i + 1)]] = [[(v ∗ − 2−m χ1)+ ≤ 2−m i]] for each i ∈ N. But letting m → ∞, we have

R

|T u × v| ≥ limm→∞

R

u × (v ∗ − 2−m χ1)+ =

R

u × v∗

because hu × (v ∗ − 2−m χ1)+ im∈N is a non-decreasing sequence with supremum u × v ∗ . In view of the reductions in (a) above, this is enough to complete the proof. 373Q Corollary Let (A, µ ¯) be a measure algebra, (B, ν¯) a semi-finite measure algebra, u ∈ Mµ¯1,∞ and 0,∞ v ∈ Mν¯ . Then

R

R

R

u∗ × v ∗ = sup{ |T u × v| : T ∈ Tµ¯,¯ν } = sup{ T u × v : T ∈ Tµ¯,¯ν }.

proof There is a non-decreasing sequence hcn in∈N in B such that ν¯cn < ∞ for every n and v ∗ = supn∈N (v × χcn )∗ . P P For each rational q > 0, we can find a countable non-empty set Bq ⊆ B such that b ⊆ [[|v| > q]], ν¯b < ∞ for every b ∈ Bq , supb∈Bq ν¯b = ν¯[[|v| > q]]

S (because (B, ν¯) is semi-finite). Let hbn in∈N be a sequence running over q∈Q,q>0 Bq and set cn = supi≤n bi , vn = v × χcn for each n. Then h|vn |in∈N and hvn∗ in∈N are non-decreasing and supn∈N vn∗ ≤ v ∗ in L0 (AL ). But in fact supn∈N vn∗ = v ∗ , because µ ¯L [[v ∗ > q]] = µ ¯[[|v| > q]] = supn∈N µ ¯[[vn > q]] = supn∈N µ ¯L [[vn∗ > q]] = µ ¯L [[supn∈N vn∗ > q]] for every rational q > 0, by 373Da. Q Q R R For each n ∈ N we have a Tn ∈ Tµ¯,¯ν such that Tn u × vn = u∗ × vn∗ (373P). Set Sn w = Tn w × χcn for n ∈ N, w ∈ Mµ¯1,∞ ; then every Sn belongs to Tµ¯,¯ν , so Z Z Z sup{ T u × v : T ∈ Tµ¯,¯µ } ≥ sup Sn u × v = sup Tn u × vn n∈N n∈N Z Z ∗ ∗ = sup u × vn = u∗ × v ∗ n∈N Z Z ≥ sup{ |T u × v| : T ∈ Tµ¯,¯µ } ≥ sup{ T u × v : T ∈ Tµ¯,¯µ } by 373J, as usual.

373S

Decreasing rearrangements

435

373R Order-continuous operators: Proposition Let (A, µ ¯) be a measure algebra, (B, ν¯) a local(0) izable measure algebra, and T0 ∈ T (0) = Tµ¯,¯ν . Then there is a T ∈ T × = Tµ¯×,¯ν extending T0 . If (A, µ ¯) is semi-finite, T is uniquely defined. proof (a) Suppose first that T0 ∈ T (0) is non-negative, regarded as a member of L∼ (Mµ¯1,0 ; Mν¯1,0 ). In this case T0 has an extension to an order-continuous positive linear operator T : Mµ¯1,∞ → L0 (B) defined by saying that T w = sup{T0 u : u ∈ Mµ¯1,0 , 0 ≤ u ≤ w} for every w ≥ 0 in Mµ¯1,∞ . P P I use 355F. Mµ¯1,0 is a solid 1,∞ linear subspace of Mµ¯ . T0 is order-continuous when its codomain is taken to be Mν¯1,0 , as noted in 371Gb, and therefore if its codomain is taken to be L0 (B), because M 1,0 is a solid linear subspace in L0 , so the embedding is order-continuous. If w ≥ 0 in Mµ¯1,∞ , let γ ≥ 0 be such that u1 = (w − γχ1)+ is integrable. If u ∈ Mµ¯1,0 and 0 ≤ u ≤ w, then (u − γχ1)+ ≤ u1 , so T0 u = T0 (u − γχ1)+ + T0 (u ∧ γχ1) ≤ T0 u1 + γχ1 ∈ L0 (B). Thus {T0 u : u ∈ Mν¯1,0 , 0 ≤ u ≤ w} is bounded above in L0 (B), for any w ≥ 0 in Mµ¯1,∞ . L0 (B) is Dedekind complete, because (B, ν¯) is localizable, so sup{T0 u : 0 ≤ u ≤ w} is defined in L0 (B); and this is true for every w ∈ (Mµ¯1,∞ )+ . Thus the conditions of 355F are satisfied and we have the result. Q Q (b) Now suppose that T0 is any member of T (0) . Then T0 has an extension to a member of T × . P P |T0 |, = 21 (|T0 | + T0 ) and T0− = 12 (|T0 | − T0 ), taken in L∼ (Mµ¯1,0 ; Mν¯1,0 ), all belong to T (0) (371G), so have extensions S, S1 and S2 to order-continuous positive linear operators from Mµ¯1,∞ to L0 (B) as defined in (a). Now for any w ∈ L1µ¯ , T0+

kSwk1 = k|T0 |wk1 ≤ kwk1 , ∞

and for any w ∈ L (A), |Sw| ≤ S|w| = sup{|T0 |u : u ∈ Mµ¯1,0 , 0 ≤ u ≤ w} ≤ kwk∞ χ1, so kSwk∞ ≤ kwk∞ . Thus S ∈ T ; similarly, S1 and S2 can be regarded as operators from Mµ¯1,∞ to Mν¯1,∞ , and as such belong to T . Next, for w ≥ 0 in Mµ¯1,∞ , S1 w + S2 w = sup{T0+ u : u ∈ Mµ¯1,0 , 0 ≤ u ≤ w} + sup{T0− u : u ∈ Mµ¯1,0 , 0 ≤ u ≤ w} = sup{T0+ u + T0− u : u ∈ Mµ¯1,0 , 0 ≤ u ≤ w} = Sw. But this means that S = S1 + S2 ≥ |S1 − S2 | and T = S1 − S2 ∈ T , by 373Bc; while of course T extends T0+ − T0− = T0 . Finally, because S1 and S2 are Q order-continuous, T ∈ L× (Mµ¯1,∞ ; Mν¯1,∞ ), so T ∈ T × . Q (c) If (A, µ ¯) is semi-finite, then Mµ¯1,0 is order-dense in Mµ¯1,∞ (because it includes L1µ¯ , which is order-dense in L0 (A)); so that the extension T is unique, by 355F(iii). 373S Adjoints in T (0) : Theorem Let (A, µ ¯) and (B, ν¯) be measure algebras, and T any member of R R (0) Then there is a unique operator T 0 ∈ Tν¯,¯µ such that a T 0 (χb) = b T (χa) for every a ∈ Af , b ∈ Bf , R R R and now u × T 0 v = T u × v whenever u ∈ Mµ¯1,0 , v ∈ Mν¯1,0 are such that u∗ × v ∗ < ∞. (0) Tµ¯,¯ν .

proof (a) For each v ∈ Mν¯1,0 we can define T 0 v ∈ Mµ¯1,0 by the formula

R

T 0v = a

R

R

T (χa) × v

R for every a ∈ Af . P P Set θa = T (χa) × v for each a ∈ Af ; because (χa)∗ × v ∗ < ∞, the R integral is defined and finite (373J). Of course θ : Af → R is additive because χ is additive and T , × and are linear. Also limt↓0 supµ¯a≤t |θa| ≤ limt↓0 1 t

Rt 0

limt→∞ supµ¯a≤t |θa| ≤ limt→∞

v ∗ = 0,

1 t

Rt 0

v∗ = 0

436

Linear operators between function spaces

373S

because v ∈ Mν¯1,0 , so v ∗ ∈ Mµ¯1,0 . By 373Ha, there is a unique T 0 v ∈ Mµ¯1,0 such that L f a∈A .Q Q

R a

T 0 v = θa for every (0)

(b) Because the formula uniquely determines T 0 v, we see that T 0 : Mν¯1,0 → Mµ¯1,0 is linear. Now T 0 ∈ Tν¯,¯µ . P P (i) If v ∈ L1ν¯ , then (because T 0 v ∈ Mµ¯1,0 ) |T 0 v| = supa∈Af |T 0 v| × χa, and Z Z Z Z kT 0 vk1 = |T 0 v| = sup |T 0 v| = sup ( T 0 v − T 0 v) a∈Af

Z

b,c∈Af

a

c

(χb − χc)∗ × v ∗

T (χb − χc) × v ≤ sup

= sup

b,c∈Af

b,c∈Af

Z =

b

Z

v ∗ = kvk1 .

(ii) Now suppose that v ∈ L∞ (B) ∩ Mν¯1,0 , and set γ = kvk∞ . ?? If a = [[|T 0 v| > γ]] 6= 0, then T 0 v 6= 0 so v 6= 0 and γ > 0 and µ ¯a < ∞, because T 0 v ∈ Mµ¯1,0 . Set b = [[(T 0 v)+ > γ]], c = [[(T 0 v)− > γ]]; then Z Z Z Z γµ ¯a < |T 0 v| = T 0 v − T 0 v = T (χb − χc) × v a

b

c

≤ γkT (χb − χc)k1 ≤ γkχb − χck1 = γ µ ¯a, which is impossible. X X Thus [[|T 0 v| > γ]] = 0 and kT 0 vk∞ ≤ γ = kvk∞ . (0) Q Putting this together with (i), we see that T 0 ∈ Tν¯,¯µ . Q (0)

(c) Let |T | be the modulus of T in L∼ (Mµ¯1,0 ; Mν¯1,0 ), so that |T | ∈ Tµ¯,¯µ , by 371Gb. If u ≥ 0 in Mµ¯1,0 , v ≥ 0 R sequence in S(Af )+ with supremum in Mν¯1,0 are such that u∗ × v ∗ < ∞, let R hun in∈N be a non-decreasing R u. In this case |T |u = supn∈N |T |un , so |T |u × v = supn∈N |T |un × v and |

R

as n → ∞, because

R

At the same time, because

R

Tu × v −

u × |T 0 v| ≤

R

R

|

R

R

T un × v| ≤

|T |u × v ≤

u × T 0v −

R

R

R

|T |(u − un ) × v → 0

u∗ × v ∗ < ∞.

un × T 0 v| ≤

R

(u − un ) × |T 0 v| → 0

u∗ × v ∗ < ∞. So

T u × v = limn→∞

R

T un × v = limn→∞

R

un × T 0 v =

R

u × T 0 v,

the middle equality being valid because each un is a linear of characteristic functions. R combination R Because T Rand T 0 are linear, it follows at once that u × T 0 v = T u × v whenever u ∈ Mµ¯1,0 , v ∈ Mν¯1,0 are such that u∗ × v ∗ < ∞. (d) Finally, to see that T 0 is uniquely defined by the formula in the statement of the theorem, observe that this surely defines T 0 (χb) for every b ∈ Bf , by the remarks in (a). Consequently it defines T 0 on S(Bf ). (0) Since S(Bf ) is order-dense in Mν¯1,0 , and any member of Tν¯,¯µ must belong to L× (Mν¯1,0 ; Mµ¯1,0 ) (371Gb), the restriction of T 0 to S(Bf ) determines T 0 (355J). 373T Corollary Let (A, µ ¯) and (B, ν¯) be localizable measure algebras. Then for any T ∈ Tµ¯×,¯ν there is a R R R unique T 0 ∈ Tν¯×,¯µ such that u×T 0 v = T u×v whenever u ∈ Mµ¯1,∞ , v ∈ Mν¯1,∞ are such that u∗ ×v ∗ < ∞. R (0) (0) proof The restriction T ¹Mµ¯1,0 belongs to Tµ¯,¯ν (373Bb), so there is a unique S ∈ Tν¯,¯µ such that u × Sv = R R T u × v whenever u ∈ Mµ¯1,0 , v ∈ Mν¯1,0 are such that u∗ × v ∗ < ∞ (373S). Now there is a unique T 0 ∈ Tν¯×,¯µ R R R extending S (373R). If u ≥ 0 in Mµ¯1,∞ , v ≥ 0 in Mν¯1,∞ are such that u∗ ×v ∗ < ∞, then u×T 0 v = T u×v. P P If T ≥ 0, then both are

R

sup{ u0 × T 0 v0 : u0 ∈ Mµ¯1,0 , v ∈ Mν¯1,0 , 0 ≤ u0 ≤ u, 0 ≤ v0 ≤ v}

373Xi

Decreasing rearrangements

437

because both T and T 0 are (order-)continuous. In general, we can apply the same argument to T + and T − , taken in L∼ (Mµ¯1,∞ ; Mν¯1,∞ ), since these belong to Tµ¯×,¯ν , by 373B and 355H, and we shall surely have R R T 0 = (T + )R0 − (T − )0 . Q Q As in 373S, it follows that u × T 0 v = T u × v whenever u ∈ Mµ¯1,∞ , v ∈ Mν¯1,∞ are such that u∗ × v ∗ < ∞. 373U Corollary Let (A, µ ¯) and (B, ν¯) be localizable measure algebras, and π : A → B an ordercontinuous measure-preserving Boolean homomorphism. Then the associated map T ∈ Tµ¯×,¯ν (373Bd) has an R adjoint P ∈ Tν¯×,¯µ defined by the formula a P (χb) = ν¯(b ∩ πa) for a ∈ Af , b ∈ Bf . proof The adjoint P = T 0 must have the property that

R

a

P (χb) =

R

χa × P (χb) =

R

T (χa) × χb =

R

χ(πa) × χb = ν¯(πa ∩ b)

for every a ∈ Af , b ∈ Bf . To see that this defines P uniquely, let S ∈ Tν¯×,¯µ be any other operator with the same property. By 373Hb, S(χb) = P (χb) for every b ∈ Bf , so S and P agree on S(Bf ). Because both P and S are supposed to belong to L× (Mν¯1,∞ ; Mµ¯1,∞ ), and S(Bf ) is order-dense in Mν¯1,∞ , S = P , by 355J. 373X Basic exercises (a) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a ring homomorphism such that ν¯πa ≤ µ ¯a for every a ∈ A. (i) Show that there is a unique T ∈ Tµ¯,¯ν such that T (χa) = χ(πa) for every a ∈ A, and that T is a Riesz homomorphism. (ii) Show that T is (sequentially) order-continuous iff π is. > (b) Let (A, µ ¯) and (B, µ ¯) be measure algebras, and φ : R → R a convex function such that φ(0) ≤ 0. ¯ u) ≤ T (φ(u)) ¯ ¯ Show that if T ∈ Tµ¯,¯ν and T ≥ 0, then φ(T whenever u ∈ Mµ¯1,∞ is such that φ(u) ∈ Mµ¯1,∞ . (Hint: 233J, 365Rb.) (c) Let (A, µ ¯) be a measure algebra. Show that if w ∈ L∞ (A) and kwk∞ ≤ 1 then u 7→ u × w : Mµ¯1,∞ → Mµ¯1,∞ belongs to Tµ¯×,¯µ . (d) Let (A, µ ¯) and (B, ν¯) be measure algebras. Show that if hai ii∈I , hbi ii∈I are disjoint families in A, B respectively, and hTi ii∈I is any family in Tµ¯,¯ν , and either I is countable or B is Dedekind complete, then we have an operator T ∈ Tµ¯,¯ν such that T u × χbi = Ti (u × χai ) × χbi for every u ∈ Mµ¯1,∞ ,¯ ν , i ∈ I. > (e) Let I, J be sets and write µ = µ ¯, ν = ν¯ for counting measure on I, J respectively. Show that there is a natural one-to-one correspondence between Tµ¯×,¯ν and the set of matrices haij ii∈I,j∈J such that P P j∈J |aij | ≤ 1 for every i ∈ I. i∈I |aij | ≤ 1 for every j ∈ J, >(f ) Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces, with measure algebras (A,R µ ¯) and (B, ν¯), and product measure λ on X ×RY . Let h : X × Y → R be a measurable function such that |h(x, y)|dx ≤ 1 for ν-almost every y ∈ Y and |h(x, y)|dy ≤ 1 for µ-almost every x ∈ X. Show that R there is a corresponding T ∈ Tµ¯×,¯ν defined by writing T (f • ) = g • whenever f ∈ L1 (µ) + L∞ (µ) and g(y) = h(x, y)f (x)dx for almost every y. > (g) Let µ beR Lebesgue measure on R, and (A, µ ¯) its measure algebra. Show that for any µ-integrable function h with |h|dµ ≤ 1 we have a corresponding T ∈ Tµ¯×,¯µ defined by setting T (f • ) = (h ∗ f )• whenever g ∈ L1 (µ) + L∞ (µ), writing h ∗ f for the convolution of h and f (255E). Explain how this may be regarded as a special case of 373Xf. > (h) Let (Ω, Σ, µ) be a probability space and X ∈ L0 (µ) a non-negative real-valued random variable on Ω; let νX be its distribution (271C). Write u = X • ∈ L0 (µ) ∼ ¯) is the measure algebra of = L0µ¯ , where (A, µ ∗ ∗ (Ω, Σ, µ). Show that µ ¯L [[u > α]] = Pr(X > α) for every α, so that each of u , νX is uniquely determined by the other. (i) Let (A, µ ¯) and (B, ν¯) be measure algebras, and π : A → B a measure-preserving Boolean homomorphism; let T : Mµ¯1,∞ → Mν¯1,∞ be the corresponding operator (373Bd). Show that (T u)∗ = u∗ for every u ∈ Mµ¯1,∞ .

438

Linear operators between function spaces

373Xj

(j) Let (A, µ ¯) be a totally finite measure algebra, and A a subset of L1µ¯ . Show that the following are equiveridical: (i) A is uniformly integrable; (ii) {u∗ : u ∈ A} is uniformly integrable in L1µ¯L ; (iii) Rt limt↓0 supu∈A 0 u∗ = 0. (l) Let (A, µ ¯) be a measure algebra, and A ⊆ (Mµ¯0 )+ a non-empty downwards-directed set. Show that ∗ (inf A) = inf u∈A u∗ in L0 (AL ). (k) Let (A, µ ¯) be a measure algebra. Show that kuk1,∞ =

R1 0

u∗ for every u ∈ M 1,∞ (A, µ ¯).

(m) Let (A, µ ¯) and (B, ν¯) be measure algebras, and φ a Young’s function (369Xc). Write Uφ,¯µ ⊆ L0 (A), 0 Uφ,¯ν ⊆ L (B) for the corresponding Orlicz spaces. (i) Show that if T ∈ Tµ¯,¯ν and u ∈ Uφ,¯µ , then T u ∈ Uφ,¯ν and kT ukφ ≤ kukφ . (ii) Show that u ∈ Uφ,¯µ iff u∗ ∈ Uφ,¯µL , and in this case kukφ = ku∗ kφ . > (n) Let (A, µ ¯) be a measure algebra and (B, ν¯) a totally finite measure algebra. Show that if A ⊆ L1µ¯ is uniformly integrable, then {T u : u ∈ A, T ∈ Tµ¯,¯ν } is uniformly integrable in L1ν¯ . (o) (i) Give examples of u, v ∈ L1 (AL ) such that (u + v)∗ 6≤ u∗ + v ∗ . (ii) Show that if (A, µ ¯) is any Rt Rt measure algebra and u, v ∈ Mµ¯0,∞ , then 0 (u + v)∗ ≤ 0 u∗ + v ∗ for every t ≥ 0. (p) Let (A, µ ¯) and (B, ν¯) be two measure algebras. For u ∈ Mµ¯1,0 , w ∈ Mν¯∞,1 set

R

(0)

ρuw (S, T ) = | (Su − T u) × w| for all S, T ∈ T (0) = Tµ¯,¯ν . The topology generated by the pseudometrics ρuw is the very weak operator topology on T (0) . Show that T (0) is compact in this topology. (0)

(q) Let (A, µ ¯) and (B, ν¯) be measure algebras and let u ∈ Mµ¯1,0 . (i) Show that B = {T u : T ∈ Tµ¯,¯ν } is compact for the topology Ts (Mν¯1,0 , Mν¯∞,1 ). (ii) Show that any non-decreasing sequence in B has a supremum in L0 (B) which belongs to B. (r) Let (A, µ ¯) and (B, ν¯) be measure algebras, and u ∈ Mµ¯1,0 , v ∈ Mν¯1,0 . Show that the following are Rt Rt (0) equiveridical: (i) there is a T ∈ Tµ¯,¯ν such that T u = v; (ii) 0 u∗ ≤ 0 v ∗ for every t ≥ 0. (s) Let (A, µ ¯) and (B, ν¯) be measure algebras. Suppose that u1 , u2 ∈ Mµ¯1,∞ and v ∈ Mν¯1,∞ are such Rt ∗ Rt that 0 v ≤ 0 (u1 + u2 )∗ for every t ≥ 0. Show that there are v1 , v2 ∈ Mν¯1,∞ such that v1 + v2 = v and Rt ∗ Rt ∗ v ≤ 0 ui for both i, every t ≥ 0. 0 i • • ∞ R > (t) Set g(t) = t/(t + 1) for t ≥ 0, and set v = g , u = χ[0, 1] ∈ L (AL ). Show that T u × v for every T ∈ Tµ¯L ,¯µL .

(0)

R

u∗ × v ∗ = 1 >

(0)

(u) Let (A, µ ¯) and (B, ν¯) be measure algebras, and for T ∈ Tµ¯,¯ν define T 0 ∈ Tν¯,¯µ as in 373S. Show that T = T. 00

(0)

(0)

(v) Let (A, µ ¯) and (B, ν¯) be measure algebras, and give Tµ¯,¯ν , Tν¯,¯µ their very weak operator topologies (0) (0) (373Xp). Show that the map T 7→ T 0 : Tµ¯,¯ν → Tν¯,¯µ is an isomorphism for the convex, order and topological structures of the two spaces. (By the ‘convex structure’ of a convex set C in a linear space I mean the operation (x, y, t) 7→ tx + (1 − t)y : C × C × [0, 1] → C.) 373Y Further exercises (a) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1]. Set u = f • and v = g • in L0 (A), where f (t) = t, g(t) = 1 − 2|t − 21 | for t ∈ [0, 1]. Show that u∗ = v ∗ , but that there is no measure-preserving Boolean homomorphism π : A → A such that Tπ v = u, writing Tπ : L0 (A) → L0 (A) for the operator induced by π, as in 364R. (Hint: show that {[[v > α]] : α ∈ R} does not τ -generate A.)

373 Notes

Decreasing rearrangements

439

(b) Let (A, µ ¯) be a totally finite homogeneous measure algebra of uncountable Maharam type. Let u, v ∈ (Mµ¯1,∞ )+ be such that u∗ = v ∗ . Show that there is a measure-preserving automorphism π : A → A such that Tπ u = v. Rt Rt (c) Let u, v ∈ Mµ¯1,∞ be such that u = u∗ , v = v ∗ and 0 v ≤ 0 u for every t ≥ 0. Show that there is L Rt Rt a non-negative T ∈ Tµ¯L ,¯µL such that T u = v and 0 T w ≤ 0 w for every w ≥ 0 in M 1,∞ . Show that any such T must belong to Tµ¯×L ,¯µL . (d) Let (A, µ ¯) and (B, ν¯) be measure algebras, and u ∈ Mµ¯1,∞ . (i) Suppose that w ∈ S(Bf ). Show directly (without quoting the result of 373O, but R R possibly using some of the ideas of the proof) that for every γ < u∗ ×w∗ there is a T ∈ Tµ¯,¯ν such that T u×w ≥ γ. (ii) Suppose that (B, ν¯) is localizable and that R R v ∈ Mν¯1,∞ \ {T u : T ∈ Tµ¯,¯ν }. Show that there is a w ∈ S(Bf ) such that v × w > sup{ T u × w : T ∈ Tµ¯,¯ν }. (Hint: use 373M and the Hahn-Banach theorem in the following form: if U is a linear space with the topology Ts (U, V ) defined by a linear subspace V of L(U ; R), C ⊆ U is a non-empty closed convex set, and v ∈ U \ C, then there is an f ∈ V such that f (v) > supu∈C f (u).) (iii) Hence prove 373O for localizable (B, ν¯). (iv) Now prove 373O for general (B, ν¯). (e) (i) Define v ∈ L∞ (AL ) as in 373Xt. Show that there is no T ∈ Tµ¯×L ,¯µL such that T v = v ∗ . (ii) Set h(t) = 1 + max(0, sint t ) for t > 0, w = h• ∈ L∞ (AL ). Show that there is no T ∈ Tµ¯×L ,¯µL such that T w∗ = w. (f ) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1]. Show that Tµ¯,¯µL = Tµ¯×,¯µL can be identified, as convex ordered space, with Tµ¯×L ,¯µ , and that this is a proper subset of Tµ¯L ,¯µ . (g) Show that the adjoint operation of 373T is not as a rule continuous for the very weak operator topologies of Tµ¯×,¯ν , Tν¯×,¯µ . 373 Notes and comments 373A-373B are just alternative expressions of concepts already treated in 371F371H. My use of the simpler formula Tµ¯,¯ν symbolizes my view that T , rather than T (0) or T × , is the most natural vehicle for these ideas; I used T (0) in §§371-372 only because that made it possible to give theorems which applied to all measure algebras, without demanding localizability (compare 371Gb with 373Bc). The obvious examples of operators in T are those derived from measure-preserving Boolean homomorphisms, as in 373Bd, and their adjoints (373U). Note that the latter include conditional expectation operators. In return, we find that operators in T share some of the characteristic properties of the operators derived from Boolean homomorphisms (373Bb, 373Xb, 373Xm). Other examples are multiplication operators (373Xc), operators obtained by piecing others together (373Xd) and kernel operators of the type described in 373Xe-373Xf, including convolution operators (373Xg). (For a general theory of kernel operators, see §376 below.) Most of the section is devoted to the relationships between the classes T of operators and the ‘decreasing rearrangements’ of 373C. If you like, the decreasing rearrangement u∗ of u describes the ‘distribution’ of |u| (373Xh); but for u ∈ / M 0 it loses some information (373Xt, 373Ye). It is important to be conscious that 0 even when u ∈ L (AL ), u∗ is not necessarily obtained by ‘rearranging’ the elements of the algebra AL by a measure-preserving automorphism (which would, of course, correspond to an automorphism of the measure space ([0, ∞[ , µL ), by 344C). I will treat ‘rearrangements’ of this narrower type in the next section; for the moment, see 373Ya. Apart from this, the basic properties of decreasing rearrangements are straightforward enough (373D-373F). The only obscure area concerns the relationship between (u + v)∗ and u∗ , v ∗ (see 373Xo). In 373G I embark on results involving both decreasing rearrangements and operators in T , leading to the characterization of the sets {T u : T ∈ T } in 373O. In one direction this is easy, and is the content of 373G. In the other direction it depends on a deeper analysis, and the easiest method seems to be through studying the ‘very weak operator topology’ on T (373K-373L), even though this is an effective tool only when one of the algebras involved is localizable (373L). A functional analyst is likely to feel that the method is both natural and illuminating; but from the point of view of a measure theorist it is not perfectly satisfactory, because it is essentially non-constructive. While it tells us that there are operators T ∈ T acting in the required ways, it gives only the vaguest of hints concerning what they actually look like.

440

Linear operators between function spaces

373 Notes

Of course the very weak operator topology is interesting in its own right; and see R t also 373Xp-373Xq. Rt The proof of 373O can be thought of as consisting of three steps. Given that 0 v ∗ ≤ 0 u∗ for every t, then I set out to show that v is expressible as T1 v ∗ (parts (c)-(d) of the proof), that v ∗ is expressible as T2 u∗ (part (g)) and that u∗ is expressible as T3 u (parts (e)-(f)), each Ti belonging to an appropriate T . In all three steps the general case follows easily from the case of u, v ∈ S(A), S(B). If we are willing to use a more sophisticated version of the Hahn-Banach theorem than those given in 3A5A and 363R, there is an alternative route (373Yd). I note that the central step above, from u∗ to v ∗ , can be performed with an order-continuous T2 (373Yc), but that in general neither of the other steps can (373Ye), so that we cannot use T × in place of T here. A companion result to 373O, in that it also shows thatR {T u : T ∈ T } is large enough to reach natural bounds, is 373P; given u and v, we can find T such that T u × v is as large as possible. In this form the result is valid only for v ∈ M (0) (373Xt). But if we do not demand that the supremum should be attained, we can deal with other v (373Q). We already know that every operator in T (0) is a difference of order-continuous operators, just because M 1,0 has an order-continuous norm (371Gb). It is therefore not surprising that members of T (0) can be extended to members of T × , at least when the codomain Mν¯1,∞ is Dedekind complete (373R). It is also very natural to look for a correspondence between Tµ¯,¯ν and Tν¯,¯µ , because if T ∈ Tµ¯,¯ν we shall surely have an adjoint operator (T ¹L1µ¯ )0 from (L1ν¯ )∗ to (L1µ¯ )∗ , and we can hope that this will correspond to some member of Tν¯,¯µ . But when we come to the details, the normed-space properties of a general member of T are not enough (373Yf), and we need some kind of order-continuity. For members of T (0) this is automatically present (373S), and now the canonical isomorphism between T (0) and T × gives us an isomorphism between Tµ¯×,¯ν and Tν¯×,¯µ when µ ¯ and ν¯ are localizable (373T).

374 Rearrangement-invariant spaces As is to be expected, many of the most important function spaces of analysis are symmetric in various ways; in particular, they share the symmetries of the underlying measure algebras. The natural expression of this is to say that they are ‘rearrangement-invariant’ (374E). In fact it turns out that in many cases they have the stronger property of ‘T -invariance’ (374A). In this section I give a brief account of the most important properties of these two kinds of invariance. In particular, T -invariance is related to a kind of transfer mechanism, enabling us to associate function spaces on different measure algebras (374B-374D). As for rearrangement-invariance, the salient fact is that on the most important measure algebras many rearrangement-invariant spaces are T -invariant (374K, 374M). 374A T -invariance: Definitions Let (A, µ ¯) be a measure algebra. Recall that I write Mµ¯1,∞ = L1µ¯ + L∞ (A) ⊆ L0 (A), Mµ¯∞,1 = L1µ¯ ∩ L∞ (A), Mµ¯0,∞ = {u : u ∈ L0 (A), inf α>0 µ ¯[[|u| > α]] < ∞}, (369N, 373C). (a) I will say that a subset A of Mµ¯1,∞ is T -invariant if T u ∈ A whenever u ∈ A and T ∈ T = Tµ¯,¯µ (definition: 373Aa). (b) I will say that an extended Fatou norm τ on L0 is T -invariant or fully symmetric if τ (T u) ≤ τ (u) whenever u ∈ Mµ¯1,∞ and T ∈ T . (c) As in §373, I will write (AL , µ ¯L ) for the measure algebra of Lebesgue measure on [0, ∞[, and u∗ ∈ Mµ¯0,∞ L for the decreasing rearrangement of any u belonging to any Mµ¯0,∞ (373C).

374C

374B

Rearrangement-invariant spaces

441

The first step is to show that the associate of a T -invariant norm is T -invariant.

Theorem Let (A, µ ¯) be a semi-finite measure algebra and τ a T -invariant extended Fatou norm on L0 (A). τ Let L be the Banach lattice defined from τ (369G), and τ 0 the associate extended Fatou norm (369H-369I). Then (i) Mµ¯∞,1 ⊆ Lτ ⊆ Mµ¯1,∞ ; R (ii) τ 0 is also T -invariant, and u∗ × v ∗ ≤ τ (u)τ 0 (v) for all u, v ∈ Mµ¯0,∞ . P Take any u ∈ L0 (A) \ Mµ¯0,∞ . There is surely some w > 0 in Lτ , proof (a) I check first that Lτ ⊆ Mµ¯0,∞ . P and we can suppose that w = χa for some a of finite measure. Now, for any n ∈ N, (|u| ∧ nχ1)∗ = nχ1 ≥ nw∗ in L0 (AL ), because µ ¯[[|u| > n]] = ∞. So there is a T ∈ Tµ¯,¯µ such that T (|u| ∧ nχ1) = nw, by 373O, and τ (u) ≥ τ (|u| ∧ nχ1) ≥ τ (T (|u| ∧ nχ1)) = τ (nw) = nτ (w). As n is arbitrary, τ (u) = ∞. As u is arbitrary, Lτ ⊆ Mµ¯0,∞ . Q Q R ∗ (b) Next, u × v ∗ ≤ τ (u)τ 0 (v) for every u, v ∈ Mµ¯0,∞ . P P If u ∈ Mµ¯1,∞ , then Z

Z u∗ × v ∗ = sup{

|T u × v| : T ∈ Tµ¯,¯µ }

(373Q) ≤ sup{τ (T u)τ 0 (v) : T ∈ Tµ¯,¯µ } = τ (u)τ 0 (v). Generally, setting un = |u| ∧ nχ1, hu∗n in∈N is a non-decreasing sequence with supremum u∗ (373Db, 373Dh), so

R

u∗ × v ∗ = supn∈N

R

Q u∗n × v ∗ ≤ supn∈N τ (un )τ 0 (v) = τ (u)τ 0 (v). Q

P If A = {0}, this is trivial. Otherwise, take u ∈ Lτ . There is surely (c) Consequently, Lτ ⊆ Mµ¯1,∞ . P 0 some non-zero a such that τ (χa) < ∞; now, setting v = χa,

R µ¯a 0

u∗ =

R

u∗ × v ∗ ≤ τ (u)τ 0 (v) < ∞

Q by (b) above. But this means that u∗ ∈ Mµ¯1,∞ , so that u ∈ Mµ¯1,∞ (373F(b-ii)). Q (d) Next, τ 0 is T -invariant. P P Suppose that v ∈ Mµ¯0,∞ , T ∈ Tµ¯,¯µ , u ∈ L0 (A) and τ (u) ≤ 1. Then u ∈ Mµ¯1,∞ , by (c), so

R

|u × T v| ≤

R

u∗ × v ∗ ≤ τ (u)τ 0 (v) ≤ τ 0 (v),

using 373J for the first inequality. Taking the supremum over u, we see that τ 0 (T v) ≤ τ 0 (v); as T and v are arbitrary, τ 0 is T -invariant. Q Q 0

(e) Finally, putting (d) and (c) together, Lτ ⊆ Mµ¯1,∞ , so that Lτ ⊇ Mµ¯∞,1 , using 369J and 369O. 374C For any T -invariant extended Fatou norm on L0 (AL ) there are corresponding norms on L0 (A) for any semi-finite measure algebra, as follows. Theorem Let θ be a T -invariant extended Fatou norm on L0 (AL ), and (A, µ ¯) a semi-finite measure algebra. (a) There is a T -invariant extended Fatou norm τ on L0 (A) defined by setting τ (u) = θ(u∗ ) if u ∈ Mµ¯0,∞ , = ∞ if u ∈ L0 (A) \ Mµ¯0,∞ . (b) Writing θ0 , τ 0 for the associates of θ and τ , we now have

442

Linear operators between function spaces

374C

τ 0 (v) = θ0 (v ∗ ) if v ∈ Mµ¯0,∞ , = ∞ if v ∈ L0 (A) \ Mµ¯0,∞ . (c) If θ is an order-continuous norm on the Banach lattice Lθ , then τ is an order-continuous norm on Lτ . proof (a)(i) The argument seems to run better if I use a different formula to define τ : set

R

τ (u) = sup{ |u × T w| : T ∈ Tµ¯L ,¯µ , w ∈ L0 (AL ), θ0 (w) ≤ 1} for u ∈ L0 (A). (By 374B, w ∈ Mµ¯1,∞ whenever θ0 (w) ≤ 1, so there is no difficulty in defining T w.) Now τ (u) = θ(u∗ ) for every u ∈ Mµ¯0,∞ . P P (α) If w ∈ L0 (AL ) and θ0 (w) ≤ 1, then w ∈ Mµ¯1,∞ , by 374B(i), so L there is an S ∈ Tµ¯L ,¯µL such that Sw = w∗ (373O). Accordingly θ0 (w∗ ) ≤ θ0 (w) (because θ0 is T -invariant, by 374B); now

R

|u × T w| ≤

R

u∗ × w∗ ≤ θ(u∗ )θ0 (w∗ ) ≤ θ(u∗ )θ0 (w) ≤ θ(u∗ );

as w is arbitrary, τ (u) ≤ θ(u∗ ). (β) If w ∈ L0 (AL ) and θ0 (w) ≤ 1, then Z

Z ∗

(u∗ )∗ × w∗

|u × w| ≤ (373E)

Z

Z u × w = sup{ |u × T w| : T ∈ Tµ¯L ,¯µL } ∗

=

∗

(373Q) ≤ τ (u). But because θ is the associate of θ0 (369I(ii)), this means that θ(u∗ ) ≤ τ (u). Q Q (ii) Now τ is an extended Fatou norm on L0 (A). P P Of the conditions in 369F, (i)-(iv) are true just R because τ (u) = supv∈B |u × v| for some set B ⊆ L0 . As for (v) and (vi), observe that if u ∈ Mµ¯∞,1 then u∗ ∈ Mµ¯∞,1 (373F(b-iv)), so that τ (u) = θ(u∗ ) < ∞, by 374B(i), while also L u 6= 0 =⇒ u∗ 6= 0 =⇒ τ (u) = θ(u∗ ) > 0. ¯) is semi-finite), 369F(v)-(vi) As Mµ¯∞,1 is order-dense in L0 (A) (this is where I use the hypothesis that (A, µ are satisfied, and τ is an extended Fatou norm. Q Q (iii) τ is T -invariant. P P Take u ∈ Mµ¯1,∞ and T ∈ Tµ¯,¯µ . There are S0 ∈ Tµ¯L ,¯µ and S1 ∈ Tµ¯,¯µL such that S0 u∗ = u, S1 T u = (T u)∗ (373O); now S1 T S0 ∈ Tµ¯L ,¯µL (373Be), so τ (T u) = θ((T u)∗ ) = θ(S1 T S0 u∗ ) ≤ θ(u∗ ) = τ (u) because θ is T -invariant. Q Q (iv) We can now return to the definition of τ . I have already remarked that τ (u) = θ(u∗ ) if u ∈ Mµ¯0,∞ . For other u, we must have τ (u) = ∞ just because τ is a T -invariant extended Fatou norm (374B(i)). So the definitions in the statement of the theorem and (i) above coincide. (b) We surely have τ 0 (v) = ∞ if v ∈ L0 (A) \ Mµ¯0,∞ , by 374B, because τ 0 , like τ , is a T -invariant extended Fatou norm. So take v ∈ Mµ¯0,∞ . (i) If u ∈ L0 (A) and τ (u) ≤ 1, then

R

|v × u| ≤

R

v ∗ × u∗ ≤ θ0 (v ∗ )θ(u∗ ) = θ0 (v ∗ )τ (u) ≤ θ0 (v ∗ );

as u is arbitrary, τ 0 (v) ≤ θ0 (v ∗ ). (ii) If w ∈ L0 (AL ) and θ(w) ≤ 1, then

374D

Rearrangement-invariant spaces

Z

Z

443

Z

∗

∗

|v × w| ≤

∗

v × w = sup{

|v × T w| : T ∈ Tµ¯L ,¯µ }

(373Q) ≤ sup{τ 0 (v)τ (T w) : T ∈ Tµ¯L ,¯µ } = sup{τ 0 (v)θ((T w)∗ ) : T ∈ Tµ¯L ,¯µ } ≤ sup{τ 0 (v)θ(ST w) : T ∈ Tµ¯L ,¯µ , S ∈ Tµ¯,¯µL } (because, given T , we can find an S such that ST w = (T w)∗ , by 373O) ≤ sup{τ 0 (v)θ(T w) : T ∈ Tµ¯L ,¯µL } ≤ τ 0 (v). As w is arbitrary, θ0 (v ∗ ) ≤ τ 0 (v) and the two are equal. This completes the proof of (b). (c)(i) The first step is to note that Lτ ⊆ Mµ¯0 . P P?? Suppose that u ∈ Lτ \Mµ¯0 , that is, that µ ¯[[|u| > α]] = ∞ • ∗ 0 ∞ θ for some α > 0. Then u ≥ αχ1 in L (AL ), so L (AL ) ⊆ L . For each n ∈ N, set vn = χ [n, ∞[ . Then ∗ vn = v0 , so we can find a Tn ∈ Tµ¯L ,¯µL such that Tn vn = v0 (373O), and θ(vn ) ≥ θ(v0 ) for every n. But as hvn in∈N is a decreasing sequence with infimum 0, this means that θ is not an order-continuous norm. X XQ Q (ii) Now suppose that A ⊆ Lτ is non-empty and downwards-directed and has infimum 0. Then inf u∈A µ ¯[[u > α]] = 0 for every α > 0 (put 364Nb and 321F together). But this means that B = {u∗ : u ∈ A} must have infimum 0; since B is surely downwards-directed, inf v∈B θ(v) = 0, that is, inf u∈A τ (u) = 0. As A is arbitrary, τ is an order-continuous norm. 374D

What is more, every T -invariant extended Fatou norm can be represented in this way.

Theorem Let (A, µ ¯) be a semi-finite measure algebra, and τ a T -invariant extended Fatou norm on L0 (A). Then there is a T -invariant extended Fatou norm θ on L0 (AL ) such that τ (u) = θ(u∗ ) for every u ∈ Mµ¯0,∞ . proof I use the method of 374C. If A = {0} the result is trivial; assume that A 6= {0}. (a) Set

R

θ(w) = sup{ |w × T v| : T ∈ Tµ¯,¯µL , v ∈ L0 (A), τ 0 (v) ≤ 1} for w ∈ L0 (AL ). Note that

R

θ(w) = sup{ w∗ × v ∗ : v ∈ L0 (A), τ 0 (v) ≤ 1} , by 373J and 373Q again. for every w ∈ Mµ¯0,∞ L θ is an extended Fatou norm on L0 (AL ). P P As in 374C, the conditions 369F(i)-(iv) are Relementary. If w > 0 in L0 (AL ), take any v ∈ L0 (A) such that 0 < τ 0 (v) ≤ 1; then w∗ × v ∗ 6= 0 so θ(w) ≥ w∗ × v ∗ > 0. So 369F(v) is satisfied. As for 369F(vi), if w > 0 in L0 (AL ), take a non-zero a ∈ A of finite measure such that α = τ (χa) < ∞. Let β > 0, b ∈ AL be such that 0 < µ ¯L b ≤ µ ¯a and βχb ≤ w; then θ(χb) = supτ 0 (v)≤1

R

(χb)∗ × v ∗ ≤ supτ 0 (v)≤1

R

(χa)∗ × v ∗ ≤ τ (χa) < ∞

by 374B(ii). So θ(βχb) < ∞ and 369F(vi) is satisfied. Thus θ is an extended Fatou norm. Q Q (b) θ is T -invariant. P P If T ∈ Tµ¯L ,¯µL and w ∈ Mµ¯1,∞ , then L θ(T w) = supτ 0 (v)≤1

R

(T w)∗ × v ∗ ≤ supτ 0 (v)≤1

R

w∗ × v ∗ = θ(w)

by 373G and 373I. Q Q (c) θ(u∗ ) = τ (u) for every u ∈ Mµ¯0,∞ . P P We have τ (u) = supτ 0 (v)≤1

R

|u × v| ≤ supτ 0 (v)≤1

using 369I, 373E and 374B. So θ(u∗ ) = supτ 0 (v)≤1 by the remark in (a) above. Q Q

R

R

u∗ × v ∗ ≤ τ (u),

u∗ × v ∗ = τ (u)

444

Linear operators between function spaces

374E

374E

I turn now to rearrangement-invariance. Let (A, µ ¯) be a measure algebra.

