Topological Riesz Spaces and Measure Theory

Topological Riesz Spaces and Measure Theory

Topological Riesz Spaces and Measure Theory D.H.FREMLIN Lecturer in Mathematics, University of Essex

WE

CAMBRIDGE AT THE UNIVERSITY PRESS 1974

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www. Cambridge. org Information on this title: www.cambridge.org/9780521201704 © Cambridge University Press 1974 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1974 This digitally printed version 2008 A catalogue recordfor this publication is available from the British Library Library of Congress Catalogue Card Number: 72—95410 ISBN 978-0-521-20170-4 hardback ISBN 978-0-521-09031-5 paperback

Contents Acknowledgements Preface Prerequisites 1 Riesz spaces 11 Partially ordered sets 12 Partially ordered linear spaces 13 Lattices 14 Riesz spaces 15 Archimedean Riesz spaces 16 Linear maps between Riesz spaces 17 Order-dense Riesz subspaces 18 The countable sup property IX Examples for Chapter 1 2 Topological Riesz spaces 21 Compatible topologies 22 Locally solid topologies 23 Fatou topologies 24 Lebesgue topologies 25 Complete metrizable topologies 26 L-spaces and M-spaces 2X Examples for Chapter 2 3 Dual spaces 31 The space E~ 32 The space E x 33 Perfect Riesz spaces 4 Riesz spaces on Boolean rings 41 Boolean rings 42 The space 5(91) 43 The space L°°(9l) 44 The space L # 45 Ring homomorphisms 4X Examples for Chapter 4

page vii ix

xiii 1 4 6 9 17 20 26 30 33

36 37 43 53 61 68 75 82 84 87 91 97 107 111 114 119

CONTENTS 5 Measure algebras 51 Measure rings 52 The space L\%) 53 Maharam algebras 54 Measure-preserving ring homomorphisms 5X Examples for Chapter 5

page 126 130 136 139 142

6 Measure spaces 61 Definitions and basic properties 62 Measurable functions; the space L° 63 Integration 64 Maharam measure spaces 65 Banach function spaces 6X Examples for Chapter 6

145 151 160 169 176 185

7 Representation of linear functionals 71 Sequentially smooth functionals 72 Smooth functionals: quasi-Radon measure spaces 73 Radon measures and Riesz' theorem 7X Examples for Chapter 7

193 202 210 216

8 Weak compactness 81 Weak compactness in E~ 82 Weak compactness in £ x 83 Weak compactness in L-spaces 8X Examples for Chapter 8

219 226 234 247

Appendix Al Linear topological spaces A2 Spaces of continuous functions

250 255

References

258

Index of special symbols

261

Index

263

VI

Acknowledgements The gestation of this book has extended over three appointments: as a lecturer at United College, The Chinese University of Hong Kong; as a Central Electricity Generating Board Junior Research Fellow at Churchill College, Cambridge; and as a lecturer at the University of Essex. I am grateful for the material support of all these institutions, and for the moral support of my colleagues there. Almost every mathematician I have spoken to in the last six years has influenced the book in some way; but I should like to give special thanks to Professor W. A. J. Luxemburg, who introduced me to its subject matter; to Dr N. N. Chan, who suggested that it should be written; and to Dr F. Smithies, who gently moulded it into publishable form. Finally I must mention Miss M. Mitchell, who typed the bulk of the MS, and the editors of the Cambridge University Press, for their skill and patience in transforming an idiosyncratic manuscript into print.

VII

Preface This book is addressed to functional analysts who would like to understand better the application of their subject to the older discipline of measure theory. The relationship of the two subjects has not always been easy. Measure theory has been the source of many examples for functional analysis; and these examples have been leading cases for some of the most important developments of the general theory. Such a stimulation is, of course, entirely welcome. But there have in addition been several cases in which special results in measure theory have been applied to prove general theorems in analysis. The ordinary functional analyst feels inadequately prepared for these applications, and is exasperated by the intrusion of a large body of knowledge in an unfamiliar style into his own concerns. My aim therefore is to identify those concepts in measure theory which have most affected functional analysis, and to integrate them into the latter subject in a way consistent with its own structure and habits of thought. The most powerful idea is undoubtedly that of Riesz space, or vector lattice. The principal Banach spaces which measure theory has contributed to functional analysis all have natural partial orderings which render them Riesz spaces. Many of their special properties can be related to the ways in which their orderings, their linear structures and their topologies are connected. For a clear understanding of the difference, for instance, between an L1 space and an L00 space, there is no substitute for an abstract analysis of their properties as ordered linear topological spaces. The other point at which measure theory has had an impact on functional analysis is in the representation of linear functionals as integrals, and the consequent deduction of surprising properties. I think that the analyst's instinctive rejection of such methods is a perfectly sound reaction. The difficulty is to provide techniques powerful enough to act as substitutes. Here again I believe that a thorough understanding of some quite simple topological Riesz spaces can see us through most of the difficulties. Now as soon as we begin to look at measure theory with a sceptical and abstract eye, a number of peculiarities strike us. The first is the ix

PREFACE relative unimportance of measure spaces themselves. They are used to construct spaces of equivalence classes of functions which then take on a life of their own; and quite different measure spaces can give rise to isomorphic function spaces. So it is natural to ask just what it is about a measure space that determines the associated function spaces. The answer is readily to hand: it is a construction called the measure algebra. Two measure spaces with isomorphic measure algebras will have isomorphic L 1 spaces, isomorphic L00 spaces, and so on; and to a large extent the converse is true. I think that the most convincing way of demonstrating these facts is to exhibit methods of constructing the function spaces (or, rather, isomorphic copies of them) directly from a measure algebra. These constructions are not particularly simple; in this book they take up Chapter 4 and the first half of Chapter 5. However, given some intuitive grasp of the nature of Riesz spaces, they are fairly straightforward, and provide invaluable insights. For instance, another curiosity of traditional measure theory is the unimportance of any notion of homomorphism between measure spaces. This distinguishes it from all comparably abstract branches of mathematics. I believe that this deficiency occurs because the natural and important homomorphisms of the theory are between measure algebras and not between measure spaces at all. In §§45 and 54 I discuss such homomorphisms and their effects on the constructions I have set up. The reader may be forgiven for wondering at this stage just how much he needs to know to cope with this book. A basic knowledge of functional analysis is essential. It would be possible to go a fair way with normed spaces alone; but this would shut off many of the most interesting ideas, and I have written on the assumption that I may call on the fundamental concepts of the theory of linear topological spaces. As for measure theory, in a formal sense I require none; I give every definition from that of measure space onwards. In an informal sense, I cannot pretend that this book is a genuine alternative to the traditional presentation of the elementary theory. §§61-3, while formally complete, are far too sketchy to be satisfactory as a first introduction. So if you are uncertain of your grounding in the subject, I refer you to the detailed advice in the Prerequisites section. I have preferred, in ordering the material of this book, to arrange results by the contexts in which they apply; thus propositions which refer to arbitrary Riesz spaces go into § 14 and those which refer only to Archimedean Riesz spaces go into § 15. Now this is not, of course,

PREFACE

a natural order for learning the subject. The beginner will certainly wish in the first instance to restrict his attention to Archimedean Riesz spaces, and leave any generalizations to one side for the moment. I think my decision is justified; it means that the hypotheses of successive theorems do not usually vary in arbitrary details, and are consequently much easier to recall precisely. But it makes a page-by-page progression inappropriate. To assist the reader to avoid spending too much time on material which will not be immediately essential, I have used asterisks to mark sections and propositions which can at first be passed over. They will be given references in the text when they come to be needed. Most of this book is taken up with fulfilling the objects I described at the beginning of this preface. But I think that topological Riesz spaces are inherently fascinating, and I have laid Chapter 2 out as a survey of their elementary properties; and in Chapter 8 I try to take a special subject up to the area of present research work. University of Essex January 1973

D. H. FREMLIN

XI

Prerequisites ZOBN'S LEMMA, TRANSFINITE INDUCTION, AND THE AXIOM OF CHOICE

The best introduction I know to this subject is in the appendix of KELLEY; HALMOS N.S.T. is also perfectly satisfactory. I shall apply the axiom of choice, and the principle of dependent choice for sequences, without comment, though when syntax allows I signal with such a word as * choose' or ' choice'. Zorn's lemma is used most often in the form If X is any collection of sets such that for any subset <W of X which is totally ordered by ordinary inclusion, we have \}®f = {*:3 YeW,teY}eX, then X has a maximal element. (This is nearly what KELLEY calls the 'maximal principle'.) When employing this form, I shall generally say merely 'by Zorn's lemma, X has a maximal element', and leave it to the reader to verify that the hypotheses on X are satisfied - supposing, of course, that the verification is straightforward. But if the lemma is to be applied in the more general form If X is a partially ordered set such that every totally ordered subset has an upper bound in X9 then X has a maximal element, then I shall give more details. FUNCTIONAL ANALYSIS I can recommend BOURBAKI V, ROBERTSON ROBERTSON, KELLEY & NAMIOKA, or SCHAEFER T.V.S. ; any of these

& is

adequate for the principal results. For details see §A1. AS I explained in the Preface, I make no formal requirements in this field. But a knowledge of the elementary theory up to and including the Radon-Nikodym theorem would certainly be helpful. The first point is that although abstract measure theory is not particularly difficult in its early stages, the only easy examples of measure spaces are essentially trivial. There is no substitute for Lebesgue measure on the real line as a leading example. So at some

MEASURE THEORY

xiii

PREREQUISITES stage the reader should make himself familiar with a proof that it exists. Secondly, several of the most important ideas in this book appeared first in measure theory. Pre-eminent among these are the arguments leading up to the Radon-Nikodym theorem. I give this theorem in § 63 with a proof which refers to almost every preceding chapter. In fact my proof is very closely related to one of the traditional ones, but its component ideas have been scattered. I think that it is illuminating to see how they can be joined more closely together in a direct proof. I shall give a concordance in the proper place. Meanwhile, I do not want the reader to feel that he must take a course in measure theory before proceeding further. Perhaps the books suggested below should be regarded as a starred section; optional on first reading, but likely to be useful later. R. G. Bartle, The Elements of Integration (Wiley, 1966) H. L. Royden, Real Analysis (Macmillan, 1963) H. Widom, Lectures on Measure and Integration (van Nostrand Reinhold, 1969) J. H. Williamson, Lebesgue Integration (Holt, Rinehart & Winston, 1962).

XIV

1. Riesz spaces The prerequisites for this chapter are few; an acquaintance with linear spaces is the principal requirement. A Riesz space is simply a linear space over the field R of real numbers which has a special kind of partial ordering, and all we need to know about partial orderings will be covered in §§11 and 13. But the theory of Riesz spaces is already rich, and some of the work in §§16 and 17 is far from trivial. It does, however, have to be taken seriously. These are the basic results which will enable us to handle Riesz spaces with assurance and facility.

11 Partially ordered sets This section is little more than a list of definitions. As such I suggest that it should be read carefully once, together with the associated examples; but that there is no need to consciously memorize anything. You will find the index perfectly reliable. Actually the concepts here have applications to every branch of mathematics, and they will mostly be familiar in everything but name. I think it is amusing and instructive to seek such applications out and consciously appreciate them. Now to work: 11A Definition A partially ordered set is a pair (A, ^ ) where A is a set and ^ is a binary relation on A such that: (i) a ^ a for every aeA; (ii) if a and b belong to A and a < b and b ^ a, then a = b; (iii) if a, b and c belong to A and a ^ b and b < c, then a ^ c. In this context we write a ^ b for b < a (that is, ^ will always be the inverse relation to a for a ^ b and a #= 6. For examples see 1XA and IXC. 11B Suprema and infima Let (A, , ^ rr D {{a:aei?, a ^ b}:beB}. !F(B \) is sometimes called the filter of sections of the directed setS. B ^('B is directed downwards'), B ]> a and ^(2? •[) are defined in the same way, but upside down. 11D Functions Let (A, ^ ) and (B, B is increasing if a ^ 6 =>/a =^/6. When / is a sequence (flOweN say> I shall write (# n ) we nt to mean that it is an increasing sequence, that is, that an =^ an+\ for every %GN. Similarly, n J5 is order-continuous if sup/[O] = fa

whenever G\a and C is not empty

inf/[O] = fa

whenever C\a and C is not empty.

PARTIALLY ORDERED SETS

[11

(Of course, if/is increasing, then/[C] is directed whenever G is.) Note that it is only directed sets which must have their sups and infs preserved in this way. *There are two more concepts which will be useful to us in special circumstances. An increasing function/: A -> B is order-continuous on the left if it satisfies the first half of the condition for ordercontinuity, that is, if /[C]f/a whenever G\a and ( 7 + 0 . A n d / is

sequentially order-continuous if

*11E Order-closed sets j ? c i 5 l shall write

whenever

neM ta

whenever

neH |6.

If (A, ^ ) is a partially ordered set and

B, C * 0 , C\a in A), B, 0 + 0 , C\a in A). Then B c JB and B c ®B. B is order-closed if JB = B = ^ 5 . *11F Order-bounded sets Let(^4, ^ ) be a partially ordered set. A set B c J. is order-bounded if it has both upper and lower bounds in A, i.e. if B c: [&? c] = {a: 6 ^ a ^ c} for some b and c in A. *11G Products Let((^4t, < t )X e I be an indexed family of partially ordered sets. Let A be the cartesian product ]JLeIAL, and define a relation ^ on A by a ^ 6iff a(t) ^ &(*) for every tel. Then (^4, < ) is a partially ordered set, the product of the family {(AL, ^ t))LeI. If, for each LEI, nL: A -> AL is the canonical map, then nL is increasing and order-continuous. *11H Exercises (a) Let (A, < ) be a partially ordered set. Then there is a topology on A for which the closed sets are precisely the order-closed sets of A, [See IXC] (b) Let (A, B is order-continuous iff/~1[C] is orderclosed in A for every order-closed C c B.

11]

RIESZ SPACES

Notes and comments We shall see a greatest lower bound or a least upper bound on every other page of this book; so it will be as well to grasp firmly the formal definitions in 11B. I generally take these formalities seriously; I want to know exactly what I mean by such expressions as inf 0. I cannot emphasize too strongly the fact that the least upper bound of a set depends on the partially ordered set in which it lies. There would be formal advantages in writing

to mean 'the least upper bound of B taken in the partially ordered set (A, < ) ' . Of course this would be absurdly cumbersome; but the abbreviation' sup B' must always be recognized as such. In 1XBI give a simple example of the danger. There's a more elaborate one in 4XF, and the distinctions become very important in Chapter 7. The same warning, of course, applies to the notations J and 3) in HE. There is a perfect symmetry in the notion of partially ordered set; if < is a partial ordering, so is its inverse ^ . Consequently all the associated definitions are doubled, as in 11B and 11C. I want to call attention to the concept of the filter of sections ^(B f), defined in 11C. I think that this filter is almost the most important thing associated with a directed set. You may have seen it already in the correspondence between net-convergence and filter-convergence in topological spaces [KELLEY 2L]. Note the technical point that ^{B\) is a filter on the whole space A, not on 2?. For another look at the material of this section, see BOURBAKI I, chapter in, §1.

12 Partially ordered linear spaces In this short section I discuss the simplest way in which a linear space structure and a partial ordering can be related. Although we shall very rarely have any reason to consider partially ordered linear spaces which are not Riesz spaces [§ 14], I think that the results here are clarified by being placed in their natural context. Here, and everywhere in this book, all linear spaces have the real numbers for their underlying field. There do exist applications in which it is more convenient to have the complex numbers; but I think that these are best approached by way of the real case. Otherfieldsare so far merely curiosities from the standpoint of this theory.

PARTIALLY ORDERED LINEAR SPACES

[12

12A Definition A partially ordered linear space is a quadruple ( £ , + , . , < ) where ( £ , + , - ) is a linear space over the field R of real numbers and ^ is a partial ordering on E such that (i) if x ^ y, thenx + z ^ y + z for every zeE; (ii) If x ^ 0 in E, then ax ^ 0 whenever a ^ 0 in R. 12B Positive cones From 12A(i), we see that x ^ yoO ^y-x. So ^ is determined entirely by £+ = {x: xe E, x ^ 0}, the positive cone of E. Given a linear space E over R and a set P ^ E, there is a partial ordering < on E such that (E, ^ ) is a partially ordered linear space and P = E+iS P n ( - P ) = {0} [for 11A (i) and (ii)], P + P^P aPc?

[for llA(iii)], V a ^ 0 [for 12A(ii)].