(a) I will say that a subset A of L0 = L0 (A) is rearrangement-invariant if Tπ u ∈ A whenever u ∈ A and π : A → A is a measure-preserving Boolean automorphism, writing Tπ : L0 → L0 for the isomorphism corresponding to π (364R). (b) I will say that an extended Fatou norm τ on L0 is rearrangement-invariant if τ (Tπ u) = τ (u) whenever u ∈ L0 and π : A → A is a measure-preserving automorphism. 374F Remarks (a) If (A, µ ¯) is a semi-finite measure algebra and π : A → A is a sequentially ordercontinuous measure-preserving Boolean homomorphism, then Tπ ¹Mµ¯1,∞ belongs to Tµ¯,¯µ ; this is obvious from the definition of M 1,∞ = L1 + L∞ and the basic properties of Tπ (364R). Accordingly, any T invariant extended Fatou norm τ on L0 (A) must be rearrangement-invariant, since (by 374B) we shall have τ (u) = τ (Tπ (u)) = ∞ when u ∈ / Mµ¯1,∞ . Similarly, any T -invariant subset of Mµ¯1,∞ will be rearrangementinvariant. (b) I seek to describe cases in which rearrangement-invariance implies T -invariance. This happens only for certain measure algebras; in order to shorten the statements of the main theorems I introduce a special phrase. 374G Definition I say that a measure algebra (A, µ ¯) is quasi-homogeneous if for any non-zero a, b ∈ A there is a measure-preserving Boolean automorphism π : A → A such that πa ∩ b 6= 0. 374H Proposition Let (A, µ ¯) be a semi-finite measure algebra. Then the following are equiveridical: (i) (A, µ ¯) is quasi-homogeneous; (ii) either A is purely atomic and every atom of A has the same measure or there is a κ ≥ ω such that the principal ideal Aa is homogeneous, with Maharam type κ, for every a ∈ A of non-zero finite measure. proof (i)⇒(ii) Suppose that (A, µ ¯) is quasi-homogeneous. α) Suppose that A has an atom a. In this case, for any b ∈ A \ {0} there is an automorphism π of (α (A, µ ¯) such that πa ∩ b 6= 0; now πa must be an atom, so πa = πa ∩ b and πa is an atom included in b. As b is arbitrary, A is purely atomic; moreover, if b is an atom, then it must be equal to πa and therefore of the same measure as a, so all atoms of A have the same measure. β ) Now suppose that A is atomless. In this case, if a ∈ A has finite non-zero measure, Aa is homo(β geneous. P P?? Otherwise, there are non-zero b, c ⊆ a such that the principal ideals Ab , Ac are homogeneous and of different Maharam types, by Maharam’s theorem (332B, 332H). But now there is supposed to be an automorphism π such that πb ∩ c 6= 0, in which case Ab , Aπb , Aπb∩c and Ac must all have the same Maharam type. X XQ Q Consequently, if a, b ∈ A are both of non-zero finite measure, the Maharam types of Aa , Aa∪b and Ab must all be the same infinite cardinal κ. (ii)⇒(i) Assume (ii), and take a, b ∈ A \ {0}. If a ∩ b 6= 0 we can take π to be the identity automorphism and stop. So let us suppose that a ∩ b = 0. α) If A is purely atomic and every atom has the same measure, then there are atoms a0 ⊆ a, b0 ⊆ b. (α Set πc = c if c ⊇ a0 ∪ b0 or c ∩ (a0 ∪ b0 ) = 0, = c 4 (a0 ∪ b0 ) otherwise. Then it is easy to check that π is a measure-preserving automorphism of A such that πa0 = b0 , so that πa ∩ b 6= 0. β ) If Ac is Maharam-type-homogeneous with the same infinite Maharam type κ for every non-zero (β c of finite measure, set γ = min(1, µ ¯a, µ ¯b) > 0. Because A is atomless, there are a0 ⊆ a, b0 ⊆ b with µ ¯ a0 = µ ¯b0 = γ (331C). Now Aa0 , Ab0 are homogeneous with the same Maharam type and the same

374J

Rearrangement-invariant spaces

445

magnitude, so by Maharam’s theorem (331I) there is a measure-preserving isomorphism π0 : Aa0 → Ab0 . Define π : A → A by setting πc = (c \ (a0 ∪ b0 )) ∪ π0 (c ∩ a0 ) ∪ π0−1 (c ∩ b0 ); then it is easy to see that π is a measure-preserving automorphism of A and that πa ∩ b 6= 0. Remark We shall return to these ideas in Chapter 38. In particular, the construction of π from π0 in the ←−−−− last part of the proof will be of great importance; in the language of 381G, π = (a0 π0 b0 ). 374I Corollary Let (A, µ ¯) be a quasi-homogeneous semi-finite measure algebra. Then (a) whenever a, b ∈ A have the same finite measure, the principal ideals Aa , Ab are isomorphic as measure algebras; (b) there is a subgroup Γ of the additive group R such that (α) µ ¯a ∈ Γ whenever a ∈ A and µ ¯a < ∞ (β) whenever a ∈ A, γ ∈ Γ and 0 ≤ γ ≤ µ ¯a then there is a c ⊆ a such that µ ¯c = γ. proof If A is purely atomic, with all its atoms of measure γ0 , set Γ = γ0 Z, and the results are elementary. If A is atomless, set Γ = R; then (a) is a consequence of Maharam’s theorem, and (b) is a consequence of 331C, already used in the proof of 374H. 374J Lemma Let (A, µ ¯) be a quasi-homogeneous semi-finite measure algebra and u, v ∈ Mµ¯0,∞ . Let Aut be the group of measure-preserving automorphisms of A. Then

R

u∗ × v ∗ = supπ∈Aut

R

|u × Tπ v|,

where Tπ : L0 (A) → L0 (A) is the isomorphism corresponding to π. f proof (a) Suppose first that u, v are non-negativeP and belong to S(AP ), where Af is the ring {a : µ ¯a < ∞}, n m β χb where α ≥ . . . αm ≥ 0, α χa , v = as usual. Then they can be expressed as u = j j 0 i i j=0 i=0 β0 ≥ . . . ≥ βn ≥ 0, a0 , . . . , am are disjoint and of finite measure, and b0 , . . . , bn are disjoint and of finite measure. Extending each list by a final term having a coefficient of 0, if need be, we may suppose that supi≤m ai = supj≤n bj . Let (t0 , . . . , ts ) enumerate in ascending order the set Pk Pk ¯bj : k ≤ n}. ¯ai : k ≤ m} ∪ { j=0 µ {0} ∪ { i=0 µ Pn Pm ¯bj . For 1 ≤ r ≤ s let k(r), ¯ai = j=0 µ Then every tr belongs to the subgroup Γ of 374Ib, and ts = i=0 µ P Pl(r) Pk(r) ¯bj . Then µ ¯ai = k(r)=i tr − tr−1 , ¯ai , tr ≤ j=0 µ l(r) be minimal subject to the requirements tr ≤ i=0 µ so (using 374Ib) we can find a disjoint family hcr i1≤r≤s such that cr ⊆ ak(r) and µ ¯cr = tr − tr−1 for each r. Similarly, there is a disjoint family hdr i1≤r≤s such that dr ⊆ bl(r) and µ ¯dr = tr − tr−1 for each r. Now the principal ideals Acr , Adr are isomorphic for every r, by 374Ia; let πr : Adr → Acr be measure-preserving isomorphisms. Define π : A → A by setting

πa = (a \ sup1≤r≤s dr ) ∪ sup1≤r≤s πr (a ∩ dr ); because supr≤s cr = supi≤m ai = supj≤n bj = supr≤s dr , π : A → A is a measure-preserving automorphism. Now Ps u = r=1 αk(r) χcr , u∗ = so

R

u × Tπ v =

Ps r=1

Ps r=1

v=

v∗ =

•

αk(r) χ [tr−1 , tr [ , αk(r) βl(r) µ ¯cr =

Ps

Ps r=1

r=1

βl(r) χdr ,

Ps r=1

•

βl(r) χ [tr−1 , tr [ ,

αk(r) βl(r) (tr − tr−1 ) =

R

u∗ × v ∗ .

(b) Now take any u0 , v0 ∈ Mµ¯0,∞ . Set A = {u : u ∈ S(Af ), 0 ≤ u ≤ |u0 |},

B = {v : v ∈ S(Af ), 0 ≤ v ≤ |v0 |}.

446

Linear operators between function spaces

374J

Then A is an upwards-directed set with supremum |u0 |, because (A, µ ¯) is semi-finite, so {u∗ : u ∈ A} is an ∗ ∗ upwards-directed set with supremum |u0 | = u0 (373Db, 373Dh). Similarly {v ∗ : v ∈ B} is upwards-directed ∗ and has supremum v0∗ , so × v ∗ : u ∈ A, v ∈ B} is upwards-directed and supremum u∗0 × v0∗ . R {u R has ∗ ∗ ∗ Consequently, if γ < u0 × v0 , there are u ∈ A, v ∈ B such that γ ≤ u × v ∗ . Now, by (a), there is a π ∈ Aut such that γ≤

R

u × Tπ v ≤

R

|u0 | × Tπ |v0 | =

because Tπ is a Riesz homomorphism. As γ is arbitrary,

R

u∗0 × v0∗ ≤ supπ∈Aut

R

R

|u0 × Tπ v0 |

|u0 × Tπ v0 |.

But the reverse inequality is immediate from 373J. 374K Theorem Let (A, µ ¯) be a quasi-homogeneous semi-finite measure algebra, and τ a rearrangement -invariant extended Fatou norm on L0 = L0 (A). Then τ is T -invariant. proof Write τ 0 for the associate of τ . Then 374J tells us that for any u, v ∈ Mµ¯0,∞ ,

R

u∗ × v ∗ = supπ∈Aut

R

|Tπ u × v| ≤ supπ∈Aut τ (Tπ u)τ 0 (v) = τ (u)τ 0 (v),

writing u∗ , v ∗ for the decreasing rearrangements of u and v, and Aut for the group of measure-preserving automorphisms of (A, µ ¯). But now, if u ∈ Mµ¯1,∞ and T ∈ Tµ¯,¯µ , Z |T u × v| : τ 0 (v) ≤ 1}

τ (T u) = sup{ (by 369I)

Z ≤ sup{

u∗ × v ∗ : τ 0 (v) ≤ 1}

(by 373J) ≤ τ (u). As T , u are arbitrary, τ is T -invariant. 374L Lemma Let R(A, µ ¯) be a quasi-homogeneous semi-finite measure algebra. Suppose that u, v ∈ + ∗ (Mµ¯0,∞ ) are such that u × v ∗ = ∞. Then there is a measure-preserving automorphism π : A → A such R that u × Tπ v = ∞. proof I take three cases separately. (a) Suppose that A is purely atomic. Then it is surely infinite, since otherwise infinite. Let γ be the common measure of its atoms. For each n ∈ N, set αn = inf{α : α ≥ 0, µ ¯[[u > α]] ≤ 2n γ}. Then hαn in∈N is non-increasing, and µ ¯[[u > αn ]] ≤ 2n γ ≤ µ ¯[[u ≥ αn ]]. We can therefore choose a sequence h˜ an in∈N in A inductively so that [[u > αn ]] ⊆ a ˜n ⊆ [[u ≥ αn ]],

µ ¯a ˜n = 2n γ,

a ˜n ⊆ a ˜n+1 ,

for each n. Now if αn+1 ≤ α < αn , 2n γ < µ ¯[[u > α]] ≤ 2n+1 γ. So if we set £ £• • u ˜ = kuk∞ χ [0, γ[ ∨ supn∈N αn χ 2n γ, 2n+1 γ , then u∗ ≤ u ˜ in L∞ (AL ). Set an = a ˜n+1 \ a ˜n for each n; then han in∈N is disjoint and µ ¯an = 2n γ, for each n.

an ⊆ [[u ≥ αn+1 ]]

R

u∗ × v ∗ could not be

374L

Rearrangement-invariant spaces

447

Similarly, we can find a non-increasing sequence hβn in∈N in [0, ∞[ and a disjoint sequence hbn in∈N in A such that µ ¯bn = 2n γ, for each n, while

bn ⊆ [[v ≥ βn+1 ]]

£ £• • v ∗ ≤ v˜ = kvk∞ χ [0, γ[ ∨ supn∈N αn χ 2n γ, 2n+1 γ

in L∞ (AL ). R Now we are supposing that u∗ × v ∗ = ∞, so we must have ∞=

R

u ˜ × v˜ = γkuk∞ kvk∞ +

P∞ n=0

2n γαn βn .

Because 2n+1 αn+1 βn+1 ≤ 2 · 2n αn βn for each n, we must have P∞ 2n+1 α2n+1 β2n+1 = ∞. n=0 2 At this point, recall that we are dealing with a purely atomic algebra in which every S atom has measure γ. Let An , Bn be the sets of atoms included in an , bn for each n, and A = n∈N An ∪ Bn . Then #(An ) = #(Bn ) = 2n for eachSn. We therefore have S a bijection φ : A → A such that φ[B2n ] = A2n for every n. (The point is that A \ n∈N A2n and A \ n∈N B2n are both countably infinite.) Define π : A → A by setting πc = (c \ sup A) ∪ supa∈A,a ⊆ c φa. Then π is well-defined (because A is countable), and it is easy to check that it is a measure-preserving Boolean automorphism (because it is just a permutation of the atoms); and πb2n = a2n for every n. Consequently

R

u × Tπ v ≥

P∞

n=0

α2n+1 β2n+1 µ ¯a2n =

P∞

n=0

22n γα2n+1 β2n+1 = ∞.

So we have found a suitable automorphism. (b) Next, consider the case in which (A, µ ¯) is atomless and of finite magnitude γ. Of course γ > 0. For each n ∈ N set αn = inf{α : α ≥ 0, µ ¯[[u > α]] ≤ 2−n γ}. Then

£ £• u∗ ≤ supn∈N αn+1 χ 2−n−1 γ, 2−n γ .

Also µ ¯[[u > αn ]] ≤ 2−n γ ≤ µ ¯[[u ≥ αn ]] for each n, so we can choose inductively a decreasing sequence h˜ an in∈N such that [[u > αn ]] ⊆ a ˜n ⊆ [[u ≥ αn ]] −n

and µ ¯a ˜n = 2 γ for each n. Set an = a ˜n \ a ˜n+1 ; then han in∈N is disjoint and µ ¯an = 2−n−1 γ, an ⊆ [[u ≥ αn ]] for each n. In the same way, we can find hβn in∈N , hbn in∈N such that hbn in∈N is a£ disjoint sequence in A, µ ¯bn = 2−n−1 γ £ −n−1 ∗ −n • and bn ⊆ [[v ≥ βn ]] for each n, and v ≤ supn∈N βn+1 χ 2 γ, 2 γ . Now all the principal ideals Aan , Abn are homogeneous and of the same Maharam type, so there are measure-preserving isomorphisms πn : Abn → Aan . Define π : A → A by setting πc = supn∈N πn (c ∩ an ); then π is a measure-preserving automorphism of A, and πbn = an for each n. Since u × χan ≥ αn χan , v × χbn ≥ βn χbn for each n,

R

but on the other hand, So

R

R

u × Tπ v ≥

u∗ × v ∗ ≤

P∞

P∞

n=0

2−n−1 γαn βn ;

2−n−1 γαn+1 βn+1 ≤ 2 n=0

R

u × Tπ v.

u × Tπ v = ∞.

(c) Thirdly, consider the case in which A is atomless and not totally finite; take κ to be the common Maharam type of all the principal ideals Aa where 0 < µ ¯a < ∞. In this case, set

448

Linear operators between function spaces

αn = inf{α : µ ¯[[u > α]] ≤ 2n },

374L

βn = inf{α : µ ¯[[v > α]] ≤ 2n }

for each n ∈ Z. This time

£ £• u∗ ≤ supn∈Z αn χ 2n , 2n+1 ,

£ £• v ∗ ≤ supn∈Z βn χ 2n , 2n+1 .

There are disjoint families han in∈Z , hbn in∈Z such that µ ¯an = µ ¯bn = 2n for each n and u ≥ supn∈Z αn+1 χan ,

v ≥ supn∈Z βn+1 χbn .

(This time, start by fixing a ˜0 such that µ ¯a ˜0 = 1 and [[u > α0 ]] ⊆ a ˜0 ⊆ [[u ≥ α0 ]], and choose a ˜n+1 ⊇ a ˜n for n ≥ 0, a ˜n−1 ⊆ a ˜n for n ≤ 0.) Set d∗ = supn∈Z an ∪ supn∈Z bn . Then d1 = d∗ \ supn∈Z a2n ,

d2 = d∗ \ supn∈Z b2n

both have magnitude ω and Maharam type κ. So there is a measure-preserving isomorphism π ˜ : Ad2 → Ad1 (332J). At the same time, for each n ∈ Z there is a measure-preserving isomorphism πn : Ab2n → Aa2n . So once again we can assemble these to form a measure-preserving automorphism π : A → A, defined by the formula πc = (c \ d∗ ) ∪ π ˜ (c ∩ d2 ) ∪ supn∈Z πn (c ∩ b2n ). Just as in (a) and (b) above, while

R

u × Tπ v ≥

R

P∞ n=−∞

u∗ × v ∗ ≤

22n α2n+1 β2n+1 ,

P∞ n=−∞

2n αn βn

is infinite. Because 22n+2 α2n+2 β2n+2 + 22n+1 α2n+1 β2n+1 ≤ 6 · 22n α2n+1 β2n+1 for every n,

R

u × Tπ v ≥

1 6

R

u∗ × v ∗ = ∞.

Thus we have a suitable π in any of the cases allowed by 374H. 374M Proposition Let (A, µ ¯) be a quasi-homogeneous localizable measure algebra, and U ⊆ L0 = L (A) a solid linear subspace which, regarded as a Riesz space, is perfect. If U is rearrangement-invariant and Mµ¯∞,1 ⊆ U ⊆ Mµ¯1,∞ , then U is T -invariant. 0

proof Set V = {v : u × v ∈ L1 for every u ∈ U }, so that V is a solid linear subspace of L0 which can be identified with U × (369C), and U becomes {u : u × v ∈ L1 for every v ∈ V }; note that Mµ¯∞,1 ⊆ V ⊆ Mµ¯1,∞ (using 369Q). R If u ∈ UR + , v ∈ V + and π : A → A is a measure-preserving automorphism, then Tπ u ∈ U , so v×Tπ u < ∞; by 374L, u∗ × v ∗ is finite. But this means that if u ∈ U , v ∈ V and T ∈ Tµ¯,¯µ ,

R

|T u × v| ≤

R

u∗ × v ∗ < ∞.

As v is arbitrary, T u ∈ U ; as T and u are arbitrary, U is T -invariant. 374X Basic exercises > (a) Let (A, µ ¯) be a measure algebra and A ⊆ Mµ¯1,∞ a T -invariant set. (i) Show that A is solid. (ii) Show that if A is a linear subspace and not {0}, then it includes Mµ¯∞,1 . (iii) Rt Rt Show that if u ∈ A, v ∈ Mµ¯0,∞ and 0 v ∗ ≤ 0 u∗ for every t > 0, then v ∈ A. (iv) Show that if (B, ν¯) is any other measure algebra, then B = {T u : u ∈ A, T ∈ Tµ¯,¯ν } and C = {v : v ∈ Mν¯1,∞ , T v ∈ A for every T ∈ Tν¯,¯µ } are T -invariant subsets of Mν¯1,∞ , and that B ⊆ C. Give two examples in which B ⊂ C. Show that if (A, µ ¯) = (AL , µ ¯L ) then B = C. >(b) Let (A, µ ¯) be a measure algebra. Show that the extended Fatou norm k kp on L0 (A) is T -invariant for every p ∈ [1, ∞]. (Hint: 371Gd.)

374Yb

Rearrangement-invariant spaces

449

(c) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras, and φ a Young’s function (369Xc). Let τφ , τ˜φ be the corresponding Orlicz norms on L0 (A), L0 (B). Show that τ˜φ (T u) ≤ τφ (u) for every u ∈ L0 (A), T ∈ Tµ¯,¯ν . (Hint: 369Xn, 373Xm.) In particular, τφ is T -invariant. (d) Show that if (A, µ ¯) is a semi-finite measure algebra and τ is a T -invariant extended Fatou norm on L0 (A), then the Banach lattice Lτ defined from τ is T -invariant. (e) Let (A, µ ¯) be a semi-finite measure algebra and τ a T -invariant extended Fatou norm on L0 which is an order-continuous norm on Lτ . Show that Lτ ⊆ Mµ¯1,0 . (f ) Let θ be a T -invariant extended Fatou norm on L0 (AL ) and (A, µ ¯), (B, ν¯) two semi-finite measure algebras. Let τ1 , τ2 be the extended Fatou norms on L0 (A), L0 (B) defined from θ by the method of 374C. Show that τ2 (T u) ≤ τ1 (u) whenever u ∈ Mµ¯1,∞ and T ∈ Tµ¯,¯ν . 1 R |u| µ ¯a a

>(g) Let (A, µ ¯) be a semi-finite measure algebra, not {0}, and set τ (u) = sup0 t]])q/p dt.

450

Linear operators between function spaces

374Yb

(ii) Show that we have an extended Fatou norm k kp,q on L0 (A) defined by setting ¢1/q ¡ R∞ kukp,q = p 0 tq−1 (¯ µ[[|u| > t]])q/p dt 1/q

for every u ∈ L0 (A). (Hint: use 374Xj with w = wpq , k k = k kq .) (iii) Show that if (B, ν¯) is another measure algebra and T ∈ Tµ¯,¯ν , then kT ukp,q ≤ kukp,q for every u ∈ Mµ¯1,∞ . (iv) Show that k kp,q is an order-continuous norm on Lk kp,q . (c) Let (A, µ ¯) be a homogeneous measure algebra of uncountable Maharam type, and u, v ≥ 0 in Mµ¯0 ∗ such that u = v ∗ . Show that there is a measure-preserving automorphism π of A such that Tπ u = v, where Tπ : L0 (A) → L0 (A) is the isomorphism corresponding to π. (d) In L0 (AL ) let u be the equivalence class of the function f (t) = te−t . Show that there is no Boolean automorphism π of AL such that Tπ u = u∗ . (Hint: show that AL is τ -generated by {[[u∗ > α]] : α > 0}.) (e) Let (A, µ ¯) be a quasi-homogeneous semi-finite measure algebra and C ⊆ L0 (A) a solid convex orderclosed rearrangement-invariant set. Show that C ∩ Mµ¯1,∞ is T -invariant. 374 Notes and comments I gave this section the title ‘rearrangement-invariant spaces’ because it looks good on the Contents page, and it follows what has been common practice since Luxemburg 67b; but actually I think that it’s T -invariance which matters, and that rearrangement-invariant spaces are significant largely because the important ones are T -invariant. The particular quality of T -invariance which I have tried to bring out here is its transferability from one measure algebra (or measure space, of course) to another. This is what I take at a relatively leisurely pace in 374B-374D and 374Xf, and then encapsulate in 374Xh and 374Ya. The special place of the Lebesgue algebra (AL , µ ¯L ) arises from its being more or less the simplest algebra over which every T -invariant set can be described; see 374Xa. I don’t think this section is particularly easy, and (as in §373) there are rather a lot of unattractive names in it; but once one has achieved a reasonable familiarity with the concepts, the techniques used can be seen to amount to half a dozen ideas – non-trivial ideas, to be sure – from §§369 and 373. From §369 I take concepts of duality: the symmetric relationship between a perfect Riesz space U ⊆ L0 and the representation of its dual (369C-369D), and the notion of associate extended Fatou norms (369H-369K). From §373 I take the idea of ‘decreasing rearrangement’ and theorems guaranteeing the existence of useful members of Tµ¯,¯ν (373O-373Q). The results of the present section all depend on repeated use of these facts, assembled in a variety of patterns. There is one new method here, but an easy one: the construction of measure-preserving automorphisms by joining isomorphisms together, as in the proofs of 374H and 374J. I shall return to this idea, in greater generality and more systematically investigated, in §381. I hope that the special cases here will give no difficulty. While T -invariance is a similar phenomenon for both extended Fatou norms and perfect Riesz spaces (see 374Xh, 374Ya), the former seem easier to deal with. The essential difference is I think in 374B(i); with a T -invariant extended Fatou norm, we are necessarily confined to M 1,∞ , the natural domain of the methods used here. For perfect Riesz spaces we have examples like RN ∼ = L0 (PN) and its dual, the space of eventually-zero sequences (374Xk); these are rearrangement-invariant but not T -invariant, as I have defined it. This problem does not arise over atomless algebras (374Xl). I think it is obvious that for algebras which are not quasi-homogeneous (374G) rearrangement-invariance is going to be of limited interest; there will be regions between which there is no communication by means of measure-preserving automorphisms, and the best we can hope for is a discussion of quasi-homogeneous components, if they exist, corresponding to the partition of unity used in the proof of 332J. There is a special difficulty concerning rearrangement-invariance in L0 (AL ): two elements can have the same decreasing rearrangement without being rearrangements of each other in the strict sense (373Ya, 374Yd). The phenomenon of 373Ya is specific to algebras of countable Maharam type (374Yc). You will see that some of the labour of 374L is because we have to make room for the pieces to move in. 374J is easier just because in that context we can settle for a supremum, rather than an actual infinity, so the rearrangement needed (part (a) of the proof) can be based on a region of finite measure.

375C

Kwapien’s theorem

451

375 Kwapien’s theorem In §368 and the first part of §369 I examined maps from various types of Riesz space into L0 spaces. There are equally striking results about maps out of L0 spaces. I start with some relatively elementary facts about positive linear operators from L0 spaces to Archimedean Riesz spaces in general (375A-375D), and then turn to a remarkable analysis, due essentially to S.Kwapien, of the positive linear operators from a general L0 space to the L0 space of a semi-finite measure algebra (375I), with a couple of simple corollaries. 375A Theorem Let A be a Dedekind σ-complete Boolean algebra and W an Archimedean Riesz space. If T : L0 (A) → W is a positive linear operator, it is sequentially order-continuous. proof (a) The first step is to observe that if hun in∈N is any non-increasing sequence in L0 = L0 (A) with infimum 0, and ² > 0, then {n(un − ²u0 ) : n ∈ N} is bounded above in L0 . P P For k ∈ N set ak = supn∈N [[n(un − ²u0 ) > k]]; set a = inf k∈N ak . ?? Suppose, if possible, that a 6= 0. Because un ≤ u0 , n(un − ²u0 ) ≤ nu0 for every n and a ⊆ a0 ⊆ [[u0 > 0]] = [[²u0 > 0]] = supn∈N [[²u0 − un > 0]]. So there is some m ∈ N such that a0 = a ∩ [[²u0 − um > 0]] 6= 0. Now, for any n ≥ m, any k ∈ N, a0 ∩ [[n(un − ²u0 ) > k]] ⊆ [[²u0 − um > 0]] ∩ [[um − ²u0 > 0]] = 0. But a0 ⊆ supn∈N [[n(un − ²u0 ) > k]], so in fact a0 ⊆ supn≤m [[n(un − ²u0 ) > k]] = [[v > k]], where v = supn≤m n(un − ²u0 ). And this means that inf k∈N [[v > k]] ⊇ a0 6= 0, which is impossible. X X Accordingly a = 0; by 364Ma, {n(un − ²u0 ) : n ∈ N} is bounded above. Q Q (b) Now suppose that hun in∈N is a non-increasing sequence in L0 with infimum 0, and that w ∈ W is a lower bound for {T un : n ∈ N}. Take any ² > 0. By (a), {n(un − ²u0 ) : n ∈ N} has an upper bound v in L0 . Because T is positive, 1 n

1 n

w ≤ T un = T (un − ²u0 ) + T (²u0 ) ≤ T ( v) + T (²u0 ) = T v + ²T u0 for every n ≥ 1. Because W is Archimedean, w ≤ ²T u0 . But this is true for every ² > 0, so (again because W is Archimedean) w ≤ 0. As w is arbitrary, inf n∈N T un = 0. As hun in∈N is arbitrary, T is sequentially order-continuous (351Gb). 375B Proposition Let A be an atomless Dedekind σ-complete Boolean algebra. Then L0 (A)× = {0}. proof ?? Suppose, if possible, that h : L0 (A) → R is a non-zero order-continuous positive linear functional. Then there is a u > 0 in L0 such that h(v) > 0 whenever 0 < v ≤ u (356H). Because A is atomless, there is a disjoint sequence han in∈N such that an ⊆ [[u > 0]] for each n, so that un = u × χan > 0, while um ∧ un = 0 if m 6= n. Now however v = supn∈N n(h(un ))−1 un is defined in L0 , by 368K, and h(v) ≥ n for every n, which is impossible. X X 375C Theorem Let A be a Dedekind complete Boolean algebra, W an Archimedean Riesz space, and T : L0 (A) → W an order-continuous Riesz homomorphism. Then V = T [L0 (A)] is an order-closed Riesz subspace of W . proof The kernel U of T is a band in L0 = L0 (A) (352Oe), and must be a projection band (353I), because L0 is Dedekind complete (364O). Since U + U ⊥ = L0 , T [U ] + T [U ⊥ ] = V , that is, T [U ⊥ ] = V ; since U ∩ U ⊥ = {0}, T is an isomorphism between U ⊥ and V . Now suppose that A ⊆ V is upwards-directed and has a least upper bound w ∈ W . Then B = {u : u ∈ U ⊥ , T u ∈ A} is upwards-directed and T [B] = A. The point is that B is bounded above in L0 . P P?? If not, then {u+ : u ∈ B} cannot be bounded above, so there 0 is a u0 > 0 in L such that nu0 = supu∈B nu0 ∧ u+ for every n ∈ N (368A). Since B ⊆ U ⊥ , u0 ∈ U ⊥ and T u0 > 0. But now, because T is an order-continuous Riesz homomorphism,

452

Linear operators between function spaces

375C

nT u0 = supu∈B T (nu0 ∧ u+ ) = supv∈A nT u0 ∧ v + ≤ w+ for every n ∈ N, which is impossible. X XQ Q Set u∗ = sup B; then T u∗ = sup A = w and w ∈ V . As A is arbitrary, V is order-closed. 375D Corollary Let W be an Archimedean Riesz space and V an order-dense Riesz subspace which is isomorphic to L0 (A) for some Dedekind complete Boolean algebra A. Then V = W . proof Apply 375C to an isomorphism T : L0 (A) → V to see that V is order-closed in W . 375E I come now to the deepest result of this section, concerning positive linear operators from L0 (A) to L0 (B) where B is a measure algebra. I approach through a couple of lemmas which are striking enough in their own right. The following temporary definition will be useful. Definition Let A and B be Boolean algebras. I will say that a function φ : A → B is a σ-subhomomorphism if φ(a ∪ a0 ) = φ(a) ∪ φ(a0 ) for all a, a0 ∈ A, inf n∈N φ(an ) = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0. Now we have the following easy facts. 375F Lemma Let A and B be Boolean algebras and φ : A → B a σ-subhomomorphism. (a) φ(0) = 0, φ(a) ⊆ φ(a0 ) whenever a ⊆ a0 , and φ(a) \ φ(a0 ) ⊆ φ(a \ a0 ) for every a, a0 ∈ A. (b) If µ ¯, ν¯ are measures such that (A, µ ¯) and (B, ν¯) are totally finite measure algebras, then for every ² > 0 there is a δ > 0 such that ν¯φ(a) ≤ ² whenever µ ¯a ≤ δ. proof (a) This is elementary. Set every an = 0 in the second clause of the definition 375E to see that φ(0) = 0. The other two parts are immediate consequences of the first clause. (b) (Compare 232B, 327Bb.) ?? Suppose, if possible, otherwise. Then for every n ∈ N there is an an ∈ A such that µ ¯an ≤ 2−n and ν¯φ(an ) ≥ ². Set cn = supi≥n ai for each n; then hcn in∈N is non-increasing and has infimum 0 (since µ ¯cn ≤ 2−n+1 for each n), but ν¯φ(cn ) ≥ ² for every n, so inf n∈N φcn cannot be 0. X X 375G Lemma Let (A, µ ¯) and (B, ν¯) be totally finite measure algebras and φ : A → B a σ-subhomomorphism. Then for every non-zero b0 ∈ B there are a non-zero b ⊆ b0 and an m ∈ N such that b ∩ inf j≤m φ(aj ) = 0 whenever a0 , . . . , am ∈ A are disjoint. proof (a) Suppose first that A is atomless and that µ ¯1 = 1. 1 Set ² = 51 ν¯b0 and let m ≥ 1 be such that ν¯φ(a) ≤ ² whenever µ ¯a ≤ m . We need to know that 1 m 1 1 1 (1 − m ) ≤ 2 ; this is because (if m ≥ 2) ln m − ln(m − 1) ≥ m , so m ln(1 − m ) ≤ −1 ≤ − ln 2. Set C = {inf j≤m φ(aj ) : a0 , . . . , am ∈ A are disjoint}. ?? Suppose, if possible, that b0 ⊆ sup C. Then there are c0 , . . . , ck ∈ C such that ν¯(b0 ∩ supi≤k ci ) ≥ 4². For each i ≤ k choose disjoint ai0 , . . . , aim ∈ A such that ci = inf j≤m φ(aij ). Let D be the set of atoms of the finite subalgebra of A generated by {aij : i ≤ k, j ≤ m}, so that D is a finite partition of unity in A, and every aij is the join of the members of D it includes. Set p = #(D), and for each d ∈ D take a maximal 1 1 }, so that µ ¯(d \ sup Ed ) < pm ; set disjoint set Ed ⊆ {e : e ⊆ d, µ ¯e = pm S d∗ = 1 \ sup( d∈D Ed ) = supd∈D (d \ sup Ed ), 1 1 1 so that µ ¯d∗ is a multiple of pm and is less than m . Let E ∗ be a disjoint set of elements of measure pm with S 1 ∗ ∗ union d , and take E = E ∪ d∈D Ed , so that E is a partition of unity in A, µ ¯e = pm for every e ∈ E, and aij \ d∗ is the join of the members of E it includes for every i ≤ k, j ≤ m. Set

K = {K : K ⊆ E, #(K) = p},

M = #(K) =

(mp)! . p!(mp−p)!

375H

Kwapien’s theorem 1 m

For every K ∈ K, µ ¯(sup K) = R

453

so ν¯φ(sup K) ≤ ². So if we set P v = K∈K χφ(sup K),

v ≤ ²M . On the other hand, ν¯φ(d∗ ) ≤ ²,

ν¯(b0 ∩ supi≤k ci ) ≥ 4², so ν¯b1 ≥ 3², where Accordingly

R

b1 = b0 ∩ supi≤k ci \ φ(d∗ ). v ≤ 31 M ν¯b1 and b2 = b1 ∩ [[v < 12 M ]]

is non-zero. Because b2 ⊆ b1 , there is an i ≤ k such that b2 ∩ ci 6= 0. Now b2 ∩ ci ⊆ ci \ φ(d∗ ) = inf j≤m φ(aij ) \ φ(d∗ ) ⊆ inf j≤m φ(aij \ d∗ ). But every aij \ d∗ is the join of the members of E it includes, so b2 ∩ ci ⊆ inf φ(aij \ d∗ ) ⊆ inf φ(sup{e : e ∈ E, e ⊆ aij }) j≤m

j≤m

= inf sup{φ(e) : e ∈ E, e ⊆ aij } j≤m

= sup{ inf φ(ej ) : e0 , . . . , em ∈ E and ej ⊆ aij for every j}. j≤m

So there are e0 , . . . , em ∈ E such that ej ⊆ aij for each j and b3 = b2 ∩ inf j≤m φ(ej ) 6= 0. Because ai0 , . . . , aim are disjoint, e0 , . . . , em are distinct; set J = {e0 , . . . , em }. Then whenever K ∈ K and K∩J 6= ∅, b3 ⊆ φ(sup K). So let us calculate the size of K1 = {K : K ∈ K, K ∩ J 6= ∅}. This is M−

(mp−m−1)! p!(mp−p−m−1)!

¡ (mp−p)(mp−p−1)...(mp−p−m) ¢ =M 1− mp(mp−1)...(mp−m)

¡

≥M 1−(

mp−p m+1 ¢ ) mp

1 2

≥ M.

But this means that b3 ⊆ [[v ≥ 21 M ]], while also b3 ⊆ [[v < 12 M ]]; which is surely impossible. X X Accordingly b0 6⊆ sup C, and we can take b = b0 \ sup C. (b) Now for the general case. Let A be the set of atoms of A, and set d = 1 \ sup A. Then Ad is atomless, so there are a non-zero b1 ⊆ b0 and an n ∈ N such that b1 ∩ inf j≤n φ(aj ) = 0 whenever a0 , . . . , an ∈ Ad are disjoint. P P If µ ¯d > 0 this follows from (a), if we apply it to φ¹ Ad and (¯ µd)−1 µ ¯¹ Ad . If µ ¯d = 0 then we can just take b1 = b0 , n = 0. Q Q Let δ > 0 be such that ν¯φ(a) < ν¯b1 whenever µ ¯a ≤ δ. Let A1 ⊆ A be a finite set such that µ ¯(sup A1 ) ≥ µ ¯(sup A) − δ, and set r = #(A), d∗ = sup(A \ A1 ). Then µ ¯d∗ ≤ δ so b = b1 \ φ(d∗ ) 6= 0. Try m = n + r. If a0 , . . . , am are disjoint, then at most r of them can meet sup A1 , so (re-ordering if necessary) we can suppose that a0 , . . . , an are disjoint from sup A1 , in which case aj \ d∗ ⊆ d for each j ≤ m. But in this case (because b ∩ φ(d∗ ) = 0) b ∩ inf j≤m φ(aj ) ⊆ b ∩ inf j≤n φ(aj ) = b ∩ inf j≤n φ(aj ∩ d) = 0 by the choice of n and b1 . Thus in the general case also we can find appropriate b and m. 375H Lemma Let (A, µ ¯) and (B, ν¯) be totally finite measure algebras and φ : A → B a σ-subhomomorphism. Then for every non-zero b0 ∈ B there are a non-zero b ⊆ b0 and a finite partition of unity C ⊆ A such that a 7→ b ∩ φ(a ∩ c) is a ring homomorphism for every c ∈ C. proof By 375G, we can find b1 , m such that 0 6= b1 ⊆ b0 and b1 ∩ inf j≤m φ(aj ) = 0 whenever a0 , . . . , am ∈ A are disjoint. Do this with the smallest possible m. If m = 0 then b1 ∩ φ(1) = 0, so we can take

454

Linear operators between function spaces

375H

b = b1 , C = {1}. Otherwise, because m is minimal, there must be disjoint c1 , . . . , cm ∈ A such that b = b1 ∩ inf 1≤j≤m φ(cj ) 6= 0. Set c0 = 1 \ sup1≤j≤m cj , C = {c0 , c1 , . . . , cm }; then C is a partition of unity in A. Set πj (a) = b ∩ φ(a ∩ cj ) for each a ∈ A, j ≤ m. Then we always have πj (a ∪ a0 ) = πj (a) ∪ πj (a0 ) for all a, a0 ∈ A, because φ is a subhomomorphism. To see that every πj is a ring homomorphism, we need only check that πj (a ∩ a0 ) = 0 whenever a ∩ a0 = 0. (Compare 312H(iv).) In the case j = 0, we actually have π0 (a) = 0 for every a, because b ∩ φ(c0 ) = b1 ∩ inf 0≤j≤m φ(cj ) = 0 by the choice of b1 and m. When 1 ≤ j ≤ m, if a ∩ a0 = 0, then πj (a) ∩ πj (a0 ) = b1 ∩ inf 1≤i≤m,i6=j φ(cj ) ∩ φ(a) ∩ φ(a0 ) is again 0, because a, a0 , c1 , . . . , cj−1 , cj+1 , . . . , cm are disjoint. So we have a suitable pair b, C. 375I Theorem Let A be any Dedekind σ-complete Boolean algebra and (B, ν¯) a semi-finite measure algebra. Let T : L0 (A) → L0 (B) be a positive linear operator. Then we can find B, hAb ib∈B such that B is a partition of unity in B, each Ab is a finite partition of unity in A, and u 7→ T (u × χa) × χb is a Riesz homomorphism for every b ∈ B, a ∈ Ab . proof (a) Write B ∗ for the set of potential members of B; that is, the set of those b ∈ B such that there is a finite partition of unity A ⊆ A such that Tab is a Riesz homomorphism for every a ∈ A, writing Tab (u) = T (u × χa) × χb. If I can show that B ∗ is order-dense in B, this will suffice, since there will then be a partition of unity B ⊆ B ∗ . (b) So let b0 be any non-zero member of B; I seek a non-zero member of B ∗ included in b0 . Of course there is a non-zero b1 ⊆ b0 withR ν¯b1 < ∞. Let γ > 0 be such that b2 = b1 ∩ [[T (χ1) ≤ γ]] is non-zero. Define R µ : A → [0, ∞[ by setting µa = b2 T (χa) for every a ∈ A. Then µ is countably additive, because χ, T and are all additive and sequentially order-continuous (using 375A). Set N = {a : µa = 0}; then N is a σ-ideal of A, and (C, µ ¯) is a totally finite measure algebra, where C = A/N and µ ¯a• = µa for every a ∈ A (just as in 321H). (c) We have a function φ from C to the principal ideal Bb2 defined by saying that φa• = b2 ∩ [[T (χa) > 0]] for every a ∈ A. P P If a1 , a2 ∈ A are such that a•1 = a•2 in C, this means that a1 4 a2 ∈ N ; now [[T (χa1 ) > 0]] 4 [[T (χa2 ) > 0]] ⊆ [[|T (χa1 ) − T (χa2 )| > 0]] ⊆

[[T (|χa1 − χa2 |) > 0]] = [[T χ(a1 4 a2 ) > 0]]

R

is disjoint from b2 because b2 T χ(a1 4 a2 ) = 0. Accordingly b2 ∩ [[T (χa1 ) > 0]] = b2 ∩ [[T (χa2 ) > 0]] and we Q can take this common value for φ(a•1 ) = φ(a•2 ). Q (d) Now φ is a σ-subhomomorphism. P P (i) For any a1 , a2 ∈ A we have [[T χ(a1 ∪ a2 ) > 0]] = [[T (χa1 ) > 0]] ∪ [[T (χa2 ) > 0]] because T (χa1 ) ∨ T (χa2 ) ≤ T χ(a1 ∪ a2 ) ≤ T (χa1 ) + T (χa2 ). So φ(c1 ∪ c2 ) = φ(c1 ) ∪ φ(c2 ) for all c1 , c2 ∈ C. (ii) If hcn in∈N is a non-increasing sequence in C with infimum 0, choose an ∈ A such that a•n = cn for each n, and set a ˜n = inf i≤n ai \ inf i∈N ai for each n; then a ˜•n = cn so φ(cn ) = [[T (χ˜ an ) > 0]] for each n, while h˜ an in∈N is non-increasing and inf n∈N a ˜n = 0. ?? Suppose, if possible, an ) > 0]]) ≥ 2² for every n ∈ N. For each n, that b0 = inf n∈N φ(cn ) 6= 0; set ² = 21 ν¯b0 . Then ν¯(b2 ∩ [[T (χ˜ take αn > 0 such that ν¯(b2 ∩ [[T (χ˜ an ) > αn ]]) ≥ ². Then u = supn∈N nαn−1 χ˜ an is defined in L0 (A) (because −1 supn∈N [[nαn−1 χ˜ an > k]] ⊆ a ˜m if k ≥ maxi≤m iαi , so inf k∈N supn∈N [[nαn−1 χ˜ an > k]] = 0). But now ν¯(b2 ∩ [[T u > n]]) ≥ ν¯(b2 ∩ [[T (χ˜ an ) > αn ]]) ≥ ² for every n, so inf n∈N [[T u > n]] 6= 0, which is impossible. X X Thus inf n∈N φ(cn ) = 0; as hcn in∈N is arbitrary, φ is a σ-subhomomorphism. Q Q (e) By 375H, there are a non-zero b ∈ Bb2 and a finite partition of unity C ⊆ C such that d 7→ b ∩ φ(d ∩ c) is a ring homomorphism for every c ∈ C. There is a partition of unity A ⊆ A, of the same size as C, such that C = {a• : a ∈ A}. Now Tab is a Riesz homomorphism for every a ∈ A. P P It is surely a positive linear

375Xc

Kwapien’s theorem

455

operator. If u1 , u2 ∈ L0 (A) and u1 ∧ u2 = 0, set ei = [[ui > 0]] for each i, so that e1 ∩ e2 = 0. Observe that ui = supn∈N ui ∧ nχei , so that [[Tab ui > 0]] = supn∈N [[Tab (ui ∧ nχei ) > 0]] ⊆ [[Tab (χei ) > 0]] = b ∩ [[T χ(ei ∩ a) > 0]] for both i (of course Tab , like T , is sequentially order-continuous). But this means that [[Tab u1 > 0]] ∩ [[Tab u2 > 0]] ⊆ b ∩ [[T χ(e1 ∩ a) > 0]] ∩ [[T χ(e2 ∩ a) > 0]] = b ∩ φ(e•1 ∩ a• ) ∩ φ(e•2 ∩ a• ) = 0 because a• ∈ C, so d → 7 φ(d ∩ a• ) is a ring homomorphism, while e•1 ∩ e•2 = 0. So Tab u1 ∧ Tab u2 = 0. As u1 and u2 are arbitrary, Tab is a Riesz homomorphism (352G(iv)). Q Q (f ) Thus b ∈ B ∗ . As b0 is arbitrary, B ∗ is order-dense, and we’re home. 375J Corollary Let A be a Dedekind σ-complete Boolean algebra and U a Dedekind complete Riesz space such that U × separates the points of U . If T : L0 (A) → U is a positivePlinear operator, there is a ∞ 0 sequence hTn iP n∈N of Riesz homomorphisms from L (A) to U such that T = n=0 Tn , in the sense that n 0 T u = supn∈N i=0 Ti u for every u ≥ 0 in L (A). proof By 369A, U can be embedded as an order-dense Riesz subspace of L0 (B) for some localizable measure algebra (B, ν¯); being Dedekind complete, it is solid in L0 (B) (353K). Regard T as an Q operator from L0 (A) 0 0 to L (B), and take B, hAb ib∈B as in 375I. Note that L (B) can be identified with b∈B L0 (Bb ) (364S, 322K). For each b ∈ B let fb : Ab → N be an injection. If b ∈ B and n ∈ fb [Ab ], set Tnb (u) = χb × T (u × χa); otherwiseP set Tnb = 0. Then Tnb : L0 (A) → L0 (Bb ) is a Riesz homomorphism; because Ab is a finite partition ∞ of unity, n=0 Tnb u = χb × T u for every u ∈ L0 (A). But this means that if we set Tn u = hTnb uib∈B , Q ∼ L0 (B) Tn : L0 (A) → b∈B L0 (Bb ) = P∞ is a Riesz homomorphism for each n; and T = n=0 Tn . Of course every Tn is an operator from L0 (A) to U because |Tn u| ≤ T |u| ∈ U for every u ∈ L0 (A). 375K Corollary (a) If A is a Dedekind σ-complete Boolean algebra, (B, ν¯) is a semi-finite measure algebra, and there is any non-zero positive linear operator from L0 (A) to L0 (B), then there is a non-trivial sequentially order-continuous ring homomorphism from A to B. (b) If (A, µ ¯) and (B, ν¯) are homogeneous probability algebras and τ (A) > τ (B), then L∼ (L0 (A); L0 (B)) = {0}. proof (a) It is probably quickest to look at the proof of 375I: starting from a non-zero positive linear operator T : L0 (A) → L0 (B), we move to a non-zero σ-subhomomorphism φ : A/N → B and thence to a non-zero ring homomorphism from A/N to B, corresponding to a non-zero ring homomorphism from A to B, which is sequentially order-continuous because it is dominated by φ. Alternatively, quoting 375I, we have a non-zero Riesz homomorphism T1 : L0 (A) → L0 (B), and it is easy to check that a 7→ [[T (χa) > 0]] is a non-zero sequentially order-continuous ring homomorphism. (b) Use (a) and 331J. 375X Basic exercises (a) Let A be a Dedekind complete Boolean algebra and W an Archimedean Riesz space. Let T : L0 (A) → W be a positive linear operator. Show that T is order-continuous iff T χ : A → W is order-continuous. (b) Let A be an atomless Dedekind σ-complete Boolean algebra and W a Banach lattice. Show that the only order-continuous positive linear operator from L0 (A) to W is the zero operator. (c) Let A be a Dedekind complete Boolean algebra and W an Archimedean Riesz space. Let T : L0 (A) → W be an order-continuous Riesz homomorphism such that T [L0 (A)] is order-dense in W . Show that T is surjective.

456

Linear operators between function spaces

375Xd

(d) Let A and B be Boolean algebras and φ : A → B a σ-subhomomorphism as defined in 375E. Show that φ is sequentially order-continuous. > (e) Let A be the measure algebra of Lebesgue measure on [0, 1] and G the regular open algebra of R. (i) Show that there is no non-zero positive linear operator from L0 (G) to L0 (A). (Hint: suppose T : L0 (G) → L0 (A) were such an operator. Reduce to the case T (χ1) ≤ χ1. Let an R hbn in∈N enumerate R order-dense subset of G (316Yn). For each n ∈ N take non-zero b0n ⊆ bn such that T (χb0n ) ≤ 2−n−2 T (χ1) and consider T χ(supn∈N b0n ).) (ii) Show that there is no non-zero positive linear operator from L0 (A) to L0 (G). (Hint: suppose T : L0 (A) → L0 (G) were such an operator. For each n ∈ N choose an ∈ A, αn >P0 such that µ ¯an ≤ 2−n and if bn ⊆ [[T (χ1) > 0]] then bn ∩ [[T (χan ) > αn ]] 6= 0. Consider T u where ∞ −1 u = n=0 nαn χan .) (f ) In 375J, show that for any u ∈ L0 (A) inf n∈N supm≥n [[|T u −

Pm i=0

Ti u| > 0]] = 0.

> (g) Prove directly, without quoting 375E-375K, that if A is a Dedekind σ-complete Boolean algebra then every positive linear functional from L0 (A) to R is a finite sum of Riesz homomorphisms. 375Y Further exercises (a) Show that the following are equiveridical: (i) there is a purely atomic probability space (X, Σ, µ) such that Σ = PX and µ{x} = 0 for every x ∈ X; (ii) there are a set X and a Riesz homomorphism f : RX → R which is not order-continuous; (iii) there are a Dedekind complete Boolean algebra A and a positive linear operator f : L0 (A) → R which is not order-continuous; (iv) there are a Dedekind complete Boolean algebra A and a sequentially order-continuous Boolean homomorphism π : A → {0, 1} which is not order-continuous; (v) there are a Dedekind complete Riesz space U and a sequentially order-continuous Riesz homomorphism f : U → R which is not order-continuous; *(vi) there are an atomless Dedekind complete Boolean algebra A and a sequentially order-continuous Boolean homomorphism π : A → {0, 1} which is not order-continuous. (Compare 363S.) (b) Give an example of an atomless Dedekind σ-complete Boolean algebra A such that L0 (A)∼ 6= {0}. (c) Let A, B be Dedekind σ-complete Boolean algebras of which B is weakly σ-distributive. Let T : L0 (A) → L0 (B) be a positive linear operator. Show that a 7→ [[T (χa) > 0]] : A → B is a σ-subhomomorphism. (d) Let (A, µ ¯) be a localizable measure algebra and U a Riesz space such that U × separates the points of U . Suppose that T : L0 (A) → U is an order-continuous positive linear operator. Show that T [L0 (A)] is order-closed. (e) Let A and B be Dedekind complete Boolean algebras, and φ : A → B an σ-subhomomorphism such that φ1A = 1B . Show that there is a sequentially order-continuous Boolean homomorphism π : A → B such that πa ⊆ φa for every a ∈ A. (f ) Let G be the regular open algebra of R, and L0 = L0 (G). Give an example of a non-zero positive linear operator T : L0 → L0 such that there is no non-zero Riesz homomorphism S : L0 → L0 with S ≤ T . 375Z Problem Let G be the regular open algebra of R, and L0 = L0 (G). If T : L0 → L0 is a positive linear operator, must T [L0 ] be order-closed? 375 Notes and comments Both this section, and the earlier work on linear operators into L0 spaces, can be regarded as describing different aspects of a single fact: L0 spaces are very large. The most explicit statements of this principle are 368E and 375D: every Archimedean Riesz space can be embedded into a Dedekind complete L0 space, but no such L0 space can be properly embedded as an order-dense Riesz subspace of any other Archimedean Riesz space. Consequently there are many maps into L0 spaces (368B). But by the same token there are few maps out of them (375B, 375Kb), and those which do exist have a variety of special properties (375A, 375I).

376A

Kernel operators

457

The original version of Kwapien’s theorem (Kwapien 73) was the special case of 375I in which A is the Lebesgue measure algebra. The ideas of the proof here are mostly taken from Kalton, Peck & Roberts 84. I have based my account on the concept of ‘subhomomorphism’ (375E); this seems to be an effective tool when B is weakly (σ, ∞)-distributive (375Yc), but less useful in other cases. The case B = {0, 1}, L0 (B) ∼ = R is not entirely trivial and is worth working through on its own (375Xg). I mention 375C for the sake of its corollary 375D, but have to admit that I am not sure it is the best possible result. For the cases of principal interest to a measure theorist, there is something a good deal stronger (375Yd). Further questions concern possible relaxations of the hypotheses of 375B. One has a straightforward resolution (375Yb); others can be reduced to the Banach-Ulam problem by the techniques of 363S (375Ya).

376 Kernel operators The theory of linear integral equations is in large part the theory of operators T defined from formulae of the type (T f )(y) =

R

k(x, y)f (x)dx

for some function k of two variables. I make no attempt to study the general theory here. However, the concepts developed in this book make it easy to discuss certain aspects of such operators defined between the ‘function spaces’ of measure theory, meaning spaces of equivalence classes of functions, and indeed allow us to do some of the work in the abstract theory of Riesz spaces, omitting all formal mention of measures (376D, 376H, 376P). I give a very brief account of two theorems characterizing kernel operators in the abstract (376E, 376H), with corollaries to show the form these theorems can take in the ordinary language of integral kernels (376J, 376N). To give an idea of the kind of results we can hope for in this area, I go a bit farther with operators with domain L1 (376Mb, 376P, 376S). I take the opportunity to spell out versions of results from §253 in the language of this volume (376B376C). 376A Kernel operators To give an idea of where this section is going, I will try to describe the central idea in a relatively concrete special case. Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces; you can take them both to be X ×Y. R [0, 1] with Lebesgue measure if you like. Let λ be the product measure on 1 ∞ If k ∈ L (λ), then k(x, y)dx is defined for almost every y, by Fubini’s theorem; so if f ∈ L (µ) then R g(y) = k(x, y)f (x)dx is defined for almost every y. Also

R

g(y)dy =

R

k(x, y)f (x)dxdy

is defined, because (x, y) 7→ k(x, y)f (x) is λ-virtually measurable, defined λ-a.e. and is dominated by a multiple of the integrable function k. Thus k defines a function from L∞ (µ) to L1 (ν). Changing f on a set of measure 0 will not change g, so we can think of this as an operator from L∞ (µ) to L1 (ν); and of course we can move immediately to the equivalence class of g in L1 (ν), so getting an operator Tk from L∞ (µ) to L1 (ν). This operatorRis plainly linear; also it is easy to check that ±Tk ≤ T|k| , so that Tk ∈ L∼ (L∞ (µ); L1 (ν)), and that kTk k ≤ |k|. Moreover, changing k on a λ-negligible set does not change Tk , so that in fact we can speak of Tw for any w ∈ L1 (λ). I think it is obvious, even before investigating them, that operators representable in this way will be important. We can immediately ask what their properties will be and whether there is any straightforward way of recognising them. We can look at the properties of the map w 7→ Tw : L1 (λ) → L∼ (L∞ (µ); L1 (ν)). And we can ask what happens when L∞ (µ) and L1 (ν) are replaced by other function spaces, defined by extended Fatou norms or otherwise. Theorems 376E and 376H answer questions of this kind. R It turns out that the formula g(y) = k(x, y)f (x)dx gives rise to a variety of technical problems, and it is much easier to characterize T u in terms of its action on the dual. In the language of the special case above, if h ∈ L∞ (ν), then we shall have

R

k(x, y)f (x)h(y)d(x, y) =

R

g(y)h(y)dy;

458

Linear operators between function spaces

376A

R since g • ∈ L1 (ν) is entirely determined by the integrals g(y)h(y)dy as h runs over L∞ (ν), we can define R the operator T in terms of the functional (f, h) 7→ k(x, y)f (x)h(y)d(x, y). This enables us to extend the results from the case of σ-finite spaces to general strictly localizable spaces; perhaps more to the point in the present context, it gives them natural expressions in terms of function spaces defined from measure algebras rather than measure spaces, as in 376E. Before going farther along this road, however, I give a couple of results relating the theorems of §253 to the methods of this volume. 376B The canonical map L0 × L0 → L0 : Proposition Let (A, µ ¯) and (B, ν¯) be semi-finite measure ¯ their localizable measure algebra free product (325E). Then we have a bilinear map algebras, and (C, λ) (u, v) 7→ u ⊗ v : L0 (A) × L0 (B) → L0 (C) with the following properties. (a) For any u ∈ L0 (A), v ∈ L0 (B), α ∈ R, [[u ⊗ χ1B > α]] = [[u > α]] ⊗ 1B ,

[[χ1A ⊗ v > α]] = 1A ⊗ [[v > α]]

where for a ∈ A, b ∈ B I write a ⊗ b for the corresponding member of A ⊗ B (315M), identified with a subalgebra of C (325Dc). (b)(i) For any u ∈ L0 (A)+ , the map v 7→ u ⊗ v : L0 (B) → L0 (C) is an order-continuous multiplicative Riesz homomorphism. (ii) For any v ∈ L0 (B)+ , the map u 7→ u ⊗ v : L0 (A) → L0 (C) is an order-continuous multiplicative Riesz homomorphism. (c) In particular, |u ⊗ v| = |u| ⊗ |v| for all u ∈ L0 (A), v ∈ L0 (B). (d) For any u ∈ L0 (A)+ and v ∈ L0 (B)+ , [[u ⊗ v > 0]] = [[u > 0]] ⊗ [[v > 0]]. proof The canonical maps a 7→ a ⊗ 1B , b 7→ 1A ⊗ b from A, B to C are order-continuous Boolean homomorphisms (325Da), so induce order-continuous multiplicative Riesz homomorphisms from L0 (A) and L0 (B) to L0 (C) (364R); write u ˜, v˜ for the images of u ∈ L0 (A), v ∈ L0 (B). Observe that |˜ u| = |u|∼ , ∼ ∼ ∼ |˜ v | = |v| and (χ1A ) = (χ1B ) = χ1C . Now set u ⊗ v = u ˜ × v˜. The properties listed in (a)-(c) are just a matter of putting the definition in 364Ra together with the fact that L0 (C) is an f -algebra (364E). As for [[u ⊗ v > 0]] = [[˜ u × v˜ > 0]], this is (for non-negative u, v) just [[˜ u > 0]] ∩ [[˜ v > 0]] = ([[u > 0]] ⊗ 1B ) ∩ (1A ⊗ [[v > 0]]) = [[u > 0]] ⊗ [[v > 0]]. 376C For L1 spaces we have a similar result, with additions corresponding to the Banach lattice structures of the three spaces. Theorem Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras with localizable measure algebra free product ¯ (C, λ). ¯ and (a) If u ∈ L1µ¯ = L1 (A, µ ¯) and v ∈ L1ν¯ = L1 (B, ν¯) then u ⊗ v ∈ L1λ¯ = L1 (C, λ)

R

u⊗v =

R R u

v,

ku ⊗ vk1 = kuk1 kvk1 .

(b) Let W be a Banach space and φ : L1µ¯ × L1ν¯ → W a bounded bilinear map. Then there is a unique bounded linear operator T : L1λ¯ → W such that T (u⊗v) = φ(u, v) for all u ∈ L1µ¯ and v ∈ L1ν¯ , and kT k = kφk. (c) Suppose, in (b), that W is a Banach lattice. Then (i) T is positive iff φ(u, v) ≥ 0 for all u, v ≥ 0; (ii) T is a Riesz homomorphism iff u 7→ φ(u, v0 ) : L1µ¯ → W , v 7→ φ(u0 , v) : L1ν¯ → W are Riesz homomorphisms for all v0 ≥ 0 in L1ν¯ and u0 ≥ 0 in L1µ¯ . proof (a) I refer to the proof of 325D. Let (X, Σ, µ) and (Y, T, ν) be the Stone spaces of (A, µ ¯) and (B, ν¯) ¯ can be identified with the measure algebra of the c.l.d. product measure λ on X × Y (321K), so that (C, λ) (part (a) of the proof of 325D), and L1µ¯ , L1ν¯ , L1λ¯ can be identified with L1 (µ), L1 (ν) and L1 (λ) (365B). Now ¯ if f ∈ L0 (µ) and g ∈ L0 (ν) then f ⊗ g ∈ L0 (λ) (253Cb), and it is easy to check that (f ⊗ g)• ∈ L0 (λ) corresponds to f • ⊗ g • as defined in 376B. (Look first at the cases in which one of f , g is a constant function 1 with value R1.) RBy 253E, we have a canonical map (f • , g • ) 7→ (f ⊗ g)• from L1 (µ) × L1 (ν) R R to L (λ), R with R f ⊗ g = f g (253D); so that if u ∈ L1µ¯ and v ∈ L1ν¯ we must have u ⊗ v ∈ L1λ¯ , with u ⊗ v = u v. As in 253E, it follows that ku ⊗ vk1 = kuk1 kvk1 .

376D

Kernel operators

459

(b) In view of the situation described in (a) above, this is now just a translation of the same result about L1 (µ), L1 (ν) and L1 (λ), which is Theorem 253F. (c) Identifying the algebraic free product A ⊗ B with its canonical image in C (325Dc), I write (A ⊗ B)f ¯ < ∞}, so that (A ⊗ B)f is a subring of C. Recall that any member of A ⊗ B is for {c : c ∈ A ⊗ B, λc expressible as supi≤n ai ⊗ bi where a0 , . . . , an are disjoint (315Na); evidently this will belong to (A ⊗ B)f iff µ ¯ai · ν¯bi is finite for every i. The next fact to lift from previous theorems is in part (e) of the proof of 253F: the linear span M of {χ(a ⊗ b) : a ∈ Af , b ∈ Bf } is norm-dense in L1λ¯ . Of course M can also be regarded as the linear span of {χc : c ∈ (A ⊗ B)f }, or S(A ⊗ B)f . (Strictly speaking, this last remark relies on 361J; the identity map from (A ⊗ B)f to C induces an injective Riesz homomorphism from S(A ⊗ B)f into S(C) ⊆ L0 (C). To see that χc ∈ M for every c ∈ (A ⊗ B)f , we need to know that c can be expressed as a disjoint union of members of A ⊗ B, as noted above.) (i) If T is positive then of course φ(u, v) = T (u ⊗ v) ≥ 0 whenever u, v ≥ 0, since u ⊗ v ≥ 0. On the other hand, if φ is non-negative on U + × V + , then, in particular, T χ(a ⊗ b) = φ(χa, χb) ≥ 0 whenever µ ¯a· ν¯b < ∞. Consequently T (χc) ≥ 0 for every c ∈ (A⊗B)f and T w ≥ 0 whenever w ≥ 0 in M ∼ = S(A⊗B)f , as in 361Ga. Now this means that T |w| ≥ 0 whenever w ∈ M . But as M is norm-dense in L1λ¯ , w 7→ T |w| is continuous and W + is closed, it follows that T |w| ≥ 0 for every w ∈ L1λ¯ , that is, that T is positive. (ii) If T is a Riesz homomorphism then of course u 7→ φ(u, v0 ) = T (u ⊗ v0 ), v 7→ φ(u0 , v) = T (u0 ⊗ v) are Riesz homomorphisms for v0 , u0 ≥ 0. On the other hand, if all these maps are Riesz homomorphisms, then, in particular, T χ(a ⊗ b) ∧ T χ(a0 ⊗ b0 ) = φ(χa, χb) ∧ φ(χa0 , χb0 ) ≤ φ(χa, χb + χb0 ) ∧ φ(χa0 , χb + χb0 ) = φ(χa ∧ χa0 , χb + χb0 ) = 0 whenever a, a0 ∈ Af , b, b0 ∈ Bf and a ∩ a0 = 0. Similarly, T χ(a ⊗ b) ∧ T χ(a0 ⊗ b0 ) = 0 if b ∩ b0 = 0. But this means that T χc ∧ T χc0 = 0 whenever c, c0 ∈ (A ⊗ B)f and c ∩ c0 = 0. P P Express c, c0 as supi≤m ai ⊗ bi , 0 0 0 0 supj≤n aj ⊗ bj where ai , aj , bi , bj all have finite measure. Now if i ≤ m, j ≤ n, (ai ∩ a0j ) ⊗ (bi ∩ b0j ) = (ai ⊗bi ) ∩ (a0j ⊗b0j ) = 0, so one of ai ∩ a0j , bi ∩ b0j must be zero, and in either case T χ(ai ⊗bi )∧T χ(a0j ⊗b0j ) = 0. Accordingly m n X X T χc ∧ T χc0 ≤ ( T χ(ai ⊗ bi )) ∧ ( T χ(a0j × b0j )) i=0

≤

X

j=0

T χ(ai ⊗ bi ) ∧ T χ(a0j ⊗ b0j ) = 0,

i≤m,j≤n

using 352Fa for the second inequality. Q Q This implies that T ¹M must be a Riesz homomorphism (361Gc), that is, T |w| = |T w| for all w ∈ M . Again because M is dense in L1λ¯ , T |w| = |T w| for every w ∈ L1λ¯ , and T is a Riesz homomorphism. 376D Abstract integral operators: Definition The following concept will be used repeatedly in the theorems below; it is perhaps worth giving it a name. Let U be a Riesz space and V a Dedekind complete Riesz space, so that L× (U ; V ) is a Dedekind complete Riesz space (355H). If f ∈ U × and v ∈ V write Pf v u = f (u)v for each u ∈ U ; then Pf v ∈ L× (U ; V ). P P If f ≥ 0 in U × and v ≥ 0 in V × then Pf v is a positive linear operator from U to V which is order-continuous because if A ⊆ U is non-empty, downwards-directed and has infimum 0, then (as V is Archimedean) inf u∈A Pf v (u) = inf u∈A f (u)v = 0. Of course (f, g) 7→ Pf g is bilinear, so Pf v ∈ L× (U ; V ) for every f ∈ U × , v ∈ V . Q Q Now I call a linear operator from U to V an abstract integral operator if it is in the band of L× (U ; V ) generated by {Pf v : f ∈ U × , v ∈ V }.

460

Linear operators between function spaces

376D

The first result describes these operators when U , V are expressed as subspaces of L0 (A), L0 (B) for measure algebras A, B and V is perfect. 376E Theorem Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras, with localizable measure algebra ¯ and U ⊆ L0 (A), V ⊆ L0 (B) order-dense Riesz subspaces. Write W for the set of free product (C, λ), those w ∈ L0 (C) such that w × (u ⊗ v) is integrable for every u ∈ U , v ∈ V . Then we have an operator w 7→ Tw : W → L× (U ; V × ) defined by setting Tw (u)(v) =

R

w × (u ⊗ v)

for every w ∈ W , u ∈ U and v ∈ V . The map w 7→ Tw is a Riesz space isomorphism between W and the band of abstract integral operators in L× (U ; V × ). × proof (a) The first thing to check is that the formula offered does R define a member Tw (u) of V for any w ∈ W, u ∈ U. P P Of course Tw (u) isR a linear operator because is linear and ⊗ and × are bilinear. It belongs to V ∼ because, writing g(v) = |w|×(|u|⊗v), g is a positive linear operator and |Tw (u)(v)| ≤ g(|v|) for every v. (I am here using 376Bc to see that |w × (u ⊗ v)| = |w| × (|u| ⊗ |v|).) Also g ∈ V × because R 0 0 v 7→ |u| ⊗ v, w 7→ |w| × w and are all order-continuous; so Tw (u) also belongs to V × . Q Q

(b) Next, for any given w ∈ W , the map Tw : U → V × is linear (again because ⊗ and × are bilinear). It is helpful to note that W is a solid linear subspace of L0 (C). Now if w ≥ 0 in W , then Tw ∈ L× (U ; V × ). P P If u, v ≥ 0 then u ⊗ v ≥ 0, w × (u ⊗ v) ≥ 0 and Tw (u)(v) ≥ 0; as v is arbitrary, Tw (u) ≥ 0 whenever u ≥ 0; as u is arbitrary, Tw is positive. If A ⊆ U is non-empty, downwards-directed and has infimum 0, then Tw [A] is downwards-directed, and for any v ∈ V + (inf Tw [A])(v) = inf u∈A Tw (u)(v) = inf u∈A

R

w × (u ⊗ v) = 0

because u 7→ u ⊗ v is order-continuous. So inf Tw [A] = 0; as A is arbitrary, Tw is order-continuous. Q Q For general w ∈ W , we now have Tw = Tw+ − Tw− ∈ L× (U ; V × ). (c) Ths shows that w 7→ Tw is a map from W to L× (U ; V × ). Running through the formulae once again, it is linear, positive and order-continuous; this last because, given a non-empty downwards-directed C ⊆ W with infimum 0, then for any u ∈ U + , v ∈ V + R

(because 0.

(inf w∈C Tw )(u)(v) ≤ inf w∈C

R

w × (u ⊗ v) = 0

and × are order-continuous); as v is arbitrary, (inf w∈C Tw )(u) = 0; as u is arbitrary, inf w∈C Tw =

(d) R All this is easy, being nothing but a string of applications of the elementary properties of ⊗, × and . But I think a new idea is needed for the next fact: the map w 7→ Tw : W → L× (U ; V × ) is a Riesz homomorphism. P P Write D for the set of those d ∈ C such that Tw ∧ Tw0 = 0 whenever w, w0 ∈ W + , [[w > 0]] ⊆ d and [[w0 > 0]] ⊆ 1C \ d. (i) If d1 , d2 ∈ D, w, w0 ∈ W + , [[w > 0]] ⊆ d1 ∪ d2 and [[w0 > 0]] ∩ (d1 ∪ d2 ) = 0, then set w1 = w × χd1 , w2 = w − w1 . In this case [[w1 > 0]] ⊆ d1 ,

[[w2 > 0]] ⊆ d2 ,

so Tw1 ∧ Tw0 = Tw2 ∧ Tw0 = 0,

Tw ∧ Tw0 ≤ (Tw1 ∧ Tw0 ) + (Tw2 ∧ Tw0 ) = 0.

0

As w, w are arbitrary, d1 ∪ d2 ∈ D. Thus D is closed under ∪ . (ii) The symmetry of the definition of D means that 1C \ d ∈ D whenever d ∈ D. (iii) Of course 0 ∈ D, just because Tw = 0 if w ∈ W + and [[w > 0]] = 0; so D is a subalgebra of C. (iv) If D ⊆ D is non-empty and upwards-directed, with supremum c in C, and if w, w0 ∈ W + are such that [[w > 0]] ⊆ c, [[w0 > 0]] ∩ c = 0, then consider {w × χd : d ∈ D}. This is upwards-directed, with supremum w; so Tw = supd∈D Tw×χd , because the map q 7→ Tq is order-continuous. Also Tw×χd ∧ Tw0 = 0 for every d ∈ D, so Tw ∧ Tw0 = 0. As w, w0 are arbitrary, c ∈ D; as D is arbitrary, D is an order-closed subalgebra of C. (v) If a ∈ A and w, w0 ∈ W + are such that [[w > 0]] ⊆ a ⊗ 1B , [[w0 > 0]] ∩ (a ⊗ 1B ) = 0, then any u ∈ U + is expressible as u1 + u2 where u1 = u × χa, u2 = u × χ(1A \ a). Now Tw (u2 )(v) =

R

w × (u2 ⊗ v) =

R

w × χ(a ⊗ 1B ) × (u ⊗ v) × χ((1A \ a) ⊗ 1B ) = 0

376E

Kernel operators

461

for every v ∈ V , so Tw (u2 ) = 0. Similarly, Tw0 (u1 ) = 0. But this means that (Tw ∧ Tw0 )(u) ≤ Tw (u2 ) + Tw0 (u1 ) = 0. As u is arbitrary, Tw ∧ Tw0 = 0; as w and w0 are arbitrary, a ⊗ 1B ∈ D. (vi) Now suppose that b ∈ B and that w, w0 ∈ W + are such that [[w > 0]] ⊆ 1A ⊗ b, [[w0 > 0]] ∩ (1A ⊗ b) = 0. If u ∈ U + , v ∈ V + then (Tw ∧ Tw0 )(u)(v) ≤

R

w × (u ⊗ (v × χ(1B \ b))) +

R

w0 × (u ⊗ (v × χb)) = 0.

As u, v are arbitrary, Tw ∧ Tw0 = 0; as w and w0 are arbitrary, 1A ⊗ b ∈ D. (vii) This means that D is an order-closed subalgebra of C including A ⊗ B, and is therefore the whole of C (325D(c-ii)). (viii) Now take any w, w0 ∈ W such that w ∧ w0 = 0, and consider c = [[w > 0]]. Then [[w0 > 0]] ⊆ 1C \ c and c ∈ D, so Tw ∧ Tw0 = 0. This is what we need to be sure that w 7→ Tw is a Riesz homomorphism (352G). Q Q (e) The map w 7→ Tw is injective. P P (i) If w > 0 in W , then consider A = {a : a ∈ A, ∃ u ∈ U, χa ≤ u}, 0

B = {b : b ∈ B, ∃ v ∈ V, χb ≤ v}.

0

Because U and V are order-dense in L (A) and L (B) respectively, RA and B are order-dense in A and B. R Also both are upwards-directed. So supa∈A,b∈B a ⊗ b = 1C and 0 < w = supa∈A,b∈B a⊗b w. Take a ∈ A, R b ∈ B such that a⊗b w > 0; then there are u ∈ U , v ∈ V such that χa ≤ u, χb ≤ v, so that Tw (u)(v) ≥

R

a⊗b

w>0

and Tw > 0. (ii) For general non-zero w ∈ W , we now have |Tw | = T|w| > 0 so Tw 6= 0. Q Q Thus w 7→ Tw is an order-continuous injective Riesz homomorphism. ˜ for {Tw : w ∈ W }, so that W ˜ is a Riesz subspace of L× (U ; V × ) isomorphic to W , and (f ) Write W × × c ˜ is order-dense in W c. P W for the band it generates in L (U ; V ). Then W P Suppose that S > 0 in ⊥⊥ ⊥ c=W ˜ ˜ , so there is a w ∈ W such that S ∧ Tw > 0. Set w1 = w ∧ χ1C . Then W (353Ba). Then S ∈ /W w = supn∈N w ∧ nw1 , so Tw = supn∈N Tw ∧ nTw1 and R = S ∧ Tw1 > 0. Set U1 = U ∩ L1 (A, µ ¯). Because U is an order-dense Riesz subspace of L0 (A), U1 is an order-dense Riesz 1 1 subspace of Lµ¯ = L (A, µ ¯), therefore also norm-dense. Similarly V1 = V ∩ L1 (B, ν¯) is a norm-dense Riesz subspace of L1ν¯ = L1 (B, ν¯). Define φ0 : U1 × V1 → R by setting φ0 (u, v) = R(u)(v) for u ∈ U1 , v ∈ V1 . Then φ0 is bilinear, and |φ0 (u, v)| = |R(u)(v)| ≤ |R(u)|(|v|) ≤ R(|u|)(|v|) ≤ Tw1 (|u|)(|v|) Z Z = w1 × (|u| ⊗ |v|) ≤ |u| ⊗ |v| = kuk1 kvk1 for all u ∈ U1 , v ∈ V1 , because 0 ≤ R ≤ Tw1 in L× (U ; V × ). Because U1 , V1 are norm-dense in L1µ¯ , L1ν¯ respectively, φ0 has a unique extension to a continuous bilinear map φ : L1µ¯ × L1ν¯ → R. (To reduce this to standard results on linear operators, think of R as a function from U1 to V1∗ ; since every member of V1∗ has a unique extension to a member of (L1ν¯ )∗ , we get a corresponding function R1 : U1 → (L1ν¯ )∗ which is continuous and linear, so has a unique extension to a continuous linear operator R2 : L1µ¯ → (L1ν¯ )∗ , and we set φ(u, v) = R2 (u)(v).) ¯ ∗ such that h(u ⊗ v) = φ(u, v) for every u ∈ L1 , v ∈ L1 . By 376C, there is a unique h ∈ (L1λ¯ )∗ = L1 (C, λ) µ ¯ ν ¯ ¯ Because (C, λ) is localizable, this h corresponds to a w0 ∈ L∞ (C) (365Jc), and

R

w0 × (u ⊗ v) = h(u ⊗ v) = φ0 (u, v) = R(u)(v)

for every u ∈ U1 , v ∈ V1 . Because U1 is norm-dense in L1µ¯ , U1+ is dense in (L1µ¯ )+ , and similarly V1+ is dense in (L1ν¯ )+ , so U1+ × V1+ is dense in (L1µ¯ )+ × (L1ν¯ )+ ; now φ0 is non-negative on U1+ × V1+ , so φ (being continuous) is non-negative on (L1µ¯ )+ ×(L1ν¯ )+ . By 376Cc, h ≥ 0 in (L1λ¯ )∗ and w0 ≥ 0 in L∞ (C). In the same way, because φ0 (u, v) ≤ Tw (u)(v) for u ∈ U1+ and v ∈ V1+ , w0 ≤ w1 ≤ w in L0 (C), so w0 ∈ W . We have Tw0 (u)(v) =

R

w0 × (u ⊗ v) = R(u)(v)

for all u ∈ U1 , v ∈ V1 . If u ∈ U1+ , then Tw0 (u) and R(u) are both order-continuous, so must be identical, since V1 is order-dense in V . This means that Tw0 and R agree on U1 . But as both are themselves order-continuous

462

Linear operators between function spaces

376E

linear operators, and U1 is order-dense in U , they must be equal. ˜ is quasi-order-dense in W c , therefore order-dense Thus 0 < Tw0 ≤ S in L× (U ; V × ). As S is arbitrary, W (353A). Q Q ˜ is an injective Riesz homomorphism, we have an inverse map Q : (g) Because w 7→ Tw : W 7→ W ˜ → L0 (C), setting Q(Tw ) = w; this is a Riesz homomorphism, and it is order-continuous because W is W solid in L0 (C), so that the embedding W ⊆ L0 (C) is order-continuous. By 368B, Q has an extension to ˜ : W c → L0 (C). Because Q(S) > 0 whenever S > 0 in W ˜, an order-continuous Riesz homomorphism Q ˜ c ˜ ˜ c Q(S) > 0 whenever S > 0 in W , so Q is injective. Now Q(S) ∈ W for every S ∈ W . P P It is enough ˜ ˜ w ) : w ∈ W, Tw ≤ S} = sup C, where to look at non-negative S. In this case, Q(S) must be sup{Q(T C = {w : Tw ≤ S} ⊆ W . Take u ∈ U + , v ∈ V + . Then {w × (u ⊗ v) : w ∈ C} is upwards-directed, because C is, and supw∈C

R

w × (u ⊗ v) = supw∈C Tw (u)(v) ≤ S(u)(v) < ∞.

˜ ˜ So Q(S) × (u ⊗ v) = supw∈C w × (u ⊗ v) belongs to L1λ¯ (365Df). As u and v are arbitrary, Q(S) ∈ W. Q Q ˜ =W c and Q ˜ = Q, that is, that w 7→ Tw : W 7→ W c is a Riesz space (h) Of course this means that W isomorphism. c as the band Z of abstract integral operators in (i) I have still to check on the identification of W × × × L (U ; V ). Write Pf g (u) = f (u)g for f ∈ U , g ∈ V and u ∈ U . ×

Set U # = {u : u ∈ L0 (A), u × u0 ∈ L1µ¯ for every u0 ∈ U }, V # = {v : v ∈ L0 (B), v × v 0 ∈ L1ν¯ for every v 0 ∈ V }. R From 369C we know that if we set fu (u0 ) = u × u0 for u ∈ U # and u0 ∈ U , then fu ∈ U × for every u ∈ U # , and u 7→Rfu is an isomorphism between U # and an order-dense Riesz subspace of U × . Similarly, setting gv (v 0 ) = v × v 0 for v ∈ V # and v 0 ∈ V , v 7→ gv is an isomorphism between V # and an order-dense Riesz subspace of V × . If u ∈ U # , v ∈ V # then

R

(u ⊗ v) × (u0 ⊗ v 0 ) =

R

R

R

(u × u0 ) ⊗ (v × v 0 ) = ( u × u0 )( v × v 0 ) = fu (u0 )gv (v 0 )

for every u0 ∈ U , v 0 ∈ V , so u ⊗ v ∈ W and Tu⊗v = Pfu gv . Now take f ∈ (U × )+ and g ∈ (V × )+ . Set A = {u : u ∈ U # , u ≥ 0, fu ≤ f } and B = {v : v ∈ V # , v ≥ 0, gv ≤ g}. These are upwards-directed, so C = {u ⊗ v : u ∈ A, v ∈ B} is upwards-directed in L0 (C). Because {fu : u ∈ U # } is order-dense in U × , f = supu∈A fu ; by 355Ed, f (u0 ) = supu∈A fu (u0 ) for every u0 ∈ U + . Similarly, g(v 0 ) = supv∈B fv (v 0 ) for every v 0 ∈ V + . ?? Suppose, if possible, that C is not bounded above in L0 (C). Because C and L0 (C) are Dedekind complete, c = inf n∈N supu∈A,v∈B [[u ⊗ v ≥ n]] must be non-zero (364Ma). Because U and V are order-dense in L0 (A), L0 (B) respectively, 1A = sup{[[u0 > 0]] : u0 ∈ U }, and there are u0 ∈ U + , v 0 ∈ V + any n ∈ N,

1B = sup{[[v 0 > 0]] : v 0 ∈ V }, R such that c ∩ [[u0 > 0]] ⊗ [[v 0 > 0]] 6= 0, so that c u0 ⊗ v 0 > 0. But now, for

376E

Kernel operators

f (u0 )g(v 0 ) ≥

463

fu (u0 )gv (v 0 ) Z = sup (u ⊗ v) × (u0 ⊗ v 0 ) u∈A,v∈B Z ≥ sup ((u ⊗ v) ∧ nχc) × (u0 ⊗ v 0 ) u∈A,v∈B Z = sup ((u ⊗ v) ∧ nχc) × (u0 ⊗ v 0 ) sup

u∈A,v∈B

(because w 7→

R

u∈A,v∈B 0

0

w × (u ⊗ v ) is order-continuous) Z Z = (nχc) × (u0 ⊗ v 0 ) = n u0 ⊗ v 0 , c

which is impossible. X X Thus C is bounded above in L0 (C), and has a supremum w ∈ L0 (C). If u0 ∈ U + , v 0 ∈ V + then Z

Z 0

0

w × (u ⊗ v ) =

sup

(u ⊗ v) × (u0 ⊗ v 0 )

u∈A,v∈B

=

sup u∈A,v∈B

fu (u0 )gv (v 0 ) = f (u0 )g(v 0 ) = Pf g (u0 )(v 0 ).

Thus w ∈ W and ˜ ⊆W c. Pf g = Tw ∈ W c for every f ∈ U × , And this is true for any non-negative f ∈ U × , g ∈ V × . Of course it follows that Pf g ∈ W c is a band, it must include Z. g ∈ V × ; as W c ⊆ Z. P c ∩ Z ⊥ = {0}. Take any T > 0 in (j) Finally, W P Since Z = Z ⊥⊥ , it is enough to show that W c . There are u0 ∈ U + , v 0 ∈ V + such that T (u0 )(v 0 ) > 0. So there is a v ∈ V # such that 0 ≤ gv ≤ T (u0 ) W 0 0 0 0 R0 and gv (v00 ) > 0, that is, v × v00 > 0. Because V is order-dense in L0 (B), there is a v10 ∈ V such that 0 < v10 ≤ v00 × χ[[v > 0]], so that 0
0]] ⊆ [[v > 0]]. Now consider the functional u0 7→ h(u0 ) = T (u0 )(v10 ) : U → R. This belongs to (U × )+ and h(u00 ) > 0, R # 0 so there is a u ∈ U such that 0 ≤ fu ≤ h and fu (u0 ) > 0. This time, u × u00 > 0 so (because U is order-dense in L0 (A)) there is a u01 ∈ U such that h(u01 ) > 0 and [[u01 > 0]] ⊆ [[u > 0]]. We can express T as Tw where w ∈ W + . In this case, we have

R

w × (u01 ⊗ v10 ) = T (u01 )(v10 ) = h(u01 ) > 0,

so 0 6= [[w > 0]] ∩ [[u01 ⊗ v10 > 0]] = [[w > 0]] ∩ ([[u01 > 0]] ⊗ [[v10 > 0]]) ⊆

[[w > 0]] ∩ ([[u > 0]] ⊗ [[v > 0]]) = [[w > 0]] ∩ [[u ⊗ v > 0]],

and w ∧ (u ⊗ v) > 0, so Tw ∧ Pfu gv = Tw ∧ Tu⊗v = Tw∧(u⊗v) > 0. c ∩ Z ⊥ = {0} and W c ⊆ Z ⊥⊥ = Z. Q Thus T ∈ / Z ⊥ . Accordingly W Q c Since we already know that Z ⊆ W , this completes the proof.