In particular, P = {0} will do [cf. 1XA]. 12C Lemma Let £ be a partially ordered linear space, xe E, A, B^ E. Then (a) sup (x + A) = x + sup A if either side exists. (b) sup ( — A) = — inf^4 if either side exists. (c) sup (A+B) = sup A + sup B if the right-hand side exists. (d) If a ^ 0, sup (ocA) = asup^l if the right-hand side exists. Proof of (c) Apply (a) twice, as follows: sup A + sup B = sup (A -}- sup B)

= sup {x + sup B: x eA} , yeB} = sup(,4+J3). Notes and comments The correspondence between an ordering and a positive cone [12B] is extremely important; it is one of the easiest ways of defining partial orderings on linear spaces. In 12C I prove only (c), because the other parts are direct consequences of the definitions. In a partially ordered linear space E, the map 2/h->o; + 2 / : E - > E i s a n order-automorphism for every x e £, and

12]

RIESZ SPACES

therefore must preserve suprema and infima. The same applies to the map y\-*ay: E->E for every oc > 0. On the other hand, the map yi-^ — y:E->E is order-reversing, that is, x ^ yo — y^—x, so it exchanges suprema for infima. Although I shall have no space to discuss them, many of the ideas of this book, set out for Riesz spaces, have extensions to partially ordered linear spaces. Compatible topologies [§21] are an obvious example; Lebesgue topologies [§24] are another. Some of these developments may be found in PERESSINI. 13 Lattices Once again we must have a section consisting mostly of definitions. I maintain a careful separation between general remarks on lattices and those concerning linear spaces because I wish to apply the former to Boolean lattices in § 41. Although it is possible to regard a lattice as a set with two binary algebraic operations, governed by certain identities, I prefer to take it as a special kind of partially ordered set. This makes it easier to think of the suprema and infima of infinite sets, which is something we must do continually. The important sections below are A-E, though G should be examined because it attaches a rather unusual meaning to the word 'disjoint'. 13A Definitions A lattice is a partially ordered set (A, x = sxip{y:yeF, 0 ^ y < x}. A solid linear subspace is always locally order-dense. A band or normal subspace of £ is a solid linear subspace F such that whenever i g f and sup A exists in E, then supAeF; i.e. an order-closed solid linear subspace [HE]. The intersection of any set of Riesz subspaces is a Riesz subspace; the intersection of any set of solid linear subspaces is a solid linear subspace; the intersection of any set of bands is a band. The range of a Riesz homomorphism is a Riesz subspace; the kernel 12

RIESZ SPACES

[14

of a Riesz homomorphism is a solid linear subspace; the kernel of an order-continuous Riesz homomorphism is a band. 14G Quotient spaces: proposition Let £ be a Riesz space and F a solid linear subspace of £. Then the linear space quotient E/F can be given a partial ordering such that it is a Riesz space and the canonical map 0: £ -> E/F is a Riesz homomorphism. Proof Let P = {<j>x: x e £+} s £/F. (a) Given X G £ , ^ 6 P O x r e F . P If areF, then Conversely, if <j>x e P, there is a y e £ + such that 02/ = 0#, i.e. y — xeF. Now — a < y ~~ x> s o and #- e F as .F is solid. Q (b) Now P satisfies the conditions of 12B. P If <j>x e P n ( - P), then x~eF by (a) above. But also 0( — #)eP, so #+ = ( — #)~eF, and a; = a ; + - r G F . Thus P n ( - P ) = {0}. But the other conditions are trivially satisfied. Q (c) So E/F can be given a partial ordering under which it is a partially ordered linear space and P = (£/F)+. Now if a; e £, sup {0#, 0} exists in E/F and is equal to x, 0}. Q (d) By the exercise in 14A, E/F is a Riesz space, and 0 is a Riesz homomorphism by 14Eb. Remarks Thus not only is the kernel of every Riesz homorphism a solid linear subspace, but every solid linear subspace is the kernel of some Riesz homomorphism. For this reason some authors call solid linear subspaces ' ideals'. Although it will not be used in this book, the result of 14Lb is very important in understanding the nature and significance of bands. 13

14]

RIESZ SPACES

*14H Products Let {Et)ieI be an indexed family of Riesz spaces. On YlieIEt there are natural linear space and lattice structures [13H] which render it a Riesz space. The canonical projections to each EL are all order-continuous Riesz homomorphisms. *14I Order units Two concepts we shall want to use later are the following. Let £ be a Riesz space. An order unit of E (sometimes called a strong order unit) is an e e E+ such that for every xeE there is an ne N such that x ^ ne. A weak order unit of E is an ee E+ such that x = supneN# A ne V xe E. *14J At various points of the book we shall want to use the following results. Lemma Let £ be a Riesz space, (a) If x, y and z belong to £ + , (b) If (xi)imzt)

= p(sup m6H inf i>m a; i -ag

[using 23C for the first time]

[using 23C for the second time]. Now however, if k > m > n, m-X

k-1

m-1

so

|^n-i fc1

and

P t ^ - i n ^ ^ i ) < S P(^+i-^) ^ 2~n+1>

whence the result. *23E Notation In the rest of this section we shall have occasion to use some manipulations of filters which, while fairly straightforward, seem to have no generally accepted description. The principle I use is the following. If X is a (non-empty) set, &> a filter on X, Y another set, a n d / : I - > 7 a function, t h e n / ^ is the filter on Y i.e. the filter on Y with base

If X and Y are topological spaces a n d / i s continuous, t h e n / ^ ->ft whenever J*" -> t. It X and Y are uniform spaces and / is uniformly continuous, t h e n / J ^ is Cauchy whenever iF is. 46

FATOU TOPOLOGIES

[23

Two special cases we shall be concerned with are: (i) Let £ be a linear space, p: £ -> R a function, u G £, and &> a filter on £. Then />(J^ - u) is the filter on R {A:{x:p(x-u)eA}e&r},

Ve>0 3 , | p ( ) | (ii) Let £ be a Riesz space, we £, and #* a filter on £. Then is the filter on £ uis defined similarly. *23F Lemma Let £ be a Riesz space and p a Fatou pseudo-norm on £. Then (a) F = {u :pu = 0} is a band in £. (b) If xeE, {u:p(x — u) = 0} is an order-closed [definition: H E ] sublattice of £. (c) If !F is a filter on £, ^4 = {^ :/>(#* — w) -> 0} is an order-closed sublattice of £ (perhaps 0 ) . Proof (a) is a simple computation; F is solid and closed under addition, so is a solid linear subspace; now 23C shows that it is a band. Any translate of a band is an order-closed sublattice, which proves (b). As for (c), it is easy to see that if x e A, then A = x + F, reducing it to (b). *23G Lemma Let £ be a Dedekind cr-complete Riesz space and p a Fatou pseudo-norm on £. Let SF be a filter on £ which is '/)-0auchy', that is, ^piu-v) = 0. Suppose that A is a member of 3F which is order-bounded and closed under the lattice operation A. Then there is a ueJQiA such that Proof

Choose a sequence (An}ne1x in !F such that p{x-y) ^ 2-n V x,yeAn

V neN.

For each weN, choose an element

47

23]

TOPOLOGICAL RIESZ SPACES

Then {un:neN} n e N . So if

is order-bounded and p(un-un+1)

< 2rn for every

then p(un — u)^ 2~n+1 for every n e N by 23D, and p(x-u)

< 3.2~" V #e^l w V

neN,

But as J. is closed under A, infk>i>mUi eQiA

VmeN,

*23H Lemma Let £ be a Dedekind complete Riesz space with a Fatou topology. Let B S>B such that ^ -> x0. Proof If p is a continuous Fatou pseudo-norm on E, C(p) is closed under A [23Fc]. So if w(/>) = inf G(p), C(p)^w(p) and w{p)e@B. But also, again by 23Fc, p(0r-w{p))-+O. Now if /?! and p 2 are both continuous Fatou pseudo-norms on E, is also a continuous Fatou pseudo-norm on £, and so w(pl) < MPi+/>2). SO D = {t(;(/o): p is a continuous Fatou pseudo-norm on £} f; a s D c ^JS a n d B i s bounded above, D\x0 say, andxQe*f@B. Moreover, if p0 is any continuous Fatou pseudo-norm on £, D o = {t0) («^" — ^(/> +p0)) -> 0 for every continuous Fatou pseudonorm p. So by 23Fc again, P o ^ ~~ xo) -> 0; as p 0 is arbitrary, !F -> a:0 [using 23B]. 48

FATOU TOPOLOGIES

[23

231 Definition Let £ be a Riesz space. A linear space topology on E is Levi if every set A c= E which is bounded (in the linear topological space sense) and directed upwards, has an upper bound (in the ordering of £). *23J Lemma Let E be a Dedekind complete Riesz space with a Levi Hausdorff Fatou topology. Suppose that 3F is a Cauchy filter on E such that 3* t\z [notation: 23E(ii)] converges for each zeE. Then IF converges. Proof For each z e £, let x(z) be the limit of !F A z. (a) If zx ^ z2, x{z1) ^ x(z2). P & A zx = ^ A (z2 A z±) = (J^ A Z2) A zx and the function u i—• u A zx is continuous; so a ^ ) = lim (J^ A z2 A zx) = zx A lim (J^ A z2) = zx A a;(z2). Q (b) Set ^4 = {x(z):zeE+}. Then u4f, by (a) above. But also A is bounded. P Let U be any solid neighbourhood of 0 in £. Let V be a balanced neighbourhood of 0 such that V+V £ ?7, and let Be^ be such that B — B^ V. Let w e J5; there is an a e R + such that u e a V and then B^u+V ^aV+V £ ( a + i ) ( 7 + F ) c (a + l)?7. But for any «(2f) e {yAziyeB}

as C7 is solid. So ^L is absorbed by U. As C7 is arbitrary, A is bounded. Q So A is bounded above and x0 = sup J. exists. (c) Now !F -> #0. P Let p be a continuous Fatou pseudo-norm on E, and let e > 0. Let Be3F be such that p(u — v) ^ e V%, andfixw0 e JS. Now we know that A f x0, so [using (a) above], and by 23C there is a zx ^ wj}" such that But we know that

x(zx)E{UAZ1:UEB},

so there is a % e JS such that

49

23]

TOPOLOGICAL RIESZ SPACES

Now

l ^ i A ^ — uo\ = \u± AZX — u0Azx\ ^ \ux — uo\,

so/?(% Azx-u0)

^ p(u±-u0)

^ e, and

p(xo-uo)

So

P(#o~^) O e

V

;

as e and p are arbitrary, ^" -> x0 [using 23B again]. Q *23K Nakano's theorem Let E be a Dedekind complete Riesz space with a Levi Hausdorff Fatou topology. Then E is complete (as a uniform space). Proof Let & be a Cauchy filter on £. Then for any y,zeE,(&rvy)Az is Cauchy and contains the order-bounded set [y Az,z], so it converges [23G, 23H, setting A = B = [y AZ,Z]]. So #" vy converges [23J]. So J^ converges [23J, upside down], *23L Proposition Let E be a Riesz space with a Hausdorff Fatou topology. Let A c E be a solid set such that JA c A. Then ^4 is closed. Proof (a) Let O be the set of continuous Fatou pseudo-norms on E. ForpeO, set F(p) = {z:zeE,pz = 0}. Then F(p) is a band in E [23Fa]. Set where for each p e O F(p)d = {U:UEE,

\U\ A \Z\ = 0

V ^

d

as in 15F. We recall that each F(p) is a band [15Fa]. Also, if px andp 2 belong to O, so does Pi+p2, and F(p1+p2)d = (F(Pl) n F(pt))* 2 F(^)^ u Fta)«, from which it follows that F is a solid linear subspace of E. (b) In fact, F is order-dense. P Suppose that w > 0 in E. As the topology on E is Hausdorff, there is a neighbourhood U of 0 such that w $ U; by 23B, there is a p e O such that /w > 0. So w $ F(p) = F(p)dd [15Fb; recalling that by 22Eb E is Archimedean]. Let z e F(/>)d be such 50

FATOU TOPOLOGIES

[23

that u = \z\ A w > 0. Then ueF and 0 < u ^ w. As w is arbitrary, Fis order-dense by the criterion of 15E. Q (c) Let x be any member of A. Suppose that peO, yeF(p)d, and that 0 < y 0 and y — we F(p)d, y — w 0. Let neN be such t h a t 2~n < \p{y — w).

Since every lower bound of {yi :i ^ n} is less than or equal to w, ^ inf {ym — z:m^n, by 15C. So mfm>n{ym-w)+ So

z ^ yi V i ^ n} = 0

= 0, i.e.

y - w = sup m ^(i/ -ymvw)

= sup m>n Bupm>i>w(y -

and Vi) < So indeed JBf 2/ and yeJA c ^4. (d) Thus D = {y:yeF,0^y^

P(y-w). X

\x\) ^ A.

But as F is order-dense, Z)f |#| so |a;| G^4. As .4 is solid, ^ G 4 ; which proves the proposition. *23M Lemma Let E be any Riesz space. Then %S{E~, E) and %S(EX, E) are Levi topologies on E~ and £ x respectively. Proof Suppose that A is a non-empty subset of E~, directed upwards and bounded for %S(E~, E). Then sup {fx :feA} < 00 for every x e £ + . By 16Db, sup A exists in Er, and 4. is bounded above in E~. If 4 c Ex, then sup 4 G £ X because Ex is a band in E~; so 4. is bounded above infix. 51

23]

TOPOLOGICAL RIESZ SPACES

*23N Exercises (a) Let £ be a Riesz space with a locally convex Fatou topology %. Then % is defined by the continuous Fatou seminorms on E. (b) Let £ be a Riesz space with a Fatou topology. Let A be a closed totally bounded set in £. Then A is order-closed. (c) Let £ be a Riesz space with a Fatou topology. Then 0 has a neighbourhood basis consisting of solid order-closed sets. (d) A locally solid Levi linear space topology on an Archimedean Riesz space is Hausdorff. (e) Let £ be a Riesz space with a Levi linear space topology such that order-bounded sets are bounded. Then £ has the 'boundedness property5 i.e. If B c E+ is such that for every sequence (xn}nelx in B and every sequence OeZ. 24B Theorem Let £ be a Riesz space with a Lebesgue linear space topology. (a) I f 4 c £ , / i c l and 2A c A. (b) 0 has a base of neighbourhoods U such that JTJ £ [7. (c) If 0 ci J.10 in £, and C7 is a neighbourhood of 0, then there is an xeA such that [ — x,x] c JJ. (d) If 0 c 4 | 0 in £, then J ^ l ) -> 0. Proofs

(a) and (b) are easy.

(c) Let F be a closed neighbourhood of 0 such that V—V^ U. Then there is an xe A such that [0,x] c F. P ? Otherwise, for each XEA, set Ax = Aft [0, x] and Bx = {z:3yeAx,

y ^ z ^ x}.

Now ^ J , 0, so for each w e [0, a], {w;Vj/ :yeAx} | w, and we3)Bx c JS^. Thus [0, a:] c JJ^. But we are supposing that [0, #] $ V, and F is closed, so Bx $ F. Set C = {Z:ZEE\V,

Now (i) V ^e-4 (ii) V zeC

3 a,

3 zeC, z ^ x, 3 yEA,y ^ z,

(iii) ^4^0. It follows that Cj 0. But 0$C. X Q Now [ - x, x] = [0, x] - [0, #] c Y -Y 0}! 0 in £ because E is Archimedean [15D]. So for any neighbourhood U of 0 there is an a > 0 such that [0,oc(y-x)] c [7, i.e. [ 0 , y - a ] c a" 1 ?/ [24Bc]. Thus [0, y — x] is bounded. So £ c: a; + [0, y — x] must be bounded. 24E Corollary If £ is a Riesz space with a compatible Lebesgue linear space topology under which it is complete as a uniform space, it is Dedekind complete. Proof

21Ba, 24Da, 21Bd.

24F Corollary If £ is an Archimedean Riesz space with a Lebesgue linear space topology, then £ ' c fix. Proof L e t / e £ ' . Then/is bounded on order-bounded sets [24Db], s o / e E~[16C]. Consider | / | in £~. If 0 c A\ 0 in £, and e > 0, then U = {y:yeE, \fy\ < e}is a neighbourhood of 0, so there is an xeA such that [ - x, x] c C7 [24Bc]. Now |/1 (») < e [16Ea]. As e is arbitrary, infX€A | / | (x) = 0; as J. is arbitrary, | / | is order-continuous [14Ec], i.e. | / | efix.S o / e £ x , as required [16H]. 24G Proposition Let £ be a Riesz space and % a locally convex locally solid linear space topology on £ such that £ ' c £x. Then % is Lebesgue. Proof Suppose that 0 ^ A^OinE. Then 0 belongs to the closure of A for the weak topology %S(E, £'). P Suppose (fi}i 0 for 2 . Proof (a) If % is Lebesgue, it satisfies the conditions (i) and (ii) by 24F and 241. (b) The point is that condition (i) above implies both (i) and (ii) of 241. If E' c £x, then order-bounded sets are %S{E, £>l>ounded, therefore ^-bounded. At the same time, if A c Ex has polar A0 c £, then JA° c A0. P Suppose that 0 c JS c ^o a n ( j ^ a t £ f ^ in £. Then 0, so for any/e-4, as-y) = 0, and |/a;| ^ sup^^l/^l ^ 1. A s / i s arbitrary, XG^4°. Q SO if ?7 is any closed absolutely convex neighbourhood of 0 for %,JTJx £ such that x = inf n6H sup n = s u p ^ i n f ^ a v Then for %. Notes and comments Lebesgue topologies are a perpetual source of surprises. The early results of this section show how quite simple arguments can have unexpected consequences; moreover, subject to suitable modifications, many of them can be generalized to much wider contexts. The subtlest appears to be 24Bc; this is the foundation of the most interesting results. 24F and 24G seem to sort out locally solid locally convex Lebesgue topologies, which are of course the most important ones. Note that the condition 'locally solid' in 24G cannot be dispensed with [8XB]. However, the general condition 241 seems to add something even in this case. In § 82 this condition is tied in with duality theory in some general remarks on locally convex Lebesgue topologies. 24H is a very powerful result. Apart from its application to 241 above, it has a striking consequence in 81H. Perhaps it would be profitable to study spaces satisfying the conditions of 24H in their own right. The locally solid ones are of course dealt with by 24Lf. Of course there are many important examples of Lebesgue topologies. Some are given in 1XD, 1XF, 26B, 2XB, 2XC, 2XF, 2XG, 63K, 6XF, 6X1 and 82H. *The name 'Lebesgue topology' is suggested by 24Lg, which is an abstract version of Lebesgue's Dominated Convergence Theorem [63Md]. It corresponds to 'condition A(ii)' of LUXEMBURG & ZAANEN B.F.S. [note x, § 33]. NAKANO L.T. [§ 6] gives a definition of' continuous 60

LEBESGUE TOPOLOGIES

[24

topology' which is equivalent to my' sequentially Lebesgue topology' [83 Ka], but in a context which ensures that it will be actually Lebesgue [NAKANO L.T., theorem 6.2]. 25 Complete metrizable topologies As in other branches of functional analysis, Banach spaces have a preeminent position in the theory of topological Riesz spaces. Banach lattices, that is, Riesz spaces endowed with Riesz norms under which they are complete, were the first topological Riesz spaces to be studied, and remain the most important examples. One of the purposes of this chapter is to exhibit their properties in such a way that it is clear which aspects of their structure are involved at each point. In the present section I give those results which are consequences of the closed graph theorem and therefore apply to all complete metrizable linear space topologies. We find that a Riesz space structure on a Banach space, if related to the topology at all, is very strongly related; the same is true, of course, of other kinds of algebraic structure. If we restrict our attention to spaces in which the topology is compatible, they seem to divide naturally into three types: (i) not locally solid; (ii) locally solid, not Lebesgue; (iii) locally solid and Lebesgue. The first class will not concern us, though I gave an example in 2XD. The other two are both important. Since we know from 25A that the topology can be regarded as defined by the Riesz space structure, it is natural to ask what Riesz space properties correspond to the classification above. The division between (i) and the others can be expressed directly in terms of a concept, 'uniform completeness', which will be useful elsewhere; so I include this result [25K]. In the locally convex case, the three classes are distinguished very simply by the nature of the dual space [25G,25M]. 25A Proposition Let £ be a Riesz space. Then there is at most one compatible complete metrizable linear space topology on E, and for any such topology (U n E+) — (U ft E+) is a neighbourhood of 0 whenever U is. Proof Let % and %' be two compatible complete metrizable linear space topologies on E. Choose sequences (U^)ne^ and (Vn}nex of balanced sets in E such that {Un: n e N} is a neighbourhood basis at 0 for X, 61

25]

TOPOLOGIGAL RIESZ SPACES

{Vn: n e N} is a neighbourhood basis at 0 for %', Un+1 + Un+1^Un Set

Wn =

and Vn+1 + Vn+lc:Vn V neN.

{Un(\VnnE*-)-(Un(\VnnE+).

Then for every n e N, T^ is balanced, absorbent [because E + — £+ = £], Wn+1 + Wn+X s TTn, and Tfn+1 a l s o converging to 0, such that xn = yn-zn for every n e N . Proof Let (Uk}ke1Si be a decreasing sequence of sets forming a neighbourhood basis at 0 in £, and let (nkykE^ be a strictly increasing sequence in N such that xie(Uk n E+) — (Uk 0 E+) whenever i ^ nk. Now, for nk ^ i < nk+1, choose yi and zi in Uk n £ + such that xi = yi — zi. (For i < n0, choose ^ and ^ arbitrarily such that xi = yi — z^) Then (y^)ie^ and F is continuous. Proof Let (xn}n€ix be a sequence converging to 0 in £. Let neN be a subsequence of Fis a continuous linear map. To show that it is in L~(E; F), I shall use the approach of 16F. Let xe E+ and write Ax = {Zi R such that

H i = 2*6x1^(01 V ^ °- ^ W i s complete under || \r p. 100; or use 25Na], so it is an L-space. There is a natural duality between lx(X) and /°°(X) given by

[TAYLOR,

the sum being absolutely convergent for every x e V-(X) and y e l°°(X). This identifies /°°(X), as normed Riesz space, with [26C; the identification l°°(X) = I 1 ^ ) ' is proved by KOTHE, 14.7.8, and TAYLOR, theorem 4.32-A; or we may use 65B/6XD to prove that At the same time it identifies P(Z) with /°°(X)x. P This is a consequence of 33F. Alternatively, it can be proved directly as follows. (a) Suppose that x > 0 in P-(X) and that 0 c B\ 0 in /°°(Z). Then 0 = inf B in Rx, i.e. in%(0:2/e5} = 0 V teX. Fix y 0 G ^

an(

i ^ et ^ > 0. Then there is a finite I ^ X such that

Since, for each tel, there is a, yeB such that i/(£) < e, and since there is a yx e B such that ^x ^ y0 and j/1(^) ^ e for every £ e / . Now

As e is arbitrary, m£yeB(x, y} = 0. Thus x acts on Z°°(X) as a member of /CO(X)X. Since Z°°(X) certainly separates its points, /1(X) may be regarded as embedded in /°°(X)X. (b) Now suppose t h a t / ^ 0 in t°(X)\ For each teX, let et be the member of V°(X) given by et(t) = 1, et( where e is the unit of /°°(X). So

As / is arbitrary

Stex^W

e

^fe 2/) f° r ©very yelQO(X). We know that gr(et) = f(et) for every i e l . But suppose that y is any member of /°°(X)+. Consider * = { S t « i y ( * K : J s Z f /finite}. Then/z = grz for every zeB, and 5 f i/. So, a s / a n d g are both ordercontinuous, , « /«/ = sup 0ei? /« = sup^^flfz = gy. As ?/ is arbitrary, / = g. Thus every positive element of /°°(iL)x can be represented by an element of lx{X)\ it follows that the duality between /°°(X) and induces a linear space isomorphism between /°°(X)X and V-{X). (c) Finally, we see that for xeP-(X)9

x^Oo

x(t) ^ 0 V teX o (x,et) ^ 0 V teX

so that the ordering on P-(X) induced by its embedding in /°°(X)X is correct, and the isomorphism is a Riesz space isomorphism. Q (The argument above is of course a particularly easy special case of thatin65A.) Hence l\X) is perfect [33A, 33F]. 2XG The space co(X) Let X be any non-empty set. Let cQ(X) be the set of all functions x: X -> R such that

{t:teX,\x(t)\ >e} is finite for every e > 0. Then cQ(X) is a solid linear subspace of /°°(X) [2XA above]. It is easy to see that co(X) is closed under || J^, so (co(X), I || oo) is a Dedekind complete Riesz space with a Fatou norm under which it is complete. The norm topology is Lebesgue; but it will be Levi only if X is finite. (So co(X) does not satisfy the hypotheses of Nakano's theorem, 23K.) Because /°°(X) is an M-space, so is co(X) [26Ac]. 77

2X]

TOPOLOGICAL RIESZ SPACES

*Note that a subset of cQ(X) is relatively compact for || H^ iff it is order-bounded; cf. 26Hf. [See also 2XF, 4XCb.] 2XD

Let E be the set of sequences x e RN such that ||a|| =m&x(\x(O)\,su-pie}x\x(i+l)-x(i)\)

< oo.

Then E is an order-dense Riesz subspace of RN (for if x e E then \x\ e E and || \x\ || ^ ||aj||). || || is a norm on E under which £ is a Banach space isomorphic (as Banach space) with f°(N). The topology % derived from || || is compatible (since for each neN \x{n)\ ^ (n+l)\\x\\

V XEE,

so the coordinate map x\->x(n) is continuous). £ has an order unit e givenby

e(n) = n + l V » 6 N .

But the interval [ — e, e] is not bounded, so % is an example of a compatible complete metrizable linear space topology which is not locally solid. Since [ — e,e] includes the unit ball, % must be Levi; but of course it is not Lebesgue [24Db]. Exercise

Show that the lattice operations on E are not continuous.

*2XE An example for 26E Recall that the space I2 = Z2(N), defined by 19 c J /2 = {x: is a Hilbert space under the inner product Clearly it is also a solid linear subspace of RN, and its norm is a Riesz norm. Moreover, it is easy to see (adapting the arguments of 2XB, or using 65A) that the inner product identifies I2 with (/2)x, and therefore that the norm topology is Lebesgue [25M]; alternatively, it is equally easy to show directly that the norm topology is Lebesgue, and therefore that /2 - (/2)' = (/2)x [25M]. [See also 6XG.] Now let us examine H = L~(/2; / 2 ). We know that H c L(/2; I2), the space of continuous linear endomorphisms of /2 [25D]. The norm of L(/2; /2) induces a metrizable linear space topology on H. The positive cone of H i s 78

{T:TeL(P;

I2), Tx > 0 V ^ O } ,

EXAMPLES FOR CHAPTER 2 [2X which must be closed in L(P; I2), and therefore in H, as the positive cone of l% itself is closed. Next, if

because the norm on I2 is a Riesz norm. So the interval [0, T] is bounded in H; it follows that all order-bounded sets are bounded. However, given neN, there is a 2n x 2n matrix An = (ocnij} such that \ani:}\ = 1 for all i, j < 2n and Yi]Coinikanjk = 2n i£i=j,

0 otherwise.

P Set Ao = (1) and

K for each neN.Q

Consider An as a linear transformation of /2, given by ( A * * ) (t) = i:j 0 there is a finite I ^ X such [This is 83F below.] This makes it obvious that the solid convex hull of A is still relatively weakly compact, from which it follows that ^(Z00,11) is locally solid. [See also 82H.] So it must be Lebesgue [24G, 82F]. A ^(Z 00 , Z1)-bounded set must be £S(Z°°, /^-bounded; by the uniform boundedness theorem (Z00 being (I1)'), it is || ^-bounded, i.e. orderbounded. Thus £&(Z°°, I1) must be Levi. [See also 33B.] I t is also well known that Z00 is complete under %k(lco, I1); this could be proved as a corollary of Nakano's theorem [23K], but the usual proof is from the fact that Z00 is the dual of the Banach space Z1 [see A IE]. Thus Sfc(Z°°, I1) is a complete HausdorfF Lebesgue Levi Fatou topology. *2XG The topology %k(P-, c0) Now let us consider the Mackey topology on Z1 = lx(X) derived from the duality between Z1 and c0 = co(X) [2XC]. I t is coarser than the Lebesgue topology derived from || || x [2XB], so it must be Lebesgue. It has the same bounded sets as %S{11, c0), which (because I1 is the dual of c0) has the same bounded sets as || || x; so, like || l^, it must be Levi. Finally, it is locally solid. P First method Let B be the unit ball of Z00 = Z°°(Z). Then B can be identified with [— 1, l ] x . Now it is easy to see that the product topology of B is the topology on B induced by %s{lco, Z1), under which B is compact. Consider the map (x9y)\-+ xy: B x B -> B, where we define (xy)(t) = x(t)y(t) for every teX. Then this is continuous for the product topology. So if A c B is compact for %s(lcc, Z1), {xy:xeA,yeB} is also compact. But this is precisely the solid hull of A in Z00. Now suppose that U c: I1 is a neighbourhood of 0 for S^Z1, c0). Then its polar U° c c0 is compact for %s{c0,11) or %s(lcoy I1). So certainly U° is bounded for ||fl^,and there is an a > 0 such that <xU° c B. Now the 80

EXAMPLES FOR CHAPTER 2

[2X

solid hull C of ocU° is compact, andD = arxG, which is the solid hull of U° itself, is compact, But D is still included in c0. We must now observe that by Krein's theorem [AlG], the convex hull of D is relatively compact for %s(cQ, I1). So D°, the polar of D in I1, is a neighbourhood of 0 for %k(P-9 c0); but of course it is solid, so it is the required solid neighbourhood of 0 included in U. Second method I t is in fact not difficult to show that a set A c c0 is relatively %s(c0, /^-compact iff: (i) A is bounded for || H^; (ii) whenever (^>^eN is a sequence of distinct points in X, = 0.

From this it follows at once that the solid hull of a relatively compact set in c0 is relatively compact. The rest of the work proceeds as above. Q Now we can say that I1 is complete under S^/ 1 , c0), either by Nakano's theorem [23K], or by AlE. (The results above have been generalized by MOORE & REBER.) Further reading for Chapter 2 The principal omission from this chapter has been the concept of 'normal cone'; at the same time, generalizations to the theory of partially ordered topological linear spaces have hardly been considered. Both of these subjects are tackled by SCHAEFER T.V.S. and PERESSINI. For further ideas on these lines, and for results concerning complete metrizable partially ordered linear spaces, see JAMESON. There is an alternative introduction to the elementary theory of L- and if-spaces in KELLY & NAMIOKA, with a few additional results. For more about normed Lebesgue and Fatou topologies, see LUXEMBURG & ZAANEN B.F.S. [notes x-xvi].

81

3 . Dual spaces In the theory of linear topological spaces, the concept of a dual space is of paramount importance. In this short chapter I supplement this theory with a brief discussion of the dual spaces E~ and £ x which we have already seen, and which are fundamental to the ideas of the rest of the book. The most important new results concern the canonical evaluation map from a Riesz space £ to the space of linear functionals on £x [32B below]. In the third section I give an equally abbreviated note on 'perfect' Riesz spaces, which include the majority of the spaces which interest us. Further general results concerning spaces in duality will be derived in the course of Chapter 8, while studying weak compactness.

31 The space £~ This is the basic dual space associated with a given Riesz space £. Recall that so far we know that E~ is the set of linear functionals on E which are bounded on order-bounded sets, and that it is a Dedekind complete Riesz space [16C, 16D]. The importance of the results below is that they apply to the evaluation map from E into G* for any solid linear subspace G of E~; their principal application is of course in the case G = £ x . 31A Let us begin with a list of definitions, repeating some given in Chapter 1. Definitions (a) Let E be any linear space over R. Then £* will denote the linear space of all linear maps from £ to R. (b) Let £ be a Riesz space. Then E~ is the Dedekind complete Riesz space L~(£; R) [16B, 16D]. (c) Let £ be a Riesz space. Then £ x is the band L x (£; R) in £~ [16G, 16H]. (d) Let £ be a linear topological space over R. Then £' is the linear space of continuous linear maps from £ to R. (e) Suppose that £ is a linear space over R and that F is a linear 82

THE SPACE JET

subspace of £*. For any xeE,I given by

[31

shall write £ for the member of F*

£(/)=/(*)

V feF.

31B Lemma Let £ be a Riesz space; suppose t h a t / e E~+ and that a; e £ + . Then there is a g e E~ such that 0 ^ g ^ f, gx = fx, and gy = 0 whenever a; A 2/ = 0 in £. Proof

Define gr: E+ -> R by jf2 = sup {f(z Aax):oce R+} ^ fz,

as/is increasing. Then without difficulty (using nothing stronger than 14Ja), we can see that g is linear on £+ (I mean that it satisfies the conditions of 16A), so extends to become a member of JE*. The rest of its properties are immediate [using 14Kb for the last]. 31C Proposition Let £ be a Riesz space and G a solid linear subspace of E~. Then the canonical map x\->$: £->G* is a Riesz homomorphism from £ into G x . Proof TfxeE+, £G G x by 16Ec [forif 0 c A],0in G, then A),0 in £~, so m£feA%{f) = mff€Afx = 0]. As G x is a linear subspace of G*, £ G G x for every # e £. The map a:h-> ^ is plainly linear and increasing. Finally, suppose that x A y = 0 in £ and that/G G+. Then there is a gr G £~ such that 0 ^ g ^ f, gx = fx, and gy = 0 [31B]. But G is solid in E~, so ^ G G . Now

(* A 0) (/) $ is a Riesz homomorphism. 31D Corollary

Let E be a Riesz space and/G £~+. Then

r, \g\ G* is a Riesz homomorphism from E to G x . (b) Let E and F be Riesz spaces and T: E-+F a Riesz homomorphism which is onto. Define T: F* -> £* by (T'gr) 0*0 =fl^s)for every xe E and #eF*. Then T'[F~] c £~ and T':F~ -> E~ is a oneto-one Riesz homomorphism. Notes and comments Of course 3IB is the central idea of this section. In effect, this asserts that E~ contains enough functionals to make 31C and 31D true. 31D and 16E together indicate a curious symmetry in the relationship between E and £~.

32 The space £ x This seems to be the most interesting of the Riesz space duals of a given Riesz space E, especially in view of the remarkable topological properties discussed in § 82 below. At the same time it turns out that, in particular cases, the ' natural' dual of £ is £ x as often as it is E~ [see, for instance, 2XA, 52E, 6XD]. Of course they are frequently identical [as in 25M], which leads to particularly powerful results. In this section I give the fundamental embedding theorem 32B and an elementary, but useful, characterization of members of £ x [32D]. 32A I begin with a vital technical result. Lemma Let £ be an Archimedean Riesz space. Let / > 0 in £ x . Then there is an x > 0 in £ such that: (i) fx > 0; (ii) 0 < y < x=>fy > 0; (iii) if g A / = 0 in £ x , gx = 0. Proof

(a) Let

F = {z:zeE,f(\z\) = 0}. Then F is a band in £ [cf. 23Fa]. But F + £ because/ # 0. So Fd * {0} [definition: 15F], because [by 15F] F = Fdd. Now if x > 0 in Fd, then 0 < y ^ x => x Ay = y > 0 => y$F => fy > 0, and of course fx > 0. 84

THE

SPACE £x

[32

(b) Now suppose that g A / = 0 in Ex; of course g > 0. Let e > 0. By 16Eb we may choose, for each weN, an element zn of E such that 0 ^ zn ^ # and For each weN, let wn = infi<w2$. Then and

g{x-wn) ^ S^nflrfc-z*)
$: E->GX is a Riesz homomorphism. It is order-continuous because, if 0 c A\ 0 in J3, = mf^/z

= 0 V/eG+

and infjg^^ = 0 by 16Ec. So we are left considering the Riesz subspace {$ : x e E} in Gx. I shall use 15E to show that it is order-dense. (b) Perhaps I should first remark that G, and similarly G x , are Archimedean, because they are Riesz subspaces of the Dedekind complete Riesz spaces E~ and G~. Suppose that <j> > 0 in G x . Then there is a n / e G+ such that (i) / > 0, (ii) if tjr A $5 = 0 in G x , ifrf = 0 [32A]. Now there is an x e £+ such that /# > 0. Since £(/) = fx > 0, $ A ^ > 0. Because G x is Archimedean, inf{a£:a> 0} = 0 [15D], and there is an a > 0 such that > 0. 85

32]

DUAL SPACES

Let geG+ be such that (i)frg> 0, (ii) 0iJrh>0, (iii) if 6 A f = 0 in G x , #0 = 0 [32A again]. Let y e E+ be such that (i) gy > 0, (ii) 0gz>0, (iii) if g A h = 0 in G, hy = 0 [32A once more]. Now so $ A ^ > 0, by the choice of ^; as i/r < &, $A& > 0; by 14Kb, P A a£ > 0. Set z = y A a#; then £ = 0 A a£ by 31C, so 2 > 0. (c) Infact, 2 ^ cj). P ? Otherwise, (2 - 0)+j > 0, so there is an hoeG+ such that (i) (2 - 0)+ (fe0) > 0 (ii) 0 < h ^ h0 => (2 - 0)+ (A) > 0 [32A, for the last time]. Now examine But ^ A (a£ - §S A ^) + = 0, so (a£ - ^ A ^) + (^) = 0, by the choice of g. So h0 A g = 0, by the choice of ^0. Now, however, ho(y) = 0, by the choice of y. So (S-0)+(fto) ^ HK) = ho(z) < ho(y) = 0, which contradicts the orginal requirements on hQ. X Q (d) Thus we have 0 < 2 < <j>. As (j> is arbitrary, {$: we£} is orderdense in Gx, by 15E. 32G Corollary Let £ be a Dedekind complete Riesz space and G a solid linear subspace of JEX. Then the map x\-*& takes £ onto an order-dense solid linear subspace of Gx. Proof

Immediate from 32B and 17D.