464

Linear operators between function spaces

376F

376F Corollary Let (A, µ ¯) and (B, ν¯) be localizable measure spaces, with localizable measure algebra ¯ Let U ⊆ L0 (A), V ⊆ L0 (B) be perfect order-dense solid linear subspaces, and T : U → V free product (C, λ). a linear operator. Then the following are equiveridical: (i) T is an abstract integral operator; R R (ii) there is a w ∈ L0 (C) such that w × (u ⊗ v 0 ) is defined and equal to T u × v 0 whenever u ∈ U and v 0 ∈ L0 (B) is such that v 0 × v is integrable for every v ∈ V . proof Setting V # = {v 0 : v 0 ∈ L0 (B), v × v 0 ∈ L1 for every v ∈ V }, we know that we can identify V # with V × and V with (V # )× (369C). So the equivalence of (i) and (ii) is just 376E applied to V # in place of V . 376G Lemma Let U be a Riesz space, V an Archimedean Riesz space, T : U → V a linear operator, f ∈ (U ∼ )+ and e ∈ V + . Suppose that 0 ≤ T u ≤ f (u)e for every u ∈ U + . Then if hun in∈N is a sequence in U such that limn→∞ g(un ) = 0 whenever g ∈ U ∼ and |g| ≤ f , hT un in∈N order*-converges to 0 in V (definition: 367A). proof Let Ve be the solid linear subspace of V generated by e; then T u ∈ Ve for every u ∈ U . We can identify Ve with an order-dense and norm-dense Riesz subspace of C(X), where X is a compact Hausdorff space, with e corresponding to χX (353M). For x ∈ X, set gx (u) = (T u)(x) for every u ∈ U ; then 0 ≤ gx (u) ≤ f (u) for u ≥ 0, so |gx | ≤ f and limn→∞ (T un )(x) = 0. As x is arbitrary, hT un in∈N order*-converges to 0 in C(X), by 367L, and therefore in Ve , because Ve is order-dense in C(X) (367F). But Ve , regarded as a subspace of V , is solid, so 367F also tells us that hT un in∈N order*-converges to 0 in V . 376H Theorem Let U be a Riesz space and V a weakly (σ, ∞)-distributive Dedekind complete Riesz space (definition: 368N). Suppose that T ∈ L× (U ; V ). Then the following are equiveridical: (i) T is an abstract integral operator; (ii) whenever hun in∈N is an order-bounded sequence in U + and limn→∞ f (un ) = 0 for every f ∈ U × , then hT un in∈N order*-converges to 0 in V ; (iii) whenever hun in∈N is an order-bounded sequence in U and limn→∞ f (un ) = 0 for every f ∈ U × , then hT un in∈N order*-converges to 0 in V . proof For f ∈ U × , v ∈ V and u ∈ U set Pf v (u) = f (u)v. Write Z ⊆ L× (U ; V ) for the space of abstract integral operators. (a)(i)⇒(iii) Suppose that T ∈ Z + , and that hun in∈N is an order-bounded sequence in U such that limn→∞ f (un ) = 0 for every f ∈ U × . Note that {Pf v : f ∈ U ×+ , v ∈ V + } is upwards-directed, so that T = sup{T ∧ Pf v : f ∈ U ×+ , v ∈ V + } (352Va). Take u∗ ∈ U + such that |un | ≤ u∗ for every n, and set w = inf n∈N supm≥n T um , which is defined because |T un | ≤ T u∗ for every n. Now w ≤ (T − Pf v )+ (u∗ ) for every f ∈ U ×+ , v ∈ V + . P P Setting T1 = T ∧ Pf v , w0 = (T − Pf v )+ (u∗ ) we have T un − T1 un ≤ |T − T1 |(u∗ ) = (T − Pf v )+ (u∗ ) = w0 for every n ∈ N, so T un ≤ w0 + T1 un . On the other hand, 0 ≤ T1 u ≤ f (u)v for every u ∈ U + , so by 376G we must have inf n∈N supm≥n T1 um = 0. Accordingly w ≤ w0 + inf n∈N supm≥n T1 um = w0 . Q Q But as inf{(T − Pf v )+ : f ∈ U ×+ , v ∈ V + } = 0, w ≤ 0. Similarly (or applying the same argument to h−un in∈N ), supn∈N inf n∈N T un ≥ 0 and hT un in∈N order*-converges to zero. For general T ∈ Z, this shows that hT + un in∈N and hT − un in∈N both order*-converge to 0, so hT un in∈N order*-converges to 0, by 367C. As hun in∈N is arbitrary, (iii) is satisfied. (b)(iii)⇒(ii) is trivial. (c)(ii)⇒(i) ?? Now suppose, if possible, that (ii) is satisfied, but that T ∈ / Z. Because L× (U ; V ) is Dedekind complete (355H), Z is a projection band (353I), so T is expressible as T1 + T2 where T1 ∈ Z, T2 ∈ Z ⊥ and T2 6= 0. At least one of T2+ , T2− is non-zero; replacing T by −T if need be, we may suppose that T2+ > 0.

376I

Kernel operators

465

Because T2+ , like T , belongs to L× (U ; V ), its kernel U0 is a band in U , which cannot be the whole of U , and there is a u0 > 0 in U0⊥ . In this case T2+ u0 > 0; because T2+ ∧ (T2− + |T1 |) = 0, there is a u1 ∈ [0, u0 ] such that T2+ (u0 − u1 ) + (T2− + |T1 |)(u1 ) 6≥ T2+ u0 , so that T u1 ≥ T2 u1 − |T1 |(u1 ) 6≤ 0 and T u1 6= 0. Now this means that the sequence (T u1 , T u1 , . . . ) is not order*-convergent to zero, so there must be some f ∈ U × such that (f (u1 ), f (u1 ), . . . ) does not converge to 0, that is, f (u1 ) 6= 0; replacing f by |f | if necessary, we may suppose that f ≥ 0 and that f (u1 ) > 0. By 356H, there is a u2 such that 0 < u2 ≤ u1 and g(u2 ) = 0 whenever g ∈ U × and g ∧ f = 0. Because 0 < u2 ≤ u0 , u2 ∈ U0⊥ and v0 = T2+ u2 > 0. Consider Pf v0 ∈ Z. Because T2 ∈ Z ⊥ , T2+ ∧ Pf v0 = 0; set S = Pf v0 + T2− , so that T2+ ∧ S = 0. Then inf u∈[0,u2 ] T2+ (u2 − u) + Su = 0,

supu∈[0,u2 ] T2+ u − Su = v0

(use 355Ec for the first equality, and then subtract both sides from v0 ). Now Su ≥ f (u)v0 for every u ≥ 0, so that for any ² > 0 supu∈[0,u2 ],f (u)≥² T2+ u − Su ≤ (1 − ²)v0 and accordingly supu∈[0,u2 ],f (u)≤² T2+ u = v0 , since the join of these two suprema is surely at least v0 , while the second is at most v0 . Note also that v0 = supu∈[0,u2 ],f (u)≤² T2+ u = sup0≤u0 ≤u≤u2 ,f (u)≤² T2 u0 = sup0≤u0 ≤u2 ,f (u0 )≤² T2 u0 . For k ∈ N set Ak = {u : 0 ≤ u ≤ u2 , f (u) ≤ 2−k }. We know that Bk = {supu∈I T2 u : I ⊆ Ak is finite} is an upwards-directed set with supremum v0 for each k. Because V is weakly (σ, ∞)-distributive, we can find a sequence hvk0 ik∈N such that vk0 ∈ Bk for every k and v1 = inf k∈N vk0 > 0. For each k let Ik ⊆ Ak be a finite set such that vk0 = supu∈Ik T2 u. Because each Ik is finite, we can build a sequence hu0n in∈N in [0, u2 ] enumerating each in turn, so that limn→∞ f (u0n ) = 0 (since f (u) ≤ 2−k if u ∈ Ik ) while supm≥n T2 u0m ≥ v1 for every n (since {u0m : m ≥ n} always includes some Ik ). Now hT2 u0n in∈N does not order*-converge to 0. However, limn→∞ g(u0n ) = 0 for every g ∈ U × . P P Express |g| as g1 + g2 where g1 belongs to the band of U × generated by f and g2 ∧ f = 0 (353Hc). Then g2 (u0n ) = g2 (u2 ) = 0 for every n, by the choice of u2 . Also g1 = supn∈N g1 ∧ nf (352Vb); so, given ² > 0, there is an m ∈ N such that (g1 − mf )+ (u2 ) ≤ ² and (g1 − mf )+ (u0n ) ≤ ² for every n ∈ N. But this means that |g(u0n )| ≤ |g|(u0n ) ≤ ² + mf (u0n ) for every n, and lim supn→∞ |g(u0n )| ≤ ²; as ² is arbitrary, limn→∞ g(u0n ) = 0. Q Q 0 Now, however, part (a) of this proof tells us that hT1 un in∈N is order*-convergent to 0, because T1 ∈ Z, while hT u0n in∈N is order*-convergent to 0, by hypothesis; so hT2 u0n in∈N = hT u0n − T1 u0n in∈N order*-converges to 0. X X This contradiction shows that every operator satisfying the condition (ii) must be in Z. 376I

The following elementary remark will be useful for the next corollary and also for Theorem 376S.

Lemma Let (X, Σ, µ) be a σ-finite measure space and U an order-dense solid linear subspace of L0 (µ). Then there is a non-decreasing sequence hXn in∈N of measurable subsets of X, with union X, such that χXn• ∈ U for every n ∈ N. proof Write A for the measure algebra of µ, so that L0 (µ) can be identified with L0 (A) (364Jc). A = {a : a ∈ A \ {0}, χa ∈ U } is order-dense in A, so includes a partition of unity hai ii∈I . Because µ is σ-finite, A is • ccc (322G) and I isScountable, so we can S take I to be a subset of N. Choose Ei ∈ Σ such that Ei = ai for i ∈ I; set E = X \ i∈I Ei , Xn = E ∪ i∈I,i≤n Ei for n ∈ N.

466

Linear operators between function spaces

376J

376J Corollary Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces, with product measure λ on X × Y . Let U ⊆ L0 (µ), V ⊆ L0 (ν) be perfect order-dense solid linear subspaces, and T : U → V a linear operator. Write U = {f : f ∈ L0 (µ), f • ∈ U }, V# = {h : h ∈ L0 (ν), h• × v ∈ L1 for every v ∈ V }. Then the following are equiveridical: (i) T is an abstract integral operator; 0 (ii) there R is a k ∈ L (λ) such that (α) |k(x, y)f (x)h(y)|d(x, y) 0. Let hXn in∈N be a non-decreasing sequence of sets of finite measure covering X, and set an = Xn• for each n. For each n, inf u∈A [[u > 2−n ]] = 0, so we can find u ˜n ∈ A such that µ ¯(an ∩ [[˜ un > 2−n ]]) ≤ 2−n . Set un = inf i≤n u ˜i for each n; then hun in∈N is non-increasing and has infimum 0; also, [0, un ] meets A for each n, so that v0 ≤ sup{T u : 0 ≤ u ≤ un } for each n. Because V is weakly (σ, ∞)-distributive, we can find + a sequence hIS > 0. n in∈N of finite sets such that In ⊆ [0, un ] for each n and v1 = inf n∈N supu∈In (T u) 0 0 Enumerating n∈N In as hun in∈N , as in part (d) of the proof of 376H, we see that hun in∈N is order-bounded

376M

Kernel operators

467

and limn→∞ f (u0n ) = 0 for every f ∈ U × (indeed, hu0n in∈N order*-converges to 0 in U ), while hT u0n in∈N 6→ 0 in V . X XQ Q Similarly, T − is order-continuous, so T ∈ L× (U ; V ). Accordingly T is an abstract integral operator by condition (ii) of 376H. 376K As an application of the ideas above, I give a result due to N.Dunford (376N) which was one of the inspirations underlying the theory. Following the method of Zaanen 83, I begin with a couple of elementary lemmas. Lemma Let U and V be Riesz spaces. Then there is a Riesz space isomorphism T 7→ T 0 : L× (U ; V × ) → L× (V ; U × ) defined by the formula (T 0 v)(u) = (T u)(v) for every u ∈ U , v ∈ V . If for f ∈ U × , g ∈ V × we write Pf g (u) = f (u)g for every u ∈ U , then Pf g ∈ L× (U ; V × ) and Pf0 g = Pgf in L× (V ; U × ). Consequently T is an abstract integral operator iff T 0 is. proof All the ideas involved have already appeared. For positive T ∈ L× (U ; V × ) the functional (u, v) 7→ (T u)(v) is bilinear and order-continuous in each variable separately; so (just as in the first part of the proof of 376E) corresponds to a T 0 ∈ L× (V ; U × ). The map T 7→ T 0 : L× (U ; V × )+ → L× (V ; U × )+ is evidently an additive, order-preserving bijection, so extends to an isomorphism between L× (U ; V × ) and L× (V ; U × ) given by the same formula. I remarked in part (i) of the proof of 376E that every Pf g belongs to L× (U ; V × ), and the identification Pf0 g = Pgf is just a matter of checking the formulae. Of course it follows at once that the bands of abstract integral operators must also be matched by the map T 7→ T 0 . 376L Lemma Let U be a Riesz space with an order-continuous norm. If w ∈ U + there is a g ∈ U × such that for every ² > 0 there is a δ > 0 such that kuk ≤ ² whenever 0 ≤ u ≤ w and g(u) ≤ δ. proof (a) As remarked in 356D, U ∗ = U ∼ = U × . Set A = {v : v ∈ U and there is an f ∈ (U × )+ such that f (u) > 0 whenever 0 < u ≤ |v|}. Then v 0 ∈ A whenever |v 0 | ≤ |v| ∈ A and v + v 0 ∈ A for all v, v 0 ∈ A (if f (u) > 0 whenever 0 < u ≤ |v| and f 0 (u) > 0 whenever 0 < u ≤ |v 0 |, then (f + f 0 )(u) > 0 whenever 0 < u ≤ |v + v 0 |); moreover, if v0 > 0 in U , there is a v ∈ A such that 0 < v ≤ v0 . P P Because U × = U ∗ separates the points of U , there is a g > 0 in × U such that g(v0 ) > 0; now by 356H there is a v ∈ ]0, v0 ] such that g is strictly positive on ]0, v], so that v ∈ A. Q Q But this means that A is an order-dense solid linear subspace of U . (b) In fact w ∈ A. P P w = sup B, where B = A ∩ [0, w]. Because B is upwards-directed, w ∈ B (354Ea), and there is a sequence hu0n in∈N in B converging to w for the norm. For each n, choose fn ∈ (U × )+ such that fn (u) > 0 whenever 0 < u ≤ u0n . Set P∞ 1 f = n=0 n fn 2 (1+kfn k)

∗

×

in U = U . Then whenever 0 < u ≤ w there is some n ∈ N such that u ∧ u0n > 0, so that fn (u) > 0 and f (u) > 0. So f witnesses that w ∈ A. Q Q (c) Take g ∈ (U × )+ such that g(u) > 0 whenever 0 < u ≤ w. This g serves. P P?? Otherwise, there is some ² > 0 such that for every n ∈ N we can find a un ∈ [0, w] with g(un ) ≤ 2−n and kun k ≥ ². Set vn = supi≥n ui ; then 0 ≤ vn ≤ w, g(vn ) ≤ 2−n+1 and kvn k ≥ ² for every n ∈ N. But hvn in∈N is nondecreasing, so v = inf n∈N vn must be non-zero, while 0 ≤ v ≤ w and g(v) = 0; which is impossible. X X Q Q Thus we have found an appropriate g. 376M Theorem (a) Let U be a Banach lattice with an order-continuous norm and V a Dedekind complete M -space. Then every bounded linear operator from U to V is an abstract integral operator. (b) Let U be an L-space and V a Banach lattice with order-continuous norm. Then every bounded linear operator from U to V × is an abstract integral operator.

468

Linear operators between function spaces

376M

proof (a) By 355Kb and 355C, L× (U ; V ) = L∼ (U ; V ) ⊆ B(U ; V ); but since norm-bounded sets in V are also order-bounded, {T u : |u| ≤ u0 } is bounded above in V for every T ∈ B(U ; V ) and u0 ∈ U + , and B(U ; V ) = L× (U ; V ). I repeat ideas from the proof of 376H. (I cannot quote 376H directly as I am not assuming that V is weakly (σ, ∞)-distributive.) ?? Suppose, if possible, that B(U ; V ) is not the band Z of abstract integral operators. In this case there is a T > 0 in Z ⊥ . Take u1 ≥ 0 such that T u1 6= 0. Let f ≥ 0 in U × be such that for every ² > 0 there is a δ > 0 such that kuk ≤ ² whenever 0 ≤ u ≤ u1 and f (u) ≤ δ (376L). Set v0 = T u1 . Then, just as in part (d) of the proof of 376H, supu∈[0,u1 ],f (u)≤δ T u = v0 for every δ > 0. But there is a δ > 0 such that kT kkuk ≤ 12 kv0 k whenever 0 ≤ u ≤ u1 and f (u) ≤ δ; in which case k supu∈[0,u1 ],f (u)≤δ T uk ≤ 12 kv0 k, which is impossible. X X Thus Z = B(U ; V ), as required. (b) Because V has an order-continuous norm, V ∗ = V × = V ∼ ; and the norm of V ∗ is a Fatou norm with the Levi property (356Da). So B(U ; V ∗ ) = L× (U ; V × ), by 371C. By 376K, this is canonically isomorphic to L× (V ; U × ). Now U × = U ∗ is an M -space (356Pb). By (a), every member of L× (V ; U × ) is an abstract integral operator; but the isomorphism between L× (V ; U × ) and L× (U ; V × ) matches the abstract integral operators in each space (376K), so every member of B(U ; V ∗ ) is also an abstract integral operator, as claimed. 376N Corollary: Dunford’s theorem Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces and T : L1 (µ) → Lp (ν) a bounded linear operator, where 1 < p ≤ ∞. Then there is a measurable function R k : X × Y → R such that T f • = gf• , where gf (y) = k(x, y)f (x)dx almost everywhere, for every f ∈ L1 (µ). p proof Set q = p−1 if p is finite, 1 if p = ∞. We can identify Lp (ν) with V × , where V = Lq (ν) ∼ = Lp (ν)× (366Dc, 365Jc) has an order-continuous norm because 1 ≤ q < ∞. By 376Mb, T is an abstract integral operator. By 376F/376J, T is represented by a kernel, as claimed.

376O

Under the right conditions, weakly compact operators are abstract integral operators.

Lemma Let U be a Riesz space, and W a solid linear subspace of U ∼ . If C ⊆ U is relatively compact for the weak topology Ts (U, W ) (3A5E), then for every g ∈ W + and ² > 0 there is a u∗ ∈ U + such that g(|u| − u∗ )+ ≤ ² for every u ∈ C. proof Let Wg be the solid linear subspace of W generated by g. Then Wg is an Archimedean Riesz space with order unit, so Wg× is a band in the L-space Wg∗ = Wg∼ (356Na), and is therefore an L-space in its own right (354O). For u ∈ U , h ∈ Wg× set (T u)(h) = h(u); then T is an order-continuous Riesz homomomorphism from U to Wg× (356F). Now Wg is perfect. P P I use 356K. Wg is Dedekind complete because it is a solid linear subspace of the Dedekind complete space U ∼ . Wg× separates the points of W because T [U ] does. If A ⊆ Wg is upwardsdirected and suph∈A φ(h) is finite for every φ ∈ Wg× , then A acts on Wg× as a set of bounded linear functionals which, by the Uniform Boundedness Theorem (3A5Ha), is uniformly bounded; that is, there is some M ≥ 0 such that suph∈A |φ(h)| ≤ M kφk for every φ ∈ Wg× . Because g is the standard order unit of Wg , we have kφk = |φ|(g) and |φ(h)| ≤ M |φ|(g) for every φ ∈ Wg× , h ∈ A. In particular, h(u) ≤ |h(u)| = |(T u)(h)| ≤ M |T u|(g) = M T u(g) = M g(u) +

for every h ∈ A, u ∈ U . But this means that h ≤ M g for every h ∈ A and A is bounded above in Wg . Thus all the conditions of 356K are satisfied and Wg is perfect. Q Q Accordingly T is continuous for the topologies Ts (U, W ) and Ts (Wg× , Wg×× ), because every element φ of Wg×× corresponds to a member of Wg ⊆ W , so 3A5Ec applies. Now we are supposing that C is relatively compact for Ts (U, W ), that is, is included in some compact set C 0 ; accordingly T [C 0 ] is compact and T [C] is relatively compact for Ts (Wg× , Wg×× ). Since Wg× is an L-space, T [C] is uniformly integrable (356Q); consequently (ignoring the trivial case C = ∅) there are φ0 , . . . , φn ∈ T [C] such that k(|φ| − supi≤n |φi |)+ k ≤ ² for every φ ∈ T [C] (354Rb), so that (|φ| − supi≤n |φi |)+ (g) ≤ ² for every φ ∈ T [C].

376R

Kernel operators

469

Translating this back into terms of C itself, and recalling that T is a Riesz homomorphism, we see that there are u0 , . . . , un ∈ C such that g(|u| − supi≤n |ui |)+ ≤ ² for every u ∈ C. Setting u∗ = supi≤n |ui | we have the result. 376P Theorem Let U be an L-space and V a perfect Riesz space. If T : U → V is a linear operator such that {T u : u ∈ U, kuk ≤ 1} is relatively compact for the weak topology Ts (V, V × ), then T is an abstract integral operator. proof (a) For any g ≥ 0 in V × , Mg = supkuk≤1 g(|T u|) is finite. P P By 376O, there is a v ∗ ∈ V + such ∗ + ∗ that g(|T u| − v ) ≤ 1 whenever kuk ≤ 1; now Mg ≤ g(v ) + 1. Q Q Considering kuk−1 u, we see that g(|T u|) ≤ Mg kuk for every u ∈ U . Next, we find that T ∈ L∼ (U ; V ). P P Take u ∈ U + . Set Pn Pn B = { i=0 |T ui | : u0 , . . . , un ∈ U + , i=0 ui = u} ⊆ V + . Then B is upwards-directed. (Cf. 371A.) If g ≥ 0 in V × , n X

sup g(v) = sup{ v∈B

i=0 n X

≤ sup{

n X

g(|T ui |) : Mg kui k :

i=0

i=0 n X

ui = u} ui = u} = Mg kuk

i=0

is finite. By 356K, B is bounded above in V ; and of course any upper bound for B is also an upper bound for {T u0 : 0 ≤ u0 ≤ u}. As u is arbitrary, T is order-bounded. Q Q Because U is a Banach lattice with an order-continuous norm, T ∈ L× (U ; V ) (355Kb). (b) Since we can identify L× (U ; V ) with L× (U ; V ×× ), we have an adjoint operator T 0 ∈ L× (V × ; U × ), as in 376K. Now if g ≥ 0 in V × and hgn in∈N is a sequence in [0, g] such that limn→∞ gn (v) = 0 for every v ∈ V , hT 0 gn in∈N order*-converges to 0 in U × . P P For any ² > 0, there is a v ∗ ∈ V + such that g(|T u| − v ∗ )+ ≤ ² whenever kuk ≤ 1; consequently kT 0 gn k = sup (T 0 gn )(u) = sup gn (T u) kuk≤1

kuk≤1

∗

≤ gn (v ) + sup gn (|T u| − v ∗ )+ kuk≤1 ∗

≤ gn (v ) + sup g(|T u| − v ∗ )+ ≤ gn (v ∗ ) + ² kuk≤1

for every n ∈ N. As limn→∞ gn (v ∗ ) = 0, lim supn→∞ kT 0 gn k ≤ ²; as ² is arbitrary, hkT 0 gn kin∈N → 0. But as U × is an M -space (356Pb), it follows that hT 0 gn in∈N order*-converges to 0. Q Q By 368Pc, V is weakly (σ, ∞)-distributive. By 376H, T 0 is an abstract integral operator, so T also is, by 376K. 376Q Corollary Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces and T : L1 (µ) → L1 (ν) a weakly compact bounded linear operator. Then there is a function k : X × Y → R such that T f • = gf• , where R gf (y) = k(x, y)f (x)dx almost everywhere, for every f ∈ L1 (µ). proof This follows from 376P and 376J, just as in 376N. 376R So far I have mentioned actual kernel functions k(x, y) only as a way of giving slightly more concrete form to the abstract kernels of 376E. But of course they can provide new structures and insights. I give one result as an example. The following lemma is useful. Lemma Let (X, Σ, µ) be a measure space, (Y, T, ν) a σ-finite measure space, and λ the c.l.d. product measure on X × Y . Suppose that k is a λ-integrable real-valued function. Then for any ² > 0 there is a finite partition E0 , . . . , En of X into measurable sets such that kk − k1 k1 ≤ ², where

470

Linear operators between function spaces

k1 (x, y) =

1 µEi

376R

Z k(t, y)dt whenever x ∈ Ei , 0 < µEi < ∞ Ei

and the integral is defined in R, = 0 in all other cases. proof Once again I refer to the proof of 253F: there are sets H0 , . . . , Hr of finite P measure in X, sets r F0 , . . . , Fr of finite measure in Y , and α0 , . . . , αr such that kk − k2 k1 ≤ 12 ², where k2 = j=0 αi χ(Hj × Fj ). R Let E0 , . . . , En be the partition of X generated by {Hi : i ≤ r}. Then for any i ≤ n, Ei ×Y |k − k1 | is R defined and is at most 2 Ei ×Y |k − k2 |. P P If µEi = 0, this is trivial, as both are zero. If µEi = ∞, then again the result is elementary, since both k 1 and k2 are zero on Ei × Y . So let us suppose that 0 < µEi < ∞. R In this case Ei k(t, y)dt must be defined for almost every y, by Fubini’s theorem. So k1 is defined almost everywhere on Ei × Y , and

R

Ei ×Y

|k − k1 | =

R R Y

Now take some fixed y ∈ Y such that β=

1 µEi

Ei

|k(x, y) − k1 (x, y)|dxdy.

R Ei

k(t, y)dt

is defined. Then β = k1 (x, y) for every x ∈ Ei . For every x ∈ Ei , we must have k2 (x, y) = α where R P α = {αj : Ei ⊆ Hj , y ∈ Fj }. But in this case, because Ei k(x, y) − β dx = 0, we have

R

Ei

If β ≥ α,

max(0, k(x, y) − β)dx =

R Ei

if β ≤ α,

R Ei

R

max(0, k(x, y) − β)dx ≤

max(0, β − k(x, y))dx ≤

in either case, 1 2

R Ei

Ei

max(0, β − k(x, y))dx =

R Ei

R Ei

1 2

max(0, k(x, y) − α)dx ≤

max(0, α − k(x, y))dx ≤

|k(x, y) − k1 (x, y)|dx ≤

R Ei

R

Ei

R Ei

R Ei

|k(x, y) − k1 (x, y)|dx.

|k(x, y) − k2 (x, y)|dx; |k(x, y) − k2 (x, y)|dx;

|k(x, y) − k2 (x, y)|dx.

This is true for almost every y, so integrating with respect to y we get the result. Q Q Now, summing over i, we get

R

|k − k1 | ≤ 2

R

|k − k2 | ≤ ²,

as required. 376S Theorem Let (X, Σ, µ) be a complete locally determined measure space, (Y, T, ν) a σ-finite measure space, and λ the c.l.d. product measure on X × Y . Let τ be an extended Fatou norm on L0 (ν) and 0 write Lτ for {g : g ∈ L0 (ν), τ 0 (g • ) < ∞}, where τ 0 is the associate extended Fatou norm of τ (369H-369I). 0 Suppose that k ∈ L0 (λ) is such that k × (f ⊗ g) is integrable whenever f ∈ LR1 (µ) and g ∈ LRτ . Then we have a corresponding linear operator T : L1 (µ) → Lτ defined by saying that (T f • ) × g • = k × (f ⊗ g) 0 whenever f ∈ L1 (µ), g ∈ Lτ . For x ∈ X set kx (y) = k(x, y) whenever this is defined. Then kx ∈ L0 (ν) for almost every x; set vx = kx• ∈ L0 (ν) for such x. In this case x 7→ τ (vx ) is measurable and defined and finite almost everywhere, and kT k = ess supx τ (vx ). Remarks The discussion of extended Fatou norms in §369 regarded them as functionals on spaces of the form L0 (A). I trust that no-one will be offended if I now speak of an extended Fatou norm on L0 (ν), with 0 the associated function spaces Lτ , Lτ ⊆ L0 , taking for granted the identification in 364Jc. Recall that (f ⊗ g)(x, y) = f (x)g(y) for x ∈ dom f , y ∈ dom g (253B).

376S

Kernel operators

471

By ‘ess supx τ (vx )’ I mean inf{M : M ≥ 0, {x : vx is defined and τ (vx ) ≤ M } is conegligible} (see 243D). R 0 proof (a) To see that the formula (f, g) 7→ k × (f ⊗ g) gives rise to an operator in L× (U ; (Lτ )× ), it is perhaps quickest to repeat the argument of parts (a) and (b) of the proof of 376E. (We are not quite in a position to quote 376E, as stated, because the localizable measure algebra free product there might be strictly larger than the measure algebra of λ; see 325B.) R The first step, of course, is to note that changing f or g on a negligible set does not affect the integral k × (f ⊗ g), so that we have a bilinear functional on 0 L1 × Lτ ; and the other essential element is the fact that the maps f • 7→ (f ⊗ χY )• , g • 7→ (χX ⊗ g)• are order-continuous (put 325A and 364Rc together). 0 By 369K, we can identify (Lτ )× with Lτ , so that T becomes an operator in L× (U ; Lτ ). Note that it must be norm-bounded (355C). (b) By 376I, there is a non-decreasing sequence hYn in∈N of measurable sets in Y , covering Y , such that 0 χYn ∈ Lτ for every n. Set X0 = {x : x ∈ X, kx ∈ L0 (ν)}. Then X0 is conegligible in P Let E ∈ Σ be R X. P any set of finite measure. Then for any n ∈ N, k × (χE ⊗ χYn ) is integrable, that is, E×Yn k is defined and R finite; so by Fubini’s theorem Yn kx is defined and finite for almost every x ∈ E. Consequently, for almost every x ∈ E, kx × χYn ∈ L0 (ν) for every n ∈ N, that is, kx ∈ L0 (ν), that is, x ∈ X0 . Thus E \ X0 is negligible for every set E of finite measure. Because µ is complete and locally determined, X0 is conegligible. Q Q This means that vx and τ (vx ) are defined for almost every x. (c) τ (vx ) ≤ kT k for almost every x. P P Take any E ∈ Σ of finite measure, and n ∈ N. Then k × χ(E × Yn ) is integrable. For each r ∈ N, there is a finite partition Er0 , . . . , Er,m(r) of E into measurable sets such that R |k − k (r) | ≤ 2−r , where E×Yn Z 1 k (r) (x, y) = k(t, y)dt whenever y ∈ Yn , x ∈ Eri , µEri > 0 µEri

Eri

and the integral is defined in R, = 0 otherwise (r)

(r)

(376R). Now k (r) is also integrable over E × Yn , so kx ∈ L0 (ν) for almost every x ∈ E, writing kx (y) = (r) (r) (r) (r) k (r) (x, y), and we can speak of vx = (kx )• for almost every x. Note that kx = kx0 whenever x, x0 belong to the same Eri . 0 0 (r) If µEri > 0, then vx must be defined for every x ∈ Eri . If v 0 ∈ Lτ is represented by g ∈ Lτ then Z Z k × (χEri ⊗ (g × χYn )) = k(t, y)g(y)d(t, y) Eri ×Yn Z Z = µEri k (r) (x, y)g(y)dy = µEri vx(r) × v 0 for any x ∈ Eri . But this means that µEri

R

(r)

vx × v 0 =

R

• T (χEri ) × v 0 × χYn•

0

for every v 0 ∈ Lτ , so (r)

vx =

1 • T (χEri ) × χYn• , µEri

(r)

τ (vx ) ≤

1 • kT kkχEri k1 µEri (r)

= kT k

for every x ∈ Eri . RThis is true whenever µEri > 0, so in fact τ (vx ) ≤ kT k for almost every x ∈ E. P Because r∈N E×Yn |k − k (r) | < ∞, we must have k(x, y) = limr→∞ k (r) (x, y) for almost every (x, y) ∈ E × Yn . Consequently, for almost every x ∈ E, k(x, y) = limr→∞ k (r) (x, y) for almost every y ∈ Yn , that is,

472

Linear operators between function spaces

376S

(r)

hvx ir∈N order*-converges to vx × χYn• (in L0 (ν)) for almost every x ∈ E. But this means that, for almost every x ∈ E, (r)

τ (vx × χYn• ) ≤ lim inf r→∞ τ (vx ) ≤ kT k (369Mc). Now τ (vx ) = limn→∞ τ (vx × χYn• ) ≤ kT k for almost every x ∈ E. As in (b), this implies (since E is arbitrary) that τ (vx ) ≤ kT k for almost every x ∈ X. Q Q (d) I now show that x 7→ τ (vx ) is measurable. P P Take γ ∈ [0, ∞[ and set A = {x : x ∈ X0 , τ (vx ) ≤ γ}. ˜ y) = k(x, y) when x ∈ G Suppose that µE < ∞. Let G be a measurable envelope of A ∩ E (132Ed). Set k(x, 0 and (x, y) ∈ dom k, 0 otherwise. If f ∈ L1 (µ) and g ∈ Lτ , then

R

˜ y)f (x)g(y)d(x, y) = k(x,

R

k(x, y)f (x)g(y)d(x, y) =

G×Y

R

G

f (x)

R

Y

k(x, y)g(y)dydx

is defined. R ˜ y)g(y)|dy. Then h is finite almost everywhere and measurable. For For x ∈ X0 , set h(x) = |k(x, x ∈ A ∩ E, R R ˜ y)g(y)|dy = |vx × g • | ≤ γτ 0 (g • ). |k(x, So the measurable set G0 = {x : h(x) ≤ γτ 0 (g • )} includes A ∩ E, and µ(G \ G0 ) = 0. Consequently |

R

˜ y)f (x)g(y)d(x, y)| ≤ k(x,

R

G

|f (x)|h(x)dx ≤ γkf k1 τ 0 (g • ),

0

and this is true for every f ∈ L1 (µ), g ∈ Lτ . Now we have an operator T˜ : L1 (µ) → Lτ defined by the formula

R

R

0 k˜ × (f ⊗ g) when f ∈ L1 (ν) and g ∈ Lτ , R 0 and the formula just above tells us that | T˜u × v 0 | ≤ γkuk1 τ 0 (v 0 ) for every u ∈ L1 (ν), v 0 ∈ Lτ ; that is, τ (T˜u) ≤ γkuk1 for every u ∈ L1 (µ); that is, kT˜k ≤ γ. But now (c) tells us that τ (˜ vx ) ≤ γ for almost every ˜ x ∈ X, where v˜x is the equivalence class of y 7→ k(x, y), that is, v˜x = vx for x ∈ G ∩ X0 , 0 for x ∈ X \ G. So τ (vx ) ≤ γ for almost every x ∈ G, and G \ A is negligible. But this means that A ∩ E is measurable. As E is arbitrary, A is measurable; as γ is arbitrary, x 7→ τ (vx ) is measurable. Q Q

(T˜f • ) × g • =

(e) Finally, the ideas in (d) show that kT k ≤ ess supx τ (vx ). P P Set M = ess supx τ (vx ). If f ∈ L1 (µ) τ0 and g ∈ L , then

R

|k(x, y)f (x)g(y)|d(x, y) ≤

R

|f (x)|τ (vx )τ 0 (g • )dx ≤ M kf k1 τ 0 (g • );

as g is arbitrary, τ (T f • ) ≤ M kf k1 ; as f is arbitrary, kT k ≤ M . Q Q 376X Basic exercises > (a) Let µ be Lebesgue measure on R. Let h be a µ-integrable real-valued function with khk1 ≤ 1,R and set k(x, y) = h(y − x) whenever this is defined. Show that if f is in either L1 (µ) or L∞ (µ) then g(y) = k(x, y)f (x)dx is defined for almost every y ∈ R, and that this formula gives rise to an operator T ∈ Tµ¯×,¯µ as defined in 373A. (Hint: 255H.) (b) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras with localizable measure algebra free product ¯ and take p ∈ [1, ∞]. Show that if u ∈ Lp (A, µ ¯ and (C, λ), ¯) and v ∈ Lp (B, ν¯) then u ⊗ v ∈ Lp (C, λ) ku ⊗ vkp = kukp kvkp . > (c) Let U , V , W be Riesz spaces, of which V and W are Dedekind complete, and suppose that T ∈ L× (U ; V ) and S ∈ L× (V ; W ). Show that if either S or T is an abstract integral operator, so is ST . (d) Let h be a Lebesgue integrable function on R, and f a square-integrable function. Suppose that R hfn in∈N is a sequence of measurable functions such that (α) |fn | ≤ f for every n (β) limn→∞ E fn = 0 for every measurable set E of finite measure. Show that limn→∞ (h ∗ fn )(y) = 0 for almost every y ∈ R, where h ∗ fn is the convolution of h and fn . (Hint: 376Xa, 376F.)

376Xn

Kernel operators

473

(e) Let U and V be Riesz spaces, of which V is Dedekind complete. Suppose that W ⊆ U ∼ is a solid linear subspace, and that T belongs to the band in L∼ (U ; V ) generated by operators of the form u 7→ f (u)v, where f ∈ W and v ∈ V . Show that whenever hun in∈N is an order-bounded sequence in U such that limn→∞ f (un ) = 0 for every f ∈ W , then hT un in∈N order*-converges to 0 in V . (f ) Let (A, µ ¯) be a semi-finite measure algebra and U ⊆ L0 = L0 (A) an order-dense Riesz subspace such × that U separates the points of U . Let hun in∈N be an order-bounded sequence in U . Show that the following are equiveridical: (i) limn→∞ f (|un |) = 0 for every f ∈ U × ; (ii) hun in∈N → 0 for the topology of convergence in measure on L0 . (Hint: by 367T, condition (ii) is intrinsic to U , so we can replace (A, µ ¯) by a localizable algebra and use the representation in 369D.) (g) Let U be a Banach lattice with an order-continuous norm, and V a weakly (σ, ∞)-distributive Riesz space. Show that for T ∈ L∼ (U ; V ) the following are equiveridical: (i) T belongs to the band in L∼ (U ; V ) generated by operators of the form u 7→ f (u)v where f ∈ U ∼ , v ∈ V ; (ii) hT un in∈N order*-converges to 0 in V whenever hun in∈N is an order-bounded sequence in U + which is norm-convergent to 0; (iii) hT un in∈N order*-converges to 0 in V whenever hun in∈N is an order-bounded sequence in U which is weakly convergent to 0. (h) Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces, with product measure λ on X × Y , and measure algebras (A, µ ¯), (B,Rν¯). Suppose that k ∈ L0 (λ). Show that the following are equiveridical: (i)(α) 1 if f ∈ L (µ) then gf (y) = k(x, y)f (x)dx is defined for almost every yR and gf ∈ L1 (ν) (β) there is an operator T ∈ Tµ¯×,¯ν defined by setting T f • = gf• for every f ∈ L1 (µ); (ii) |k(x, y)|dy ≤ 1 for almost every R x ∈ X, |k(x, y)|dx ≤ 1 for almost every y ∈ Y . > (i) Let µ be Lebesgue measureRon R. Give an example of a measurable function k : [0, 1]2 → R such that, for any f ∈ L2 (µ), gf (y) = k(x, y)f (x)dx is defined for every y and kgf k2 = kf k2 , but k is not integrable, so the linear isometry on L2 = L2 (µ) defined by k does not belong to L∼ (L2 ; L2 ). (Hint: 371Ye.) (j) Let (X, Σ, µ) be a σ-finite measure space and (Y, T, ν) a complete locally determined measure space. Let U ⊆ L0 (µ), V ⊆ L0 (ν) be solid linear subspaces, of which V is order-dense; write V # = {v : v ∈ L0 (ν), v × v 0 is integrable for every v 0 ∈ V }, U = {f : f ∈ L0 (ν), f • ∈ U }, V = {g : g ∈ L0 (ν), g • ∈ V }, V# = {h : h ∈ L0 (ν), h• ∈ V # }. Let λ be the c.l.d. product measure on X × Y , and k ∈ LR0 (λ) a function such that k×(f ⊗g) is integrable for every f ∈ U, g ∈ V. (i) Show that for any f ∈ U, hf (y) = k(x, y)f (x)dx × # is defined for almost every y ∈ Y , and that hf ∈ V# . (ii) Show R that we haveRa map T ∈ L (U ; V ) defined either by writing T f • = h•f for every f ∈ U or by writing (T f • ) × g • = k × (f ⊗ g) for every f ∈ U, g ∈ V. (k) Let (X, Σ, µ), (Y, T, ν) and (Z, Λ, λ) be σ-finite measure spaces, and U , V , W perfect order-dense solid linear subspaces of L0 (µ), L0 (ν) and L0 (λ) respectively. Suppose that T : U → V and S : V → W are abstract integral operators corresponding to kernels k1 ∈ L0 (µ × ν), k2 ∈ L0 (ν × λ), writing µ × ν for the (c.l.d. or primitive) product measure on R X × Y . Show that ST : U → W is represented by the kernel k ∈ L0 (µ × λ) defined by setting k(x, z) = k1 (x, y)k2 (y, z)dy whenever this integral is defined. (l) Let U be a perfect Riesz space. Show that a set C ⊆ U is relatively compact for Ts (U, U × ) iff for every g ∈ (U × )+ , ² > 0 there is a u∗ ∈ U such that g(|u| − u∗ )+ ≤ ² for every u ∈ C. (Hint: 376O and the proof of 356Q.) > (m) Let µ be Lebesgue measure on [0, 1], and ν counting measure on [0, 1]. Set k(x, y) = 1 if x = y, 0 otherwise. Show that 376S fails in this context (with, e.g., τ = k k∞ ). (n) Suppose, in 376Xj, that U = Lτ for some extended Fatou norm on L0 (µ) and that V = L1 (ν), so that V # = L∞ (ν). Set ky (x) = k(x, y) whenever this is defined, uy = ky• whenever ky ∈ L0 (ν). Show that uy ∈ Lτ for almost every y ∈ Y , and that the norm of T in B(Lτ ; L∞ ) is ess supy τ 0 (uy ). (Hint: do the case of totally finite Y first.)