32D The following is useful for proving that functionals belong to £ x . I t works only for Archimedean spaces. Lemma Let E be an Archimedean Riesz space, and let/e E* be such that i n f ^ \fx\ = 0 whenever 0 7ieN i n

E+

V xeE 3 weN, \x\ ^ xn. Then £ is perfect. Notes and comments The original' vollkommene Raume', studied in a series of papers by G. Kothe, were sequence spaces (that is to say, subspaces of RN); they were defined in a way which we can now recognize [see 6XD below] as the form £ x , for an arbitrary solid linear subspace £ of RN; so Kothe's first theorem was that £ x x x = £ x . His most important results are repeated in KOTHE, § 30. They were taken up by DIEUDOKNE, who generalized them to include spaces of equivalence classes of measurable functions, as in § 65 below. The abstract approach above comes from LUXEMBURG & ZAANEN B.F.S. [note vn, § 28]; the two systems are linked by FREMLIN A.K.S. II, in which I gave a representation theorem showing that an arbitrary perfect Riesz space is isomorphic to one of Dieudonne's 'espaces de Kothe'. Examples for Chapter 3 I give no special examples for this chapter. Whenever we encounter a Riesz space £, one of our first tasks is to seek to identify £~ and £ x . The latter is generally easier, especially 89

33]

DUAL SPACES

when we can use 65A or 65B. Important examples from other chapters include the elementary sequence spaces [2XA-2XC], function spaces [§ 65], and, for some insight into less familiar cases, 8XA-8XC. Further reading for Chapter 3 Much the most important reference for this work is LTJXEMBUBG & ZAANEN B.F.S. [notes vi-x].

90

4 . Riesz spaces on Boolean rings The main purpose of this book is to show how the abstract theory of Riesz spaces may be applied to the study of function spaces arising in measure theory. In Chapters 1-3 I have set out the most important concepts needed for this approach. But in this chapter I propose to open up another line of attack. My eventual aim is to describe the relationship between the measure algebra of a measure space and its function spaces. In order to do this, I demonstrate methods of constructing Riesz spaces from Boolean algebras which, when applied to measure algebras, will produce isomorphic copies of the basic function spaces L 1 and L00. At the same time I shall apply the concepts of the first three chapters to describe the properties of these Riesz spaces. The technical problems encountered along the way are considerable. An intuitive understanding, however, of the basic constructions S and L00, is not hard to attain; this is because some relatively easy examples already offer most of the principal aspects of the theory [4XA-4XD]. The construction L # is essentially deeper, and requires faith in Chapters 1-3 to be meaningful at all; its real significance will not appear until § 62. One of the advantages of this method is that all the constructions are functors, and behave reasonably when the right kind of homomorphism is applied. This will enable us to see deeper into one of the more puzzling aspects of measure theory; details are in §§45, 54 and 61E. For readers who have not encountered measure algebras before, I recommend taking § 61 in parallel with this chapter.

41 Boolean rings The theory of Boolean rings is an abstract framework for the study of rings of sets. If X is any set, then its power set 0*X, with the operations of symmetric difference and intersection, forms a Boolean ring [4XA]. Conversely, any Boolean ring is isomorphic to a 'ring of sets', i.e. a subring of some SPX [41D]. A ring of sets has a natural ordering and X in further constructions, so it is important to note now that they are canonical. 41F The ordering of a Boolean ring may define a relation s on 91 by

If 91 is a Boolean ring, we

a c b o ab = a. Now we see that if ^ X is the Stone representation of 91, then a c j o ^ c ^ , From this it follows at once that £ is a partial ordering under which 91 is a lattice. The lattice operations on 91 are given by inf {a, 6} = a&, sup{a,6} = a + b + ab. Of course, these facts can readily be verified without appeal to Stone's theorem. 91 always has a least element, which is 0; it has a greatest element iff it has a multiplicative identity, and they are then the same. 41G Notation If 91 is a Boolean ring, I shall use the symbols n, U and \ for the operations on 91 corresponding to n, U and \, that is, a n b = inf (a, b} = ab, a\)b = sup{a, b} = a + b + ab, a\b = a + ab, 93

41]

RIESZ SPACES ON BOOLEAN RINGS

as well as £ defined above and its inverse 2 . 1 n a Boolean ring, the words sup and inf will always refer to the ordering c. So, for instance, 21 is Dedekind complete or cr-complete iff it is complete in the sense of 13B with respect to this ordering. If 21 is a Boolean ring, a subring 93 of 21 (i.e. a subset containing 0 and closed under addition and multiplication) is a Boolean ring in its own right. The ordering on 93 induced by the ordering on 21 coincides with the ordering on 93 derived from its own Boolean ring structure. As 93 is a sublattice of 21, the lattice operations on 93 are also those derived from 21. It is arbitrary, but convenient, to use the phrase 'Boolean algebra' to mean a Boolean ring with a 1, a multiplicative identity. A subalgebra of a Boolean algebra is now a subring containing 1. *If 21 is a Boolean ring, a a-subring of 21 is a subring which is a c-sublattice in the sense of 13F, i.e. one which is closed under countable sups and infs in so far as these exist in 21. A o-subalgebra, or o-ideal, of 21 is a (r-subring which is also a subalgebra or ideal respectively. *41H A lemma on suprema and infima Let 21 be a Boolean ring, and A a non-empty subset of 21. (a) If the set B of upper bounds of A is not empty, then inf{& + a : a e ^ , beB} = inf{6\a:ae^, beB} = 0. (b) Given a0 e 21, A f a0 iff a0 is an upper bound for A and {a + ao:aeA} = {aQ\a:aeA}^0. (c) Given a0 e 21, A ! a0 iff a0 is a lower bound for A and ao:aeA} = {a\a o :aei}|O. Proof I shall appeal to the correspondence between u, n, \ and U, fl, \ to justify intuitively the following arguments. They can, of course, be validated formally from the definitions. We observe first that, for a, b e 21, a^b

o a + 6 = b\a.

(a) If c is a lower bound for {b\a:beB, aeA}, then fix boeB. As c^bo\a V aeA,cna = 0 V a e A, So bo\c is an upper bound for A. As A #= 0 , c g bo\c and c = 0. (b) If a0 is an upper bound for A, then the map b h-> ao\b is an orderreversing involution on {6: b £ a0}. So A f a0 iff {ao\a: a e A} 10. 94

BOOLEAN RINGS

[41

(c) If a0 is a lower bound for A, then A | iff B = {a\a0: a e A} 4,. Now c is a lower bound for 5 iff c u a0 is a lower bound for A and c n a0 = 0. So if a0 = in£A, then for any lower bound c of B,c ^ aQ and coao = 0, so c = 0. Conversely, if 0 = inf B, then for any lower bound d of A, d\a0 = 0 and d £ a0.

*41I Corollary Let 31 be a Boolean ring. Then 91 is Dedekind cr-complete iff every countable non-empty set in 9t has a greatest lower bound. Proof For now if A ^ % is a non-empty countable set with an upper .. r r , .. bound aa, A u sup J. = ao\inl {ao\a: a e A). 41J Ideals and quotients of Boolean rings Observe first that if 91 is a Boolean ring, then a set / c 91 is an ideal (written / < 91) iff (i) OeJ; (ii) a,bel=>a[)bel; (iii) a c bel=>ael. For (iii) is precisely the condition that /9I £ / , and now (ii) is equivalent to / + / c: / . Obviously, if / EI shall write i> for the corresponding linear map from 5(91) to £. Thus v = vx* 42E The Riesz space structure of 5(91) Let 91 be a Boolean ring, and let 5 = 5(91) be denned as in 42B. (a) 5 is a Riesz subspace of V*(X) [2XA], where X is the Stone space of 91. (b) Every non-zero member x of 5 can be expressed in the form where (a?)i 0 o o^ > 0 V i < n,

(c) Every non-zero member x of 5+ can be expressed in the form x

= where (bjyj 0 for each j < n. In this case, *(d) Suppose that a, /?eR, a, 6e9t, and 0 < a ^ ^ fyb. Then 0 < a < /? and 0 4= a s 6. Proof Let $S: 91 -> ^ X be the Stone representation of 91. Let 5 0 be the set of those functions x: X -> R such that (i) x[X] is finite, (ii) for each a #= 0, {£: #(£) = a} G ^[91]. Then it is clear that 5 0 is a linear subspace of Kx (because ^[91] is a subring of &X), that xae^o f° r each a e 91, and that each non-zero member x of 5 0 can be expressed in the form of (b) (the oct being j ust the non-zero values taken by the function x). Consequently 5 0 = 5. But So is clearly a Riesz subspace of /°°(X), which proves (a). The identities in (b) are now elementary. For (c), express x = S i < w a i / t a i a s in (b), but with the sum so arranged that 0 < a 0 < ... < ocn_1. Set JSO = a 0 and pi = o^ — a^_x for 1 < i < n. Set bj = sup J < i < r i a i for j < n, so that xbj = Tij^i R is order-continuous. Conversely, suppose that *> is order-continuous. To show that v is order-continuous, we must examine an arbitrary non-empty set A ^ 0 in 5. Let e > 0. F i x # 0 e A . By 42Ec, there is an ocoeR+ and an a0e91 such that #0 < ao%ao. Let ^ > 0 be such that Sva0 ^ e. For each x e S+, define bx e 21 as follows. Let ^: 21 -> ^ Z be the Stone representation of 31. Then xeRx. Let

Ex = {t:x(t)>8}. Then Ex^ X and in fact ^ e ^[21] (see the characterization of S given in the proof of 42E). So there is a unique bxe 21 such that <j)bx = JS

whenever 0 < a; < a;0. Observe also that

so that B = {6X: xeA, x < ^ 0 }\. If 6 is a lower bound for B in 21, then Sxbx < x whenever

xeA, x ^ x0.

As .410, Sxb = 0; as 5 > 0, 6 = 0 [42Ed]. Thus 5 1 0 in 21. As v is ordercontinuous, there is an xeA such that x ^ x0 and a o ^ ^ e. Now vx < ^ K A + % 0 ) = ^vbx + Sva0 ^ 2e. As e is arbitrary, infxeJpx = 0; as A is arbitrary, v is order-continuous. 102

THE SPACE S(%) [42 42K Completely additive functionals Let 21 be a Boolean ring. An additive functional v\ 2i -> R is called completely additive if infae^ \va\ = 0 whenever

0 c: A \ 0 in51.

(The reason for the phrase 'completely additive' lies in 42Rj.) 42L Theorem Let 21 be a Boolean ring. Then an additive functional v: 21 -> R is completely additive iff v e 5(2l)x. Proof (a) If v: 21 -> R is completely additive, it is locally bounded. P Use 421. ? Otherwise, there is a decreasing sequence neN in 21 such that | van\ ^ n for each neN. Let B be the set of lower bounds of {an:neN}; then Bf. By the argument of 41Ha, C = {an\b:neN, beB}^0 Let

in2l.

D = {an\b: beB, n > \vb\ + 1} c C.

Then D is cofinal with C [i.e. V c e C 3rfeD, d s c] so D j 0. But | ^ | ^ 1 for every deD. X Q (b) If y:2l->R is completely additive, y+:2t->R is ordercontinuous. P The argument is similar to that of 24Bc. Suppose that 0 0 such that v+a > e for every a e A. Then for every a e A there is a b c a such that z>6 > e. But So there is a C E . 4 such t h a t c^a y(& \jc) = vb + v(c\b) ^ e.

and |^(c\6)| ^ vb — e, so t h a t

Now let B = {b : b G 21, vb ^ e, 3 a,ceA such that c c ^ g a } , Since for every ae^4 there is a beB such that 6 c a, 5 ^ 0 . But inf6eB |^61 ^ e > 0, which is impossible. X Thus inf ae ^p + a must be 0; as A is arbitrary, y+ is order-continuous [42Ja]. Q (c) Now if v\ 21 -> R is completely additive, 0eS(2l)~ by (a) and 42H, and D+ = (z^+)A. But v+ is order-continuous by (b), so (v+)A is ordercontinuous [42Jc], and 0+G5(21) X . Similarly, v~= (-*>)+ = ( - ^ and

VGSX.

(d) Conversely, if j> e 5 X , and 0 c A \ 0 in 21, as |0| and ^ are both order-continuous. So y is completely additive. 103

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RIESZ SPACES ON BOOLEAN RINGS

*42M Countably additive functionals A slightly weaker condition than 'completely additive' is 'countably additive'. If 21 is a Boolean ring, an additive functional v: 21 -> R is countably additive if (yan}nen -> ® whenever (ctn)ne^ 0 in 21. [See also 42Rj below.] Clearly, a completely additive functional is countably additive, and an increasing additive functional is countably additive iff it is sequentially order-continuous. *42N Proposition Let 21 be a Boolean ring and v: 21 -> R a locally bounded countably additive functional. Then v is the difference of two increasing countably additive functionals. Proof We know that v+: 21->R is an additive functional [42H]. Suppose that neN ^ 0 in 21. ? Suppose, if possible, that Since certainly (y+an}n€x | , there must be an e > 0 such that v+an > e for every neN. Now for each WGN there is a b £ an such that vb > e. Since (v(b n #m))meN -* 0> there must be an m e N such that and now v(b\am) ^ e. So v+(an\am) ^ e.

Consequently, we can find a strictly increasing sequence (n(i)}iels such that +

K#))

> e,

i.e.

for every i e N. But now he V

which is impossible. X As (an)neji is arbitrary, this shows that v+ is countably additive. But similarly v~ = (— v)+ is countably additive, and v = v+ — v~ [using 42H, or otherwise]. *42O Proposition Let 21 be a Boolean ring and v\ 21->R an increasing additive functional. Then v is countably additive iff i>: 5(21) -> R is sequentially order-continuous. Proof The argument is exactly the same as that of 42Jc; the only difference is that arbitrary directed sets are replaced throughout by monotonic sequences. 104

THE SPACE 5(31)

[42

*42P These arguments can be elaborated along the lines of 42L to set up a one-to-one correspondence between locally bounded countably additive functionals v\ 91 -> R and linear functional/: 5(91) -> R such that (fxn}nelSt -> 0 whenever (^n>7ieN ^ 0 in 5(91) [see the remark at the end of §32, and 83K]. The principal difference is that, unlike completely additive functionals, countably additive functionals need not be locally bounded [42L, proof, part (a), and 4XG]. However, the most important situations are covered by the following lemma. *42Q Lemma Let 91 be a Dedekind c-complete Boolean ring, and v\ 91 -> R a countably additive functional. Then v is locally bounded. Proof ? Otherwise, there is a decreasing sequence (an)nex in 91 such {an\a:neN}^0 that (\van\}n€}i^> oo [421]. Let a = infnelxan. Then [4lHc]. But now {v{an\a))n^ -> 0, i.e. n6l -> va, which is impossible. X 42R Exercises (a) Let 91 be a Boolean ring, £ a Riesz space, and v: 91-> £ an additive function. Then v: 5(91) -> E is a Riesz homomorphism iff v is a lattice homomorphism. (b) Let 91 be a Boolean ring, E an Archimedean Riesz space, and v: 91 -» E an increasing additive function. Then 0: 5(91) -> £ is ordercontinuous iff v is order-continuous. *(c) Let 91 be a Boolean ring, E a normed linear space over R, and v\ 91 -» £ an additive function. Then i>: 5(91) -> E is continuous for the norm || H^ on 5(91)iff v is bounded. (d) Let 91 be a Boolean ring, E a Dedekind complete Riesz space, and v\ 91 -> E an additive function, (i) Then v\ 5(91) -» £ belongs to L~(S; E) iff *>+a = sup {V6:6 ^ a) exists in E for every a e 91, and in this case j>+ = (i;+)A. *(ii) And now i>eL x (5; 15)iffirdaGA\va\ = 0 in £ whenever 0 0 3 # > 0 such that /ia < S=> \va\ ^ e. [Hint: show that the set of v satisfying the condition is a band in S~ containing ft. Now use 15F and a lemma analogous to (g) above, with A replacing v.] Under these conditions, we say that v is absolutely continuous with respect to fi *(i) Let 91 be a Dedekind cr-complete Boolean ring, and /i and v locally bounded real-valued additive functionals on 91 of which /A is countably additive and increasing. Then v is in the band of 5(9l)~ generated by ju, iff it is countably additive and fia = 0 => va = 0. *(j) Let 91 be a Boolean ring and v: 91 -> R an additive function. Then (i) v is countably additive if 2nGNmw, exists and is equal to ^(sup^eua^) whenever (anyneJSS is a disjoint sequence in 91 such that sup neN a n exists (ii) v is completely additive iff TiaeAva exists and is equal to j^(sup^l) whenever A is a disjoint set in 91 such that sup A exists. [Hint for (ii): show first that if v satisfies the condition, it is locally bounded, and v+ also satisfies the condition.] Notes and comments The basic property of 5(91) is the first universal mapping theorem, 42C. It is clear that this defines 5(91), and the function x> UP to linear space isomorphism. What is remarkable is that S is endowed with a natural partial order for which the next universal mapping theorem, 42Fc, is true; that this natural order is a lattice order [42Ea]; and that the remaining universal mapping theorems [42Jc/42Rb, 420, 42Ra, 42Rc] are all true. In fact we have six distinct theorems based on a single construction. 106

THE

SPACE S(%)

[42

In 16C we saw how functions which were the difference of increasing linear maps could be characterized; this result corresponds exactly to 42H. Similarly, functionals which are the difference of order-continuous increasing additive functionals are associated with the completely additive functionals [42L, corresponding to 32D]. In the analysis here of completely and countably additive functionals we are beginning the Radon-Nikodym theorem. The reason for studying both is that in the most important applications we are presented with countably additive functionals on rings of sets; but that (as I shall endeavour to show) their most striking properties are a result of the fact that they can be regarded as completely additive functionals on quotient rings. *Of the arguments above, two are familiar from elementary measure theory. 42C is just the theorem that an additive measure on a ring of sets defines an integral on the space of 'simple functions' [cf. 4XB]. 420 is the theorem that a countably additive measure defines a sequentially order-continuous integral; the same methods yield the more thoroughgoing 42Jc. *I ought to remark there that the method of constructing 5(91) given in 42B is far from the only one. Indeed, an essentially simpler algebraic construction, not using Stone's theorem, quickly sets up a space for which 42C is true. From this point, the positive cone of 5(91) can be defined as the cone generated by {#a: a e 91}, and the other results will follow. However, such an abstract approach introduces substantial additional complications into the proofs; and the temporary avoidance of the axiom of choice seemed an inadequate compensation. 43 The space L°°(9l) If we complete 5(91) with respect to the norm || fl^, we obtain the space L°°(9l). This is a Banach lattice, in fact an M-space. It is a model for the L00 spaces of ordinary measure theory, and shares many of their properties. The main question to which we shall address ourselves here is the Dedekind completeness of L°°(9l) [43D]. 43A Definition Let 91 be a Boolean ring and X its Stone space. Then 5(91) is a Riesz subspace of/°°(Z) [42Ea]. Let £^(91) be the closure of 5(91) in /°°(X) for ||fl*,.Then L°°(9l) is a closed Riesz subspace of /°°(-X"), because the lattice operations on /°°(JL) are continuous. Under || || a,, L°°(9l) is an M-space, because /°°(.X) is. For examples, see 4XB-4XF and 43Ec below. 107

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43B Lemma Let 21 be a Boolean ring and X its Stone space, (a) If 1: X -> R is the function with constant value 1, then for every yeL™ and a ^ 0. (b) If XELCC+, there is a sequence (xn}nGjj( in 5(9t)+ such that <Xn>neV t * and neH ">

x f

° r || || ••

(c) 5(91) is super-order-dense in L°°(9t). *(d) If x eL°°+ and 5 > 0, then {a:aeSt, ##« ^ x} is bounded above in 91. Proof (a) Observe first that if y e 5 = 5(91) and a ^ 0, ?/ A a l e 5. P For if 2/ = 2* where {ai)i /°° is continuous. (b) Given x e L°°+, there must be a sequence neN in 5+ such that | a; —yJloo < 2>~n for every weN. Set ^ = (^-3.2-^l)+ = ^ - 2 / n A 3 . 2 - M e 5 for each WGN. Since t; also, ||a?n-a;||oo < 4 - 2 ~ 7 1 for each n, so n€H->«. By

^ n > n e P N to:.

(c) By definition, 5 is super-order-dense in L00. *(d) We know that there is a y e 5+ such that ||x - y\\ „ < S. Express V a® YiKnPjXbj where (bj}:j is the Stone representation of 91. Now x

=> #(0 ^ ^ V tefia 0 V

So {a: 5^a ^ #} is bounded above by 60. 108

THE SPACE L°°(3l) 43C Corollary norm on L°°(2t).

[43

For any Boolean ring SI, the norm || ||«, is a Fatou

Proof We know that || || & is a Riesz norm. So suppose that 0 c A \ x in L°°+. Let a = sup^^H^/Ha,. Then x A aleL 0 0 [43Ba], and clearly x A a l is an upper bound for A. So x = # A a l and [[scl^ ^ a. As J. is arbitrary, this shows that || |oo is a Fatou norm. 43D Theorem Let 21 be a Boolean ring. Then (a) L°°(2l) is Dedekind complete iffSl is; (b) L°°(2l) is Dedekind cr-complete iff 91 is. Proof The proofs of the two halves of the theorem are formally independent, but almost identical; so I shall give a proof of (a), with occasional words in brackets which may be inserted to give a proof of (b). (i) Suppose first that L00 = L°°(2l) is Dedekind (cr-) complete and that A £ 21 is a non-empty (countable) set. Then { p : a e i } has an infimum x in L00. Let yeS = 5(21) be such that 0 ^ y ^ x and \x"~ y\oo < 1 [using 43Bb]. Then there is a /? > 0 and a 60 e 21 such that [using 42Ec; if y = 0, set 60 = 0]. Now, ifaeA, J$xb0 ^ y ^ x ^ xa> ^J 42Ed, b0 £ a; thus 60 is a lower bound for A. Conversely, if 6 is a lower bound for A, ^6 is a lower bound for {xa: as A), so yb < #; now

So 6\6O = 0, i.e. b £ &0. As & is arbitrary, 60 = inf^l; as A is arbitrary, 21 is Dedekind (cr-) complete [13Ca or 411]. (ii) Now suppose that 21 is Dedekind (cr-) complete, and that A £ L°°+ is a non-empty (countable) set with an upper bound weL™ say. For each x e A, we can choose a non-empty (countable) set Bx c 5 + such that x = supl?^ [43Bb]. Now B = \JxeABx is a non-empty (countable) set bounded above by w, and A and i? have the same upper bounds in L00. To save space, let us write an = 2~~n||w\\^ for eachneN, and let G be the set of upper bounds of B in L00. We can construct inductively a sequence (yn}nejn in 5+ with the following properties: 109

43]

RIESZ SPACES ON BOOLEAN RINGS

( Vn\

(P) ll^ + i-^IU 0 whenever (xn)nelx J, 0 in L . (c) Let 21 be a Boolean ring. Then L°°(2t)~ can be identified, as normed Riesz space, with 5(21)', which is a solid linear subspace of (d) Let 21 be a Boolean ring and/GL°°(2ir.Then/+^ = (/y)+, where / + is taken in L°°(2{)~ and (/#) + is defined as in 42H. (e) Given a Boolean ring 21, define j:L# (21) -> R by J / = ||/+|| - 1 | / - | | for every/eL # . Then in the sense that V e > 0 3 6e2l

JJ/

such that

|/C\;a)-j/| < e V a ^ b.

Notes and comments The fundamental idea here is part (iii) of the proof of 44Bb;if A: 21 -> R is a bounded order-continuous increasing additive functional, then there is a corresponding order-continuous increasing linear functional on LG0(2l). This is a fairly straightforward extension of 42Jc, in which arbitrary order-continuous increasing additive functionals were associated with order-continuous increasing linear functionals on 5(21). Now the bounded completely additive functionals are identified with members of LooX, just as general completely additive functionals are identified with members of Sx [42L]. 113

44]

RIESZ SPACES ON BOOLEAN RINGS

The value of this refinement lies in the fact that, because L00 is an M-space, Ifl = Lcox is an L-spaee. Thus we shall be able to apply some of our most powerful results about Biesz spaces to the space of bounded completely additive functionals. These expressions represent Ift as a subspace of 5 X c 5~. I t is important to note that L # is actually a solid linear subspace of 5~ (though not in general a band). For L# is a solid linear subspace of 00 JL°°~ = X, ', which can be identified with 5', which is a solid linear subspace of S~ by 22D. The point of this is that the lattice operations on L # can be calculated in the same way as those on 5~, as in 42Rg or 44Cd. When the ring 21 has a 1, then obviously every locally bounded additive functional on 21 is actually bounded; so S~ = 5" £ L°°~ and 5 X ~ Lft. There are further results concerning L # in § 83. It is difficult to give convincing examples of Ifi spaces, because simple Boolean rings give rise to trivial cases [see 4XC; also 4XFj]. In fact, as Kakutani's theorem shows, all Lft spaces can be based on measure algebras [see the end-note to § 26]. So further examples must await Chapters 5 and 6. 45 Ring homomorphisms An important aspect of the constructions of the last three sections is their behaviour relative to homomorphisms between the underlying rings. This is particularly significant because the rings which are most important to us are defined as quotient rings, that is, as homomorphic images. The fundamental results are that a homomorphism from 21 to 93 gives rise to maps from 5(21) to 5(93), from L°°(2l) to L°°(93), and from L#(93) to L#(2l). In the language of category theory, 5 and L00 are covariant functors, while L # is contravariant. If 93 is a quotient of 21, 5(93) and L°°(93) appear as quotients of 5(21) and L°°(2l) respectively; this is the result we need to identify the L00 space of an ordinary measure algebra with the usual function space. 45A Definition Let 21 and 93 be Boolean rings. A map n: 21 -> 93 is a ring homomorphism if n(ab) = na.nb and n(a-\-b) = na + nb for all a, 6 G 21. In this case, the formulae of 41G show that 7r(a[)b) = n(a + b + ab) = na + nb+ na.nb = na[)nb, n(a n b) = n(ab) = na.nb = nan nb, n(a\b) = n(a + ab) = na + na.nb = na\nb 114

RING HOMOMORPHISMS

[45

for all a,fee31, so that TT is a lattice homomorphism; also, of course, TT(O) = 0. 45B Proposition Let 21 and 93 be Boolean rings, and TT: 21 -> 93 a ring homomorphism. Then there is a unique Riesz homomorphism na r TT: 5(21) -> 5(93) defined by TT(X^) = X( ) f° every ae% and now ||^#|| oo < || #|| oo for every x e 5(21). So TT has a unique extension to a norm-decreasing Riesz homomorphism TT: L°°(2l) -> L^ffi). Proof The map a\-+x(na): 21 -> 5(93) is easily seen to be additive; so there is a unique linear map fr: 5(21) -> 5(93) such that TT(XCI) = x(na) for every a G 21 [42C]. Now if (ai)i L°°(2t)'. But in order to ensure that7r'[L#(93)] c L#(2I), we must impose an order-continuity condition on?r. Proposition Let 21 and 33 be Boolean rings, and TT: 21 -> 93 an ordercontinuous ring homomorphism. Then there is an order-continuous increasing linear map TT': L#(93) ->* L#(2t) given by (rr'g) (x) = jr^s)

V a e L»(«),

L#(9l) is a norm-preserving Riesz homomorphism. (e) Let ieZ be a family of Boolean rings with product 91. Then 91 is Boolean [4IK], For each tel, let nL: 91 -> 9lt be the canonical ordercontinuous ring homomorphism. Then the associated Riesz homomorphisms nL: L°°(9l) -> L°°(9lJ induce a normed-Riesz-space isomorphism between L°°(9l) and the solid linear subspace {x:\\x\\ =sup t e Z 1 of the Riesz space product Iliei^CSlt)- *At the same time, the normpreserving Riesz homomorphisms n[: £ # (9l t ) -> L#(9l) induce an isomorphism between L#(9l) and the solid linear subspace

Notes and comments The results of this section are essentially algebraic. They rely on a careful analysis of functions on 5-spaces which is then extended to functions on L°°-spaces. So they all seem natural, despite the complexity of some of the proofs. 4X Examples for Chapter 4 Since Boolean rings usually appear as rings of sets or their quotients, these are the important cases to analyse. When 91 is a ring of sets, we have particularly accessible representations of 5(91) and L°°(9l) [4XB]. From these we can derive representations of 5(91//) and L°°(9l//), where / is an ideal of 91, as quotients of 5(91) and L°°(9l) respectively, as 119

4X]

RIESZ SPACES ON BOOLEAN RINGS

explained in 45Dc. These again are especially simple when 91 is a o*-algebra of sets [4XE]. An interesting special case is the algebra of regular open sets in the real line [4XF]. 4XG is an instructive counterexample. 4XA The algebra 0>X. Let X be any set, and 0>X its power set. On 0>X define A by AAB = (A\B)[)(B\A)

V i,BcI

Then {£PX, A, (1) is a Boolean ring [41 A]; its zero is the empty set. (The direct verification of the ring postulates is lengthy but elementary. Alternatively, we may identify SPX with {0, i}x, saying that a set i c j corresponds to the function with value 1 on A, 0 on X\A. In this case, {SPX, A, n) is isomorphic to (Z 2 ) x , where Z2 is the field with two elements 0 and 1. The extra Boolean postulate, that A f) A = A for every A ^ X, is trivial.) In fact SPX is a Boolean algebra [41G]; its multiplicative identity is X. The lattice operations on SPX are just (J and n. &X is Dedekind complete; if A c SPX, sup A = \JA. *For each teX, the set Jt — {A: A c X,t$A} is a maximal ideal of SPX. So the map t H->^ is an embedding of X in the Stone space of £PX. Note however that (unless X is finite) SPX has many more maximal ideals than these. 4XB Rings of sets If 7 is a set, a subring of ^ 7 is a family 21 of subsets of 7 containing 0 and closed under A and n; clearly, this is the same as being closed under U and \. Suppose that 91 is a subring of ^ 7 . Then S = 5(21) is isomorphic, as normed Riesz space, to the linear subspace E of /°°( 7) generated by the characteristic functions of members of 21. P S is defined as the linear subspace of l°°(X) generated by the characteristic functions of members of 0[2t], where l a : 2t-> /°°(7) is clearly additive; so, by 42C, there is a linear map T: 5-> Z°°(7) such that T{%a) = la for every ae2l. If we examine the description of members of S given in 42Eb, it is clear that T is a norm-preserving Riesz homomorphism, so S ^ ^[5] as normed Riesz space. Also, since S is spanned by {^a: ae2l}, T[S] must be the linear span of {la: ae 21}, which is E. Q 120

EXAMPLES FOR CHAPTER 4

[4X

This isomorphism means that we can identify S with E; since the set Y is already to hand, while the Stone space X of 31 must be constructed with the axiom of choice, this is a much more convenient representation of S. It follows also that L°°(9I) [43A] can be identified with the || H*,completion of E, which is [isomorphic to] the closure of E in ^(Y). Note that this is indeed a Riesz space isomorphism as well as a normed space isomorphism, because the lattice operations are continuous. *As a rule, Lf (91) is not of much interest in this case. If 91 contains all finite subsets of X, then L#(9I) can be identified with l\X). One proof follows the argument of 2XB; another observes that L°°(9l) is order-dense in /°°(X), and applies 17B and the result of 2XB; another uses 65B. 4XC 0>X and 0>fX (a) If 91 = 0>X, then 4XB gives us the identifications S{£PX) ~ {x: xe'Rx, x takes finitely many values},

[2XB]. Thus the Dedekind completeness of f°°(X) [2XA] is related to the Dedekind completeness of 0*X by 43Da. (b) For any set X, let 0*fX be the family of finite subsets of X. Then (PfX is an ideal of &X\ in its own right, it is a Dedekind complete Boolean ring. Now

S{0>fX) - so(X) = {x:xeRx, {t:x(t) * 0} is finite}, L»(^,Z) s co(X), L#(0>fX) - l\X) [as in 2XC]. 4XD o-algebras of sets Suppose, in 4XB, that 91 is a cr-subalgebra of 0>X [i.e. X e9t and 91 is closed under countable unions and intersections, as well as complementation]. In this case, the representation of L°°(9l) as a subspace of R x expresses L°°(9l) as V aeR}. P (i) Just as in the proof of 42E, 5(91), which we have identified as the linear subspace of R x generated by the characteristic functions of members of 91, is {x: x G E, x takes finitely many values}. 121

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(ii) If x GL°°(21), then there is a sequence 0, so t h a t x(t) = limn_^O0xn(t) for every teX. Now for any a G R,

{*:*(*) > a} = which must belong to 91. So #e JE. (iii) Conversely, if xe E, then for each neN define xn by xn{t) = 2-n[2nx(t)]