474

Linear operators between function spaces

376Y

376Y Further exercises (a) Let U , V and W be linear spaces (over any field F ) and φ : U × V → W a bilinear map. Let W0 be the linear subspace of W generated by φ[U × V ]. Show that the following are equiveridical: (i) for every linear space Z over F and every bilinear ψ : U × V → Z, there is a (unique) linear operator T : W0 → Z such that Pn T φ = ψ (ii) whenever u0 , . . . , un ∈ U are linearly independent and v0 , . . . , vn ∈ V are non-zero, φ(ui , vi ) 6= 0 (iii) whenever u0 , . . . , un ∈ U are non-zero and i=0P n v0 , . . . , vn ∈ V are linearly independent, i=0 φ(ui , vi ) 6= 0 (iv) for any Hamel bases hui ii∈I , hvj ij∈J of U and V , hφ(ui , vj )ii∈I,j∈J is a Hamel basis of W0 (v) for some pair hui ii∈I , hvj ij∈J of Hamel bases of U and V , hφ(ui , vj )ii∈I,j∈J is a Hamel basis of W0 . ¯ their localizable measure algebra free (b) Let (A, µ ¯), (B, ν¯) be semi-finite measure algebras, and (C, λ) 0 0 0 product. Show that ⊗ : L (A) × L (B) → L (C) satisfies the equivalent conditions of 376Ya. (c) Let (X, Σ, µ) and (Y, T, ν) be semi-finite measure spaces and λ the c.l.d. product measure on X × Y . Show that the map (f, g) 7→ f ⊗ g : L0 (µ) × L0 (ν) → L0 (λ) induces a map (u, v) 7→ u ⊗ v : L0 (µ) × L0 (ν) → L0 (λ) possessing all the properties described in 376B and 376Ya. (d) Let (A, µ ¯) be the measure algebra of {0, 1}ω1 with its usual measure, and hbξ iξ (a) Let X be a set and Σ an algebra of subsets of X containing all singleton sets. Show that Aut Σ can be identified with the group of bijections φ : X → X such that φ[E] ∈ Σ, φ−1 [E] ∈ Σ for every E ∈ Σ. > (b) Let A be a Boolean algebra and G any subgroup of Aut A. Let H be the set of those π ∈ Aut A such that for every non-zero a ∈ A there are a non-zero b ⊆ a and a φ ∈ G such that πc = φc for every c ⊆ b. Show that H is a full subgroup of Aut A, the smallest full subgroup of A including G. > (c) Let A be a Dedekind complete Boolean algebra and G any subgroup of A. Show that an element π of Aut A belongs to the full subgroup of Aut A generated by G iff there are a partition of unity hai ii∈I in A and a family hπi ii∈I in G such that πa = πi a whenever i ∈ I and a ⊆ ai . (d) For a Boolean algebra A, let us say that a subgroup G of Aut A is countably full if whenever hai ii∈I is a countable partition of unity in A, hπi ii∈I is a family in G, and π ∈ Aut A is such that πa = πi ai whenever i ∈ I and a ⊆ ai , then π ∈ G. Show that if A is a Dedekind complete Boolean algebra and G is a countably full subgroup of Aut A, then every member of G is expressible as a product of at most eight involutions belonging to G. > (e) Let X be any set. Show that any automorphism of the Boolean algebra PX is expressible as a product of at most two involutions. (f ) Recall that in any group G, a commutator in G is an element of the form ghg −1 h−1 where g, h ∈ G. Show that if A is a Dedekind complete Boolean algebra and G is a subgroup of Aut A with many involutions then every involution in G is a commutator in G, so that every element of G is expressible as a product of finitely many commutators. (g) Give an example of a Dedekind complete Boolean algebra A such that not every member of Aut A is a product of commutators in Aut A. (h) Let A be a Dedekind complete Boolean algebra, and suppose that Aut A has many involutions. Show that if H C Aut A then every member of H is expressible as the product of at most eight involutions belonging to H. (i) Let A be a Dedekind complete Boolean algebra and G a full subgroup of Aut A with many involutions. Show that the partially ordered set H of normal subgroups of G is a distributive lattice, that is, H ∩ K1 K2 = (H ∩ K1 )(H ∩ K2 ), H(K1 ∩ K2 ) = HK1 ∩ HK2 for all H, K1 , K2 ∈ H. (j) Let A be a Dedekind complete Boolean algebra and G a full subgroup of Aut A with many involutions. Show that if H is a the normal subgroup of G generated by a finite subset of G, then it is the normal subgroup generated by a single involution. (k) Let A be a Dedekind complete Boolean algebra and G a full subgroup of Aut A with many involutions. Show (i) that there is an involution π ∈ G such that every member of G is expressible as a product of conjugates of π in G (ii) any proper normal subgroup of G is included in a maximal proper normal subgroup of G. (l) Let G be any group. Show that if π, φ ∈ G are involutions then πφ is conjugate to its inverse.

381 Notes

Automorphism groups of Boolean algebras

487

381Y Further exercises (a) Find a Dedekind σ-complete Boolean algebra A with an automorphism which cannot be expressed either as a product of finitely many involutions in Aut A, nor as a product of finitely many commutators in Aut A. (This seems to require a certain amount of ingenuity.) (b) Let X be a set and Σ a countably generated σ-subalgebra of subsets of X. (i) Show that if f : X → X is a bijection such that Σ = {E : E ⊆ X, f −1 [E] ∈ Σ}, then there are disjoint E 0 , E 00 , F 0 , F 00 , G ∈ Σ such that f −1 [E 0 ] = F 0 , f −1 [E 00 ] = F 00 , f −1 [F 00 ] = G, f −1 [F 0 ∪ G] = E 0 ∪ E 00 , and f (x) = x for every x ∈ X \ (E 0 ∪SE 00 ∪ F 0 ∪ F 00 ∪ G. (Hint: there is a sequence hEn in∈N in Σ such that En ∩ f −1 [En ] = ∅ for every n and n∈N En = {x : f (x) 6= x}.) (In particular, any member of Aut Σ has a support.) (ii) Show that every involution in Aut Σ is an exchanging involution. (c) Let X be a set and Σ a countably generated σ-subalgebra of subsets of X. Show that any member of Aut Σ is expressible as a product of at most eight involutions in Aut Σ. (d) Let A be a homogeneous Boolean algebra which is isomorphic to the simple power AN . (For instance, A could be the measure algebra of Lebesgue measure on R.) Show that any automorphism of A is the ˇ e ˇpa ´ nek & Rubin 89, Corollary 5.9a(ii).) product of at most five exchanging involutions. (Cf. St 381Z Problem In 381N, is ‘eight’ best possible? ˇ e ˇpa ´nek & Rubin 89 and Fathi 78. 381 Notes and comments The ideas above are adapted from St Lemma 381J is a form of what is sometimes called ‘Frol´ık’s theorem’, following Frol´ık 68. The two main results 381N and 381S, as written out above, both involve careful algebra. It seems to me that we can distinguish two essential methods. (i) There are arguments involving finitely many automorphisms, carefully pieced together from descriptions of their actions on different parts of the algebra, as in (a)-(c) of the proof of 381N, and the whole of the proof of 381S; similar ideas can be used in 381Xf. It is in these that I believe that the ‘cycle notation’ introduced in 381G-381I can be of value. Generally the hope is that we can use intuitions derived from the theory of permutation groups (that is, the case A = PX) to guide us. (ii) There is the argument in 381N involving a sequence of automorphisms, designed to express an automorphism π0 supported by an element a0 as the product of an automorphism θ0 supported by supk∈N ak with an automorphism θ1 supported by supk≥1 ak , so chosen that the actions of the θr on supk≥1 ak cancel ˇ e ˇpa ´ nek & Rubin out and we are left with π0 as the residue. (For an account of the origins of this idea see St 89.) Since we know of no automorphisms except those which can be derived from the original automorphism π, the method has to be to some extent constructive. The idea is that each πn+1 is not exactly a copy on an+1 of the preceding πn , but a modification of it by involutions. At each stage of the induction we have to mention an auxiliary element ψn of G in order to be sure that there will be room (in bn = ψn an ) for the next step, safely disjoint from the preceding ak . When we come to build the mutually cancelling pair θ0 , θ1 we find that they incorporate the modifiers φ1n , φ2n , φ3n , which can be assembled into the involutions φ˜ir in part (f) of the proof. I note that the assumption of ‘Dedekind completeness’ (as opposed to Dedekind σ-completeness) in 381N is used only in parts (b) and (c) of the proof, when applying Frol´ık’s theorem. Consequently we have a slight generalization possible (381Xd); but we do need the full hypothesis for the theorem as stated (381Ya). There is however a very important special case, when A is a countably generated σ-algebra, for which we have a version of Frol´ık’s theorem available for different reasons (381Yb), and can get a corresponding theorem to match 381N (381Yc). A natural question arising from 381T is: does every homogeneous Boolean algebra have a simple automorphism group? This leads into deep water. As remarked after 381T, every homogeneous Dedekind σ-complete algebra has a simple automophism group. Using the continuum hypothesis, it is possible to construct a homogeneous Boolean algebra which does not have a simple automorphism group; but as far as I am aware no such construction is known which does not rely on some special axiom outside ordinary set ˇ e ˇpa ´ nek & Rubin 89, §5. theory. See St In 381Z I ask whether the number ‘eight’ appearing in 381N is actually best possible. The argument is complex enough to make it seem that there may be room for improvement – see 381Xe and 381Yd. Ornstein

488

Automorphism groups

381 Notes

& Shields 731 present examples of automorphisms in the full subgroup G = Autµ¯ A of measure-preserving automorphisms of the Lebesgue probability algebra which are not conjugate (in G) to their inverses, and therefore cannot be expressible as the product of two involutions in G.

382 Automorphism groups of measure algebras I turn now to the group of measure-preserving automorphisms of a measure algebra, seeking to apply the results of the last section. The principal theorems are 382D, which is a straightforward special case of 381N, and 382I, corresponding to 381T. I give another example of the use of 381S to describe the normal subgroups of Autµ¯ A (382J). 382A Definition Let (A, µ ¯) be a measure algebra. I will write Autµ¯ A for the set of all measurepreserving automorphisms of A. This is a group, being a subgroup of the group Aut A of all Boolean automorphisms of A. 382B Lemma Let (A, µ ¯) be a measure algebra, and hai ii∈I , hbi ii∈I two partitions of unity in A. Assume either that I is countable or that (A, µ ¯) is localizable. Suppose that for each i ∈ I we have a measure-preserving isomorphism πi : Aai → Abi between the corresponding principal ideals. Then there is a unique π ∈ Autµ¯ A such that πc = πi c whenever i ∈ I and c ⊆ ai . Q Q proof (Compare 381B.) By 322K, we may identify A with each of the simple products i∈I Aai , i∈I Abi ; now π corresponds to the isomorphism between the two products induced by the πi . 382C Corollary If (A, µ ¯) is a localizable measure algebra, then, in the language of 381M, Autµ¯ A is a full subgroup of Aut A. 382D Theorem Let (A, µ ¯) be a localizable measure algebra. Then every measure-preserving automorphism of A is expressible as the product of at most eight measure-preserving involutions. proof This is immediate from 382C and 381N. 382E Lemma If (A, µ ¯) is a homogeneous semi-finite measure algebra, it is σ-finite, therefore localizable. proof If A = {0}, this is trivial. Otherwise there is an a ∈ A such that 0 < µ ¯a < ∞. The principal ideal Aa is ccc (322G), so A also is, and (A, µ ¯) must be σ-finite (by 322G again). 382F Lemma Let (A, µ ¯) be a homogeneous semi-finite measure algebra. (a) If hai ii∈I , hbi ii∈I are partitions of unity in A with µ ¯ai = µ ¯bi for every i, there is a π ∈ Autµ¯ A such that πai = bi for each i. (b) If (A, µ ¯) is totally finite, then whenever hai ii∈I , hbi ii∈I are disjoint families in A with µ ¯ ai = µ ¯bi for every i, there is a π ∈ Autµ¯ A such that πai = bi for each i. proof (a) By 382E, (A, µ ¯) is σ-finite, therefore localizable. For each i ∈ I, the principal ideals Aai , Abi are homogeneous, of the same measure and the same Maharam type (being τ (A) if ai 6= 0, 0 if ai = 0). Because they are ccc, they are of the same magnitude, as defined in 332G, and there is a measure-preserving isomorphism πi : Aai → Abi (332J). By 382B there is a measure-preserving automorphism π : A → A such that πd = πi d for every i ∈ I, d ⊆ ai ; and this π serves. (b) Set a∗ = 1 \ supi∈I ai , b∗ = 1 \ supi∈I bi . We must have P P µ ¯ a∗ = µ ¯1 − i∈I µ ¯ ai = µ ¯1 − i∈I µ ¯ bi = µ ¯b∗ , so adding a∗ , b∗ to the families we obtain partitions of unity to which we can apply the result of (a). 1I

am indebted to G.Hjorth for the reference.

382J

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489

382G Lemma (a) If (A, µ ¯) is an atomless semi-finite measure algebra, then Aut A and Autµ¯ A have many involutions. (b) If (A, µ ¯) is an atomless localizable measure algebra, then every element of A is the support of some involution in Autµ¯ A. proof (a) If a ∈ A \ {0}, then by 332A there is a non-zero b ⊆ a, of finite measure, such that the principal ideal Ab is (Maharam-type-)homogeneous. Now because A is atomless, there is a c ⊆ b such that µ ¯c = 21 µ ¯b (331C), so that Ac and Ab\c are isomorphic measure algebras. If θ : Ac → Ab\c is any measure-preserving ←−−−− isomorphism, then π = (c θ b \ c) is an involution in Autµ¯ A (and therefore in Aut A) supported by a. (b) Use 382C, (a) and 381R. 382H Corollary Let (A, µ ¯) be an atomless localizable measure algebra. Then (a) the lattice of normal subgroups of Aut A is isomorphic to the lattice of Aut A-invariant ideals of A; (b) the lattice of normal subgroups of Autµ¯ A is isomorphic to the lattice of Autµ¯ A-invariant ideals of A. proof Use 381S. Taking G to be either Aut A or Autµ¯ A, and I to be the family of G-invariant ideals in A, we have a map I 7→ HI = {π : π ∈ G, supp π ∈ I} from I to the family H of normal subgroups of G. Of course this map is order-preserving; 381S tells us that it is surjective; and 382Gb tells us that it is injective and its inverse is order-preserving, since if a ∈ I \ J there is a π ∈ G with supp π = a, so that π ∈ HI \ HJ . Thus we have an order-isomorphism between H and I. 382I Normal subgroups of Aut A and Autµ¯ A 381S provides the machinery for a full description of the normal subgroups of Aut A and Autµ¯ A when (A, µ ¯) is an atomless localizable measure algebra, as we know that they correspond exactly to the invariant ideals of A. The general case is complicated. But the following are simple enough. Theorem Let (A, µ ¯) be a homogeneous semi-finite measure algebra. (a) Aut A is simple. (b) If (A, µ ¯) is totally finite, Autµ¯ A is simple. (c) If (A, µ ¯) is not totally finite, Autµ¯ A has exactly one non-trivial proper normal subgroup. proof (a) This is a special case of 381T. (b)-(c) The point is that the only possible Autµ¯ A-invariant ideals of A are {0}, Af and A. P P If A is {0} or {0, 1} this is trivial. Otherwise, A is atomless. Let I C A be an invariant ideal. (i) If I 6⊆ Af , take a ∈ I with µ ¯a = ∞. By 382E, A is σ-finite, so a has the same magnitude ω as 1. By 332I, there is a partition of unity hen in∈N in A with µ ¯en = 1 for every n; setting b = supn∈N e2n , b0 = 1 \ b, we 0 see that both b and b are of infinite measure. Similarly we can divide a into c, c0 , both of infinite measure. Now by 332J the principal ideals Ab , Ab0 , Ac , A1\c are all isomorphic as measure algebras, so that there are automorphisms π, φ ∈ Autµ¯ A such that πc = b,

φc = b0 .

But this means that both b and b0 belong to I, so that 1 = b ∪ b0 ∈ I and I = A. (ii) If I ⊆ Af and I 6= {0}, take any non-zero a ∈ I. If b is any member of A, then (because A is atomless) b can be partitioned into b0 , . . . , bn , all of measure at most µ ¯a. Then for each i there is a b0i ⊆ a 0 0 such that µ ¯bi = µ ¯bi ; since this common measure is finite, µ ¯(1 \ bi ) = µ ¯(1 \ bi ). By 332J and 382Fa, there is a πi ∈ Autµ¯ A such that πi b0i = bi , so that bi belongs to I. Accordingly b ∈ I. As b is arbitrary, I = Af . Thus the only invariant ideals of A are {0}, Af and A. Q Q By 382Hb we therefore have either one, two or three normal subgroups of Autµ¯ A, according to whether µ ¯1 is zero, finite and not zero, or infinite. Remark For the Lebesgue probability algebra, (b) is due to Fathi 78. The extension to algebras of uncountable Maharam type is from Choksi & Prasad 82. 382J

The language of §352 offers a way of describing another case.

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382J

Proposition Let (A, µ ¯) be an atomless totally finite measure algebra. For each infinite cardinal κ, let eκ be the Maharam-type-κ component of A, and let K be {κ : eκ 6= 0}. Let H be the lattice of normal subgroups of Autµ¯ A. Then (i) if K is finite, H is isomorphic, as partially ordered set, to PK; (ii) if K is infinite, then H is isomorphic, as partially ordered set, to the lattice of solid linear subspaces of `∞ . proof (a) Let I be the family of Autµ¯ A-invariant ideals of A, so that H ∼ = I, by 382Hb. For a, b ∈ A, say that a ¹ b if there is some k ∈ N such that µ ¯(a ∩ eκ ) ≤ k µ ¯(b ∩ eκ ) for every κ ∈ K. Then an ideal I of A is Autµ¯ A-invariant iff a ∈ I whenever a ¹ b ∈ I. P P (α) Suppose that I is Autµ¯ A invariant and that b ∈ I, µ ¯(a ∩ eκ ) ≤ k µ ¯(b ∩ eκ ) for every κ ∈ K. Then for each κ we can find aκ1 , . . . , aκk such that a ∩ eκ = supi≤k aκi and µ ¯aκi ≤ µ ¯(b ∩ eκ ) for every i. Now there are measure-preserving automorphisms πκi of the principal ideal Aeκ such that πκi aκi ⊆ b. Setting πi d = supκ∈K πκi (d ∩ eκ ) for every d ∈ A, and ai = supκ∈K aκi , we have πi ∈ Autµ¯ A and πi ai ⊆ b, so ai ∈ I for each i; also a = supi≤k ai , so a ∈ I. (β) On the other hand, if a ∈ A and π ∈ Autµ¯ A, then µ ¯(πa ∩ eκ ) = µ ¯π(a ∩ eκ ) = µ ¯(a ∩ eκ ) for every κ ∈ K, because πeκ = eκ , so that πa ¹ a. So if I satisfies the condition, π[I] ⊆ I for every π ∈ Autµ¯ A and I ∈ I. Q Q (b) Consequently, for I ∈ I and κ ∈ K, eκ ∈ I iff there is some a ∈ I such that a ∩ aκ 6= 0, since in this case eκ ¹ a. (This is where I use the hypothesis that (A, µ ¯) is totally finite.) It follows that if K is finite, any I ∈ I is the principal ideal generated by sup{eκ : eκ ∈ I}. Conversely, of course, all such ideals are Autµ¯ A-invariant. Thus I is in a natural order-preserving correspondence with PK, and H ∼ = PK. (c) Now suppose that K is infinite; enumerate it as hκn in∈N . Define θ : A → `∞ by setting θa = h¯ µ(a ∩ eκn )/¯ µ(eκn )in∈N for a ∈ A; so that a ¹ b iff there is some k such that θa ≤ kθb, θa ≤ θ(a ∪ b) ≤ θa + θb ≤ 2θ(a ∪ b) for all a, b ∈ A, while θ(1A ) is the standard order unit 1 of `∞ . Let U be the family of solid linear subspaces of `∞ and define functions I 7→ VI : I → U , U 7→ JU : U → I by saying VI = {f : f ∈ `∞ , |f | ≤ kθa for some a ∈ I, k ∈ N}, JU = {a : a ∈ A, θa ∈ U }. The properties of θ just listed ensure that VI ∈ U , JU ∈ I for every I ∈ I, U ∈ U. Of course both I 7→ VI and U 7→ JU are order-preserving. If I ∈ I, then JVI = {a : ∃ b ∈ I, a ¹ b} = I, Finally, VJU = U for every U ∈ U . P P VJU = {f : ∃ a ∈ A, k ∈ N, |f | ≤ kθa ∈ U } ⊆ U because U is a solid linear subspace. But also, given g ∈ U , there is an a ∈ A such that µ ¯(a ∩ eκn ) = min(1, |g(n)|)¯ µ(eκn ) for every n (because A is atomless); in which case θa ≤ |g| ≤ max(1, kgk∞ )θa so a ∈ JU and g ∈ VJU . Thus U = VJU . Q Q So the functions I 7→ VI and U 7→ JU are the two halves of an order-isomorphism between I and U , and H ∼ =I∼ = U, as claimed. 382X Basic exercises >(a) Let (A, µ ¯) be a localizable measure algebra. For each infinite cardinal κ, let eκ be the Maharam-type-κ component of A. (i) Show that Autµ¯ A is a simple group iff either there is just one infinite cardinal κ such that eκ 6= 0, that eκ has finite measure and all the atoms of A (if any) have different measures or A is purely atomic and there is just one pair of atoms of the same measure or A is purely atomic and all its atoms have different measures. (ii) Show that Aut A is a simple group iff either

382Ye

Automorphism groups of measure algebras

491

(A, µ ¯) is σ-finite and there is just one infinite cardinal κ such that eκ 6= 0 and A has at most one atom or A is purely atomic and has at most two atoms. (b) Let (A, µ ¯) be a localizable measure algebra. (i) Show that Autµ¯ A is simple iff it is isomorphic to one of the groups {ι}, Z2 or Autν¯κ Bκ where κ is an infinite cardinal and (Bκ , ν¯κ ) is the measure algebra of the usual measure on {0, 1}κ . (ii) Show that Aut A is simple iff it is isomorphic to one of the groups {ι}, Z2 or Aut Bκ . (c) Show that if (A, µ ¯) is a semi-finite measure algebra of magnitude greater than c, its automorphism group Autµ¯ A is not simple. (d) Let (A, µ ¯) be an atomless localizable measure algebra. For each infinite cardinal κ write eκ for the Maharam-type-κ component of A. For π, ψ ∈ Autµ¯ A show that π belongs to the normal subgroup of Autµ¯ A generated by ψ iff there is a k ∈ N such that mag(eκ ∩ supp π) ≤ k mag(eκ ∩ supp ψ) for every infinite cardinal κ, writing mag a for the magnitude of a, and setting kζ = ζ if k > 0 and ζ is an infinite cardinal. > (e) Let (A, µ ¯) be the measure algebra of Lebesgue measure on R. For n ∈ N set en = [−n, n]• ∈ A. Let G ≤ Autµ¯ A be the group consisting of measure-preserving automorphisms π such that supp π ⊆ en for some n. Show that G is simple. (Hint: show that G is the union of an increasing sequence of simple subgroups.) (f ) Let (A, µ ¯) be an atomless totally finite measure algebra. Let H be the lattice of normal subgroups of Aut A. Show that H is isomorphic, as partially ordered set, to PK for some countable set K. (g) Let (A, µ ¯) be an atomless localizable measure algebra which is not σ-finite, and suppose that τ (Aa ) = τ (Ab ) whenever a, b ∈ A and 0 < µ ¯a ≤ µ ¯b < ∞. Let κ be the magnitude of A. (i) Show that the lattice H of normal subgroups of Autµ¯ A is well-ordered, with least member {ι}, next member {π : µ ¯(πa) < ∞ whenever µ ¯a < ∞}, and one member Hζ for each infinite cardinal ζ less than or equal to κ, setting Hζ = {π : π ∈ Autµ¯ A, mag(πa) ≤ ζ whenever mag a ≤ ζ}, where mag a is the magnitude of a. (ii) Show that the lattice H0 of normal subgroups of Aut A is wellordered, with least member {ι} and one member Hζ0 for each infinite cardinal ζ less than or equal to κ, setting Hζ0 = {π : π ∈ Aut A, mag(πa) ≤ ζ whenever mag a ≤ ζ}. 382Y Further exercises (a) Let (A, µ ¯) be an atomless totally finite measure algebra. Show that Autµ¯ A, Aut A have the same (cardinal) number of normal subgroups. (b) Let X be a set. Show that Aut PX has one normal subgroup if #(X) ≤ 1, two if #(X) = 2, three if #(X) = 3 or 5 ≤ #(X) ≤ ω, four if #(X) = 4 or #(X) = ω1 . (c) Let (A, µ ¯) be a totally finite measure algebra and π : A → A a measure-preserving homomorphism. Take any a ∈ A and set c = supn≥1 π n a. (i) Show that a ⊆ c and that c = supm≥n π m a for every n ∈ N. P∞ (ii) Set an = π n a \ sup1≤i 0 such that µ ¯(θa) = ακ µ ¯a whenever a ⊆ eκ , the Maharam-type-κ component of A. But since θeκ = eκ and µ ¯ eκ < ∞ for every κ, we must have ακ = 1 whenever eκ 6= 0; as A is atomless, X µ ¯(θa) = µ ¯(θ(a ∩ eκ )) κ is an infinite cardinal

=

X

ακ µ ¯(a ∩ eκ )

κ is an infinite cardinal

=

X

µ ¯(a ∩ eκ ) = µ ¯a

κ is an infinite cardinal

for every a ∈ A. Thus θ ∈ Autµ¯ A and q is an inner automorphism. 383P The results above are satisfying and complete in their own terms, but leave open a number of obvious questions concerning whether some of the hypotheses can be relaxed. Atoms can produce a variety of complications (see 383Ya-383Yd below). To show that we really do need to assume that our algebras are Dedekind complete or localizable, I offer the following. Example (a) There are an atomless localizable measure algebra (A, µ ¯) and an atomless semi-finite measure algebra (B, ν¯) such that Aut A ∼ = Aut B, Autµ¯ A ∼ = Autν¯ B but A and B are not isomorphic. proof Let (A0 , µ ¯0 ) be an atomless homogeneous probability algebra; for instance, the measure algebra 1 of Lebesgue measure on the unit interval. Let A be the simple product Boolean algebra Aω ¯ the 0 , and µ corresponding measure (322K); then (A, µ ¯) is an atomless localizable measure algebra. In A let I be the set {a : a ∈ A and the principal ideal Aa is ccc}; then I is an ideal of A, the σ-ideal generated by the elements of finite measure (cf. 322G). Set B = {a : a ∈ A, either a ∈ I or 1 \ a ∈ I}. Then B is a σ-subalgebra of A, so if we set ν¯ = µ ¯¹ B then (B, ν¯) is a measure algebra in its own right. The definition of I makes it plain that it is invariant under all Boolean automorphisms of A; so B is also invariant under all automorphisms, and we have a homomorphism φ 7→ q(φ) = φ¹ B : Aut A → Aut B. On the other hand, because B is order-dense in A, and A is Dedekind complete, every automorphism of B can be extended to an automorphism of A (see part (a) of the proof of 383E). So q is actually an isomorphism between Aut A and Aut B. Moreover, still because B is order-dense, q(φ) is measure-preserving iff φ is measure-preserving, so Autµ¯ A is isomorphic to Autν¯ B. But of course there is no Boolean isomorphism, let alone a measure algebra isomorphism, between A and B, because A is Dedekind complete while B is not. Remark Thus the hypothesis ‘Dedekind complete’ in 383D and 383J (and ‘localizable’ in 383M), and the hypothesis ‘homogeneous’ in 383E-383F, are essential. ¯ such that Aut C has an outer automorphism. (b) There is an atomless semi-finite measure algebra (C, λ) proof In fact we can take C to be the simple product of A and B above. I claim that the isomorphism between Aut A and Aut B gives rise to an outer automorphism of Aut C; this seems very natural, but I think there is a fair bit to check, so I take the argument in easy stages. (i) We may identify the Dedekind completion of C = A × B with A × A. For φ ∈ Aut C, we have a ˆ P corresponding φˆ ∈ Aut(A × A). Now B × A is invariant under φ. P Consider first φ(0, 1) = (a1 , b1 ) ∈ C. ∼ The corresponding principal ideal C(a1 ,b1 ) = Aa1 × Bb1 of C must be isomorphic to the principal ideal C(0,1) ∼ = Aa × Bb , = B; so that if (a, b) ∈ C and (a, b) ⊆ (a1 , b1 ), then just one of the principal ideals C(a,b) ∼ C(a1 \a,b1 \b) ∼ = Aa1 \a × Bb1 \b is ccc. But this can only happen if Aa1 is ccc and Bb1 is not; that is, if a1 and ˆ a) ⊆ (a1 , b1 ) belongs to B × A for every a ∈ A. We also find that 1 \ b1 belong to I. Consequently φ(0, φ(1, 0) = (1, 1) \ φ(0, 1) = (1 \ a1 , 1 \ b1 ) ∈ B × A.

500

Automorphism groups

383P

Now if b ∈ I, then Cφ(b,0) ∼ = C(b,0) ∼ = Ab is ccc and φ(b, 0) ∈ I × I ⊆ B × A; while φ(1 \ b, 0) = (1 \ a1 , 1 \ b1 ) \ φ(b, 0) ∈ B × A. This shows that φ(b, 0) ∈ B × A for every b ∈ B. So ˆ a) = φ(b, ˆ 0) ∪ φ(0, ˆ a) ∈ B × A φ(b, for every b ∈ B, a ∈ A. Q Q (ii) Let θ : A × A → A × A be the involution defined by setting θ(a, b) = (b, a) for all a, b ∈ A. Take ˆ −1 ∈ Aut(A × A). If c = (a, b) ∈ C, then θ−1 c = (b, a) ∈ B × A, so φ ∈ Aut C and consider ψ = θφθ −1 ˆ φθ c ∈ B × A, by (i), and ψc ∈ A × B = C. This shows that ψ¹ C is a homomorphism from C to itself. Of course ψ −1 = θφˆ−1 θ−1 has the same property. So we have a map q : Aut C → Aut C given by setting ˆ −1 ¹ C q(φ) = θφθ for φ ∈ Aut C. Evidently q is an automorphism. (iii) ?? Suppose, if possible, that q were an inner automorphism. Let χ ∈ Aut C be such that q(φ) = χφχ−1 for every φ ∈ Aut C. Then d = θφθ ˆ −1 χ ˆφˆχ ˆ−1 = q(φ) for every φ ∈ Aut C. Since G = {φˆ : φ ∈ Aut C} is a subgroup of Aut(A × A) with many involutions, the ‘uniqueness’ assertion of 383D tells us that χ ˆ = θ. But θ[C] = B × A 6= C = χ[C] = χ[C], ˆ so this cannot be. X X Thus q is the required outer automorphism of Aut C. Remark Thus the hypothesis ‘homogeneous’ in 383E, and the hypothesis ‘Dedekind complete’ in 383J, are necessary. 383Q Example Let µ be Lebesgue measure on R, and (A, µ ¯) its measure algebra. Then Autµ¯ A has an outer automorphism. P P Set f (x) = 2x for x ∈ R. Then E 7→ f −1 [E] = 21 E is a Boolean automorphism of the domain Σ of µ, and µ( 12 E) = 12 µE for every E ∈ Σ (263A, or otherwise). So we have a corresponding θ ∈ Aut A defined by setting θE • = ( 12 E)• for every E ∈ Σ, and µ ¯(θa) = 12 µ ¯a for every a ∈ A. By 383N, we have an automorphism q of Autµ¯ A defined by setting q(φ) = θφθ−1 for every measure-preserving automorphism φ. But q is now an outer automorphism of Autµ¯ A, because (by 383D) the only possible automorphism of A corresponding to q is θ, and θ is not measure-preserving. Q Q Thus the hypothesis ‘totally finite’ in 383O cannot be omitted. 383X Basic exercises (a) Let A be a Boolean algebra. Show that the following are equiveridical: (i) A is nowhere rigid; (ii) for every a ∈ A \ {0} and n ∈ N there are disjoint non-zero b0 , . . . , bn ⊆ a such that the principal ideals Abi they generate are all isomorphic; (iii) for every a ∈ A \ {0} and n ≥ 1 there is a φ ∈ Aut A, of order n, supported by a. (b) Let A be a nowhere rigid Boolean algebra. Show that its Dedekind completion is nowhere rigid. (c) Let A be an atomless homogeneous Boolean algebra and B a nowhere rigid Boolean algebra, and suppose that Aut A is isomorphic to Aut B. Show that there is an invariant order-dense subalgebra of B which is isomorphic to A.

383 Notes

Outer automorphisms

501

(d) Let A and B be nowhere rigid Boolean algebras. Show that if Aut A and Aut B are isomorphic, then b and B b are isomorphic. the Dedekind completions A (e) Find two non-isomorphic atomless totally finite measure algebras (A, µ ¯), (B, ν¯) such that Autµ¯ A and Autν¯ B are isomorphic. (This is easy.) (f ) Let (A, µ ¯) and (B, ν¯) be semi-finite measure algebras and θ : A → B a Boolean isomorphism. Show that the following are equiveridical: (i) for every φ ∈ Autµ¯ A, θφθ−1 ∈ Autν¯ B; (ii)(α) for every infinite cardinal κ there is an ακ > 0 such that ν¯(θa) = ακ µ ¯a whenever a ∈ A and the principal ideal Aa is Maharam-type-homogeneous with Maharam type κ; (β) for every γ ∈ ]0, ∞[ there is an αγ > 0 such that ν¯(θa) = αγ µ ¯a whenever a ∈ A is an atom of measure γ. (g) Let q : Aut C → Aut C be the automorphism of 383Pb. Show that q(φ) is measure-preserving whenever φ is measure-preserving, so that q¹ Autλ¯ C is an outer automorphism of Autλ¯ C. 383Y Further exercises (a) Let (A, µ ¯) and (B, ν¯) be localizable measure algebras such that Aut A ∼ = ∼ Aut B. Show that either A = B or one of A, B has just one atom and the other is atomless. ∼ Autν¯ B. Show that either (b) Let (A, µ ¯), (B, ν¯) be localizable measure algebras such that Autµ¯ A = (A, µ ¯) ∼ = (B, ν¯) or there is some γ ∈ ]0, ∞[ such that one of A, B has just one atom of measure γ and the other has none or there are γ, γ 0 ∈ ]0, ∞[ such that the number of atoms of A of measure γ is equal to the number of atoms of B of measure γ 0 , but not to the number of atoms of A of measure γ 0 . (c) Let (A, µ ¯) be a localizable measure algebra. Show that there is an outer automorphism of Aut A iff A has exactly six atoms. (d) Let (A, µ ¯) be a localizable measure algebra. For each infinite cardinal κ let eκ be the Maharam-type-κ component of A and for each γ ∈ ]0, ∞[ let Aγ be the set of atoms of A of measure γ. Show that there is an outer automorphism of Autµ¯ A iff either there is an infinite cardinal κ such that µ ¯ eκ = ∞ or there are distinct γ, δ ∈ ]0, ∞[ such that #(Aγ ) = #(Aδ ) ≥ 2 or there is a γ ∈ ]0, ∞[ such that #(Aγ ) = 6 or there are γ, δ ∈ ]0, ∞[ such that #(Aγ ) = 2 < #(Aδ ) < ω. 383 Notes and comments Let me recapitulate the results above. If A and B are Boolean algebras, any isomorphism between Aut A and Aut B corresponds to an isomorphism between A and B if either A and B are atomless and homogeneous (383E) or they are nowhere rigid and Dedekind complete (383J). If (A, µ ¯) and (B, ν¯) are atomless localizable measure algebras, then any automorphism between Autµ¯ A and Autν¯ B corresponds to an isomorphism between A and B (383M) which if µ ¯ = ν¯ is totally finite will be measure-preserving (383O). These results may appear a little less surprising if I remark that the elementary Boolean algebras PX give rise to some of the same phenomena. The automorphism group of PX can be identified with the group SX of all permutations of X, and this has no outer automorphisms unless X has just six elements. Some of the ideas of the fundamental theorem 383D can be traced through in the purely atomic case also, though of course there are significant changes to be made, and some serious complications arise, of which the most striking surround the remarkable fact that S6 does have an outer automorphism (Burnside 11, §162; Rotman 84, Theorem 7.8). I have not attempted to incorporate these in the main results. For localizable measure algebras, where the only rigid parts are atoms, the complications are superable, and I think I have listed them all (383Ya-383Yd).

502

Automorphism groups

§384 intro.

384 Entropy Perhaps the most glaring problem associated with the theory of measure-preserving homomorphisms and automorphisms is the fact that we have no generally effective method of determining when two homomorphisms are the same, in the sense that two structures (A, µ ¯, π) and (B, ν¯, φ) are isomorphic, where (A, µ ¯) and (B, ν¯) are measure algebras and π : A → A, φ : B → B are Boolean homomorphisms. Of course the first part of the problem is to decide whether (A, µ ¯) and (B, ν¯) are isomorphic; but this is solved (at least for localizable algebras) by Maharam’s theorem (see 332J). The difficulty lies in the homomorphisms. Even when we know that (A, µ ¯) and (B, ν¯) are both isomorphic to the Lebesgue measure algebra, the extraordinary variety of constructions of homomorphisms – corresponding in part to the variety of measure spaces with such measure algebras, each with its own natural inverse-measure-preserving functions – means that the question of which are isomorphic to each other is continually being raised. In this section I give the most elementary ideas associated with the concept of ‘entropy’, up to the Kolmogorov-Sinaˇı theorem (384P). This is an invariant which can be attached to any measure-preserving homomorphism on a probability algebra, and therefore provides a useful method for distinguishing non-isomorphic homomorphisms. The main work of the section deals with homomorphisms on measure algebras, but as many of the most important ones arise from inverse-measure-preserving functions on measure spaces I comment on the extra problems arising in the isomorphism problem for such functions (384T-384V). I should remark that some of the lemmas will be repeated in stronger forms in the next section. 384A Notation Throughout this section, I will use the letter q to denote the function from [0, ∞[ to R defined by saying that q(t) = −t ln t = t ln 1t if t > 0, q(0) = 0.

0

0.5

1

1.5

-0.5

The function q We shall need the following straightforward facts concerning q. (a) q is continuous on [0, ∞[ and differentiable on ]0, ∞[; q 0 (t) = −1 − ln t and q 00 (t) = − 1t for t > 0. Because q 00 is negative, q is concave, that is, −q is convex. q has a unique maximum at ( 1e , 1e ). (b) If s ≥ 0, t > 0 then q 0 (s + t) ≤ q 0 (t); consequently

Rt

q(s + t) = q(s) + 0 q 0 (s + τ )dτ ≤ q(s) + q(t) Pn Pn forPs, t ≥ 0. P It follows that q( i=0 si ) ≤ i=0 q(si ) for all s0 , . . . , sn ≥ 0 and (because q is continuous) ∞ ∞ q( i=0 si ) ≤ i=0 q(si ) for every non-negative summable series hsi ii∈N . (c) If s, t ≥ 0 then q(st) = sq(t) + tq(s); more generally, if n ≥ 1 and si ≥ 0 for i ≤ n then Qn Pn Q q( i=0 si ) = j=0 q(sj ) i6=j si . d (d) The function t 7→ q(t) + q(1 − t) has a unique maximum at ( 12 , ln 2). ( dt (q(t) + q(1 − t)) = ln 1−t t .) It 1 follows that for every ² > 0 there is a δ > 0 such that |t − 2 | ≤ ² whenever q(t) + q(1 − t) ≥ ln 2 − δ.

(e) If 0 ≤ t ≤ 21 , then q(1 − t) ≤ q(t). P P Set f (t) = q(t) − q(1 − t). Then 1 t

f 00 (t) = − +

1 1−t

=

2t−1 t(1−t)

≤0

for 0 < t ≤ 21 , while f (0) = f ( 21 ) = 0, so f (t) ≥ 0 for 0 ≤ t ≤ 12 . Q Q

384G

Entropy

503

(f ) (i) If A is a Dedekind σ-complete Boolean algebra, I will write q¯ for the function from L0 (A)+ to L (A) defined from q (364I). Note that because 0 ≤ q(t) ≤ 1 for t ∈ [0, 1], 0 ≤ q¯(u) ≤ χ1 if 0 ≤ u ≤ χ1. (ii) By (b), q¯(u + v) ≤ q¯(u) + q¯(v) for all u, v ≥ 0 in L0 (A). (Represent A as the measure algebra of a measure space, so that q¯(f • ) = (qf )• , as in 364Jb.) (iii) Similarly, if u, v ∈ L0 (A)+ , then q¯(u × v) = u × q¯(v) + v × q¯(u). 0

1 384B Lemma Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A, and ¯) → R R P : L (A, µ L (A, µ ¯) the corresponding conditional expectation operator (365R). Then q¯(u) ≤ q( u) and P (¯ q (u)) ≤ q¯(P u) for every u ∈ L∞ (A)+ . 1

proof Apply the remarks in 365Rb to −q. (¯ q (u) ∈ L∞ ⊆ L1 for every u ∈ (L∞ )+ because q is bounded on every bounded interval in [0, ∞[.) 384CP Definition Let (A, µ ¯) be a probability algebra. If A is a partition of unity in A, its entropy is H(A) = a∈A q(¯ µa), where q is the function defined in 384A. Remarks (a) In the definition of ‘partition of unity’ (311Gc) I allowed 0 to belong to the family. In the present context this is a mild irritant, and when convenient I shall remove 0 from the partitions of unity considered here (as in 384F below). But because q(0) = 0, it makes no difference; H(A) = H(A \ {0}) whenever A is a partition of unity. So if you wish you can read ‘partition of unity’ in this section to mean ‘partition of unity not containing 0’, if you are willing to make an occasional amendment in a formula. In important cases, in fact, A is of the form {ai : i ∈ I} or {ai : i ∈ I} \ {0}, where hai ii∈I is an indexed partition of unity, with Pai ∩ aj = 0 for i 6= j, but no restriction in the number of i with ai = 0; in this case, µai ). we still have H(A) = i∈I q(¯ (b) Many authors prefer to use log2 in place of ln. This makes sense in terms of one of the intuitive approaches to entropy as the ‘information’ associated with a partition. See Petersen 83, §5.1. 384D Definition Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A and A a partition of unity in A. Let P : L1 (A, µ ¯) → L1 (A, µ ¯) be the conditional expectation operator associated with B. Then the conditional entropy of A on B is R P H(A|B) = a∈A q¯(P χa), where q¯ is defined as in 384A. 384E Elementary remarks (a) In the formula R P a∈A

q¯(P χa),

we have 0 ≤ P (χa) ≤ χ1 for every a, so q¯(P χa) ≥ 0 and every term in the sum is non-negative; accordingly H(A|B) is well-defined in [0, ∞]. R (b) H(A) = H(A|{0, 1}), since if B = {0, 1} then P (χa) = µ ¯aχ1, so that q¯(P χa) = q(¯ µa). If A ⊆ B, H(A|B) = 0, since P (χa) = χa, q¯(P χa) = 0 for every a. 384F Definition If A is a Boolean algebra and A, B ⊆ A are partitions of unity, I write A ∨ B for the partition of unity {a ∩ b : a ∈ A, b ∈ B} \ {0}. (See 384Xq.) 384G Lemma Let (A, µ ¯) be a probability algebra and B a closed subalgebra. Let A ⊆ A be a partition of unity. (a) If B is another partition of unity in A, then H(A|B) ≤ H(A ∨ B|B) ≤ H(A|B) + H(B|B). (b) If B is purely atomic and D is the set of its atoms, then H(A ∨ D) = H(D) + H(A|B). (c) If C ⊆ B is another closed subalgebra of A, then H(A|C) ≥ H(A|B). In particular, H(A) ≥ H(A|B). (d) Suppose that hBn in∈N is a non-decreasing sequence of closed subalgebras of A such that B = S n∈N Bn . If H(A) < ∞ then

504

Automorphism groups

384G

H(A|B) = limn→∞ H(A|Bn ). In particular, if A ⊆ B then limn→∞ H(A|Bn ) = 0. proof Write P for the conditional expectation operator on L1 (A, µ ¯) associated with B. (a)(i) If B is infinite, enumerate it as hbj ij∈N ; if it is finite, enumerate it as hbj ij≤n and set bj = 0 for j > n. For any a ∈ A, P∞ P∞ χa = j=0 χ(a ∩ bj ), P (χa) = j=0 P χ(a ∩ bj ), n X q¯(P χa) = lim q¯( P χ(a ∩ bj )) n→∞

≤ lim

n→∞

j=0 n X

∞ X

j=0

j=0

q¯(P χ(a ∩ bj )) =

q¯(P χ(a ∩ bj ))

where all the infinite sums are to be regarded as order*-limits of the corresponding finite sums (see §367), and the middle inequality is a consequence of 384Af(ii). Accordingly Z X q¯(P χ(a ∩ b)) H(A ∨ B|B) = =

a∈A,b∈B,a∩b6=0 ∞ Z XX

XZ

a∈A j=0

a∈A

q¯(P χ(a ∩ bj )) ≥

q¯(P χai ) = H(A|B).