V teX

where [a] = s u p { m : m e Z , m < a } for each aeR. Then {*: a n 0) > a} = HmeN {*: *(*) > 2-*([2»a] + 1) - 2"-} G 21 for any a e R, so x^e 5. Since || a; — xn\^ < 2-nforeach^GN,a; 4XE Quotient rings For any Boolean ring 21 and ideal / of 21, L°°(2I//) can be identified with LOO(21)/LOO(/) [45Dc]. When 21 is a subring of ^ X , then L°°(2l) and L°°(/) are subspaces of Z°°(X). In the special case when 21 is a cr-subalgebra of SPX, as in 4XD, and / is a o*-ideal of 21 [i.e. / is an ideal closed under countable unions], then L°°(/) becomes {x:xeL™{%), {t:x(t) * 0}eI}. (For this is the solid linear subspace of Lc0(2l) generated by the characteristic functions of members of / , and it is closed for || H^. Remember that by 45Db L°°(I) is solid in L°°(2l).) Thus Lc0(2l//) becomes the space of equivalence classes in L°°(2l) under the relation x~y if {t:x(t) *4XF Algebras of regular open sets In any topological space, an open set G is regular if G = intG. The set ^ of all regular open sets is partially ordered by ^ . This ordering is induced by a Boolean ring structure on ^ under which ^ is a Dedekind complete Boolean algebra. When the underlying topological space is R, the algebra thus obtained has many facets; I shall present it as an example of an algebra on which there are no non-trivial countably additive functionals. Proofs of these remarks now follow. (a) Let X be a topological space. Suppose that E and F are closed subsets of X such that int E = int F = 0. Then int (E U F) = 0. [For X\E and X\F are dense open sets, so their intersection is dense.] 122

EXAMPLES FOR CHAPTER 4

[4X

(b) Let X be a topological space. Let 91 be the set {A:A^X,int(8A)

= 0} X,

where dA = ^4\int^. is the boundary of A. Then 91 is a subalgebra of 0>X. P If A, B c X, then 8(A [)B)^8A[) 8B. So by (a) above, 91 is closed under u. As 8(X\A) = 8A for any A c X, 91 is closed under \. Finally, 0 e 91, so 91 is a subalgebra of 0>X. Q Observe that open and closed sets all belong to 91. (c) Continuing from (b) above, let / be {A:A meN f X in ^ . So

As e is arbitrary, vJT = 0. Because v is increasing, v = 0. Q (i) I t follows at once that any countably additive functional on *3 is zero (for by 42Q and 42N, any countably additive functional on ^ is the difference of increasing countably additive functionals). At the same time we see that there can be no finite-valued measure on 3? [see 5IE]; indeed, it is easy to see that the only measure on ^ , according to the definition 51 A, is the one which takes the value oo on every non-zero element. (j) Consequently, L°°(^) is now a Dedekind complete Riesz spaceactually, an M -space with unit-such that L°°(^)x = {0} [44Bb]. Of course, L°°(#)~ = L°°(#)' [25G] is large. (k) Exercise Show that (i) there is a sequence (Gn)neV in the algebra ^(R) described above such that, for every non-empty Ge@, there is an n e N such that 0 a Gncz G (ii) any Dedekind complete Boolean algebra with this property is isomorphic to ^ . (1) Exercise Again supposing that X = R, show that L°°(9l) can be identified with the set of those bounded functions x: R->R such 124

EXAMPLES FOR CHAPTER 4 [4X that {t: x is continuous at t} is a dense G^ set. Consequently, L°°(^) is the Riesz space quotient of this by the solid linear subspace {x: x is zero on a dense Gs set}. Also, Cb(X) may be regarded as an order-dense Riesz subspace of the Dedekind complete Riesz space L°°(^). Hence, or otherwise, show that Cb(X)x = {0}. *4XG A counterexample for 420 and let 91 be

Let X be an uncountable set,

{A: A £ X, either A is finite or X\A is finite}. Then 91 is a subring of 0>X. Define v\ 91 -* R by vA — n

if A has n members, where neN;

= —n if X\A has n members. Now if (An)nelx | 0 in 91, one of the An must be finite; so in fact one of them is empty, and infweN \vAn\ = 0. Thus v is countably additive. But v is certainly not locally bounded. So we see that the condition 'Dedekind cr-complete' in 42Q is indeed necessary.

4XH Ring homomorphisms induced by functions Let X and Y be arbitrary sets, and/: X -> Y any function. Define n: SPY -> SPX by n(A) =f~x[A] for every A c 7. Then TT is an order-continuous ring homomorphism. We see that, for A ^ Y,

HxA) = X(nA) = Xif-W) = *4 o/, where here we identify %A with the characteristic function of A, as in 4XB. It follows that n: L«>(0>Y) -> L™(0>X) is given by 'fry = yof 00

for every ye/ s L*>(&Y) [see 4XCa]. Equally, identifying L#(0>X) with P(Z), TT': L#(^Z) -> L#(^7) is given by

for every ueY and a:elx{X).

125

5. Measure algebras In this chapter I shall give versions of those results in elementary measure theory which refer to measure algebras or to L1 and L00 spaces. The first two sections apply the concepts of §§ 42-4 to ' measure rings', that is, Boolean rings on which a strictly positive countably additive measure is defined. In this case, a true analogy of L 1 spaces can be found, and the correspondence between L 1 spaces and L # spaces is discussed. All measure rings of any significance are 'semi-finite', and consequently their L 1 and L # spaces can be identified; this is the basic idea of § 52. The next section deals briefly with Dedekind complete measure algebras, which seem to be central to ordinary measure theory. Finally, in § 54, the ideas of § 45 concerning homomorphisms are reviewed in the new context.

51 Measure rings In this section, we shall have only definitions and basic properties. A measure ring is a Boolean ring together with a strictly positive measure; this is an extended-real-valued functional which is additive and sequentially order-continuous on the left. The definition of measure ring which I have chosen [51 A] follows the ordinary definition of measure space [61 A] in allowing 'purely infinite5 elements, that is, non-zero elements of infinite measure such that every smaller element is either zero or also of infinite measure. It is an accident of the theory that these need not cause any inconvenience in the early stages; but fairly soon they must be outlawed, and accordingly I shall immediately introduce 'semi-finite' measure rings, that is, measure rings which have no purely infinite elements. In any measure ring, semi-finite or otherwise, the ideal of elements of finite measure is of great importance, and many of the results here are based on consideration of this ideal as a measure ring in its own right. All the most important examples of measure rings are derived from measure spaces, as in 6ID; so at this point it will be helpful to have in mind Lebesgue measure, as the simplest non-trivial example of a measure space. 126

MEASURE RINGS

[51

51A Definition A measure ring is a Boolean ring % together with a measure /i: % -» [0, oo] such that: (i) /i(a\jb) = fta+jbib if an& = 0in9l; (ii) ju,a = O o a = 0; (iii) if n6H f a in SI, then /ia = sup nels/ian. Note 'oo' here, and later, is regarded as an actual point adjoined to R+. The set [0, oo] has natural additive-semigroup and total-ordering structures, writing oo V a e [0, oo], a < oo V ae[0, oo]; and also a multiplicative-semigroup structure, writing O.oo = oo.O = 0, a.oo = oo.a = oo V a > 0, which we shall have occasion to use. Subtraction, however, is not fully defined, so we must be wary. 51B Definitions

(a) For any measure ring (91, ft), W will be {a:ae% [ia < oo}.

(b) A measure ring (%/i) is semi-finite if, whenever ae% and [ia = oo, there is a b s a such that 0 < jib < oo. 51C Proposition Let (21, fi) be a measure ring. (a) If a £ b in % jia < [ib. (b) Suppose that 0 c B c W and that B | 0 in SI. Then (i) there is a sequence (bn}ne1ii in J3 such that (bn}n€x 10; (ii) inf&ei?/*& = 0. Proof (a) For fib = fia+ju,(b\a) ^ /^a. (b) Let a = inf6eB/^6. Choose a sequence neN in B such that [icn R be a countably additive functional. Then y is completely additive. Proof Suppose that 0 R by va{b) = /i(a(\b) V 6e2I. Then, because the map 6 -> a n b: 91 f-» 21* is an order-continuous ring homomorphism, and fi: 21/->R is completely additive [51E], va is completely additive. Moreover, sup {va{b): b e 21} = fia < oo,

so there is a unique/ a GL # (2l) such that and ||/a|| = /*a [44Bb]. Thus we have a map a\-+fa: W->Lft{%). But clearly this map is additive and increasing, since if a n c = 0 in W, fa+c(Xb) = /*((« + c ) & ) = Mab + cb) = Mab) +Mcb) = fa(Xb)+UXb) for every 6e2l, and/ a + c = / a + / c - So it induces a canonical increasing linear map (j>: S(W) -> I^f(2l), given by (j>{x<x>) =fa f° r every aeW. The basic properties of ^ are contained in the two propositions 52B and 52D. 52B Proposition For any measure ring (21, fi), the canonical map <j): 5(210 -*" £#(2l) defined in 52A is a one-to-one Riesz homomorphism. For any xeS(W), l%\\ =/*(|#|)> where/? is the linear functional on S(W) corresponding to the additive functional /i: W -> R. Proof (a) If a,beW and a n 6 = 0, faAfb = 0. CG 21, A? = A;(C n 6)+#(c\6) in L°°(2l). So

P For every

0 < (fa A/ 6 ) (^e) ^ /«^(c n 6) + / 6 x ( # ) = fi(a n c n b) +/i(bn c\b) = 0, 130

THE SPACE LH^l)

[52

remembering t h a t / a A/ 6 is defined in L#(«) = L«(«)x s L«(«)~ = L~(L°°(2i); R), and applying the formula of 16Eb. As c is arbitrary, / a A/& = 0. Q Thus ^(^a) A ^(#6) = 0 whenever a n 6 = 0 in 21. But now suppose that xAy = 0 in £(9P). Then we can find a, y# > 0 in R and a,beW such that a # a ^ x < \\x\nXa, Px*> ^ V ^ \\y\Uxh [42Ec], Then (a A /?)#(a n 6) ^ x A 2/ = 0, so a n 6 = 0. Now, because <j> is increasing,

0 < 0sA0y < |MU/a A ||y||oo/6 = 0 because/^ A/ 6 = 0 [14Kb]. Since x and t/ are arbitrary, this proves that ^ is a Riesz homomorphism [14Eb]. (b) Now l|0(#a)|| = ||/o|| = fia= p>(xa) f° r any aeSl^ recalling from 52A that ||/o|| = /^a. If next xeS(W)+, then a: can be expressed as Tti where each at ^ 0; since L#(2l) is an L-space,

Finally, for arbitrary a; e

(c) It follows that is one-to-one. P For if x + 0 in AS^SI') then \x\ > 0, so there is an a > 0 and an a + 0 such that a#a ^ \x\. Now ||0||

/(||)

/

0,

so <j)x + 0. Q 52G Lemma Let 21 be a Boolean ring, and v: 21 -> R a completely additive functional. Suppose that ae2l and that va > 0. Then there is a non-zero 6 g a such that vc ^ 0 for every ccj>, Proof Define A: 21 -> R by Ac = v(a n c) for every ce21. Then it is easy to see that A is completely additive, and Xa > 0. Now consider A G 5 ( 2 1 ) X [ 4 2 L ] . Since X+(x X{xa) = Aa > 0,

X+ > 0. So by 32A there is an x > 0 in 5(21) such that X+{x) > 0, 5-s

X~(x) = 0 131

52] MEASURE ALGEBRAS

[for certainly X+ A X~ = 0]. Now by 42Ec there is a b0 e 9t and a /? > 0 such that

o

y ^

^ ii I!

t

( A ) ( A ) Now set 6 = a n 60. If c £ 6, then 0 < ;\;c < #&0, so X~(xc) = 0 and PC = Ac = A(#c) = A+(#c) ^ 0. On the other hand, vb = A60 = X(Xb0) = X+(xbQ)-A-(xb0)

> 0,

so b =)= 0 as required. ^Exercise Prove this lemma without using the construction S( ). [Hint: adapt the argument of 32A so that it refers directly to completely additive functionals on Boolean rings.] 52D Proposition Let (91, /i) be a semi-finite measure ring. Then the canonical map : S(W) -> L#(9t) defined in 52A embeds 5(81') as an order-dense Riesz subspace of L#(8l). Proof We already know that ^ is a one-to-one Riesz homomorphism [52B], so I have only to prove that ^[5] is order-dense. Suppose t h a t / > 0 in L#(9l). Then, because (%/i) is semi-finite, there is an ae W such that/(^a) > 0. P Certainly there is an aoe91 such that/(^a 0 ) > 0. Let

B = {ao\a:aeW}. Then B | in 91. ? Suppose, if possible, that B has a non-zero lower bound 60 in 91. Then there is a non-zero c £ b0 such that c e 9tA So c s ao\c, which is impossible. X Accordingly, B10 in 91. Since fX: 91 -^ R is order-continuous, inibeBf(xb) = 0. In particular, there is an a e W such that fx(ao\a) < fxao- Now

fX > fx(a n a0) = fXa0 -fx(Oo\a) > 0 as required. Q Now let a > 0 be such that/(^a) > aju,a. Set

in Lf (91) = L°°(9t)x, and let v = gx'. 91 -> R, so that v is a completely additive functional and "» = giXfl) =f(xa)-ot/ia 132

> 0.

THE SPACE L^W)

[52

By 52C, there is a non-zero b £ a such that vc ^ 0 for every c^b, i.e. /(# c ) = 9(Xc) + aMa for every c g j , Consequently, for every c e 91,

nc

) =

/ ( ^ ) > fx(c n 6) ^ a/t(c n 6) = Thus/ ^ (x,(j>(xb) in L # [44Bb]. Since a > 0 and 6 + 0 and <j> is one-to As/is arbitrary, this shows (using 15E) that ^[5] is order-dense in L # . 52E Proposition Let (91, /i) be a semi-finite measure ring. On S(W), a Riesz norm || || 1 can be defined by writing \\x\\ x = ju( \x\), where p, is the linear functional on 5(910 associated with the additive functional fi\ W -> R. Let L1(9l,/*) be the normed Riesz space completion of 5(910- Then the map 0: S(W) -> L#(9l) defined in 52A extends to a normed Riesz space isomorphism between L1(9lJ/^) andL # (9l). Consequently 5(910 i s super-order-dense in Proof (a) We know from 52B that S(W), with || ||1? can be identified with the Riesz subspace \ L1 -> h# is also a Riesz space isomorphism. Now 5 is order-dense in L1 because ^[5] is order-dense in Ifl. Consequently it is super-orderdense, because L 1 is an L-space [26Bd]. *52F L#(9t) and L#(W) If (91, /i) is any measure ring, then (W, fif) is a semi-finite measure ring, where jif is the restriction of /i to W, and of course (W)f = W. So we may apply the results above to show that 133

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), the completion of S(W) = S(Wf), is the same as L\W, fif), and is isomorphic to L#(W). Of course L1(2l, /i) can also be identified with the closure of (j)[S(W)] in L#(2I). Thus L#(W) is represented as a subspace of L#(2t). The exercise 52Hc goes into this further. 52G I conclude this section with a remark on the duality between L#(2l) and L°°(2t) when 21 carries a semi-finite measure. Proposition Let (21, /i) be a semi-finite measure ring. Then the norms on L°°(2l) andL#(2l) are dual in the sense that I/I = sup{|/x| :xeL» \\x\\a < 1} V feL#,

INU = sup{|M :feL#, \\f\\ < 1} V Proof The first part is just the definition of the norm on L # = LG0X c L00'. The second part is more interesting. Of course

for every a; e 2/°, and || ||', so defined, is a seminorm. Now if x is a nonzero member of 5(21), x can be expressed as Tn<nocix^i, where (a?)i 0. But

because a n ^ = 0 if i 4= J and ana^ = a. From this it is clear that

INI' ^ la;l = II^IU* Thus || ||' and || ^ agree on 5(21); since || ||' is

continuous with respect to || 1^, they agree on L°°(2l). *Remark We observe that the above result is a property of 21 alone; it is immaterial what the measure ji is, as long as it is semi-finite. It can be shown that this proposition characterizes those measure rings 21 on which a semi-finite measure can be defined. [See 4XFj.] *52H Exercises (a) Let 21 be a Dedekind R a countably additive functional. Then for every a e 21 there is a 6 ^ a such that vb = v+a, and v~b = 0. (If (bn)nelS( is a sequence such that bn^a and vbn ^ v+a — 2~n for each neN, set 134

THE SPACE L^W)

[52

x

(b) Let £ be a Riesz space and F a Riesz subspace of E such that the map x\->&: E -> F* is a Riesz homomorphism from E to F x . Use the method of 32B to show that the image of E is order-dense in F x . Now apply this result with E = S{W), F = L°°(2l) to prove 52D. (c) Let (%ju>) be any measure ring. Let F = {j:geL»(W)9 g(Xa) = 0 V aeW}. Then F is a band in L#(2l). Let /|AH=0

V

/

[cf. 15F]. Now the canonical map 0: 5(2l ) -> L#(9l) embeds S as an order-dense Riesz subspace of E [use the method of 52D]. Consequently <j> extends to an isomorphism between L1(2lJ/^) and E, which sets up a canonical isomorphism d: L#(W) -> E [see 52F]. At the same time, the order-continuous embedding of W in 21 induces a map n': L#(2l) -> L#(W) [45E]. Now TT'# is the identity on L#(W), and the kernel of TT' is F. Thus fr' can be thought of as the projection of L#(2l) onto E - L#(91O ^ LHSl^). with kernel F = Ed [ISP]. ^: L°°(W)X -> L°°(9l)x can also be given by (df)(x) = fsup{fy:yeL«>(W), 0 ^ y ^ x} for every ^GL°°(91)+ a n d / G L#{W)+, or by (^/) Ct«) = sup {/(^): 6 6 «/, 6 s a} for every ae% and/GL#(9l/)+. Notes and comments The reason for maintaining a careful distinction between L\%) and L#(9l), although they are canonically isomorphic in all significant cases, will appear in § 54. They are affected quite differently by homomorphisms between measure rings. The properties of L 1 can also be derived intrinsically, without considering Ift. It is easy to see that fi: S(W) -> R is strictly positive, because [i: W -> R is; so that || [^ = /«(| |) is a norm on S(W). Since of course ,. M n M H . > , ( ..

IIW + |y|||i = Hi+l|y|li for all x, y in 5(2I/), the same is true in the completion of 5, which is therefore an i-space. A less obvious point is the fact that S is orderdense in L1. This is clear from 52E, because <j)[S~\ is order-dense in IP. But it is also a consequence of the fact that, because /JL is completely additive on W [51E], fieS* [42L], so that the || Urtopology on S is 135

52]

MEASURE ALGEBRAS

Lebesgue. I t follows that S' c S\ Now S' is identified with L 1 ' as linear space, and since the positive cone of L1 is just the closure of the positive cone of 5, *S" and L 1 ' have the same ordering. Also, L 1 ' = Lx~, so L 1 can be identified with a Riesz subspace of (Lv)x = *S"X. The result now follows from 32B. However, the isomorphism between L1(9I) and L#(9I) is in itself one of the most important results of measure theory. In its more conventional forms, which will be touched on in § 63, it is the Radon-Mkodym theorem. The normal approach refers to countably additive functionals on cr-algebras of sets and uses 52Ha (the 'Hahn decomposition theorem') instead of 52C. Now an extra hypothesis must be added to ensure that, given a non-zero increasing countably additive functional v: 9t -» R, there is an a e W such that va > 0 (as in the first part of the proof of 52D). But such a hypothesis will always enable us to find a completely additive functional; see, for example, 63J.

53 Maharam algebras If the underlying Boolean ring of a semi-finite measure ring is a Dedekind complete Boolean algebra, then its L00 space turns out to be 'perfect' in the sense of § 33, that is, L00 is isomorphic to L°°xx or L # x . Thus the duality between L00 and L # or L 1 is symmetric in some respects. In many contexts, Maharam algebras appear to be the 'normal' measure rings, and all others 'unnatural'. 53A Definition A measure algebra is a measure ring (91, fi) in which 91 is an algebra, i.e. has a 1. A Maharam algebra is a semifinite measure algebra (91,/^) such that the Boolean algebra 91 is Dedekind complete. 53B Theorem Let (91, fi) be a semi-finite measure ring. Then the following are equivalent: (i) (91,/i) is a Maharam algebra; (ii) ^(91) is a Dedekind complete If-space with unit; (iii) the duality between L°°(9l) and Lx(9l), derived from the isomorphism between Lx(9l) and L°°(9l)x [52E], represents L°°(9l) as

Proof (a) (i) o (ii) We know that 91 is Dedekind complete iff L00 is Dedekind complete [43Da]. If 91 has a 1, then ^1 is the unit of 136

MAHARAM ALGEBRAS 00

[53

a

1/°°. Conversely if L has a unit e, then {a: x ^ e} is bounded above in 21 [43Bd], and its upper bound must be the 1 of 21. (b) (ii) => (Hi) We know that the canonical map from L°° to L°°xx takes L00 onto a solid linear subspace of LQ0XXJ because L00 is Dedekind complete [32C]. Now L°°xx = L#(2t)x is an If-space with unit [26C], and its unit is J, given by jf = ||/|| whenever/ ^ 0 in L # . But for/ ^ 0 in L#, Il/H = / ( # l ) [44Bb]. So J is the image of #1 in L°°xx, and the image of L00 is the whole of L°°xx. Also, we have seen in 52G that L # separates the points of L00, so L00 is isomorphic to LooXX as a Riesz space. At the same time, 52G shows that the norm of L00 is precisely that induced by its embedding in L # '; we recall from 26C that

Lf = L#x.

Thus L00 is identified with L # '. Since L # and L 1 are canonically isomorphic, we can identify L00 with L 1 '. (c) (iii) => (ii) Finally, the dual of L 1 is certainly a Dedekind complete M-space with unit [26C], so if it is isomorphic to L00 the condition (ii) must be satisfied. 53C Definition if pb\ < oo.

A measure algebra (91, fi) has finite magnitude

53D Lemma A Dedekind tr-complete measure algebra of finite magnitude is a Maharam algebra. Proof Of course any measure algebra (21, fi) of finite magnitude is semi-finite, as 21 = W. Now suppose that 0 , mfaeA/ia = 0 0. As v is strictly positive, inf n[A] = 0. This is order-continuous, and of course it is a homomorphism.

Remark The effect of this is that we can apply 54E below, which refers to order-continuous measure-preserving ring homomorphisms between semi-finite measure rings, to any measure-preserving ring homomorphism, by considering its restriction to the ideal of elements of finite measure. 54C Proposition Let (%/i) and (93, y) be measure rings, and n: 91 -> SB a measure-preserving ring homomorphism. Then n:W-^W induces a Biesz homomorphism n: 5(910 -> 5(930 [45B]. Now IITOHX = ll^Hi for every

Proof

xeS(W).

IfaeW, vn(xa) = vx{na) = v{na) = fia = fl(xa).

So 07r=/i: 5(910->R. Now

for every a; e S(W), using the definition of || || x [52E] and the fact that 7T is a Riesz homomorphism. 54D Corollary n: S(W) -> S(W) extends to a norm-preserving Riesz homomorphism n: Lx(9l) -> Proof By definition, L\%) and L x (8) are the completions of 5(810 and 5(930 respectively for the norms || l^. So n has an extension to a norm-preserving linear map from Lx(9l) to L1(93), which is a Riesz homomorphism because the lattice operations are continuous on both sides. 54E Proposition Let (9t,/0 and (93,^) be semi-finite measure rings. Let n: 91 -> 93 be an order-continuous measure-preserving ring homomorphism. Let 0: Lx(9l) -> L#(«) and ft: Lx(93) -> L#(93) be the 140

MEASURE.PRESERVING RING HOMOMORPHISMS

[54

canonical isomorphisms [52E]. Then by 54D and 45E we have a diagram:

Now this diagram commutes, i.e.

Proof

Suppose that ae 9l/. Then (TT'^TT) (#a) is given by:

= *>7r(a n 6) = /*(« n 6)

for every b e 31. So n'ft7rxa = ^ a i n ^#(9l)- But n'ljrn and ^5 are both linear, so they agree on 5(217); and they are both continuous, so they agree on L1(9l). 54F Corollary

Turning this diagram round, we can write

and see that Pnn is the identity on L1(9l). Thus if the norm-preserving Riesz homomorphism n is taken as embedding L1(3l) as a closed Riesz subspace of L1(S8), then Pn is a projection from L1(S8) onto L1(3l). i^ can be defined by saying that, for any 2/eL1(93), Pny is that member of such that (Pny, Xa) =

for every a G 21.


L#(9l) is given by (TT^) (m) = SneM«(m, n) V m e N, 2 e lx(N x N). At the same time, 5XB shows how Lx(9l) and L1(93) can be identified

and

{«/:yeR***, 2 m , n6H ? mre |y(f»,n)| < oo}.

Clearly TT: L^Sl) -> L 1 ^ ) is still given by (nx) (m,n) — x(m)

V m,«eN,a;eL 1 (2l);

and Pn: Lx(58) -> L^Sl) is given by (P.y)(m) = Vm^n^mnV^n)

V meN,

Thus, for y eL^SB), (w-P,y) ("*, n) = ^SrtrfCmiy^.«) V TO, the projection nPv: L1(93) -> L1(S3) 'averages' y down each column.

144

6. Measure spaces The object of this chapter is to show how the concepts of ordinary measure theory may be joined to the more abstract work done so far. The first step is to describe the measure algebra of a measure space [6 ID]; then § 62 and § 63 show how the L00 and L1 spaces of this measure algebra can be identified with the usual function spaces. In view of the importance of Maharam measure algebras [§53], we are naturally concerned to identify the corresponding measure spaces; the search for useful criteria involves us in consideration of a number of special kinds of measure space, which are discussed in § 64. In §65 we examine an important class of function spaces, including the IP spaces, which exemplify the results of Chapters 2 and 3 on locally convex topological Riesz spaces. 61 Definitions and basic properties The most important idea here is the construction of the measure algebra of a measure space [6ID]. The rest of the section comprises a miscellany of definitions and results which will be used later. 61A Definition A measure space is a triple (X, S, ji), where X is a set, S is a er-subalgebra of SPX [i.e. a subalgebra closed under countable unions and intersections], and ji\ 2 -> [0, oo] is a function such that: (i) (10 = 0; (ii) /i(E [}F)= fiE+jiF if E, J F G S and E[]F = 0 ; (iii) if (Enyn€$ is an increasing sequence in 2,

61B Notes (a) For a note on the use of 'oo', see 51A. (b) Observe that I allow X = 0; in this case S = SPX — { 0 } and fi must be the zero function. (c) /i is called a measure on X, and the elements of 2 are called measurable sets. 145

61] MEASURE SPACES 61C Elementary results Let (X, 2, /i) be a measure space. (a) If E, F G S and E ^F, then /iE ^ /iF. (b) HE.FGS,

(c)

If

then/J(JE7 U J1)

neN ^ &ny Sequence in 2,

Proof (a) /«JP = ju,E+/i(F\E) ^ /iE. (b) /*(# U J?7) = fiE+/i(F\E) ^ /iE+fiF. (C)

61D The measure algebra (a) Let (X, S, /i) be a measure space. Write S o for {E:EeX,/u,E = 0}. Consider S as a Boolean ring. Then S o is an ideal of S, by 6lCa and 61Cb [see 41J], and we can form the quotient 21 = S/S o ; 91 is a Boolean algebra and the canonical map i?H-»ir:Z!-^2l is a ring homomorphism. The 1 of % is X\ We recall also that

Em c F' o E\FeJl0 o ji{E\F) = 0 [41J]. Now

W =F' o /^V*7) = MJ 1 ^) = 0 => fiE = /i(E (]F)= tiF.

So fi factors through 21, that is, there is a function fi : 21 -> [0, oo] such that /6JST = /i'E' for every . E G S . Henceforth I shall allow myself to write [i instead of [i for this function on 21. (b) Now the map E i-» E': S -> 21 is sequentially order-continuous. P Suppose that (En}neln j 0 in 2. Since HneN^ belongs to S and is a lower bound for {En:neN} in S, n^eN^n = ^ - ^et a be any lower bound for {E»:neN} in S. Let JS7eS be such that ^7# = a. Then so b 61 y ^a and 6lCc fiE ^ "Lnaii{E\En) = 0 because E' ^E'n for every neN. Thus ^ = 0 and a — 0. As a is arbitrary, ne N 10 in 21. Because the map E\-^E' is a, ring homomorphism, this is enough to show that it is sequentially ordercontinuous. Q (c) It follows that (%/i) is a measure algebra. P 21 is a Boolean algebra and fi is a function from 21 to [0, oo]. (i) If a, 6 e 21 and a n b = 0, 146

DEFINITIONS AND BASIC PROPERTIES [61 let E, J^eS be such that E' = aandi^ = b.A$(E()Fy = E' o F' = 0, /i(E f)F) = O, so /i(a u 6) = ji((E [)F)')= = jiE+jiF

fi{E [) F) =

/iE+/i{F\E)

= ju,a+fib.

(ii) If ae% let S G S be such that E* = a. Then fia = 0 o [iE = 0 o i ? e 2 0 O a = 0. (iii) If neNf a in 21, choose for each neN an EneH such that E^ = an. Set # n = Ui<w-®i- Since the map E\-^E' is a lattice homomorphism, V

Let 0 = \Jn^Gn. By (b) above, 0' = sup n e H 0; = a. So Thus the three conditions of 51A are satisfied and (%/i) is a measure algebra. Q (d) It also follows from (b) that 91 is Dedekind cr-complete. P Let (ari)neK ^ e a n y sequence in 21. For each ne N, choose an EneH such that En = a n . Then H = n we N^n Gs > a n d ^ = i n f n e N ^ i n s - Because the map E\-+E* is a sequentially order-continuous lattice homomorphism, H' = infneNaw in 21. By 411, this is enough to show that 21 is Dedekind cr-complete. Q (e) Now (2t,/£) is the measure algebra of the measure space

*61E Inverse-measure-preserving functions Given two measure spaces (X,TI,JU>) and (Y,T,v), a natural class of functions to consider is given by the following definition. A function / : X -> Y is inverse-measure-preserving if f-1[F]e^ and j^f-^F] = vF for every FeT. Let 21 and 33 be the measure algebras of (X, S, jx) and (Y, T, J>) respectively. If/: X -> F is inverse-measure-preserving, then there is a sequentially order-continuous measure-preserving ring homomorphism n: 93 -> 21 given by 7r[i^'] = (/- 1 !/])* for every Fe T. P (i) We must verify first that n is well-defined. But this is a direct consequence of the fact that/" 1 : T -> S is a ring homomorphism such that / ^ [ - P J G S O for every .FeT 0 . (ii) The same argument shows that n is a ring homomorphism. Now suppose that (bn)ne^ 10 in 93. Choose Fn e T such that F'n = 6W for each 147

61]

MEASURE SPACES

n e N. Now the map J^i-^-F'iT^SBisa sequentially order-continuous lattice homomorphism, so

and v(Dn^K)

= 0- Similarly,

But /if^lCinenFJ = HHneM = °> »O infneN7T&w = 0. As weM is arbitrary, n is sequentially order-continuous. (iii) Finally, it is obvious that n is measure-preserving. Q [See also 61Jc, e.] 61F Further definitions

Let (X, 2, /i) be a measure space.

(a) Following 51Ba, write

2> ^{EiEe^/iE

< oo}.

If 91 is the measure algebra of (X, S,/0, then W = S>/So. (b) (X, S,/*) is semi-finite if its measure algebra is semi-finite [5lBb], that is, whenever EeTt and fiE = oo, there is an F c iJ such that - P G S and 0 < /iF < oo. *(c) A set .EG2 is purely infinite if /^JE7 = oo, and, for every FeZ such that F ^ E, fiF is either 0 or oo. Thus (X, 2,/0 is semi-finite iff there are no purely infinite sets in S. (d) A set A c X is negligible if there is an F e S such that A^F and /^F = 0. Now (i) any subset of a negligible set is negligible, (ii) a countable union of negligible sets is negligible [using 6lCc]. (e) Suppose that O is any property applicable to points of X. I shall write '(£) for almost every teX\ or '$(£) p.p. (t)\ or '® almost everywhere on X\ or < ' !> p.p.' to mean that X\{t: 0(0} is negligible. For instance, if #: X -» R is a function, then 'x = Op.p.' means that {t: x(t) 4= 0} is negligible. Or, if (xn)neJX is a sequence of real-valued functions on X, then L°°(93). By 45F, (TT-1)^ = (7r~17r)A, which is of course the identity on L°°(2l). Since L°°(2l) is order-dense in L°(L,JLL), VU is the identity on L°(Z,ju,). Similarly, UV is the identity on L°(T, v). Thus U is indeed an isomorphism. 62M Exercises (a) If (X, S, /i) is a measure space and (#n>neN ^s a sequence in M(S) such that (xn(t)}ne^ -> o;(^) for every ^ e l , then iu: X -> R is measurable. (b) (i) Let 21 be any Boolean ring, and Z its Stone space. Then 5(21), regarded as a linear subspace of R z , is closed under the multiplication in R z . Consequently L°°(2l) is also closed under multiplication, (ii) Let (X, 2,/^) be a measure space and 21 its measure algebra. Show that Lco(2l) is closed under the multiplication in L°(S, ju,), and that this is the unique norm-continuous bilinear multiplication on LG0(2l) such that ^ x ^ 6 = x(a n b) for all a,be21. Consequently it agrees with the multiplication on LG0(2l) defined in part (i) above. (c) Let (X, S, /i) be a measure space, (i) Show that \uxv\ = \u\x \v\ for all u,veL°(L,fi). (ii) Show that a set A ^ L° is solid i& uxveA whenever ueA and |v| ^ e. (iii) Show that uxv = Oiff|w| A \v\ = 0. *(iv) Show that an element u of L° has a multiplicative inverse iff it is a weak order unit. 158

MEASURABLE FUNCTIONS, L°

[62

*(d) Show that in 62L the Riesz homomorphism /)

is multiplicative. *(e) Let (X, Ti,ju,) be a measure space and E an Archimedean Riesz space. Let F be a super-order-dense Riesz subspace of E and a sequentially order-continuous Riesz homomorphism. Then T has a unique extension to a sequentially order-continuous Riesz homomorphism from E to L°. [Use 17Gc] *(f) Let (X, 2,/*) and (F, T, y) be measure spaces, with measure algebras 21 and 95 respectively. Let n: 21 -> 93 be a sequentially ordercontinuous ring homomorphism. Then TT: £(21) -> 5(93) is sequentially order-continuous [hint: use the technique of 42J/42O], so extends uniquely to a multiplicative sequentially order-continuous Riesz homomorphism from L°(Z,,ji) to I/°(T, v). [Use (e) above.] (g) Let (X, 2,/0 be a measure space. Suppose that 0 in R, {n(un - auo)+: w e N} is bounded above in L°. (h) Let (X, 2,/^) be a measure space. Let E be any Archimedean Riesz space and T: L°(2,/£) -> E an increasing linear map. Then T is sequentially order-continuous. [Use (g) above.] Notes and comments The most important result above is the representation of the L00 space of the measure algebra of a measure space as a solid linear subspace of the L° space. Of course it is this solid linear subspace that is normally thought of as the L00 space of a measure space; so from the ordinary point of view it is this result which justifies the emphasis I have laid on the construction L°°(2l), and also its name. If we look closer at the construction of L°, se see that it does not really depend on the measure. The original space M(2), of course, is defined by the pair (X, S). Now the solid linear subspace M o in 62P depends only on the cr-ideal S o of S, not on anything else about fi. Thus it is not surprising that L° = M/MQ is determined by the algebra 21 = S/So, without further reference to the measure [62L]. The same result is the reason for the notation I use. Since L° is determined by the algebra 2 and the function/^: S -> [0, oo], regardless of how 2 is embedded in 0>X, I write L° = L°(2,/0. Similarly, M is determined, up to a multiplicative Riesz space isomorphism, by the algebra 2. For it is easy to prove 62 J and 62K with M replacing L°, and 159