(ii) Suppose for the moment that A and B are both finite. For a ∈ A set ua = P (χa). If a, b ∈ A we 0 have 0 ≤ ua∩b ≤ ub inPL0 (B), so we may choose vP ab ∈ L (B) such that 0 ≤ vab ≤ χ1 P and ua∩b = vab × ub . = ub (because For any b ∈ B, a∈A vab = ub . Since a∈A χ(a ∩ b) = χb), so ub × a∈A ua∩b P [[|¯ q (ub )| > 0]] ⊆ [[ub > 0]], q¯(ub ) × a∈A vab = q¯(ub ). For any a ∈ A, q¯(ua ) = q¯(

X

ua∩b ) = q¯(

b∈B

X

ub × vab ) = q¯(P (

b∈B

X

χb × vab ))

b∈B

(because vab ∈ L0 (B) for every b, so P (χb × vab ) = P (χb) × vab ) X ≥ P (¯ q( χb × vab )) b∈B

(384B) = P(

X

χb × q¯(vab ))

b∈B

(because B is disjoint) =

X

ub × q¯(vab )

b∈B

(because q¯(vab ) ∈ L0 (B) for every b). Putting these together, H(A ∨ B|B) =

Z

X

q¯(ua∩b ) =

a∈A,b∈B

=

X XZ

a∈A

q¯(ub × vab )

a∈A,b∈B

Z

ub × q¯(vab ) +

a∈A,b∈B

≤

Z

X

q¯(ua ) +

XZ b∈B

X

Z vab × q¯(ub )

a∈A,b∈B

q¯(ub ) = H(A|B) + H(B|B).

384G

Entropy

505

(iii) For general partitions of unity A and B, take any finite set C ⊆ A ∨ B. Then C ⊆ {a ∩ b : a ∈ A0 , b ∈ B0 } where A0 ⊆ A and B0 ⊆ B are finite. Set A0 = A0 ∪ {1 \ sup A0 },

B 0 = B0 ∪ {1 \ sup B0 },

so that A0 and B 0 are finite partitions of unity and C ⊆ A0 ∨ B 0 . Now XZ

Z

X

q¯(P χc) ≤

q¯(P χc) = H(A0 ∨ B 0 |B) ≤ H(A0 |B) + H(B 0 |B)

c∈A0 ∨B 0

c∈C

(by (ii)) ≤ H(A0 ∨ A|B) + H(B 0 ∨ B|B) (by (i)) = H(A|B) + H(B|B). As C is arbitrary, H(A ∨ B|B) =

R

P c∈A∨B

q¯(P χc) ≤ H(A|B) + H(B|B).

P (b) It follows from 384Ab that d∈D q(¯ µ(a ∩ d)) ≥ q(¯ µa) for any a ∈ A. Now, because B is purely atomic and D is its set of atoms, P P µ ¯(a∩d) µ ¯(a∩d) P (χa) = d∈D χd, q¯(P (χa)) = d∈D q( )χd µ ¯d

µ ¯d

for every a ∈ A, H(A|B) =

P a∈A,d∈D

q(

µ ¯(a∩d) )¯ µd. µ ¯d

Putting these together, X

H(A ∨ D) =

q(¯ µ(a ∩ d)) =

a∈A,d∈D

= H(A|B) +

X a∈A,d∈D

X

q(

µ ¯(a∩d) µ ¯(a∩d) )¯ µd + q(¯ µd) µ ¯d µ ¯d

q(¯ µd) = H(A|B) + H(D).

d∈D

(c) Write PC for the conditional expectation operator corresponding to C. If a ∈ A, q¯(PC χa) = q¯(PC P χa) ≥ PC q¯(P χa) by 384B. So H(A|C) =

R

P a∈A

R R P P q¯(PC χa) ≥ a∈A PC q¯(P χa) = a∈A q¯(P χa) = H(A|B).

Taking C = {0, 1}, we get H(A) ≥ H(A|B). (d) Let Pn be the conditional expectation operator corresponding to Bn , for each n. Fix a ∈ A. Then P (χa) is the order*-limit of hPn (χa)in∈N , by L´evy’s martingale theorem (367Kb). Consequently (because q is continuous) h¯ q (Pn χa)in∈N is order*-convergent to q¯(P χa) for every a ∈ A (367I). Also, because 0 ≤ Pn χa ≤ χ1Rfor every n, R0 ≤ q¯(Pn χa) ≤ 1e χ1 for every n. By the Dominated Convergence Theorem (367J), limn→∞ q¯(Pn χa) = q¯(P χa). By 384B, we also have 0≤

R

R

for every a ∈ A and n ∈ N; since also we have |

R

q¯(Pn χa) −

R

R

q¯(Pn χa) ≤ q( Pn (χa)) = q( χa) = q(¯ µa) 0≤

R

q¯(P χa) = q(¯ µa),

q¯(P χa)| ≤ q(¯ µa) for every a ∈ A, n ∈ N.

506

Automorphism groups

384G

P Now we are supposing that H(A) is finite. Given ² > 0, we can find a finite set I ⊆ A such that µa) ≤ ², and an n0 ∈ N such that a∈A\I q(¯ R R P q¯(Pn χa) − q¯(P χa)| ≤ ² a∈I | for every n ≥ n0 ; in which case P

a∈A\I |

R

q¯(Pn χa) −

R

q¯(P χa)| ≤

P a∈A\I

q(¯ µa) ≤ ²

and |H(A|Bn ) − H(A|B)| ≤ 2² for every n ≥ n0 . As ² is arbitrary, H(A|B) = limn→∞ H(A|Bn ). 384H Corollary Let (A, µ ¯) be a probability algebra and A, B two partitions of unity in A. Then H(A) ≤ H(A ∨ B) ≤ H(A) + H(B). proof Take B = {0, 1} in 384Ga. 384I Lemma Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. If A ⊆ A is a partition of unity, then H(π[A]) = H(A). P P proof µπa) = a∈A q(¯ µa). a∈A q(¯ 384J Lemma Let (A, µ ¯) be a measure algebra. Let A be the set of its atoms. Then the following are equiveridical: (i) either A is not purely atomic or A is purely atomic and H(A) = ∞; (ii) there is a partition of unity B ⊆ A such that H(B) = ∞; (iii) for every γ ∈ R there is a finite partition of unity C ⊆ A such that H(C) ≥ γ. proof (i)⇒(ii) We need examine only the case in which A is not purely atomic. Let a ∈ A be a non-zero element such that the principal ideal Aa is atomless. By 331C we can choose inductively a disjoint sequence han in∈N such that an ⊆ a and µ ¯an = 2−n−1 µ ¯a. Now, for each n ∈ N, choose a disjoint set Bn such that n

#(Bn ) = 22 ,

n

b ⊆ an and µ ¯b = 2−2 µ ¯an for each b ∈ Bn .

Set B=

S n∈N

Bn ∪ {1 \ a}.

Then B is a partition of unity in A and H(B) ≥ =

∞ X X n=0 b∈Bn ∞ X

q(¯ µBn ) =

∞ X n=0

n

22 q

¡

µ ¯a ¢ 2n+1+2n

n ∞ ¡ 2n+1+2 ¢ X µ ¯a µ ¯a n ln ≥ 2 ln 2 = ∞. n+1 n+1 2 µ ¯a 2 n=0 n=0

(ii)⇒(iii) Enumerate B as hbi ii∈N . For each n ∈ N, Cn = {bi : i ≤ n}∪{1 \ supi≤n bi } is a finite partition of unity, and Pn limn→∞ H(Cn ) ≥ limn→∞ i=0 q(¯ µbi ) = H(B) = ∞. (iii)⇒(i) We need only consider the case in which A is purely atomic and A is its set of atoms. In this case, A ∨ C = A for every partition of unity C ⊆ A, so H(C) ≤ H(A) for every C (384H), and H(A) must be infinite. 384K Definition Let A be a Boolean algebra. If π : A → A is an order-continuous Boolean homomorphism, A ⊆ A is a partition of unity and n ≥ 1, write Dn (A, π) for the partition of unity generated by {π i a : a ∈ A, 0 ≤ i < n}, that is, {inf i 0 there 1 is an m ≥ 1 such that m αm ≤ α + ². Set M = maxj<m αj . Now, for any n ≥ m, there are k ≥ 1, j < m such that n = km + j, so that αn ≤ kαm + αj ,

1 αn n

k n

≤ αm +

M n

≤

1 αm m

+

M . n

Accordingly lim supn→∞ n1 αn ≤ α + ². As ² is arbitrary, 1 n

1 n

α ≤ lim inf n→∞ αn ≤ lim supn→∞ αn ≤ α and limn→∞ n1 αn = α is defined in [0, ∞]. Remark See also 384Yb and 385Nc below. 384M Definition Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. For any partition of unity A ⊆ A, set h(π, A) = inf n≥1 n1 H(Dn (A, π)) = limn→∞ n1 H(Dn (A, π)) (384L). Now the entropy of π is h(π) = sup{h(π, A) : A ⊆ A is a finite partition of unity}. Remarks (a) We always have h(π, A) ≤ H(D1 (A, π)) =H(A). (b) Observe that if π is the identity automorphism then Dn (A, π) = A \ {0} for every A and n, so that h(π) = 0. 384N Lemma Let (A, µ ¯) be a probability algebra and A, B two partitions of unity in A. Let π : A → A be a measure-preserving Boolean homomorphism. Then h(π, A) ≤ h(π, B) + H(A|B), where B is the closed subalgebra of A generated by B. proof We may suppose that 0 ∈ / B, since removing 0 from B changes neither Dn (B, π) nor B. For each n ∈ N, set An = π n [A], Bn = π n [B]. Let Bn = π n [B] be the closed subalgebra of A generated by Bn , and B∗n the closed subalgebra of A generated by Dn (B, π). Then H(An |Bn ) = H(A|B) for each n. P P The point is that, because B is purely atomic and B is its set of atoms, P µ ¯(a∩b) )¯ µb H(A|B) = a∈A,b∈B q( µ ¯b

as in the proof of 384Gb. Similarly, H(An |Bn ) = Accordingly, for any n ≥ 1,

P a∈A,b∈B

q(

µ ¯(π n a∩π n b) )¯ µ(π n b) µ ¯(π n b)

= H(A|B). Q Q

508

Automorphism groups

H(Dn (A, π)|B∗n ) ≤

n−1 X

384N

H(Ai |B∗n )

i=0

(by 384Ga) ≤

n−1 X

H(Ai |Bi )

i=0

(by 384Gc) = nH(A|B). Now

h(π, A) = lim

1

n→∞ n

1 n n→∞

H(Dn (A, π)) ≤ lim sup H(Dn (A, π) ∨ Dn (B, π))

(384Ga) 1 n

1 n

≤ lim sup H(Dn (B, π)) + H(Dn (A, π)|B∗n ) n→∞

(384Gb) ≤ h(π, B) + H(A|B). Remark Compare 385Nd below. 384O Lemma Let (A, µ ¯) be a probability algebra, π : A → A a measure-preserving Boolean homomorphism, and A ⊆ A a partition of unity such that H(A) < ∞. Then h(π, A) ≤ h(π). proof If A is finite, this is immediate from the definition of h(π); so suppose that A is infinite. Enumerate A as hai ii∈N . For each n ∈ N let Bn be the subalgebra of A generated by a0 , . . . , an ; then Bn has atoms Bn = {a0 , . . . , an , bn } where bn = supi>n ai . Now P∞ α H(A|Bn ) = i=n+1 q( i )βn βn

where αi = µ ¯ai , βn = µ ¯bn , because if Pn is the conditional expectation associated with Bn then Pn (χai ) = χai if i ≤ n, Pn (χai ) = βαni χbn if i > n. But ∞ X i=n+1

βn q(

αi ) βn

= =

because

P∞ i=0

∞ X i=n+1 ∞ X

q(αi ) −

αi q(βn ) βn

q(αi ) − q(βn ) → 0 as n → ∞

i=n+1

q(αi ) < ∞ and limn→∞ βn = 0. So by 384N and 384Gd we get h(π, A) ≤ h(π, Bn ) + H(A|Bn ) ≤ h(π) + H(A|Bn ) → h(π)

as n → ∞, and h(π, A) ≤ h(π). Remark Compare 385Nb below. 384P Theorem (Kolmogorov 58, Sinaˇı 59) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. (i) SupposeSthat A ⊆ A is a partition of unity such that H(A) < ∞ and the closed subalgebra of A generated by n∈N π n [A] is A itself. Then h(π) = h(π, A). (ii) Suppose that π is an automorphism, S and that A ⊆ A is a partition of unity such that H(A) < ∞ and the closed subalgebra of A generated by n∈Z π n [A] is A itself. Then h(π) = h(π, A).

384R

Entropy

509

proof I take the two arguments together. In both cases, by 384O, we have h(π, A) ≤ h(π), so I have to show that if B ⊆ A is S any finite partition of unity, then h(π, B) ≤ h(π, A). For (i), let An be S the partition of unity generated by 0≤j 0. Let φ : A1 → A2 be a measurepreserving automorphism such that π2 = φπ1 φ−1 . Because both µ1 and µ2 are complete and strictly

384Xi

Entropy

513

localizable and compact (343K), there are inverse-measure-preserving functions g1 : X1 → X2 , g2 : X2 → X1 representing φ−1 , φ respectively (343B). Now g1 g2 : X2 → X2 , g2 g1 : X1 → X1 , f2 g1 : X1 → X2 and g1 f1 : X1 → X2 represent, respectively, the identity automorphism on A2 , the identity automorphism on A1 , the homomorphism φ−1 π2 = π1 φ−1 : A2 → A1 and the homomorphism π1 φ−1 again. Next, because both µ1 and µ2 are countably separated, the sets E1 = {x : g2 g1 (x) = x}, H = {x : f2 g1 (x) = g1 f1 (x)} and E2 = {y : g1 g2 (y) = y} are all conegligible (343F). As in part (b) of the proof of 344I, g1 ¹E1 and g2 ¹E2 are the two halves of a bijection, a measure space isomorphism if E1 and E2 are given their subspace −1 measures. T Set G0 = E1 ∩ H, and for n ∈ N0 set Gn+1 = Gn ∩ f1 [Gn ]. Then every G0 n is conegligible, so 0 X1 = n∈N Gn is conegligible. Because X1 is a conegligible subset of E1 , h = g1 ¹X1 is a measure space isomorphism between X10 and X20 = g1 [X10 ], which is conegligible in X2 . Because f1 [Gn+1 ] ⊆ Gn for each n, f1 [X10 ] ⊆ X10 . Because X10 ⊆ H, g1 f1 (x) = f2 g1 (x) for every x ∈ X10 . Next, if y ∈ X20 , g2 (y) ∈ X10 , so f2 (y) = f2 g1 g2 (y) = g1 f1 g2 (y) ∈ g1 [f1 [X10 ]] ⊆ g1 [X10 ] = X20 . Accordingly we have f20 = hf10 h−1 , where fi0 = fi ¹Xi0 for both i. Thus h is an isomorphism between (X10 , f10 ) and (X20 , f20 ), and (X1 , Σ1 , µ1 , f1 ) and (X2 , Σ2 , µ2 , f2 ) are almost isomorphic. 384X Basic exercises (a) Let (A, µ ¯) be a probability algebra and A ⊆ A a partition of unity. Show that if #(A) = n then H(A) ≤ ln n. > (b) Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A and A a partition of unity in A, enumerated as han in∈N . Set a∗n = supi>n ai , An = {a0 , . . . , an , a∗n } for each n. Show that H(An |B) ≤ H(An+1 |B) for every n, and that H(A|B) = limn→∞ H(An |B). (c) Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A and A a partition of unity in A. Show that H(A|B) = 0 iff A ⊆ B. (d) Let (A, µ ¯) be a probability algebra, B a closed subalgebra of A and A a partition of unity in A. Show that H(A|B) = H(A) iff µ ¯(a ∩ b) = µ ¯a · µ ¯b for every a ∈ A, b ∈ B. (Hint: for ‘only if’, start with the case B = {0, b, 1 \ b, 1} and use 384Gc.) (e) Let (A, µ ¯) be a probability algebra and A, B two partitions of unity in A. Show that H(A ∨ B) = H(A) + H(B) iff µ ¯(a ∩ b) = µ ¯a · µ ¯b for all a ∈ A, b ∈ B. Show that H(A ∨ B) = H(A) iff every member of A is included in some member of B, that is, iff A = A ∨ B. ¯ (f ) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras, with probability algebra free product (C, λ) (325K). Suppose that πi : Ai → Ai is a measure-preserving Boolean homomorphism for each i ∈ P I, and that π : C → C is the measure-preserving Boolean homomorphism they induce. Show that h(π) = i∈I h(πi ). (Hint: use 384Gb and N384Gd to show that h(π) is the supremum of h(π, A) as A runs over the finite partitions of unity in i∈I Ai . Use this to reduce to the case I = {0, 1}. Now show that if Ai ⊆ Ai is a finite partition of unity for each i, and A = {a0 ⊗ a1 : a0 ∈ A0 , a1 ∈ A1 }, then H(A) = H(A0 ) + H(A1 ), so that h(π, A) = h(π0 , A0 ) + h(π1 , A1 ).) > (g) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving automorphism. Show that h(π −1 ) = h(π). (h) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. Show that h(π k ) = kh(π) for any k ∈ N. (Hint: if A ⊆ A is a partition of unity, h(π k , A) ≤ h(π k , Dk (A, π)) = kh(π, A).) >(i) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. (i) Suppose there is a partition of unity A ⊆ A such ¯(a ∩ πb) = µ ¯a · µ ¯b for every a ∈ A, b ∈ A (β) S that (α) µ A is the closed subalgebra of itself generated by n∈N π n [A]. Show that π is a one-sided Bernoulli shift, and that h(π) = H(A). (ii) Suppose that π is a one-sided Bernoulli shift of finite entropy. Show that there is a partition of unity satisfying (α) and (β).

514

Automorphism groups

384Xj

> (j) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1[. Fix an integer k ≥ 2, and define f : [0, 1[ → [0, 1[ by setting f (x) = , the fractional part of kx, for every x ∈ [0, 1[; let π : A → B be the corresponding homomorphism. (Cf. 372Xr.) Show that £ π is £a• one-sided Bernoulli shift and that h(π) = ln k. (Hint: in 384Xi, set A = {a0 , . . . , ak−1 } where ai = ki , i+1 for i < k.) k > (k) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1]. Set f (x) = 2 min(x, 1 − x) for x ∈ [0, 1] (see 372Xm). Show that the corresponding homomorphism π : A → A is a one-sided Bernoulli shift and that h(π) = ln 2. (Hint: in 384Xi, set A = {a, 1 \ a} where a = [0, 12 ]• .) (l) Let (A, µ ¯) be a probability algebra and π : A → A a two-sided Bernoulli shift. Show that π −1 is a twosided Bernoulli shift and that there is a measure-preserving involution φ : A → A such that π −1 = φπφ−1 , so that π is a product of two involutions in Autµ¯ (A). ¯ their probability algebra free product. (m) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras, and (C, λ) Suppose that for each i ∈ I we have a measure-preserving Boolean homomorphism πi : Ai → Ai , and that π : C → C is the measure-preserving homomorphism induced by hπi ii∈I (325Xd). (i) Show that if every πi is a one-sided Bernoulli shift so is π. (ii) Show that if every πi is a two-sided Boolean shift so is π. (n) Show that the relation ‘almost isomorphic to’ (384U) is an equivalence relation. (o) Show that the concept of ‘almost isomorphism’ described in 384U is not changed if we amend the definition to require that the subspaces X10 , X20 should be measurable. (p) Show that if (X1 , Σ1 , µ1 , f1 ) and (X2 , Σ2 , µ2 , f2 ) are almost isomorphic quadruples as described in 384U, then (A1 , µ ¯1 , π1 ) and (A2 , µ ¯2 , π2 ) are isomorphic, where for each i (Ai , µ ¯i ) is the measure algebra of (Xi , Σi , µi ) and πi : Ai → Ai is the measure-preserving Boolean homomorphism derived from fi : Xi → Xi . (q) Let (A, µ ¯) be a probability algebra, and write A for the set of partitions of unity in A not containing 0, ordered by saying that A ≤ B if every member of B is included in some member of A. (i) Show that A is a Dedekind complete lattice, and can be identified with the lattice of purely atomic closed subalgebras of A. Show that for A, B ∈ A, A ∨ B, as defined in 384F, is sup{A, B} in A. (ii) Show that H(A ∨ B) + H(A ∧ B) ≤ H(A) + H(B) for all A, B ∈ A, where ∨, ∧ are the lattice operations on A. (iii) Set A1 = {A : A ∈ A, H(A) < ∞}. For A, B ∈ A1 set ρ(A, B) = 2H(A ∨ B) − H(A) − H(B). Show that ρ is a metric on A1 (the entropy metric). (iv) Show that if π : A → A is a measure-preserving Boolean homomorphism, then |h(π, A) − h(π, B)| ≤ ρ(A, B) for all A, B ∈ A1 . (iv) Show that the lattice operations ∨, ∧ are ρ-continuous on A1 . (v) Show that H : A1 → [0, ∞[ is order-continuous. (vi) Show that if B is any closed subalgebra of A, then A 7→ H(A|B) is order-continuous on A1 . 384Y Further exercises (a) Let (A, µ ¯) be a probability algebra, and write P for the lattice of closed subalgebras of A. Show that if A is any partition of unity in A of finite entropy, then the order-preserving function B 7→ −H(A|B) : P → ]−∞, 0] is order-continuous. (b) Let (A, µ ¯) be a probability algebra, A a partition of unity in A of finite entropy, and π : A → A a measure-preserving Boolean homomorphism. Show that h(π, A) = limn→∞ H(A|Bn ), where Bn is the closed S subalgebra of A generated by 1≤i≤n π i [A]. (Hint: use 384Gb to show that H(A|Bn ) = H(Dn+1 (A, π)) − H(Dn (A, π)) and observe that limn→∞ H(A|Bn ) is defined.) (c) Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. Suppose that there is a partition of unity A of finite entropy such that the closed subalgebra of A generated S by i≥1 π i [A] is A. Show that h(π) = 0. (Hint: use 384Yb and 384Pa.) (d) Let µ be Lebesgue measure on [0, 1[, and take any α ∈ ]0, 1[. Let f : [0, 1[ → [0, 1[ be the measure space automorphism defined by saying that f (x) is to be one of x + α, x + α − 1. Let (A, µ ¯) be the measure algebra of ([0, 1[ , µ) and π : A → A the measure-preserving automorphism corresponding £to f £. Show that • h(π) = 0. (Hint: if α ∈ Q, use 384Xh; otherwise use 384Yc with A = {a, 1 \ a} where a = 0, 21 .)

385A

More about entropy

515

R 1 (e) Set X = [0, 1] \ Q, let ν be the measure on X defined by setting νE = ln12 E 1+x dx for every Lebesgue measurable set E ⊆ X, and for x ∈ X let f (x) be the fractional part < x1 > of x1 . Recall that f is inverse-measure-preserving for ν (372N). Let (A, ν¯) be the measure algebra of (X, ν) and π : A → A the homomorphism corresponding to f . Show that h(π) = π 2 /6 ln 2. (Hint: use the Kolmogorov-Sinaˇı theorem and 372Yf(v).) (f ) Consider the triplets ([0, 1[ , µ1 , f1 ) and ([0, 1], µ2 , f2 ) where µ1 , µ2 are Lebesgue measure on [0, 1[, [0, 1] respectively, f1 (x) = for each x ∈ [0, 1[, and f2 (x) = 2 min(x, 1 − x) for each x ∈ [0, 1]. Show that these structures are almost isomorphic in the sense of 384U, and give a formula for an isomorphism. (g) Let (A, µ ¯) be a probability algebra, and A1 the set of partitions of unity of finite entropy not containing 0, as in 384Xq. Show that A1 is complete under the entropy metric. (Hint: show that if hAn in∈N S is a nondecreasing sequence in A1 and supn∈N H(An ) < ∞, then the closed subalgebra of A generated by n∈N An is purely atomic.) 384 Notes and comments In preparing this section I have been heavily influenced by Petersen 83. I have taken almost the shortest possible route to Theorem 384P, the original application of the theory, ignoring both the many extensions of these ideas and their intuitive underpinning in the concept of the quantity of ‘information’ carried by a partition. For both of these I refer you to Petersen 83. The techniques described there are I think sufficiently powerful to make possible the calculation of the entropy of any of the measure-preserving homomorphisms which have yet appeared in this treatise. Of course the idea of entropy of a partition, or of a homomorphism, can be translated into the language of probability spaces and inverse-measure-preserving functions; if (X, Σ, µ) is a probability space, with measure algebra (A, µ ¯), then partitions of unity in A correspond (subject to decisions on how to treat negligible sets) to countable partitions of X into measurable sets, and an inverse-measure-preserving function from X to itself gives rise to a measure-preserving homomorphism πf : A → A; so we can define the entropy of f to be h(πf ). The whole point of the language I have sought to develop in this volume is that we can do this when and if we choose; in particular, we are not limited to those homomorphisms which are representable by inverse-measure-preserving functions. But of course a large proportion of the most important examples do arise in this way (see 384Xj, 384Xk). The same two examples are instructive from another point of view: the case k = 2 of 384Xj is (almost) isomorphic to the tent map of 384Xk. The similarity is obvious, but exhibiting an actual isomorphism is I think another matter (384Yf). I must say ‘almost’ isomorphic here because the doubling map on [0, 1[ is everywhere two-to-one, while the tent map is not, so they cannot be isomorphic in any exact sense. This is the problem grappled with in 384T-384V. In some moods I would say that a dislike of such contortions is a sign of civilized taste. Certainly it is part of my motivation for working with measure algebras whenever possible. But I have to say also that new ideas in this topic arise more often than not from actual measure spaces, and that it is absolutely necessary to be able to operate in the more concrete context.

385 More about entropy In preparation for the next two sections, I present a number of basic facts concerning measure-preserving homomorphisms and entropy. Compared with the work to follow, they are mostly fairly elementary, but the Halmos-Rokhlin-Kakutani lemma (385E) and the Shannon-McMillan-Breiman theorem (385G), in their full strengths, go farther than one might expect. 385A Periodic and aperiodic parts If X is a set and f : X → X is a bijection, then the orbits Ωx = {f n (x) : n ∈ Z} of f are described by their cardinalities, and X has a natural decomposition into the sets Xi = {x : #(Ωx ) = i} for 1 ≤ i ≤ ω. Corresponding to this is a partition of unity of a Dedekind complete Boolean algebra with an automorphism, as follows. Definition If A is a Boolean algebra, a Boolean homomorphism π : A → A is periodic, with period n ≥ 2, if A 6= {0}, π n is the identity operator and whenever b ∈ A \ {0} and 1 ≤ i < n there is a c ⊆ b such that

516

Automorphism groups

385A

π i c 6= c. π is periodic with period 1 iff it is the identity operator. π is aperiodic if for every non-zero b ∈ A, n ≥ 1 there is a c ⊆ b such that π n c 6= c. I remark immediately that if π is aperiodic, so is π n for every n ≥ 1. Note that if A = {0} then the trivial automorphism of A is counted both as aperiodic and as periodic with period 1. 385B Lemma Let A be a Dedekind complete Boolean algebra and π : A → A a Boolean homomorphism which is periodic with period n ≥ 2. Then π is a Boolean automorphism and there is an a ∈ A such that (a, πa, π 2 a, . . . , π n−1 a) is a partition of unity in A; that is (in the language of 381G) π is of the form −− (← a1−π−a−2−π−−.− .− .− π an ) where (a1 , . . . , an ) is a partition of unity in A. proof Because π n : A → A is injective and surjective, so is π, and π is a Boolean automorphism, therefore order-continuous. Set D = {d : d ∈ A, π i d ∩ d = 0 whenever 1 ≤ i < n}. Then the supremum of any upwards-directed subset of D is defined in A (because A is Dedekind complete) and belongs to D (because π i is order-continuous for every i – use 313Bc); so D has a maximal element a, by Zorn’s lemma. ?? If (a, πa, . . . , π n−1 a) is not a partition of unity in A, set b0 = 1 \ (a ∪ πa ∪ . . . ∪ π n−1 a). Then πb0 = 1 \ (πa ∪ . . . ∪ π n−1 a ∪ a) = b0 . Now we can choose b1 , . . . , bn−1 such that 0 6= bi ⊆ bi−1 and π i bi ∩ bi = 0 for 1 ≤ i < n. P P Given that bi−1 6= 0, where 1 ≤ i < n, then (by the definition of ‘periodic’ homomorphism) there is a c ⊆ bi−1 such that π i c 6= c. If c \ π i c 6= 0, take bi = c \ π i c. Otherwise, π i c \ c 6= 0 so c \ π −i c 6= 0 and we can take bi = c \ π −i c. Q Q At the end of this induction, set d = bn−1 ; then d ∩ π i d = 0 for 1 ≤ i < n, so d ∈ D; also d ∩ π i a = π i d ∩ a = 0 for 1 ≤ i < n (because π i d ⊆ π i b0 = b0 for every i), so a ∪ d ∈ D, which is supposed to be impossible. X X Thus (a, πa, . . . , π n−1 a) is a partition of unity, as required. 385C Proposition Let A be a Dedekind complete Boolean algebra and π : A → A an injective ordercontinuous Boolean homomorphism. Then there is a partition of unity hci i1≤i≤ω in A such that πci = ci for every i and π¹ Acn is periodic with period n whenever n is finite and cn 6= 0, while π¹ Acω is aperiodic. Remark As usual, I write Aa for the principal ideal of A generated by a. proof For each n ≥ 1, set Bn = {b : b ∈ A, π n c = c for every c ⊆ b},

bn = sup Bn .

Because π is a Boolean homomorphism, Bn is an ideal of A; because π is order-continuous, bn ∈ Bn , that is, Bn is the principal ideal Abn . Also πbn = bn . P P If b ∈ Bn and c ⊆ πb, then π n−1 c ⊆ π n b = b, so π n−1 (π n c) = π n (π n−1 c) = π n−1 c. But π n−1 , like π, is injective, so π n c = c. As c is arbitrary, πb ∈ Bn . As π is order-continuous, πbn = supb∈Bn πb ⊆ bn . But this means that π

i+1

i

bn ⊆ π bn for every i, so bn = π n bn ⊆ πbn ⊆ bn

and πbn = bn . Q Q Set cn = bn \ sup1≤i 0. By 331B, there is a d ∈ A such that µ ¯(c ∩ d) = δ µ ¯c for every c ∈ C. Set dk = π k d \ supi 12 ]] ∩ [[P (χa0 ) > 21 ]] ⊆ [[P (χa) + P (χa0 ) > 1]] = 0, by 364D(b-i). (b) By 384Ae, q(1−t) ≤ q(t) whenever 0 ≤ t ≤ 12 . Consequently q(t) ≤ q(min(t, 1−t)) for every t ∈ [0, 1], and q¯(u) ≤ q¯(u ∧ (χ1 − u)) whenever u ∈ L0 (A) and 0 ≤ u ≤ χ1. Fix a ∈ A for the moment. We have q¯(P (χa)) ≤ q¯(P (χa) ∧ (χ1 − P (χa)) = q¯(|P (χa) − χh(a)|). Consequently Z

Z q¯(P χa) ≤

q¯(|P (χa) − χh(a)|) ≤ q

¡

Z

¢ |P (χa) − χh(a)|

(because q is concave) = q(ρ(a, B)). Summing over a, H(A|B) =

R

P a∈A

q¯(P χa) ≤

P a∈A

q(ρ(a, B)).

(c) Set b0i = h(ai ) for each i ∈ N. Then hb0i ii∈N is disjoint. Next, for each i ∈ N, take b00i ∈ B such that b0i and µ ¯b00i = min(¯ µai , µ ¯b00i ); then hb00i ii∈N is disjoint and µ ¯b00i ≤ µ ¯ai for every i. We can therefore find a 00 partition of unity hbi ii∈N such that bi ⊇ bi and µ ¯ bi = µ ¯ai for every i. (Use 331C to choose hdi ii∈N inductively so that di ⊆ 1 \ (supj µ ¯ai , then

b00i ⊆

µ ¯(ai 4 bi ) = µ ¯(ai 4 b00i ) ≤ µ ¯(ai 4 b0i ) + µ ¯(b0i 4 b00i ) =µ ¯(ai 4 b0i ) + µ ¯b0i − µ ¯ai ≤ 2¯ µ(ai 4 b0i ) = 2ρ(ai , B). If µ ¯b0i ≤ µ ¯ai , then µ ¯(ai 4 bi ) ≤ µ ¯(ai 4 b0i ) + µ ¯(b0i 4 bi ) =µ ¯(ai 4 b0i ) + µ ¯ai − µ ¯b0i ≤ 2¯ µ(ai 4 b0i ) = 2ρ(ai , B).

526

Automorphism groups

385P

385P Lemma Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving automorphism. Suppose that B ⊆ A. For k ∈ N, let Bk be the closed subalgebra of A generated by {π j b : b ∈ B, |j| ≤ k}, and let B be the closed subalgebra of A generated by {π j b : b ∈ B, j ∈ Z}. Then S (a) B is the topological closure k∈N Bk . (b) π[B] = B. (c) If C is any closed subalgebra of A such that π[C] = C, and a ∈ Bk , then P ρ(a, C) ≤ (2k + 1) b∈B ρ(b, C). proof (a) Because hBk ik∈N is non-decreasing, and must be B.

S k∈N

Bk is a subalgebra of A, so its closure also is (323J),

(b) Of course π −1 [Bk+1 ] is a closed subalgebra of A containing π j b whenever |j| ≤ k and b ∈ B, so includes Bk ; thus π[Bk ] ⊆ Bk+1 ⊆ B for every k, and S S π[B] = π[ k∈N Bk ] ⊆ k∈N π[Bk ] ⊆ B ⊆ B because π is continuous (324Kb). Similarly, π −1 [B] ⊆ B and π[B] = B. (c) For each b ∈ B, choose cb ∈ C such that µ ¯(b 4 cb ) = ρ(cb , C) (385Oa). Set e = sup|j|≤k supb∈B π j (b 4 cb ); then µ ¯e ≤ (2k + 1)

P b∈B

µ ¯(b 4 cb ) = (2k + 1)

P b∈B

ρ(b, C).

Now B0 = {d : d ∈ A, ∃ c ∈ C such that d \ e = c \ e} is a subalgebra of A. By 314Fa, applied to the order-continuous homomorphism c 7→ c \ e : C → A1\e , {c \ e : c ∈ C} is an order-closed subalgebra of the principal ideal A1\e ; by 313Id, applied to the ordercontinuous function d 7→ d \ e : A → A1\e , B0 is order-closed. If b ∈ B and |j| ≤ k, then π j b 4 π j cb ⊆ e, so π j b ∈ B0 ; accordingly B0 ⊇ Bk . Now a ∈ Bk , so there is a c ∈ C such that a 4 c ⊆ e, and P ρ(a, C) ≤ µ ¯(a 4 c) ≤ µ ¯e ≤ (2k + 1) b∈B ρ(b, C), as claimed. 385Q Lemma Let (A, µ ¯) be a probability algebra and suppose either that A is not purely atomic or that it is purely atomic and H(D0 ) = ∞, where D0 is the set of atoms of A. Then whenever A ⊆ A is a partition of unity and H(A) ≤ γ ≤ ∞, there is a partition of unity B, refining A, such that H(B) = γ. proof (a) By 384J, there is a partition of unity D1 such that H(D1 ) = ∞. Set D = D1 ∨ A; then we still have H(D) = ∞. Enumerate D as hdi ii∈N . Choose hBk ik∈N inductively, as follows. B0 = A. Given that Bk is a partition of unity, then if H(Bk ∨ {dk , 1 \ dk }) ≤ γ, set Bk+1 = Bk ∨ {dk , 1 \ dk }; otherwise set Bk+1 = Bk . S Let B be the closed subalgebra of A generated by k∈N Bk . Note that, for each d ∈ D, {c : c ∈ A, d ⊆ c or d ∩ c = 0} is a closed subalgebra of A including every Bk , so includes B. If b ∈ B \ {0}, there is surely some d ∈ D such that b ∩ d 6= 0, so b ⊇ d; thus B must be purely atomic. Let B be the set of atoms of B. Because A = B0 ⊆ B, B refines A. (b) H(B) ≤ γ. P P For each k ∈ N, let Bk be the closed subalgebra of A generated by Bk , so that S B = k∈N Bk . Suppose that b0 , . . . , bn are distinct members of B. Then for each k ∈ N we can find disjoint b0k , . . . , bnk ∈ Bk such that µ ¯(bik 4 bi ) ≤ ρ(bi , Bk ) for every i ≤ n (385Oa). Accordingly µ ¯bi = limk→∞ µ ¯bik for each i, and Pn Pn µbi ) = limk→∞ i=0 q(¯ µbik ) ≤ supk∈N H(Bk ) ≤ γ. i=0 q(¯ As b0 , . . . , bn are arbitrary, H(B) ≤ γ. Q Q

385Yd

More about entropy

527

(c) H(B) ≥ γ. P P?? Suppose otherwise. We know that limk→∞ H({dk , 1 \ dk }) = limk→∞ q(¯ µdk ) + q(1 − µ ¯dk ) = 0. Let m ∈ N be such that H(B) + H({dk , 1 \ dk }) ≤ γ for every k ≥ m. Because B refines Bk , we must have H(Bk ∨ {dk , 1 \ dk }) ≤ H(Bk ) + H({dk , 1 \ dk }) ≤ γ, so that Bk+1 = Bk ∨ {dk , 1 \ dk } for every k ≥ m. But this means that dk ∈ B for every k ≥ m, so that P∞ γ > H(B) ≥ k=m q(¯ µdk ) = ∞, which is impossible. X XQ Q Thus B has the required properties. 385X Basic exercises > (a) Let A be a Boolean algebra, not {0}, and π : A → A an automorphism; set C = {c : πc = c}. Show that π is periodic, with period n ≥ 1, iff π¹ Ac has order n in the group Aut Ac whenever c ∈ C \ {0}. Show that π is aperiodic iff π¹ Ac has infinite order in the group Aut Ac whenever c ∈ C \ {0}. (b) In 385C, show that the family hci i1≤i≤ω is uniquely determined. > (c) Show that for a Bernoulli shift π, the Shannon-McMillan-Breiman theorem is a special case of the Pn−1 Ergodic Theorem. (Hint: wn = n1 i=0 T i w1 .) (d) Let (A, µ ¯) be a totally finite measure algebra, hBk ik∈N a non-decreasing sequence of subsets of A such that 0 ∈ B0 , and hci ii∈I a partition of unity in A. Show that P P limk→∞ i∈I ρ(ci , Bk ) = i∈I ρ(ci , B) S where B = k∈N Bk . > (e) Let (A, µ ¯) be a probability algebra, π : A → A a measure-preserving Boolean homomorphism and A a partition of unity in A. Show that h(π, Dn (A, π)) = h(π, A) = h(π, π[A]) for any n ≥ 1. (f ) Let (A, µ ¯) be a totally finite measure algebra and π : A → A a measure-preserving Boolean homomorphism. Suppose that B ⊆ A. For k ∈ N, let Bk be the closed subalgebra of A generated by {π j b : b ∈ B, j ≤ k}, and let B be the closed subalgebra of A generated by {π j b : b ∈ B, j ∈ N}. Show that S B = k∈N Bk , π[B] ⊆ B, P and that if C is any closed subalgebra of A such that π[C] ⊆ C, and a ∈ Bk , then ρ(a, C) ≤ (k+1) b∈B ρ(b, C). 385Y Further exercises (a) Give an example to show that the word ‘injective’ in the statement of 385C is essential. (b) Let (A, µ ¯) be a totally finite measure algebra and π : A → A an aperiodic measure-preserving Boolean homomorphism. Set C = {c : πc = c}. Show that whenever n ≥ 1, 0 ≤ γ < n1 and B ⊆ A is finite, there is an a ∈ A such that a, πa, π 2 a, . . . , π n−1 a are disjoint and µ ¯(a ∩ b ∩ c) = γ µ ¯(b ∩ c) for every b ∈ B, c ∈ C. (c) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism. Let P be the set of all closed subalgebras of A which are invariant under π, ordered by inclusion. Show that B 7→ h(π¹ B) : P → [0, ∞] is order-preserving and order-continuous on the left, in the sense that if Q ⊆ P is non-empty and upwards-directed then h(π¹ sup Q) = supB∈Q h(π¹ B). (d) Let (A, µ ¯) be a probability algebra, and π : A → A a measure-preserving Boolean homomorphism of finite entropy. Let P be the set of all π-invariant closed subalgebras of A. Show that B 7→ h(π¹ B) : P → [0, ∞[ is order-continuous. (Hint: if Q ⊆ P is non-empty and downwards-directed, then for any partition of unity A ⊆ A, H(A| inf Q) = supB∈Q H(A|B).)

528

Automorphism groups

385 Notes

385 Notes and comments I have taken the trouble to give sharp forms of the Halmos-Rokhlin-Kakutani lemma (385E) and the Czisz´ar-Kullback inequality (385J); while it is possible to get through the principal results of the next two sections with rather less, the formulae become better focused if we have the exact expressions available. Of course one can always go farther still (385Yb). Ornstein’s theorem in §386 (though not Sinaˇı’s, as stated there) can be deduced from the special case of the Shannon-McMillan-Breiman theorem (385G) in which the homomorphism π is a Bernoulli shift, which can be deduced from the Ergodic Theorem (385Xc). Lemma 385D is the starting point of the theory of ‘recurrence’; the next steps are in 382Yc-382Yd and 387E-387F.

386 Ornstein’s theorem I come now to the most important of the handful of theorems known which enable us to describe automorphisms of measure algebras up to isomorphism: two two-sided Bernoulli shifts (on algebras of countable Maharam type) of the same entropy are isomorphic (386I, 386K). This is hard work. It requires both delicate ²-δ analysis and substantial skill with the manipulation of measure-preserving homomorphisms. The proof is based on two difficult lemmas (386C and 386F), and includes Sinaˇı’s theorem (386E, 386L), describing the Bernoulli shifts which arise as factors of a given ergodic automorphism. 386A

The following definitions offer a language in which to express the ideas of this section.

Definitions Let (A, µ ¯) be a probability algebra and π : A → A a measure-preserving Boolean homomorphism. (a) A Bernoulli partition for π is a partition of unity hai ii∈I such that Qk ¯ai(j) µ ¯(inf j≤k π j ai(j) ) = j=0 µ whenever i(0), . . . , i(k) ∈ I. (b) If π is an automorphism, a Bernoulli partition hai ii∈I for π is (two-sidedly) generating if the closed subalgebra generated by {π j ai : i ∈ I, j ∈ Z} is A itself. (c) A factor of (A, µ ¯, π) is a triple (B, µ ¯¹ B, π¹ B) where B is a closed subalgebra of A such that π[B] = B. 386B Remarks Let (A, µ ¯) be a probability algebra, π : A → A a measure-preserving Boolean homomorphism and hai ii∈I a Bernoulli partition for π. (a) hπ k [A0 ]ik∈N is independent, where A0 is the closed subalgebra of A generated by {ai : i ∈ I}. P P Suppose that cj ∈ π j [A0 ] for j ≤ k. Then each π −j cj ∈ A0 is expressible as supi∈Ij ai for some Ij ⊆ I. Now µ ¯( inf cj ) = µ ¯( j≤k

inf π j aij )

sup

i0 ∈I0 ,... ,ik ∈Ik j≤k

=

X i0 ∈I0 ,... ,ik ∈Ik

=

k X Y j=0 i∈Ij

X

µ ¯( inf π j aij ) =

µ ¯ai =

j≤k

k Y j=0

k Y

µ ¯aij

i0 ∈I0 ,... ,ik ∈Ik j=0

µ ¯(sup ai ) = i∈Ij

k Y

µ ¯cj .

j=0

As c0 , . . . , ck are arbitrary, hπ k [A0 ]ik∈N is independent. Q Q (b) If π is an automorphism, then hπ k [A0 ]ik∈Z is independent, by 384Sf. (c) Setting A = {ai : i ∈ I} \ {0}, we have h(π, A) = H(A), as in part (a) of the proof of 384R, so h(π) ≥ H(A).

386C

Ornstein’s theorem

529

(d) If H(A) > 0, then A is atomless. P P As A contains at least two elements of non-zero measure, γ = maxa∈A µ ¯a < 1. Because hai ii∈I is a Bernoulli partition, every member of Dk (A, π) has measure at most γ k , for any k ∈ N. Thus any atom of A could have measure at most inf k∈N γ k = 0. Q Q (e) If B is any closed subalgebra of A such that π[B] ⊆ B, then h(π¹ B) ≤ h(π), just because h(π¹ B) is calculated from the action of π on a smaller set of partitions. If C+ is the closed subalgebra of A generated by {π j ai : i ∈ I, j ∈ N}, then π[C+ ] ⊆ C+ (compare 385Pb), and π¹ C+ is a one-sided Bernoulli shift with root algebra A0 and entropy H(A), so that H(A) = h(π¹ C+ ) by the Kolmogorov-Sinaˇı theorem (384P, 384R). (f ) If π is an automorphism, and C is the closed subalgebra of A generated by {π j ai : i ∈ I, j ∈ Z}, then π[C] = C (385Pb) and π¹ C is a two-sided Bernoulli shift with root algebra A0 . (g) Thus every Bernoulli partition for π gives rise to a factor of (A, µ ¯, π) which is a one-sided Bernoulli shift, and if π is an automorphism we can extend this to the corresponding two-sided Bernoulli shift. If π has a generating Bernoulli partition then it is itself a Bernoulli shift. (h) Now suppose that (B, ν¯) is another probability algebra, φ : B → B is a measure-preserving Boolean homomorphism, and hbi ii∈I is a Bernoulli partition for φ such that ν¯bi = µ ¯ai for every i. We have a unique measure-preserving Boolean homomorphism θ+ : C+ → B such that θ+ (π j ai ) = φj bi for every i ∈ I, j ∈ N. (Apply 324P.) Now θ+ π = φθ+ . (The set {a : θ+ πa = φθ+ a} is a closed subalgebra of C+ containing every π j ai .) (i) If, in (g) above, π and φ are both automorphisms, then the same arguments show that we have a unique measure-preserving Boolean homomorphism θ : C → B such that θai = bi for every i ∈ I and θπ = φθ. 386C Lemma Let (A, µ ¯) be an atomless probability algebra and π : A → A an ergodic measurepreserving automorphism. Let hai ii∈N be a partition of unity in A, of finite entropy, and hγi ii∈N a sequence of non-negative real numbers such that P∞ P∞ i=0 q(γi ) ≤ h(π), i=0 γi = 1, where q is the function of 384A. Then for any ² > 0 we can find a partition ha0i ii∈N of unity in A such that (i) {i : a0i 6= 0} is finite, P∞ (ii) i=0 |γi − µ ¯a0i | ≤ ², qP p P∞ ∞ ¯(a0i 4 ai ) ≤ ² + 6 (iii) i=0 µ µai − γi | + 2(H(A) − h(π, A)) i=0 |¯ where A = {ai : i ∈ N} \ {0}, (iv) H(A0 ) ≤ h(π, A0 ) + ² where A0 = {a0i : i ∈ N} \ {0}. p proof (a) Of course h(π, A) ≤ H(A), by 384Ma, so the square root 2(H(A) − h(π, A)) gives no difficulty. qP p ∞ 1 ²). Set β = µai − γi | + 2(H(A) − h(π, A)), δ = min( 14 , 24 i=0 |¯ P∞ There is a sequence h¯ γi ii∈N of non-negative real numbers such that {i : γ¯i > 0} is finite, i=0 γ¯i = 1, P∞ P P ∞ ∞ 2 |¯ γi −γi | ≤ 2δ 2 and i=0 q(¯ γi ) ≤ h(π). P P Take i=0P P∞k ∈ N such that i=k γi ≤ δ , and set γ¯i = γi for i < k, ∞ γ¯k = i=k γi and γ¯i = 0 for i > k; then q(¯ γk ) ≤ i=k q(γi ) (384Ab), so P∞ P∞ γi ) ≤ i=0 q(γi ) ≤ h(π), i=0 q(¯ while

P∞ i=0

|¯ γi − γi | ≤ γ¯k +

P∞ i=k

γi ≤ 2δ 2 . Q Q

P∞ P∞ Because i=0 q(¯ γi ) is finite, there is a partition of unity C in A, of finite entropy, such that i=0 q(¯ γi ) ≤ h(π, C) + 3δ; replacing C by C ∨ A if need be (note that C ∨ A still has finite entropy, by 384H), we may suppose that C refines A.

530

Automorphism groups

386C

There is a sequence hγi0 ii∈N of non-negative real numbers such that P∞ 0 2 i=0 |γi − γi | ≤ 4δ and P∞ 0 i=0 q(γi ) = h(π, C) + 3δ.

P∞ i=0

γi0 = 1, {i : γi0 > 0} is finite,

P P Take k ∈ N such that γ¯i = 0 for i > k. Take r ≥ 1 such that δ 2 ln( δr2 ) ≥ h(π, C) + 3δ and set γ˜i = (1 − δ 2 )¯ γi for i ≤ k, 1 r

= δ 2 for k + 1 ≤ i ≤ k + r, = 0 for i > k + r. Then

P∞ i=0

Pk+r i=0

|˜ γi − γ¯i | = 2δ 2 ,

P∞ i=0

|˜ γi − γi | ≤ 4δ 2 , δ2 r

r δ

q(¯ γi ) ≤ h(π, C) + 3δ ≤ δ 2 ln( 2 ) = rq( ) ≤

Now the function α 7→

Pk+r i=0

Pk+r i=0

q(˜ γi ).

q(α¯ γi + (1 − α)˜ γi ) : [0, 1] → R

is continuous, so there is some α ∈ [0, 1] such that Pk+r γi + (1 − α)˜ γi ) = h(π, C) + 3δ, i=0 q(α¯ γi + (1 − α)˜ γi for every i; of course and we can set γi0 = α¯ P∞ P∞ P∞ 0 γi − γi | ≤ 4δ 2 . Q Q γi − γi | + (1 − α) i=0 |˜ i=0 |¯ i=0 |γi − γi | ≤ α Set M = {i : γi0 6= 0}, so that M is finite. (b) Let η ∈ ]0, δ] be so small that δ

(i) |q(s) − q(t)| ≤ whenever s, t ∈ [0, 1] and |s − t| ≤ 3η, 1+#(M ) P µc, 2η)) ≤ δ, (ii) c∈C q(min(¯ 1 6

(iii) η ≤ . P µc) is finite.) (Actually, (iii) is a consequence of (i). For (ii) we must of course rely on the fact that c∈C q(¯ Let ν be the probability measure on M defined by saying that ν{i} = γi0 for every i ∈ M , and λ the product measure on M N . Define Xij : M N → {0, 1}, for i ∈ M and j ∈ N, and Yj : M N → R, for j ∈ N, by setting Xij (ω) = 1 if ω(j) = i, = 0 otherwise, 0 Yj (ω) = ln(γω(j) ) for everyω ∈ M N .

Then, for each i ∈ M , hXij ij∈N is an independent sequence of random variables, all with expectation γi0 , and hYj ij∈N is also an independent sequence of random variables, all with expectation P P∞ 0 0 0 i∈M γi ln γi = − i=0 q(γi ) = −h(π, C) − 3δ. Let n ≥ 1 be so large that (iv) µ ¯[[wn − h(π, C)χ1 ≥ δ]] < η, where 1 P 1 )χd; wn = d∈Dn (C,π) ln( n

µ ¯d

(v) ¡P ¢ 1 Pn−1 0 Pr j=0 Xij − γi | ≤ η ≥ 1 − δ, i∈M | n

386C

Ornstein’s theorem

531

¡ 1 Pn−1 ¢ Pr | j=0 Yj + h(π, C) + 3δ| ≤ δ ≥ 1 − δ; n

(vi) enδ ≥ 2,

1 n+1

≤ η,

q(

1 n ) + q( ) n+1 n+1

≤ δ;

these will be true for all sufficiently large n, using the Shannon-McMillan-Breiman theorem (385H) for (iv) and the strong law of large numbers (in any of the forms 273D, 273H or 273I) for (v). (c) There is a family hbji ij n this is trivial, because µ ¯ci ≤ ξ, by the choice of n. Otherwise, ci ∈ E. Take b ∈ Br such that µ ¯(b 4 ci ) = ρ(ci , Br ) ≤ ξ (385Oa). Let b∗ ∈ B∗ be such that θb = b∗ ∩ e˜. Then ρ(ci , B∗k ) ≤ µ ¯(ci 4 b∗ ) ≤ 1 − µ ¯e˜ + µ ¯(˜ e ∩ (ci 4 b∗ )) =

2r+1 2n+2

+µ ¯((˜ e ∩ ci ) 4 θb) =

=

2r+1 2n+2

+µ ¯(θ(ci 4 b)) ≤

2r+1 2n+2

+µ ¯(θci 4 θb)

(by (d-iv)) 2r+1 2n+2

+µ ¯(ci 4 b)

(by (d-ii)) ≤ 2ξ by the choice of n. Q Q P (f ) Set B ∗ = {b∗i : i ∈ N} \ {0}. Then H(B ∗ ) = h(π, C) ≤ h(π, B ∗ ) + η. P H(B ∗ ) = H(B) = H(C) ¯bi for every i, and we supposed from the beginning that H(C) = H(B)) (because µ ¯b∗i = µ = h(π, C) (because C is a Bernoulli partition, see 386Bc) ≤ h(π¹ B∗r ) + H(C|B∗r ) (385Nd) ≤ h(π¹ B∗ ) +

∞ X

q(ρ(ci , B∗r ))

i=0

(by the definition of h(π¹ B∗ ), and 385Ob) ≤ h(π, B ∗ ) +

∞ X

q(min(2ξ, µ ¯ci ))

i=0

(by the Kolmogorov-Sinaˇı theorem, 384P(ii), and (e) above, recalling that ξ ≤ 61 , so that q is monotonic on [0, 2ξ]) ≤ h(π, B ∗ ) + η by the choice of ξ. Q Q

386G

Ornstein’s theorem

541

(g) By 386D, applied to π¹ C and the partition hb∗i ii∈N of unity in C and the sequence hγi ii∈N = h¯ µb∗i ii∈N , ∗ we have a Bernoulli partition hdi ii∈N in C such that µ ¯di = µ ¯ bi = µ ¯bi for every i ∈ N and P∞ √ ² ¯(di 4 b∗i ) ≤ η + 6 4 2η ≤ . i=0 µ 4(2m+1)

j

Let D ⊆ C be the closed subalgebra of A generated by {π di : i ∈ N, j ∈ Z}. Then (B, π¹ B, hbi ii∈N ) is isomorphic to (D, π¹ D, hdi ii∈N ), with an isomorphism φ : B → D such that φπ = πφ and φbi = di for every i ∈ N (386Bi). (h) Set e∗ = e˜ \ sup|j|≤m,i∈N π j (di 4 b∗i ). Then φ(π j bi ) ∩ e∗ = θ(π j bi ) ∩ e∗ whenever i ≤ m and |j| ≤ m. P P φ(π j bi ) ∩ e∗ = π j (φbi ) ∩ e∗ = π j di ∩ e∗ = π j b∗i ∩ e∗ = π j b∗i ∩ e˜ ∩ e∗ = θ(π j bi ) ∩ e∗ by (d-iii), because i, |j| ≤ r ≤ n. Q Q Since b 7→ φb ∩ e∗ : A → Ae∗ , b 7→ θb ∩ e∗ : E → Ae∗ are Boolean ∗ ∗ homomorphisms, φb ∩ e = θb ∩ e for every b ∈ Bm . Now µ ¯(ci 4 φci ) ≤ ² for every i ∈ N. P P If i > n then of course µ ¯(φci 4 ci ) ≤ 2¯ µci ≤ 2ξ ≤ ². If i ≤ n, then (by the choice of m) there is a b ∈ Bm such that µ ¯(ci , b) ≤ 14 ². So φci 4 ci ⊆ (φci 4 φb) ∪ (φb 4 θb) ∪ (θb 4 θci ) ∪ (θci 4 ci ) ⊆

φ(ci 4 b) ∪ (1 \ e∗ ) ∪ θ(b 4 ci )

(using the definition of e∗ and (d-iv)) has measure at most µ ¯(ci 4 b) + µ ¯(1 \ e∗ ) + µ ¯(b 4 ci ) (by (d-ii), since b and c both belong to E) ≤ 2¯ µ(ci 4 b) + µ ¯(1 \ e˜) + (2m + 1)

∞ X

µ ¯(di 4 b∗i )

i=0

≤

² 2

+

2r+1 2n+2

+

² 4

≤ ²,

as required. Q Q 386G Lemma Let (A, µ ¯) be an atomless probability algebra and π a measure-preserving automorphism of A. Let hbi ii∈N , hci ii∈N be Bernoulli partitions for π, of the same finite entropy, and write B, C for the closed subalgebras generated by {π j bi : i ∈ N, j ∈ Z} and {π j ci : i ∈ N, j ∈ Z}. Suppose that C ⊆ B. Then for any ² > 0 we can find a Bernoulli partition hdi ii∈N for π such that (i) µ ¯ di = µ ¯ci for every i ∈ N, (ii) µ ¯(di 4 ci ) ≤ ² for every i ∈ N, (iii) writing D for the closed subalgebra of A generated by {π j di : i ∈ N, j ∈ Z}, ρ(bi , D) ≤ ² for every i ∈ N. proof (a) By 386F, there is a Bernoulli partition hb∗i ii∈N for π such that b∗i ∈ C for every i ∈ N, µ ¯b∗i = µ ¯bi 1 for every i ∈ N, and µ ¯(φci 4 ci ) ≤ 4 ² for every i ∈ N, where φ : B → C is the measure-preserving Boolean homomorphism such that φbi = b∗i for every i and πφ = φπ. Note that this implies that π −1 φ = φπ −1 , and generally that π j φ = φπ −j for every j ∈ Z; so φ[B] ⊆ C is the closed subalgebra of A generated by {φπ j bi : i ∈ N, j ∈ Z} = {π j b∗i : i ∈ N, j ∈ Z} (324L), and is invariant under the action of π and π −1 . Let m ∈ N be such that ρ(ci , Bm ) ≤ 41 ² for every i ∈ N,

542

Automorphism groups

386G

where Bm is the closed subalgebra of A generated by {π j bi : i ∈ N, |j| ≤ m} (385M). Let η ∈ ]0, π] be such that P∞ (2m + 1) i=0 min(η, 2¯ µbi ) ≤ 41 ². By 386F again, applied to π¹ C, there is a Bernoulli partition hc∗i ii∈N for π such that c∗i ∈ φ[B], µ ¯c∗i = µ ¯ci ∗ ∗ and µ ¯(ψbi 4 bi ) ≤ η for every i ∈ N, where ψ : C → C is the measure-preserving Boolean homomorphism such that ψci = c∗i for every i ∈ N and ψπ = πψ. Once again, ψ[C] will be the closed subalgebra of A generated by {ψπ j ci : i ∈ N, j ∈ Z} = {π j c∗i : i ∈ N, j ∈ Z}; because every c∗i belongs to φ[B], ψ[C] ⊆ φ[B]. (b) Now µ ¯(c∗i 4 φci ) ≤ ² for every i ∈ N. P P There is a b ∈ Bm such that µ ¯(ci 4 b) ≤ 41 ². We know that φ[Bm ] is the closed subalgebra of A generated by {φπ j bi : i ∈ N, |j| ≤ m} = {π j b∗i : i ∈ N, |j| ≤ m}, and contains φb. Because ψ(φb) 4 φb ⊆ supi∈N,|j|≤m ψ(π j b∗i ) 4 π j b∗i = sup|j|≤m π j (supi∈N ψb∗i 4 b∗i ), we have µ ¯(ψφb 4 φb) ≤ (2m + 1)

∞ X

µ ¯(ψb∗i 4 b∗i )

i=0

≤ (2m + 1)

∞ X

1 4

min(η, 2¯ µbi ) ≤ ².

i=0

But this means that µ ¯(c∗i 4 φci ) = µ ¯(ψci 4 φci ) ≤ µ ¯(ψci 4 ψφb) + µ ¯(ψφb 4 φb) + µ ¯(φb 4 φci ) ≤µ ¯(ci 4 φb) + ≤

² 4

² 4

+µ ¯(b 4 ci ) ≤ µ ¯(ci 4 φci ) + µ ¯(φci 4 φb) +

+µ ¯(ci 4 b) +

² 2

² 2

≤ ². Q Q

(c) Set di = φ−1 c∗i for each i; this is well-defined because φ is injective and c∗i ∈ φ[B]. Write D for the closed subalgebra of A or of B generated by {π j di : i ∈ N, j ∈ Z} = {φ−1 ψπ j ci : i ∈ N, j ∈ Z}; then D = φ−1 [ψ[C]], by 324L again, because φ−1 : φ[B] → B is a measure-preserving homomorphism. Then ¯ci for every i ∈ N, and hdi ii∈N is a Bernoulli partition for π. P P If i(0), . . . , i(n) ∈ N, then µ ¯ di = µ ¯c∗i = µ µ ¯( inf π j di(j) ) = µ ¯( inf π j φ−1 c∗i(j) ) = µ ¯(φ( inf π j φ−1 c∗i(j) )) j≤n j≤n j≤n Y Y j ∗ ∗ =µ ¯( inf π ci(j) )) = µ ¯ci(j) = µ ¯di(j) . Q Q j≤n

j≤n

j≤n

Next, µ ¯(ci 4 di ) = µ ¯(φci 4 φdi ) = µ ¯(φci 4 c∗i ) ≤ ² for every i, by (b). Finally, if i ∈ N, then ψb∗i belongs to ψ[C], while D = φ−1 [ψ[C]], so ρ(bi , D) = ρ(φbi , ψ[C]) ≤ µ ¯(φbi 4 ψb∗i ) = µ ¯(b∗i 4 ψb∗i ) ≤ η ≤ ². This completes the proof. 386H Lemma Let (A, µ ¯) be an atomless probability algebra and π a measure-preserving automorphism of A. Let hbi ii∈N , hci ii∈N be Bernoulli partitions for π, of the same finite entropy, and write B, C for the closed subalgebras generated by {π j bi : i ∈ N, j ∈ Z} and {π j ci : i ∈ N, j ∈ Z}. Suppose that C ⊆ B. Then for any ² > 0 we can find a Bernoulli partition hdi ii∈N for π such that (i) µ ¯di = µ ¯ci for every i ∈ N, (ii) µ ¯(di 4 ci ) ≤ ² for every i ∈ N, (iii) the closed subalgebra of A generated by {π j di : i ∈ N, j ∈ Z} is B.

386H

Ornstein’s theorem

543

proof (a) To begin with (down to the end of (c) below) suppose that A = B. Choose h²n in∈N , hδn in∈N , hrn in∈N and hhdni ii∈N in∈N inductively, as follows. Start with d0i = ci for every i, r0 = 0. Given that hdni ii∈N is a Bernoulli partition with µ ¯dni = µ ¯ci for every i, take ²n > 0 such that (2rm + 1)²n ≤ 2−n for every m ≤ n, δn > 0 such that δn ≤ 2−n−1 ²,

P∞ i=0

min(δn , 2¯ µci ) ≤ ²n ,

and use 386G to find a Bernoulli partition hdn+1,i ii∈N for π such that µ ¯dn+1,i = µ ¯ci ,

µ ¯(dn+1,i 4 dni ) ≤ δn ,

ρ(bi , D(n+1) ) ≤ 2−n−1

for every i ∈ N, where D(n+1) is the closed subalgebra of A generated by {π j dn+1,i : i ∈ N, j ∈ Z}. Let rn+1 be such that (n+1)

ρ(bi , Drn+1 ) ≤ 2−n (n+1)

for every i ∈ N, where Drn+1 is the closed subalgebra of A generated by {π j dn+1,i : i ∈ N, |j| ≤ rn+1 }. Continue. (b) For any i ∈ N,

P∞ n=0

µ ¯(dn+1,i 4 dni ) ≤

so hdni in∈N has a limit di in A. Of course µ ¯(ci 4 di ) ≤

P∞ n=0

P∞

n=0 δn

≤ ²,

µ ¯(dn+1,i 4 dni ) ≤ ²

for every i. We must have µ ¯di = limn→∞ µ ¯dni = µ ¯ ci for each i, and if i 6= j then since

P∞

di ∩ dj = limn→∞ dni ∩ dnj = 0; µ ¯ci = 1, hdi ii∈N is a partition of unity in A. For any i(0), . . . , i(k) in N, Qk Qk ¯di(j) , µ ¯(inf j≤k π j di(j) ) = limn→∞ µ ¯(inf j≤k π j dn,i(j) ) = limn→∞ j=0 µ ¯dn,i(j) = j=0 µ

i=0

so hdi ii∈N is a Bernoulli partition. (c) Let D be the closed subalgebra of A generated by {π j di : i ∈ N, j ∈ Z}. Then bj ∈ D for every j ∈ N. (m+1) (m+1) ¯(bj 4 b) ≤ 2−m . Now P P Fix m ∈ N. Then ρ(bj , Drm+1 ) ≤ 2−m , so there is a b ∈ Drm+1 such that µ ∞ X

ρ(dm+1,i , D) ≤

i=0

≤

∞ X

µ ¯(dm+1,i 4 di ) ≤

i=0 ∞ X

∞ X

∞ ∞ X X

µ ¯(dk+1,i 4 dki )

i=0 k=m+1 ∞ X

min(2¯ µci , δk ) ≤

k=m+1 i=0

²k .

k=m+1

So ρ(b, D) ≤ (2rm+1 + 1)

∞ X

ρ(dm+1,i , D)

i=0

≤

∞ X

(2rm+1 + 1)²k ≤

k=m+1

∞ X

2−k = 2−m ,

k=m+1

and ρ(bj , D) ≤ µ ¯(bj 4 b) + ρ(b, D) ≤ 2−m + 2−m = 2 · 2−m . As m is arbitrary, ρ(bj , D) = 0 and bj ∈ D. Q Q (d) This completes the proof if A = B. For the general case, apply the arguments above to (B, µ ¯¹ B, π¹ B).

544

Automorphism groups

386I

386I Ornstein’s theorem (finite entropy case) Let (A, µ ¯) and (B, ν¯) be probability algebras, and π : A → A, φ : B → B two-sided Bernoulli shifts of the same finite entropy. Then (A, µ ¯, π) and (B, ν¯, φ) are isomorphic. proof (a) Let hai ii∈N , hbi ii∈N be (two-sided) generating Bernoulli partitions in A, B respectively. By the Kolmogorov-Sinaˇı theorem, hai ii∈N and hbi ii∈N both have entropy equal to h(π) = h(φ). If this entropy is zero, then A and B are both {0, 1}, and the result is trivial; so let us assume that h(π) > 0, so that A is atomless (386Bd). (b) By Sinaˇı’s theorem (386E), there is a Bernoulli partition hci ii∈N for π such that µ ¯ci = ν¯bi for every i ∈ N. By 386H, there is a Bernoulli partition hdi ii∈N for π such that µ ¯di = µ ¯ci for every i and the closed subalgebra of A generated by {π j di : i ∈ N, j ∈ Z} is A. But now (A, µ ¯, π, hdi ii∈N ) is isomorphic to (B, ν¯, φ, hbi ii∈N ), so (A, µ ¯, π) and (B, ν¯, φ) are isomorphic. 386J Using the same methods, we can extend the last result to the case of Bernoulli shifts of infinite entropy. The first step uses the ideas of 386C, as follows. Lemma Let (A, µ ¯) be a measure algebra and π : A → A an ergodic measure-preserving automorphism. Suppose that hai ii∈I is a finite Bernoulli partition for π, with #(I) = r and µ ¯ai = 1/r for every i ∈ I, and that h(π) ≥ ln 2r. Then for any ² > 0 there is a Bernoulli partition hbij ii∈I,j∈{0,1} for π such that µ ¯(ai 4 (bi0 ∪ bi1 )) ≤ ²,

µ ¯bi0 = µ ¯bi1 =

1 2r

for every i ∈ I. proof (a) Let δ > 0 be such that

√ δ + 6 4δ ≤ ².

Let η > 0 be such that η < ln 2,

√

8η ≤ δ,

and |t − 21 | ≤ δ whenever t ∈ [0, 1] and q(t) + q(1 − t) ≥ ln 2 − 4η (384Ad). We have 1 r

H(A) = rq( ) = ln r, and µ ¯d = r−n whenever n ∈ N, d ∈ Dn (A, π). Note that A is atomless. P P?? If a ∈ A is an atom, then (because π is ergodic) supj∈N π j a = 1, and A is purely atomic, with finitely many atoms all of the same size as a; but this means that H(C) ≤ ln( µ¯1a ) for every partition of unity C ⊆ A, so that 1 n

h(π, C) = limn→∞ H(Dn (C, π), π) ≤ limn→∞

1 n

ln(

1 ) µ ¯a

=0

for every partition of unity C, and 0 = h(π) ≥ ln 2r ≥ ln 2. X XQ Q (b) There is a finite partition of unity C ⊆ A such that h(π, C) = ln 2r − η, and C refines A = {ai : i ∈ I} \ {0}. P P Because h(π) ≥ ln 2r, there is a finite partition of unity C 0 such 0 0 that h(π, C ) ≥ ln 2r − η; replacing C by C 0 ∨ A if need be, we may suppose that C 0 refines A; take such a C 0 of minimal size. Because H(C 0 ) ≥ h(π, C 0 ) > H(A), there must be distinct c0 , c1 ∈ C 0 included in the same member of A. Because A is atomless, the principal ideal generated by c1 has a closed subalgebra isomorphic, as measure algebra, to the measure algebra of Lebesgue measure on [0, 1], up to a scalar multiple of the measure; and in particular there is a family hdt it∈[0,1] such that ds ⊆ dt whenever s ≤ t, d1 = c1 and µ ¯dt = t¯ µc1 for every t ∈ [0, 1]. Let Dt be the partition of unity

386J

Ornstein’s theorem

545

(C 0 \ {c0 , c1 }) ∪ {c0 ∪ dt , c1 \ dt } for each t ∈ [0, 1]. Then h(π, D1 ) = h(π, (C 0 \ {c0 , c1 }) ∪ {c0 ∪ c1 }) < ln 2r − η, by the minimality of #(C 0 ), while h(π, D0 ) = h(π, C 0 ) ≥ ln 2r − η. Using 384N, we also have, for any s, t ∈ [0, 1] such that |s − t| ≤ 1e , h(π, Ds ) − h(π, Dt ) ≤ H(Ds |Dt ) (where Dt is the closed subalgebra generated by Dt ) ≤ q(ρ(c0 ∪ ds , Dt )) + q(ρ(c1 \ ds , Dt )) (by 385Ob, because Ds \ Dt ⊆ {c0 ∪ ds , c1 \ ds }) ≤ q(¯ µ((c0 ∪ ds ) 4 (c0 ∪ dt ))) + q(¯ µ((c1 \ ds ) 4 (c1 \ dt ))) = 2q(¯ µ(ds 4 dt )) = 2q(|s − t|¯ µc1 ) because q is monotonic on [0, |s − t|¯ µc1 ]. But this means that t 7→ h(π, Dt ) is continuous and there must be some t such that h(π, Dt ) = ln 2r − η; take C = Dt . Q Q (c) Let ξ > 0 be such that 1 6

ξ ≤ η,

ξ≤ , P c∈C

q(2ξ) + q(1 − 2ξ) ≤ η,

q(min(2ξ, µ ¯c)) ≤ η.

Let n ∈ N be such that 1 n+1

≤ ξ,

q(

1 n ) + q( ) n+1 n+1

≤ η,

µ ¯[[wn − h(π, C)χ1 ≥ η]] ≤ ξ, where wn =

1 n

P d∈Dn (C,π)

ln(

1 )χd. µ ¯d

(The Shannon-McMillan-Breiman theorem, 385H, assures us that any sufficiently large n has these properties.) (d) Let D be the set of those d ∈ Dn (C, π) such that µ ¯d ≥ (2r)−n ,

i.e.,

1 n

ln(

1 ) µ ¯d

≤ ln 2r.

Then µ ¯(sup D) ≥ 1 − ξ, by the choice of n, because h(π, C) = ln 2r − η. Note that every member of D is included in some member of Dn (A, π), because C refines A. If b ∈ Dn (A, π), then µ ¯b = r−n , so n n #({d : d ∈ D, d ⊆ b}) ≤ 2 ; we can therefore find a function f : D → {0, 1} such that f is injective on {d : d ∈ D, d ⊆ b} for every b ∈ Dn (A, π). (e) By 385E(iv), as usual, there is an a ∈ A such that a, π −1 a, . . . , π −n+1 a are disjoint and µ ¯(a ∩ d) = for every d ∈ Dn (C, π). Set

1 ¯d n+1 µ

e = supd∈D,j 0 such that (2rm + 1)²n ≤ 2−n for every m < n. We know that h(π¹ Cn+1 ) = (n + 1) ln 2 = ln(2 · 2n ). So we can apply 386J to (Cn+1 , π¹ Cn+1 ) to see that there is a Bernoulli partition hb0nτ iτ ∈{0,1}n+1 for π such that b0nτ ∈ Cn+1 ,

µ ¯b0nτ = 2−n−1

for every τ ∈ {0, 1}n+1 , µ ¯(bnσ 4 (b0n,σa 0 ∪ b0n,σa 1 )) ≤ 2−n ²n for every σ ∈ {0, 1}n . By 386H (with B = C = Cn+1 ), there is a Bernoulli partition hbn+1,τ iτ ∈{0,1}n+1 for π¹ Cn+1 such that the closed subalgebra generated by {π j bn+1,τ : τ ∈ {0, 1}n+1 , j ∈ Z} is Cn+1 , µ ¯bn+1,τ = 2−n−1 for every τ ∈ {0, 1}n+1 , and P ¯(bn+1,τ 4 b0nτ ) ≤ ²n . τ ∈{0,1}n+1 µ (n+1)

For each k ∈ N, let Bk be the closed subalgebra of Cn+1 generated by {π j bn+1,τ : τ ∈ {0, 1}n+1 , |j| ≤ k}. Since dm ∈ Cm+1 ⊆ Cn+1 for every m ≤ n, there is an rn ∈ N such that (n+1)

ρ(dm , Brn

) ≤ 2−n for every m ≤ n.

Continue. (d) Fix m ≤ n ∈ N for the moment. For σ ∈ {0, 1}m , set bnσ = sup{bnτ : τ ∈ {0, 1}n , τ extends σ}. (If n = m, then of course σ is the unique member of {0, 1}m extending itself, so this formula is safe.) Then µ ¯bnσ = 2−n #({τ : τ ∈ {0, 1}n , τ extends σ}) = 2−n 2n−m = 2−m . Next, if σ, σ 0 ∈ {0, 1}m are distinct, there is no member of {0, 1}n extending both, so bnσ ∩ bnσ0 = 0; thus hbnσ iσ∈{0,1}m is a partition of unity. If σ(0), . . . , σ(k) ∈ {0, 1}m , then

386K

Ornstein’s theorem

µ ¯( inf π j bm,σ(j) ) = µ ¯( j≤k

549

inf π j bn,τ (j) )

sup

τ (0),... ,τ (k)∈{0,1}n j≤k τ (j)⊇σ(j)∀j≤k

X

=

µ ¯( inf π j bn,τ (j) ) j≤k

τ (0),... ,τ (k)∈{0,1}n τ (j)⊇σ(j)∀j≤k

X

=

(2−n )k+1

τ (0),... ,τ (k)∈{0,1} τ (j)⊇σ(j)∀j≤k

n

= (2n−m )k+1 (2−n )k+1 = (2−m )k+1 =

k Y

µ ¯bn,σ(j) ,

j=0

so hbnσ iσ∈{0,1}m is a Bernoulli partition. (e) If m ≤ n ∈ N, then

P σ∈{0,1}m

µ ¯(bnσ 4 bn+1,σ ) ≤ 2²n

whenever m ≤ n ∈ N. P P We have X

µ ¯(bnσ 4 bn+1,σ ) ≤

σ∈{0,1}m

X

µ ¯(bnτ 4 bn+1,τ )

τ ∈{0,1}n

=

X

µ ¯(bnτ 4 (bn+1,τ a 0 ∪ bn+1,τ a 1 ))

τ ∈{0,1}n

≤

X

τ ∈{0,1}n

≤

X

X

µ ¯(bnτ 4 (b0n,τ a 0 ∪ b0n,τ a 1 )) +

µ ¯(b0nυ 4 bn+1,υ )

υ∈{0,1}n+1

2−n ²n + ²n = 2²n . Q Q

τ ∈{0,1}n

(f ) In particular, for any m ∈ N and σ ∈ {0, 1}m , P∞ P∞ ¯(bnσ 4 bn+1,σ ) ≤ n=m 2²n < ∞. n=m µ So we can define bσ = limn→∞ bnσ in A. We have µ ¯bσ = limn→∞ µ ¯bnσ = 2−m ; and if σ, σ 0 ∈ {0, 1}m are distinct, then bσ ∩ bσ0 = limn→∞ bnσ ∩ bnσ0 = 0, so hbσ iσ∈{0,1}m is a partition of unity in A. If σ(0), . . . , σ(k) ∈ {0, 1}m , then µ ¯( inf π j bσ(j) ) = lim µ ¯( inf π j bn,σ(j) ) j≤k

n→∞

= lim

n→∞

j≤k

k Y

µ ¯bn,σ(j) =

j=0

k Y

µ ¯bσ(j) ,

j=0

so hbσ iσ∈{0,1}m is a Bernoulli partition for π. If σ ∈ {0, 1}m , then bσa 0 ∪ bσa 1 = limn→∞ bn,σa 0 ∪ bn,σa 1 = limn→∞ bn,σ = bσ . S (g) Let E be the closed subalgebra of A generated by m∈N {bσ : σ ∈ {0, 1}m }. Then E is atomless and countably generated, so (E, µ ¯¹ E) is isomorphic to the measure algebra of Lebesgue measure on [0, 1]. Now Qk µ ¯(inf j≤k π j ej ) = j=0 µ ¯ej for all e0 , . . . , ek ∈ E. P P Let ² > 0. For m ∈ N, let Em be the subalgebra of E

550

Automorphism groups

386K

S generated by {bσ : σ ∈ {0, 1}m }. hEm im∈N is non-decreasing, so m∈N Em is a closed subalgebra of A, and must be E. Now the function Qk (a0 , . . . , ak ) → µ ¯(inf j≤k π j aj ) − j=0 µ ¯aj : Ak+1 → R is continuous and zero on Ek+1 for every m, by 386Bb, so is zero on Ek+1 , and in particular is zero at m (e0 , . . . , ek ), as required. Q Q By 384Sf, hπ j [E]ij∈Z is independent. S (h) Let B∗ be the closed subalgebra of A generated by {π j bσ : σ ∈ m∈N {0, 1}m , j ∈ Z}; then E ⊆ B∗ , S so B∗ is the closed subalgebra of A generated by j∈Z π j [E]. It follows from (e) that, for any m ∈ N, X X ρ(bmσ , B∗ ) ≤ µ ¯(bmσ , bσ ) σ∈{0,1}m

σ∈{0,1}m ∞ X

X

≤

µ ¯(bnσ , bn+1,σ ) ≤ 2

σ∈{0,1}m n=m (m+1)

So if b ∈ Brm

∞ X

²n .

n=m

, X

ρ(b, B∗ ) ≤ (2rm + 1)

ρ(bm+1,σ , B∗ )

σ∈{0,1}m+1

≤ 2(2rm + 1)

∞ X

²n ≤ 2

n=m+1

∞ X

2−n = 2−m+1 .

n=m+1

It follows that, whenever m ≤ n in N, (n+1)

ρ(dm , B∗ ) ≤ ρ(dm , Brn

) + 2−n+1 ≤ 2−n + 2−n+1

by the choice of rn . Letting n → ∞, we see that ρ(dm , B∗ ) = 0, that is, dm ∈ B∗ , for every m ∈ N. But this means that A0 ⊆ B∗ , by the choice of hdm im∈N . Accordingly π j [A0 ] ⊆ B∗ for every j and B∗ must be the whole of A. (i) Thus π is a two-sided Bernoulli shift with root algebra E; by 384Sc, (A, µ ¯, π) is isomorphic to (B, ν¯, φ). 386L Corollary: Sinaˇı’s theorem (general case) Let (A, µ ¯) be an atomless probability algebra, and π : A → A a measure-preserving automorphism. Let (B, ν¯) be a probability algebra of countable Maharam type, and φ : B → B a one- or two-sided Bernoulli shift with h(φ) ≤ h(π). Then (B, ν¯, φ) is isomorphic to a factor of (A, µ ¯, π). proof (a) To begin with (down to the end of (b)) suppose that φ is two-sided. Let B0 be a root algebra for φ. If B0 is purely atomic, then there is a generating Bernoulli partition hbi ii∈N for φ of entropy h(φ). By 386E, there is a Bernoulli partition hci ii∈N for π such that µ ¯ci = ν¯bi for every i. Let C be the closed subalgebra of A generated by {π j ci : i ∈ N, j ∈ Z}. Now (C, µ ¯¹ C, π¹ C) is a factor of (A, µ ¯, π) isomorphic to (B, ν¯, φ). (b) If B0 is not purely atomic, then there is still a partition of unity hbi ii∈N in B0 of infinite entropy. Again, let C be the closed subalgebra of A generated by {π j ci : i ∈ N, j ∈ Z}, where hci ii∈N is a Bernoulli partition for π such that µ ¯ci = ν¯bi for every i. Now π¹ C is a Bernoulli shift of infinite entropy and C has countable Maharam type, so 386K tells us that there is a closed subalgebra C0 ⊆ C such that hπ k [C0 ]ik∈N is independent and (C0 , µ ¯¹ C0 ) is isomorphic to the measure algebra of Lebesgue measure on [0, 1]. But (B0 , ν¯¹ B0 ) is a probability algebra of countable Maharam type, so is isomorphic to a closed subalgebra k ∗ C1 of C0 (332N). S Of kcourse hπ ∗[C1 ]ik∈N is independent, so if we take C1 to be the closed subalgebra of A generated by k∈Z π [C1 ], π¹ C1 will be a two-sided Bernoulli shift isomorphic to φ (384Sf). (c) If φ is a one-sided Bernoulli shift, then 384Sa and 384Sc show that (B, ν¯, φ) can be represented in terms of a product measure on a space X N and the standard shift operator on X N . Now this extends naturally to the standard two-sided Bernoulli shift represented by the product measure on X Z , as described in 384Sb; so that (B, ν¯, φ) becomes represented as a factor of (B0 , ν¯0 , φ0 ) where φ0 is a two-sided Bernoulli

387A

Dye’s theorem

551

shift with the same entropy as φ (since the entropy is determined by the root algebra, by 384R). By (a)-(b), (B0 , ν¯0 , φ0 ) is isomorphic to a factor of (A, µ ¯, π), so (B, ν¯, φ) also is. Remark Thus (A, µ ¯, π) has factors which are Bernoulli shifts based on root algebras of all countablygenerated types permitted by the entropy of π. 386X Basic exercises (a) Let (A, µ ¯) be a probability algebra, and π : A → A a one- or two-sided Bernoulli shift. Show that π n isSa Bernoulli shift for any n ≥ 1. (Hint: if A0 is a root algebra for π, the closed subalgebra generated by j 0, there is an n ≥ 1 such that ν[[|u| > n²]] ≤ ², in which case ν[[|αu| > ²]] ≤ ² whenever Q |α| ≤ n1 , so that τ (αu) ≤ ² whenever |α| ≤ n1 . As ² is arbitrary, limα→0 τ (αu) = 0. Q (b) Accordingly we have a metric (u, v) 7→ τ (u−v) which defines a linear space topology T on L0 (2A5B). Now let G be an open set containing 0. Then there is an ² > 0 such that H = {u : τ (u) < ²} is included in G. If v ∈ H and |u| ≤ |v|, then ν[[|u| > τ (v)]] ≤ ν[[|v| > τ (v)]] ≤ τ (v), so τ (u) ≤ τ (v) and u ∈ H ⊆ G. So T satisfies all the conditions. 393N Proposition CM5 is true iff CM1 , . . . , CM4 are true. proof CM4 ⇒ CM5 Assume CM4 . Let A, T be as in the statement of CM5 . Let S be the topology on A induced by T and the function χ : A → L0 ; that is, S = {χ−1 [G] : G ∈ T}. Then S satisfies the conditions of CM4 . P P (i) Because T is Hausdorff and χ is injective, S is Hausdorff. (ii) If 0 ∈ G ∈ S, there is an H ∈ T such that G = χ−1 [H]. Now 0 (the zero of L0 ) belongs to H, so there is an open set H1 containing 0 such that u ∈ H whenever v ∈ H1 and |u| ≤ |v|. Next, addition on L0 is continuous for T, so there is an open set H2 containing 0 such that u + v ∈ H1 whenever u, v ∈ H2 . Consider G0 = χ−1 [H2 ]. This is an open set in A containing 0, and if a, b ∈ G0 then |χ(a ∪ b)| ≤ χa + χb ∈ H2 + H2 ⊆ H1 ,

588

Measurable algebras

393N

so χ(a ∪ b) ∈ H and a ∪ b ∈ G. As G is arbitrary, ∪ is continuous at (0, 0). (iii) If han in∈N is a non-increasing sequence in A with infimum 0, u0 = supn∈N nχan is defined in L0 (use the criterion of 364Ma: inf m∈N supn∈N [[nχan > m]] = inf m∈N am+1 = 0.) If 0 ∈ G ∈ S, take H ∈ T such that G = χ−1 [H], and H1 ∈ T such that 0 ∈ H1 and u ∈ H whenever v ∈ H1 , |u| ≤ |v|. Because scalar multiplication is continuous for T, there is a k ≥ 1 such that k1 u0 ∈ H1 . For any n ≥ k, χan ≤ k1 u0 so χan ∈ H and an ∈ G. As G is arbitrary, han in∈N → 0 for S. As han in∈N is arbitrary, condition (ii) in the statement of CM4 is satisfied. Q Q Since CM4 is true, A must be a measurable algebra. CM5 ⇒ CM1 Assume that CM5 is true, and let A be a Dedekind complete Boolean algebra with a strictly positive Maharam submeasure ν. By 393M, L0 = L0 (A) has a topology satisfying the conditions of CM5 , so A is measurable. 393O The phrase ‘control measure’ derives, in fact, from none of the formulations above; it belongs to the theory of vector measures, as follows. Definitions (a) Let A be a Dedekind σ-complete Boolean algebra and U a Hausdorff linear topological space. (The idea is intended to apply, P∞ in particular, when PnA is a σ-algebra of subsets of a set.) A function θ : A → U is a vector measure if n=0 θan = limn→∞ i=0 θai is defined in U and equal to θ(supn∈N an ) for every disjoint sequence han in∈N in A. (b) In this case, a non-negative countably additive functional µ : A → [0, ∞[ is a control measure for θ if θa = 0 whenever µa = 0. 393P Now I can formulate the last of this string of statements equiveridical to CM1 . Consider the statement (CM6 ) If A is a Dedekind σ-complete Boolean algebra, U is a metrizable linear topological space, and θ : A → U is a vector measure, then θ has a control measure. 393Q Lemma If A is a Dedekind σ-complete Boolean algebra, U a Hausdorff linear topological space, and θ : A → U a vector measure, then limn→∞ θan = 0 whenever han in∈N is a non-increasing sequence in A with infimum 0. P∞ proof θan = i=n θ(ai \ ai+1 ) for every n. 393R Proposition CM6 is true iff CM1 , . . . , CM5 are true. proof CM01 ⇒ CM6 (i) Assume CM01 , and let A, U and θ be as in the statement of CM6 . Then there is a functional τ : U → [0, ∞[ such that, for u, v ∈ U , τ (u + v) ≤ τ (u) + τ (v), τ (αu) ≤ τ (u) whenever |α| ≤ 1, limα→0 τ (αu) = 0, τ (u) = 0 iff u = 0, and the topology of U is defined by the metric (u, v) 7→ τ (u − v) (2A5Cb). For a ∈ A set νa = supb ⊆ a min(1, τ (θb)). (ii) Consider the functional ν : A → [0, ∞[. (α) ν0 = τ (θ0) = τ (0) = 0. (To see that θ0 = 0 set an = 0 for every n in the definition 393Oa.) (β) If a ⊆ b then of course νa ≤ νb. (γ) If a, b ∈ A and c ⊆ a ∪ b, then min(1, τ (θc)) = min(1, τ (θ(c ∩ a) + θ(c \ a))) ≤ min(1, τ (θ(c ∩ a)) + τ (θ(c \ a))) ≤ νa + νb. As c is arbitrary, ν(a ∪ b) ≤ νa + νb. Thus ν is a submeasure. (δ) ?? Suppose, if possible, that there is a non-decreasing sequence han in∈N in A, with infimum 0, such that hν(an )in∈N does not converge to 0. Because hν(an )in∈N is non-increasing, ² = 31 inf n∈N νan is greater than 0. Now for each n ∈ N there are m > n, b such that b ⊆ an \ am and τ (θb) ≥ ². P P As νan ≥ 3² there is a c ⊆ an such that τ (θc) ≥ 2². Now hθ(am ∩ cim∈N converges to 0 in U , by 393Q, so there is an m > n such that

*393S

The Control Measure Problem

589

τ (θ(c \ am )) = τ (θc − θ(am ∩ c)) ≥ τ (θc) − τ (θ(am ∩ c)) ≥ ², and we can take b = c \ am . Q Q We may therefore choose hbk ik∈N , hn(k)ik∈N such that n(k +1) > n(k), bk ⊆ an(k) \ an(k+1) and τ (θbk ) ≥ ² P∞ for every k. But hbk ik∈N is disjoint, so we ought to be able to form k=0 θbk = θ(supk∈N bk ) in U , and limk→∞ θbk = 0 in U , that is, limk→∞ τ (θbk ) = 0. X X Thus limn→∞ νan = 0 for every non-increasing sequence han in∈N with infimum 0, and ν is a Maharam submeasure. (iii) By CM01 , there is a non-negative countably additive functional µ : A → [0, ∞[ such that νa = 0 whenever µa = 0. In particular, if µa = 0, then τ (θa) = 0 and θa = 0. So µ is a control measure for θ. As A, U and θ are arbitrary, CM6 is true. CM6 ⇒ CM1 Assume CM6 , and let A be a Dedekind complete Boolean algebra with a strictly positive Maharam submeasure ν. Give L0 = L0 (A) the topology of 393M. Then χ : A → L0 is a vector measure in the sense of 393O. P P If han in∈N is a disjoint sequence in A with supremum a, set bn = supi≤n ai , so that Pn χbn = i=0 χai for each n. We have ν(a \ bn ) → 0, so that τ (χa − χbn )) = min(1, ν(a \ bn )) → 0, P∞ where τ is the functional of 393M, and χa = i=0 χai in L0 . Q Q CM6 now assures us that there is a non-negative countably additive functional µ on A such that µa = 0 =⇒ χa = 0 =⇒ a = 0, so that µ is strictly positive and A is measurable. As A and ν are arbitrary, CM1 is true. *393S I must not go any farther without remarking that the generality of the phrase ‘metrizable linear topological space’ in CM6 is essential. If we look only at normed spaces we do not need to know anything about the Control Measure Problem, as the following theorem shows. Theorem Let A be a Dedekind σ-complete Boolean algebra, U a normed space and θ : A → U a vector measure. Then θ has a control measure. ˆ (3A5Ib), and as θ will still be a vector proof (a) Since U can certainly be embedded in a Banach space U ˆ measure when regarded as a map from A to U , we may assume from the beginning that U itself is complete. (b) θ is bounded (that is, supa∈A kθak is finite). P P?? Suppose, if possible, otherwise. Choose han in∈N inductively, as follows. a0 = 1. Given that supa ⊆ an kθak = ∞, choose b ⊆ an such that kθbk ≥ kθan k + 1. Then kθ(an \ b)k ≥ 1. Also supa ⊆ an kθak ≤ supa ⊆ an kθ(a ∩ b)k + kθ(a \ b)k, so at least one of supa ⊆ b kθak, supa ⊆ an \b kθak must be infinite. We may therefore take an+1 to be either b or an \ b and such that supa ⊆ an+1 kθak = ∞. Observe that in either case we shall have kθ(an \ an+1 )k ≥ 1. Continue. At the end of the P induction we shall have a disjoint sequence han \ an+1 in∈N such that kθ(an \ an+1 )k ≥ 1 ∞ for every n, so that n=0 θ(an \ an+1 ) cannot be defined in U ; which is impossible. X XQ Q (c) Accordingly we have a bounded linear operator T : L∞ → U , where L∞ = L∞ (A), such that T χ = θ (363Ea). Now the key to the proof is the following fact: if hun in∈N is a disjoint order-bounded sequence in (L∞ )+ , hT un in∈N → 0 in U . P P Let γ be such that un ≤ γχ1 for every n. Let ² > 0, and let k be the integer part of γ/². For n ∈ N, i ≤ k set ani = [[un > ²(i + 1)]]; then hani in∈N is disjoint for each i, and if we set Pk vn = ² i=0 χani , we get vn ≤ uP n ≤ vn + ²χ1, so kun − vn k∞ ≤ ². ∞ Because hani in∈N is disjoint, n=0 θani is defined in U , and hθani in∈N → 0, for each i ≤ k. Consequently Pk T vn = ² i=0 θani → 0 as n → ∞. But kT un − T vn k ≤ kT kkun − vn k∞ ≤ ²kT k

590

Measurable algebras

*393S

for each n, so lim supn→∞ kT un k ≤ ²kT k. As ² is arbitrary, limn→∞ kT un k = 0. Q Q (d) Consider the adjoint operator T 0 : U ∗ → (L∞ )∗ . Recall that L∞ is an M -space (363B) so that its dual is an L-space (356N). Write A = {T 0 g : g ∈ U ∗ , kgk ≤ 1} ⊆ (L∞ )∗ = (L∞ )∼ . If u ∈ L∞ , then supf ∈A |f (u)| = supkgk≤1 |(T ∗ g)(u)| = supkgk≤1 |g(T u)| = kT uk. Now A is uniformly integrable. P P I use the criterion of 356O. Of course kf k ≤ kT 0 k for every f ∈ A, so A is norm-bounded. If hun in∈N is an order-bounded disjoint sequence in (L∞ )+ , then supf ∈A |f (un )| = kT un k → 0 as n → ∞. So A is uniformly integrable. Q Q (e) Next, A ⊆ (L∞ )∼ P If f ∈ A, it is of the form T 0 g for some g ∈ U ∗ , that is, c . P f (χa) = (T 0 g)(χa) = gT (χa) = g(θa) for every a ∈ A. If now han in∈N is a disjoint sequence in A with supremum a, P∞ P∞ P∞ f (χa) = g(θ(supn∈N an )) = g( n=0 θan ) = n=0 g(θan ) = n=0 f (χan ). So f χ is countably additive. By 363K, f ∈ (L∞ )∼ Q c . Q (f ) Because A is uniformly integrable, there is for each m ∈ N an fm ≥ 0 in (L∞ )∗ such that k(|f | − fm )+ k ≤ 2−m for every f ∈ A; moreover, we can suppose that fm is of the form supi≤km |fmi | where every fmi belongs to A (354R(b-iii)), so that fm ∈ (L∞ )∼ c and µm = fm χ is countably additive. Set P∞ 1 γm = 1 + µm 1 for each m, µ = m=0 m µm ; 2 γm

then µ : A → [0, ∞[ is a non-negative countably additive functional. Now µ is a control measure for θ. P P If µa = 0, then µm a = 0, that is, fm (χa) = 0, for every m ∈ N. But this means that if g ∈ U ∗ and kgk ≤ 1, |g(θa)| = |(T 0 g)(χa)| ≤ fm (χa) + k(|T 0 g| − fm )+ k ≤ 2−m for every m, by the choice of fm ; so that g(θa) = 0. As g is arbitrary, θa = 0; as a is arbitrary, µ is a control measure for θ. Q Q 393T This concludes the list of ‘positive’ results I wish to present in this section. I now devote a few pages to significant examples of submeasures. The first example is a classic formulation (taken from Talagrand 80) which shows in clear relief some of the fundamental ways in which submeasures differ from measures. Examples (a) Fix n ≥ 1, and let I be the set {0, 1, . . . , 2n − 1}, X = [I]n , so that X is a finite set. For each i ∈ I set Ai = {a : i ∈ a ∈ X}. For E ⊆ X set νE = =

[

1 n+1

inf{#(J) : J ⊆ I, E ⊆

1 n+1

inf{#(J) : a ∩ J 6= ∅ for every a ∈ E}.

Ai }

i∈J

It is elementary to check that ν : PX → [0, ∞[ is a strictly positive submeasure, therefore (because PX is finite) a Maharam submeasure. The essential properties of ν are twofold: (i) νX = 1; (ii) for any non-negative additive functional µ such 2 . P P (i)(α) If J ⊆ I and #(J) ≤ n, there is an a ∈ [I \ J]n , that µE ≤ νE for every E ⊆ X, µX ≤ n+1 S S so that a ∈ X \ i∈J Ai and X 6⊆ i∈J Ai . This means that X cannot be covered by fewer than n + 1 of the sets Ai , so that νX must be at leastS1. (β) On the other hand, if J ⊆ I is any set of cardinal n + 1, a ∩ J 6= ∅ for every a ∈ X, so that X = i∈J Ai and νX ≤ 1. (ii) Every member of X belongs to just n of the sets Ai , so

393U

The Control Measure Problem

nµX = and µX ≤

2 n+1 .

P i∈I

µAi ≤

#(I) n+1

=

591

2n , n+1

Q Q

(b) This example shows at least that any proof of CM3A cannot work through any generally valid inequality of the form ‘if ν is a Maharam submeasure there is an additive functional µ with δν ≤ µ ≤ ν’. If we take a sequence of these spaces we can form a result which in one direction is stronger, as follows. For each n ≥ 1 set In =Q{0, . . . , 2n − 1} and define Xn = [In ]n , Ani = {a : i ∈ a ∈ Xn }, νn : PXn → [0, 1] ∞ as in (a) above. Set Z = n=1 Xn , K = {(n, i) : n ≥ 1, i ∈ In } and Cni = {x : x ∈ Z, x(n) ∈ Ani } for (n, i) ∈ K. Now define θ : PZ → [0, 1] by setting S P 1 : J ⊆ K, W ⊆ (n,i)∈J Cni } θW = inf{ (n,i)∈J n+1

for every W ⊆ Z. Then θ is an outer measure on Z, by arguments we have been familiar with since 114D. Also θZ = 1. P P (i) Because (for instance) X1 P is covered by the two sets A10 and A11 , Z is covered by C10 1 and C11 , so that θZ ≤ 1. (ii) If J ⊆ K and (n,i)∈J n+1 < 1, set Jn = {i : (n, i) ∈ J} for each n; then S #(Jn ) < n + 1, so we can choose an x(n) ∈ Xn \ i∈Jn Ani . This defines a sequence x ∈ Z such that x ∈ / Cni S for every (n, i) ∈ J, and Z 6= (n,i)∈J Cni . As J is arbitrary, θZ ≥ 1. Q Q Finally, if Σ is any subalgebra of PZ containing every Cni , and µ : Σ → [0, ∞[ a non-negative finitely additive functional such that µE ≤ θE for every E ∈ Σ, then µ = 0. P P For each n, every point of x belongs to n different Cni , just as in (a) above; so that P 2n nµZ ≤ i∈In µCni ≤ . n+1

As this is true for every n ≥ 1, µZ = 0. Q Q (c) A non-zero submeasure ν on a Boolean algebra A is called pathological if the only additive functional µ such that 0 ≤ µa ≤ νa for every a ∈ A is the zero functional. Thus the submeasure θ of (b) above is pathological, and CM3A can be read ‘an exhaustive submeasure cannot be pathological’. (d) It is important to note that Fubini’s theorem fails catastrophically for submeasures. As a simple example, consider the space (X, ν) of (a) above, for some fixed n ≥ 1. If we define θ : P(X × X) → [0, 1] by setting P∞ S θW = inf{ i=0 νEi · νFi : W ⊆ i∈N Ei × Fi } for each W ⊆ X × X, then we obtain an outer measure, just as if ν itself were a measure; but θ(X × X) ≤ 4n/(n + 1)2 is small compared with (νX)2 . P P Give {0, . . . , 2n − 1} = Z2n its usual group operation +2n of addition mod 2n. Then S S X × X ⊆ i 369Xc, 369Xd, 369Xr, 369Yc, 369Yd, 373Xm Young’s inequality 255Ym Zermelo’s Well-ordering Theorem 2A1Ka Zermelo-Fraenkel set theory 3A1A

General index

I

667

zero-dimensional topological space 311I-311K, 315Xf, 316Xo, 316Yd, 353Yc, 3A3Ad, 3A3Bd zero-one law 254S, 272O, 272Xf, 272Xg, 325Xg zero set in a topological space 313Yb, 316Yh, 324Yb, 3A3Pa Zorn’s lemma 2A1M, 3A1G a.e. (‘almost everywhere’) 112Dd a.s. (‘almost surely’) 112De Aut (in Aut A) see automorphism group of a Boolean algebra (381A); (in Autµ A) see automorphism group of a measure algebra (382A) AL (Lebesgue measure algebra) 373C B (in B(x, δ), closed ball) 261A, 2A2B B (in B(U ; V ), space of bounded linear operators) 253Xb, 253Yj, 253Yk, 2A4F, 2A4G, 2A4H, 371B-371D, 371G, 371Xd, 371Yc, 376M, 3A5H; (B(U ; U )) 395Xb c (in c(A), where A is a Boolean algebra) see cellularity (332D) c (in c(X), where X is a topological space) see cellularity (332Xd) c (the cardinal of R or PN) 2A1H, 2A1L, 343I, 343Yb, 344H, 344Yf, 382Xc, 391Yd, 391Ye C (in C(X), where X is a topological space) 243Xo, 281Yc, 281Ye, 281Yf, 352Xj, 353M, 353Xd, 354L, 354Yf, 353Yc, 363A, 367L, 367Yi, 367Yj, 367Yk, 368Ya C([0, 1]) 242 notes, 352Xg, 356Xb, 368Yf Cb (in Cb (X), where X is a topological space) 281A, 281E, 281G, 281Ya, 281Yd, 281Yg, 285Yg, 352Xj, 354Hb C ∞ (in C ∞ (X), for extremally disconnected X) 364W, 364Yl, 368G c 354Xq, 354Xs, 355Ye c 0 354Xa, 354Xd, 354Xi, 371Yc cac (‘countable antichain condition’) 316 notes ccc Boolean algebra 316Aa, 316C-316F, 316Xa-316Xj, 316Yc-316Ye, 316Yg, 316Yp, 322G, 324Yd, 325Yd, 326L, 326Xi, 331Ge, 332D, 332H, 363Yb, 364Yb, 367Yl, 368Yg, 368Yi, 391M, 391Xa, 392Ca, 392I, 393J, 393Yd, 393Ye, 394Xf, 394Yb ccc topological space 316Ab, 316B, 316Xd, 316Xe, 316Ya, 316Yd cf (in cf P ) see cofinality (3A1Fb) c.l.d. product measure §§251-253 (251F, 251W), 254Db, 254U, 254Ye, 256K, 256L, 325A, 325B, 325C, 325H, 334A, 334Xa, 342Ge, 342Id, 342Xn, 343H, 354Yl, 376J, 376R, 376S, 376Yc c.l.d. version of a measure (space) 213E, 213F-213H, 213M, 213Xb-213Xe, 213Xg, 213Xk, 213Xn, 213Yb, 214Xf, 214Xj, 232Ye, 234Yf, 241Ya, 242Yh, 244Ya, 245Yc, 251Ic, 251S, 251Wn, 251Xd, 251Xj, 251Xk, 252Ya, 322D, 322Qb, 322Xc, 322Xi, 322Yb, 324Xc, 324Xe, 342Gb, 342Ib, 342Xn, 343H, 343Ye CM∗ (formulations of the Control Measure Problem) 393A, 393H, 393J, 393L, 393P, 393Xe, 393Xg d (in d(X)) see density (331Yf) D (in Dn (A, π), where A is a subset of a Boolean algebra, and π is a homomorphism) 384K, 384L, 384M diam (in diam A) = diameter dom (in dom f ): the domain of a function f ess sup see essential supremum (243Da) E (in E(X), expectation of a random variable) 271Ab f

(in Af ) 361Ad F (in F(B ↑), F(B↓)) 323D f -algebra 241H, 241 notes, 352W, 352Xj-352Xm, 353O, 353P, 353Xd, 353Yg, 353Yh, 361Eh, 363B, 364C-364E, 367Yg Gδ set 264Xe h (in h(π)) see entropy (384M); (in h(π, A)) 384M, 384N-384P, 384Xq, 384Yb, 386C H (in H(A)) see entropy of a partition (384C); (in H(A|B)) see conditional entropy (384D) I k see split interval (343J)

668

Index

L

`1 (in `1 (X)) 242Xa, 243Xl, 246Xd, 247Xc, 247Xd, 354Xa, 356Xc `1 (= `1 (N)) 246Xc, 354M, 354Xd, 356Xl `2 244Xn, 282K, 282Xg, 355Yb, 371Ye, 376Yh, 376Ym, 376Yn `p (in `p (X)) 244Xn, 354Xa `∞ (in `∞ (X)) 243Xl, 281B, 281D, 354Ha, 354Xa, 361D, 361L `∞ (= `∞ (N)) 243Xl, 354Xj, 356Xa, 371Yd, 382J `∞ -complemented subspace 363Yd L-space 354M, 354N-354P, 354R, 354Xt, 354Yj, 356N, 356P, 356Q, 356Xm, 356Yf, 362A, 362B, 362Yj, 365C, 365Xc, 365Xd, 367Xn, 369E, 371A-371E, 371Xa, 371Xb, 371Xf, 371Ya, 376M, 376P, 376Yi, 376Yj L0 (in L0 (µ)) 121Xb, 121Ye, §241 (241A), §245, 253C, 253Ya; (in L0 (Σ)) 345Yb, 364C, 364D, 364E, 364J, 364Yi; see also L0 (241C), L0strict (241Yh), L0C (241J) L0strict 241Yh L0C (in L0C (µ)) 241J, 253L L0 (in L0 (µ)) §241 (241A), 242B, 242J, 243A, 243B, 243D, 243Xe, 243Xj, §245, 253Xe, 253Xf, 253Xg, 271De, 272H, 323Xf, 345Yb, 352Xj, 364Jc, 376Yc; (in L0C (µ)) 241J; (in L0 (A)) §364 (364A), 368A-368E, 368H, 368K, 368M, 368Qb, 368R, 368S, 368Xa, 368Xe, 368Ya, 368Yd, 368Yi, §369, 372H, §375, 376B, 376Yb, 393M, 394I; (in L0C (A)) 364Yn; see also L0 (241A, 364C) L1 (in L1 (µ)) 122Xc, 242A, 242Da, 242Pa, 242Xb; (in L1strict (µ) 242Yg, 341Ye; (in L1C (µ) 242P, 255Yn; (in L1V (µ)) 253Yf; see also L1 , k k1 L1 (in L1 (µ)) §242 (242A), 243De, 243F, 243G, 243J, 243Xf, 243Xg, 243Xh, 245H, 245J, 245Xh, 245Xi, §246, §247, §253, 254R, 254Xp, 254Ya, 254Yc, 255Xc, 257Ya, 282Bd, 327D, 341Ye, 354M, 354Q, 354Xa, 365B, 376N, 376Q, 376S, 376Yl; (in L1V (µ)) 253Yf, 253Yi, 354Yl; (in L1 (A, µ ¯) or L1µ¯ ) §365 (365A), 366Yc, 367J, 367U, 367Yt, 369E, 369N, 369O, 369P, 371Xc, 371Yb, 371Yc, 371Yd, 372B, 372C, 372E, 372G, 372Xc, 376C, 385G, 385H, 385J; see also L1 , L1C , k k1 L1C (µ) 242P, 243K, 246K, 246Yl, 247E, 255Xc; see also convolution of functions L2 (in L2 (µ)) 244Ob, 253Yj, §286; (in L2C (µ)) 284N, 284O, 284Wh, 284Wi, 284Xi, 284Xk-284Xm, 284Yg; see also L2 , Lp , k k2 L2 (in L2 (µ)) 244N, 244Yk, 247Xe, 253Xe, 355Ye, 372 notes; (in L2C (µ)) 282K, 282Xg, 284P; (in L2 (A, µ ¯)) 366K, 366L, 366Xh, 395A, 395Xb; see also L2 , Lp , k k2 Lp (in Lp (µ)) §244 (244A), 246Xg, 252Ym, 253Xh, 255K, 255Og, 255Yc, 255Yd, 255Yl, 255Ym, 261Xa, 263Xa, 273M, 273Nb, 281Xd, 282Yc, 284Xj, 286A; see also Lp , L2 , k kp Lp (in Lp (µ), 1 < p < ∞) §244 (244A), 245G, 245Xj, 245Xk, 245Yg, 246Xh, 247Ya, 253Xe, 253Xi, 253Yk, 255Yf, 354Xa, 354Yk, 366B, 376N; (in Lp (A, µ ¯) = Lpµ¯ , 1 < p < ∞) §366 (366A), 369L, 371Gd, p 372Xp, 372Xq, 372Yb, 373Bb, 373F, 376Xb;(in LC (µ), 1 < p < ∞) 354Yk;(in Lp (A, µ ¯), 0 < p < 1) 366Ya, 366Yg;see also Lp , k kp L∞ (in L∞ (µ)) 243A, 243D, 243I, 243Xa, 243Xl, 243Xn; (in L∞ (Σ)) 341Xe, 363H;see also L∞ L∞ C 243K L∞ strict 243Xb L∞ (in L∞ (µ)) §243 (243A), 253Yd, 341Xe, 352Xj, 354Hc, 354Xa, 363I, 376Xn; (in L∞ (A)) §363 (363A), 364K, 364Xh, 365I, 365J, 365K, 365Xk, 368Q, 394N; see also L∞ , L∞ C , k k∞ L∞ 243K, 243Xm C Lτ (where τ is an extended Fatou norm) 369G, 369J, 369K, 369M, 369O, 369R, 369Xi, 374Xd, 374Xi; see also Orlicz space (369Xd), Lp , M 1,∞ (369N), M ∞,1 (369N) L (in L(U ; V ), space of linear operators) 253A, 253Xa, 351F, 351Xd, 351Xe L∼ (in L∼ (U ; V ), space of order-bounded linear operators) §355 (355A), 356Xi, 361H, 361Xc, 361Yb, 363Q, 365N, 371B-371E, 371Gb, 371Xb-371Xe, 371Ya, 371Yc-371Ye, 375Kb, 376J, 376Xe, 376Ym; see also order-bounded dual (356A) ∼ L∼ c (in Lc (U ; V )) 355G, 355I, 355Yi, 376Yf; see also sequentially order-continuous dual (356A) L× (in L× (U ; V )) 355G, 355H, 355J, 355K, 355Yg, 355Yi, 355Yj, 371B-371D, 371Gb, 376D, 376E, 376H, 376K, 376Xj, 376Yf; see also order-continuous dual (356A) lim (in lim F) 2A3S; (in limx→F ) 2A3S lim inf (in lim inf n→∞ ) §1A3 (1A3Aa), 2A3Sg; (in lim inf δ↓0 ) 2A2H; (in lim inf x→F ) 2A3S lim sup (in lim supn→∞ ) §1A3 (1A3Aa), 2A3Sg; (in lim supδ↓0 ) 2A2H, 2A3Sg; (in lim supx→F f (x)) 2A3S

General index

T

669

ln+ 275Yd M (in M (A), space of bounded finitely additive functionals) 362B, 362E, 363K M -space 354Gb, 354H, 354L, 354Xq, 354Xr, 356P, 356Xj, 363B, 363O, 371Xd, 376M; see also order-unit norm (354Ga) M 0 (in M 0 (A, µ ¯) = Mµ¯0 ) 366F, 366G, 366H, 366Yb, 366Yd, 366Yg, 373D, 373P, 373Xl 0,∞ M 252Yp M 0,∞ (in M 0,∞ (A, µ ¯) = Mµ¯0,∞ ) 373C, 373D, 373E, 373F, 373I, 373Q, 373Xo, 374B, 374J, 374L M 1,0 (in M 1,0 (A, µ ¯) = Mµ¯1,0 ) 366F, 366G, 366H, 366Ye, 369P, 369Q, 369Yh, 371F, 371G, 372D, 372Ya, 373G, 373H, 373J, 373S, 373Xp, 373Xr, 374Xe M 1,∞ (in M 1,∞ (µ)) 234Yd, 244Xl, 244Xm, 244Xo, 244Yc; (in M 1,∞ (A, µ ¯) = Mµ¯1,∞ ) 369N, 369O-369Q, 369Xi-369Xk, 369Xm, 369Xq, §373, 374A, 374B, 374M M ∞,0 (in M ∞,0 (A, µ ¯)) 366Xd, 366Yc M ∞,1 (in M ∞,1 (A, µ ¯) = Mµ¯∞,1 ) 369N, 369O, 369P, 369Q, 369Xi, 369Xj, 369Xk, 369Xl, 369Yh, 373K, 373M, 374B, 374M, 374Xa, 374Ya Mσ (in Mσ (A), space of countably additive functionals) 362B, 362Xd, 362Xh, 362Xi, 362Ya, 362Yb, 363K Mτ (in Mτ (A), space of completely additive functionals) 326Yp, 327D, 362B, 362D, 362Xd, 362Xg, 362Xi, 362Ya, 362Yb, 363K N 3A1H N × N 111Fb NN 372Xi —– see also PN N see ideal of negligibles On (the class of ordinals) 3A1E p (in p(t)) 385I, 385J P (in PX) 311Ba, 311Xe, 312B, 312C, 313Ec, 313Xf, 363S, 381Xe, 382Yb; (usual measure on PX) 254J, 254Xf, 254Xq, 254Yd PN 1A1Hb, 2A1Ha, 2A1Lb, 315O, 316Yo, 324Yg, 326Yg, 374Xk; (usual measure on) 273G, 273Xd, 273Xe P(NN ) 316Yg p.p. (‘presque partout’) 112De X ∈ E) etc. 271Ad Pr(X > a), Pr(X Q (the set of rational numbers) 111Eb, 1A1Ef, 364Yg q (in q(t)) 384A, 385O R (the set of real numbers) 111Fe, 1A1Ha, 2A1Ha, 2A1Lb, 352M RX 245Xa, 256Ye, 352Xj, 375Ya, 3A3K; see also Euclidean metric, Euclidean topology RX |F see reduced power (351M) R C 2A4A R see extended real line (§135) RO (in RO(X)) see regular open algebra (314Q) S (in S(A)) 243I, §361 (361D), 363C, 363Xg, 364K, 364Xh, 365F, 368Q, 369O; (in S f ∼ = S(Af )) 242M, ∼ ∼ × 244H, 365F, 365G, 369O, 369P; (in S(A) ) 362A; (in S(A)c ) 362Ac; (in S(A) ) 362Ad; (in SC (A)) 361Xj, 361Yd S see rapidly decreasing test function (284A) S 1 (the unit circle, as topological group) see circle group S6 (the group of permutations of six elements) 383 notes sf (in µsf ) see semi-finite version of a measure (213Xc); (in µ∗sf ) 213Xf, 213Xg, 213Xk T2 topology see Hausdorff (2A3E, 3A3Aa) (0) T (0) (in Tµ¯,¯ν ) 371F, 371G, 371H, 372D, 372Xb, 372Yb, 372Yc, 373B, 373G, 373J, 373R, 373S, 373Xp, 373Xq, 373Xr, 373Xu, 373Xv T (in Tµ¯,¯ν ) 244Xm, 244Xo, 244Yc, 246Yc, §373 (373A); see also T -invariant (374A)

670

Index

T

T × (in Tµ¯×,¯ν ) §373 (373Ab), 376Xa, 376Xh T -invariant extended Fatou norm 374Ab, 374B-374D, 374Fa, 374Xb, 374Xd-374Xj, 374Yb T -invariant set 374Aa, 374M, 374Xa, 374Xi, 374Xk, 374Xl, 374Ya, 374Ye Tm see convergence in measure (245A) Ts (in Ts (U, V )) 373M, 373Xq, 376O, 3A5E; see also weak topology (2A5I), weak* topology (2A5Ig) U (in U (x, δ)) 1A2A upr (in upr(a, C)) 314V, 314Xg, 333Xa, 365Rc, 394G, 394I, 394K-349N, Var (in Var(X)) see variance (271Ac); (in VarD f , Var f ) see variation (224A) w∗ -topology see weak* topology 2A5Ig Z (the set of integers) 111Eb, 1A1Ee; (as topological group) 255Xe Z2 (the group {0, 1}) 311Bc, 311D ZFC see Zermelo-Fraenkel set theory βr (volume of unit ball in Rr ) 252Q, 252Xh, 265F, 265H, 265Xa, 265Xb, 265Xe Γ-function 225Xj, 225Xk, 252Xh, 252Yk, 252Yn, 255Xj ∆-system 2A1Pa θ-refinable see hereditarily weakly θ-refinable µ ¯L (in §373) 373C νX see distribution of a random variable (271C) π-λ Theorem see Monotone Class Theorem (136B) σ-additive see countably additive (231C, 326E) σ-algebra of sets §111 (111A), 136Xb, 136Xi, 212Xk, 314D, 314M, 314N, 314Yd, 316D, 322Ya, 326Ys, 343D, 344D, 362Xg, 363H; see also Borel σ-algebra (111G) σ-algebra defined by a random variable 272C, 272D σ-complete see Dedekind σ-complete (241Fb, 314Ab) σ-field see σ-algebra (111A) σ-finite measure algebra 322Ac, 322Bc, 322C, 322G, 322M, 323Gb, 323Ya, 324K, 325Eb, 327B, 331N, 362Xd, 367Nd, 367Q, 367Xq, 367Xs, 369Xg, 382E σ-finite measure (space) 211D, 211L, 211M, 211Xe, 212G, 213Ha, 213Ma, 214Ia, 214Ja, 215B, 215C, 215Ya, 215Xe, 215Yb, 216A, 232B, 232F, 234F, 235O, 235R, 235Xe, 235Xk, 241Yd, 243Xi, 245Eb, 245K, 245L, 245Xe, 251K, 251Wg, 252B-252E, 252H, 252P, 252R, 252Xc, 252Yb, 252Yl, 322Bc, 342Xi, 362Xh, 365Xp, 367Xr, 376I, 376J, 376N, 376S σ-generating set in a Boolean algebra 331E σ-ideal (in a Boolean algebra) 313E, 313Pb, 313Qb, 314C, 314D, 314L, 314N, 314Yd, 316C, 316D, 316Xi, 316Ye, 321Ya, 322Ya, 392Xd —– (of sets) 112Db, 211Xc, 212Xf, 212Xk, 313Ec, 322Ya, 363H σ-linked Boolean algebra 391L, 391M, 391N, 391Xj, 391Yb, 391Yd, 391Ye, 391Z, 393Xf; σ-m-linked Boolean algebra 391Yh, 393Ya σ-order complete see Dedekind σ-complete (314Ab) σ-order-continuous see sequentially order-continuous (313H) σ-subalgebra of a Boolean algebra 313E, 313F, 313G, 313Xd, 313Xe, 314Eb, 314Fb, 314Hb, 314Jb, 314Xc, 315Yc, 321G, 321Xb, 322M, 323Z, 324Xb, 326Fg, 331E, 331G, 364Xc, 366I; see also order-closed subalgebra σ-subalgebra of sets §233 (233A), 321Xb, 323Xb σ-subhomomorphism between Boolean algebras 375E, 375F-375H, 375Xd, 375Yc, 375Ye (σ, P ∞)-distributive see weakly (σ, ∞)-distributive (316G) i∈I ai 112Bd, 222Ba, 226A

special symbols

General index

671

τ (in τ (A)) see Maharam type (331Fa); (in τC (A)) see relative Maharam type (333Aa) τ -additive functional on a Boolean algebra see completely additive (326J) τ -additive measure 256M, 256Xb, 256Xc τ -generating set in a Boolean algebra 313Fb, 313M, 331E, 331F, 331G, 331Yb, 331Yc Φ see normal distribution function (274Aa) χ (in χA, where A is a set) 122Aa; (in χa, where a belongs to a Boolean ring) 361D, 361Ef, 361L, 361M, 364K; (the function χ : A → L0 (A)) 364Kc, 367R ω (the first infinite ordinal) 2A1Fa, 3A1H; (in [X] 363L; (in A f )R 131D, 214D, 235Xf; R see also subspace measure; (in u) 242Ab, 242B, 242D, 363L, 365D, 365Xa; (in A u) 242Ac; (in a u) 365D, 365Xb; see also upper integral, lower integral (133I) R R see upper integral (133I) see lower integral (133I) R R see Riemann integral (134K) ¹ (in f ¹A, the restriction of a function to a set) 121Eh | | (in a Riesz space) 241Ee, 242G, §352 (352C), 354Aa, 354Bb k ke see order-unit norm (354Ga) k k1 (on L1 (µ)) §242 (242D), 246F, 253E, 275Xd, 282Ye; (on L1 (µ)) 242D, 242Yg, 273Na, 273Xi; (on 1 L (A, µ ¯)) 365A, 365B, 365C, 385G, 385H; (on the `1 -sum of Banach lattices) 354Xb, 354Xo k k2 244Da, 273Xj, 366Yh; see also L2 , k kp k kp (for 1 < p < ∞) §244 (244Da), 246Xb, 246Xh, 246Xi, 252Ym, 252Yp, 253Xe, 253Xh, 273M, 273Nb, 275Xe, 275Xf, 275Xh, 276Ya, 366A, 366C, 366D, 366H, 366J, 366Xa, 366Xi, 366Yf, 367Xp, 369Oe, 372Xb, 372Yb, 374Xb; see also Lp , Lp , k kp,q k kp,q (the Lorentz norm) 374Yb k k∞ 243D, 243Xb, 243Xo, 244Xh, 273Xk, 281B, 354Xb, 354Xo, 356Xc, 361D, 361Ee, 361I, 361J, 361L, 361M, 363A, 364Xh; see also essential supremum (243D), L∞ , L∞ , `∞ k k1,∞ 369O, 369P, 369Xh-369Xj, 371Gc, 372D, 372F, 373F, 373Xk; see also M 1,∞ , M 1,0 k k∞,1 369N, 369O, 369Xi, 369Xj, 369Xl; see also M ∞,1 ⊗ (in f ⊗ g) 253B, 253C, 253J, 253L, 253Ya, 253Yb; (in u ⊗ v) §253 (253E); (in A ⊗ B, a ⊗ b) see free product N (315M) N (in i∈I Ai ) see free product (315H) b (in Σ⊗T) b ⊗ 251D, 251K, 251L, 251Xk, 251Ya, 252P, 252Xd, 252Xe, 253C N N c (in c Σi ) 251Wb, 251Wf, 254E, 254F, 254Mc, 254Xc, 254Xi, 343Xb Q Q i∈I Q (in i∈I αi ) 254F; (in i∈I Xi ) 254Aa # (in #(X), the cardinal of X) 2A1Kb

672

Index

special symbols

S S S T (in Tn∈N En ) 111C; (in T A) 1A1F (in n∈N En ) 111C; (in E) 1A2F 4 (in E4F , ‘symmetric difference’) 111C, 311Ba ∪ , ∩ (in a Boolean ring or algebra) 311Ga, 313Xi, 323B \ , 4 (in a Boolean ring or algebra) 311Ga, 323B ⊆ , ⊇ (in a Boolean ring or algebra) 311H, 323Xa C (in I C R) see ideal (3A2Ea) −− ←−−−− −) (in (← (← a π b), (a π b φ c) etc.) see cycle notation (381G), cyclic automorphism, exchanging involution (381G) + (in κ+ , successor cardinal) 2A1Fc; (in f + , where f is a function) 121Xa, 241Ef; (in u+ , where u belongs to a Riesz space) 241Ef, 352C; (in U + , where U is a partially ordered linear space) 351C; (in F (x+ ), where F is a real function) 226Bb − (in f − , where f is a function) 121Xa, 241Ef; (in u− , in a Riesz space) 241Ef, 352C; (in F (x− ), where F is a real function) 226Bb 2 (in 2κ ) 3A1D ∨, ∧ (in a lattice) 121Xa, 2A1Ad; (in A ∨ B, where A, B are partitions of unity in a Boolean algebra) 384F ⊥ (in A⊥ , in a Boolean algebra) 313Xo; (in A⊥ , in a Riesz space) 352O, 352P, 352Q, 352R, 352Xg; (in ⊥ V , in a Hilbert space) see orthogonal complement; see also complement of a band 0 (in U 0 ) see algebraic dual; (in T 0 ) see adjoint operator a (in z a i) 3A1H ,

∧

∨

(in f , f ) see Fourier transform, inverse Fourier transform (283A) ¯ ¯ (in h(u), where h is a Borel function and u ∈ L0 ) 241I, 241Xd, 241Xi, 245Dd, 364I, 364J, 364Xg, 364Xq, 364Yd, 364Ye, 367I, 367S, 367Xl, 367Ys {0, 1}I (usual measure on) 254J, 254Xe, 254Yc, 272N, 273Xb, 331J-331L, 332B, 332C, 332N, 332Xm, 332Xn, 341Yc, 341Yd, 341Zb, 342Jd, 343Ca, 343I, 343Xc, 343Yd, 344G, 344L, 344Xg, 345Ab, 345C-345E, 345Xa, 346C, 382Xb; (when I = N) 254K, 254Xd, 254Xj, 256Xk, 261Yd, 341Xb, 343Cb, 343H, 343M, 345Yc, 346Zb, 387H; see also PX —– (usual topology of) 311Xh, 3A3K; (when I = N) 314Ye —– (open-and-closed algebra of) 311Xh, 315Xh, 316Xq, 316Yj, 316Ym, 331Yg, 391Xd, 393A, 393F —– (regular open algebra of) 316Yj (2, ∞)-distributive lattice 367Yd ¿ (in ν ¿ µ) see absolutely continuous (232A) 4τG (in a 4τG b) 394A, 394G, 394I, 394K, 394Ma, 394Xb ∞ see infinity [ ] (in [a, b]) see closed interval (115G, 1A1A, 2A1Ab); (in f [A], f −1 [B], R[A], R−1 [B]) 1A1B; (in [X]κ , [X] ν]]) 326O, 326P [ [ (in [a, b[) see half-open interval (115Ab, 1A1A) ] ] (in ]a, b]) see half-open interval (1A1A) ] [ (in ]a, b[) see open interval (115G, 1A1A) d e (in db : ae) 394I, 394J, 394K-394M, 394Xa b c (in bb : ac) 394I, 394J, 394K-394M, 394Xa (in µ E) 234E, 235Xf ∧

∨