62]

MEASURE SPACES

therefore to prove an analogue of 62L showing that if 2 and T are cr-algebras of sets which are isomorphic as Boolean algebras, then M(2) ~ M(T). Indeed, every space M(2) can be regarded as an L°. For suppose that X is a set and 2 a cr-subalgebra of 8PX. Define fi: 2 -> [0, oo] by fiE — n if E has n members, where % G N ; = oo otherwise. Then (X,2,/^) is a measure space. Since 2 0 = {0}, 2/2 0 ~ 2 and L°(L,ju,) ~ M(2). Thus it is unsurprising that most of the results of this section [62G, 621, 62J] consist in showing that L° spaces have properties which are a matter of course for M(2) spaces. It is clear that the theorem 62K relies only on the lemma 62J and the Dedekind ^-completeness of L°. A more powerful result along the same lines is given in 64C/D. *Note that we have seen in 5XAb that any Dedekind cr-complete Boolean algebra 91 can be given a measure, so by 611 it is isomorphic to the measure algebra of a measure space. This gives us a construction of L°(9l), which by 62L is well-defined up to isomorphism. The functorial nature of the construction L° is given by 62Mf. 63 Integration The theory of integration on an arbitrary measure space (X, 2,/*) can be developed in many ways. All the customary approaches, however, proceed by defining a space 8H/0 of' integrable' real-valued functions, together with an 'integral', a sequentially order-continuous increasing linear functional J which extends the functional fi: 5(2*) -> R. On S1 a semi-norm || 1^ is defined by writing ||#||i = J|#| for every x efi1, and S 1 is found to be complete, in the sense that every Cauchy sequence has at least one limit. It can then be pointed out that

H = {x:xe21, 1x^ = 0} is a linear subspace of S1, so that the quotient St/H is a Banach space, in fact an L-space. In most systems, all integrable functions are measurable; so, in the language of §62 above, S 1 c M(2) and H = S 1 0 Mo. So the quotient QXIH can be identified with the image of S 1 in L°(2,/^). It is usually easy to show that 5(2*) is || || ^dense in 2 1 , so that its image (identified in 62H with 5(21*)) becomes dense in S1/!*; it follows at once that fiVH is isomorphic to L1(9l) as defined in § 52, where 91 is, as usual, the measure algebra of (X, 2,/*). 160

INTEGRATION

[63

In this book, however, I propose to use this work to demonstrate the power of the general theorems I have introduced so far. We know that S(W) is a super-order-dense Riesz subspace of L1(9l) [52E]; it is easy to see that the natural embedding of S(W) in L° [62H] is an ordercontinuous Riesz homomorphism. So by 62K it extends to an embedding of L1 in L°. I define &1 in reverse as

the characterization of members of S 1 requires care, but proceeds satisfactorily [63D]. We can accordingly regard the integral on S 1 as the linear functional defined by the integral on the i-space L1, i.e. jx = jx' = ||#'||i whenever

x ^ 0 in S1.

The famous convergence theorems are now consequences of the fact that L 1 is an L-space [26HJ-1, 63Ma-d]. I prove without difficulty that L1 is solid in L° [63E]; it follows that uxveL1 whenever ueL1 and veL00. Now, writing (u,v) = jux v9 we have a duality between L 1 and L00; this is of course the usual duality taken between these spaces; we have to check that it agrees with the identification of L1 with LG0X in 52E. On the basis of this, I can use previous results to establish a form of the Radon-Nikodym theorem [63J]. 63A Theorem Let (X,Tt,/i) be any measure space, with measure algebra (91, /i). The embedding of S(W) as a Riesz subspace of L°(S, JLL), described in 62H, extends uniquely to an order-continuous embedding of L1(9l) as a Riesz subspace of Proofs (a) It will help if we regard the embedding of S(W) in L° as induced by a one-to-one Riesz homomorphism T: S(W) -> L°. Since L°°(9l) is solid in L° [62H], L«>(W) is solid in L°°(8l) [45Db], and S(W) is order-dense in L°°(21O [43Bc], the embeddings S(W) ), 0 < y < |a;|} < oo; (ii) {t: t eX, x(t) + 0} has no purely infinite measurable subset [for definition, see 61Fc]. In this case, a = j\x\ djit = \\x'lv Proof

(a) Suppose that xeQ1, i.e. that x EL1, embedded in L° by

63A. Then j\x\ =j\x\ #

= \\xm\\v If yeS(J2)

0 ^ 2 / ' ^ |o; | inL°and

fiy = jy = jy' = \\y'\\i 162

and 0 ^ y ^ \x\, then

INTEGRATION

[63

using 63Ca. So a ^ ||#"||i < oo. On the other hand, let Fe^L be any purely infinite set. Then ju,(F n E) = 0 for every i?G 2', b e c a u s e / ^ n •#) must be either 0 or oo. So (xF)' A (##)' = x(E OF)' = 0 for every J2eS>. It follows that (xF)' A|M| = 0 for every ueS(W) [using 63Cb]. Now S(W) is order-dense in L1, so A = {u:ueS(W), 1

O^u^

\x\}^\x\

1

in L ; as the embedding of L in L° is order-continuous, J. f \x'\ in L°; so by the distributive law [14D], (XF A \x\)m = # F ' A |a?'| = mipu€AxF'

Thus

hu = 0.

ju,{t:teF, \x(t)\ > 0} = 0, 7

and i' cannot be a subset of {t: \x(t)\ > 0}. Thus x satisfies both conditions. (b) Conversely, suppose that x e M(S)+ satisfies the conditions. Let A be the set r Then A f and HZ ' ^Hi ^ aforevery^G^ [using 63Ca again], so v = sup A exists in L 1 [because the Zrspace L1 is Levi and Dedekind complete, by 26B]. Again, the embedding of L 1 in L° being order-continuous, v = sup.4 in L°. Since u ^ x' for every usA, v ^ x' in L°. Let ZEM(?Z)+ be such that a;" = v. Now consider j» = ^ . ^ < ? Suppose, if possible, that/*# > 0. For yeQ, set since ^ = Ur6Q^> there is a yeQ such that /d!^, > 0. Now y > 0 as 2 ^ 0, so Fy c {^: o:(^) =)= 0}. By the condition (ii), Fy is not purely infinite, and there is a measurable set H ^ Fy such that 0 < fiH < oo. Let y — yxH. Then i/ e S(Lf) and 2/ ^ x, so ^/# G^4. Thus y' < sup J. = z . But

^ : «(*) < y(t)} = ju,H > 0,

which is impossible. X So we see that /iF = 0 and x' ^ z'; thus #* = z' eL1 and Also, because the norm on the i-space L 1 is Fatou, ), 0 ^ y ^ x} = oc. Thus the result is true for x ^ 0. In general, it is clear that \x\ and 6-2

163

63]

MEASURE SPACES

x+ satisfy the conditions if x does, and now x = 2x+- \x\ eS 1 ,

while a HI K 111 == IK 11163E Corollaries (a) S 1 is solid in M(2) and L 1 is solid in L°. (b) For any x ^ 0 in fi1, / # = sup{Jy :ye5(S/), 0 ^ y < «}. Proof (a) It is obvious from the criterion in 63D that S 1 is solid in M, and it follows that L1 is solid in L°. (Or use 17Gd.) (b) This is just a restatement of the last sentence in 63D, identifying jy with fly by 63Ca. If (X, £, /i) is a semi-finite measure space, then L1 is

63F Lemma order-dense in

Proof For, if 91 is the measure algebra of (Z,S,/^) then L°°(9t) is order-dense in L° [62H], and 5(91) is order-dense in L°°(8t) [43Bc]. But, if (Z, S,/e) is semi-finite, then S(W) is order-dense in 5(91) [5lGd]. So [using 15E, or otherwise], 5(910 i s order-dense in L° and L1 is orderdense in L°. 63G The duality between L1 and L00 Let (X,S,/^) be a semifinite measure space, with measure algebra 91. In 62H and 63A we have identified both L°°(9t) and L\%) with subspaces of L°(L,/i). liueL1 00 and veL , then \v\ < IML^, where e = #X* [see 62H]. So | U X V | = \u\ X \v\ ^ \u\ X ||v||«x>C = ||v||oo|^|. 1

As L is solid [63E], uxveL1 and Now we already have a duality between L 1 and L00 given by the identification of L 1 with L°°x in 52E. This duality is continuous for the norms of L 1 and L00 and is defined by saying that (X

M

^, 6 G 91.

/

But, given a e 9l and b e 91, because %E x xF = #(# n J7) in M(S) for all E,FeX.

164

So

INTEGRATION

[63

But now it follows that (u, v} = juxv for every u e S(W) and v e S( 21), and therefore for every u e L1 and v e L00, by continuity. Thus the duality between L1 and L00, under which L 1 is a representation of L°°x, is given by (u,v} = \uxv

V w e L1, v eL00,

and this is the way in which I shall generally regard it from now on. *63H For any measure space (X, £,/£), we can identify L1(3l) with L°°(21OX [52F]. Exactly the same arguments show that the same formula (u,v) = juxv gives this duality also. 631 Notation Let (X, 2,/^) be a measure space. Suppose that

xe&(/i) and that # e 2 . Then \x x xE\ < \x\ soxx xEeZ1. We write

regarding x^ as a member of L°°(2l). 63J The Radon-Nikoctym theorem famous theorem into the light of day.

I can now bring this

First form Let (91, ji) be a measure ring. Let v: W -> R be bounded and completely additive. Then there is a unique ^eL 1 (9l) such that pa = (u, xa) for every a e W. Second form Let (X, 2,/£) be a measure space. Let v: 2 ' -> R be bounded, countably additive, and such that vE = 0 whenever ju,E = 0. Then there is an integrable function x: X -> R such that vE = \Ex for every EHJ Proof The 'first form' is just 44Bb (which finds u in L#(W)) and 52E/F (which is the isomorphism between Lx(9l) and L#(W)). The 'second form' follows. You see at once that the condition '/iE = 0 => vE = 0' is simply to ensure that p descends to a function p': W -> R given by p'(E') = PE for every Eeltf. (As usual, 91 is the measure algebra of (X, £,/£).) Then J/ is bounded and additive; in fact it is completely additive. P By 5 ID, it is enough to show that p' is countably additive. 165

63]

MEASURE SPACES

Suppose that i/(^) for every teX. Then jy = l i m ^ ^ j xn. [See 26HL] (e) Let (X, 2, /^) be a measure space and x an integrable real-valued function on X. Let FeTt be such that /iF > 0 and #(£) > 0 for every teF. Then jFx > 0. (f) Let (X, 2,/^) be a measure space, and x: X -> R an integrable function, (i) The map jPh->JF#: S->R is countably additive, (ii) For every e > 0, there is a yeSi^J) such that J |o: — 2/| ^ e. (iii) For every e > 0 there is a set EeHf and a # > 0 such that \\Fx\ < e whenever i ^ S d ^ ^ ) 8 167

63]

MEASURE SPACES

(g) Let (X, 2,/^) be a measure space of finite magnitude. Then L°°(2) c £i(/£), and J: L°°(S) -> R is continuous for || \\n. (h) Let (X, 2,/^) be a measure space, and i c : I - > R a non-negative integrable function, (i) For every a > 0, /i{t: x(t) > a} < oo. (ii) There is a sequence (%n}ne$ in 5(2') such that (xn)nelx t x in M(S); i.e. S(Zf) is super-order-dense in S 1 ^). *(i) Let (X, 2,/^) and (Y, T, p) be measure spaces. Let/: X -» Y be an inverse-measure-preserving function [6IE], (i) If #: Y-> R is measurable, so is xf: X -> R. In this case, J #di> exists i&jxfdju, exists, and they are then equal, (ii) The map x\-+xf: M(T) -> M(S) induces a sequentially order-continuous Riesz homomorphism from L°(T, y) to L°(2,/£), which includes norm-preserving maps from L 1 to L 1 and from L00 to L00. These maps are precisely the maps n described in 54D and 45B, where n is the homomorphism between the measure algebras described in 61E. See also 62Mf. *(j) Let (X,li,/i) be a measure space. For EEILJ and e > 0, let U(E,e) = {x: xeM(Ii)9 p{t: teE, \x(t)\ ^ e} ^ e} c L°(S,/^). Then {U(E, e): EeU, e > 0} is a base of neighbourhoods of 0 for the topology % on L° described in 63K. Hence, or otherwise, show that % is HausdorffifF(X, 2,/^) is semi-finite. *(k) Let (X, 2,/£) be a measure space of finite magnitude, and suppose that (xn}neV is a sequence in M(S). (i) Show that (x'n}neI[ -> 0 in L°(Z,/i) for the topology £ of 63K iff (/i{t:\xn(t)\>e})n€jx->0

V e>0.

(ii) Let / : R -> R be continuous and bounded. Show that there is a function R given by 0(a;') = jfxd/i

V »eM(S),

and that is continuous for S. Notes and comments In Chapters 4 and 5, the relationships 5(910 £ ^(Sl) and 5(910 £ S(9i) £ L°°(9l) were established for any measure ring 91. Now the effect of 62H and 63A is to show that when 91 is the measure algebra of a measure space (X, 2, /i), then all these spaces can be embedded in L0(L,ji). This is, of course, the ordinary way of looking at L 1 and L°°. So the work of 63G/H is essential, since we must be sure not only that the abstract spaces of Chapters 4 and 5 are isomorphic to the usual function spaces, but also that they are in the familiar duality. 168

INTEGRATION

[63

Most of the results in this section are, like 63G, a matter of checking that we really are in the situation we expected. But we nowfindour labour beginning to bear fruit, in a quick proof of the Radon-Nikodym theorem [63J]. The 'first form' is just a restatement of earlier results; the 'second form', the conventional Radon-Nikodym theorem, is a straightforward corollary. It is instructive to take a conventional proof of the Radon-Nikodym theorem [e.g. WILLIAMSON p. 100, theorem 6.3d, or ROYDEN chap. 11, §5] for contrast with the proof here. The 'Jordan decomposition theorem' is 42N; the 'Hahn decomposition theorem' is 52Ha. The proof proceeds very much on the lines of 52D, reading HJ for 21 throughout. It finishes with an argument based essentially on the fact that the space of bounded countably additive functionals on U is an i-space, just as in 52E. You will see that the fact that 2 is a cr-algebra of sets is vital, for the sake of the Hahn decomposition; whereas no such condition is required in § 52. In fact it is used, not in this part of the argument, but in the embedding of L 1 in L° [62K/63A]. *It is clear from 63Mj that the topology of L° can be described without reference to the theory of integration; but this seems an unnecessary refinement. It has usually been studied for measure spaces of finite magnitude [see 63Mk], and various notions of convergence have been introduced in other cases. But I think there can be no doubt that the topology of 63K is the right one for ordinary purposes.

64 Maharam measure spaces Now that we have identified the function spaces L 1 and L00, the work of § 53 can be seen in a new light. I shall define aMaharam measure space to be one with Maharam measure algebra, so that by 53B the dual of L 1 is identified with L00. We find that, because L00 is Dedekind complete, so is L°, and that this leads to striking consequences [64D,64E]. The brief discussion in 53C-53E showed that a product of measure algebras of finite magnitude is Maharam; so a direct sum of measure spaces of finite magnitude is Maharam. Such a measure space I will call decomposable. All important Maharam measure spaces are of this kind. A useful sufficient condition for a measure space to be decomposable is in 641. 64A Definition A measure space is Maharam (sometimes called localizable) if its measure algebra is Maharam, i.e. semi-finite and Dedekind complete [53A]. 169

64] MEASURE SPACES

64B Theorem If (X,H,/i) is a semi-finite measure space, with measure algebra 31, then these are equivalent: (i) (X, 2, /i) is Maharam; (ii) L°°(2t) is Dedekind complete; (iii) L°(L,fi) is Dedekind complete; (iv) the natural duality between L1(3l) and LG0(9l) represents L00 as Proof In 53B it was shown that (i), (ii), and (iv) are equivalent. (Note that of course L°°(2l) has a unit, e = (xX)\) Obviously (iii) => (ii), as L00 is a solid linear subspace of L° [62H]. Conversely, suppose that L00 is Dedekind complete and that A c (L°)+ is non-empty and bounded above. Then for each n e N.

vw = svp{uAne:ueA} 00

exists in L , because L00 is Dedekind complete; as L00 is solid in L°, #w = sup{WAneiueA} in L°. Now sup^L = sup^eNv% exists in L° because L° is Dedekind cr-complete [62G]. Thus (ii) => (iii). 64C We can now prove a stronger version of 62K, which will be useful later, as well as being intrinsically remarkable. As before, I begin with a lemma. Lemma Let (X, 2,/£) be a Maharam measure space, and let A c: L°(£, /i)+. Then either A is bounded above or there is a v > 0 such &# = supwe^i6A&# V i e N . Proof

If ^4 is not bounded above, set un = svipUEAuAne

V neN,

#

where as usual e = (^^) ; these all exist because L° is Dedekind complete [64B]. Then {un: n e N} is not bounded above, so by 62J there is 8bV > 0 such that Jcv = supweNwn A kv < supMe^it Akv ^ Jcv for every jfc e N. 64D Theorem Let (X, 2,/^) be a Maharam measure space and £ an Archimedean Eiesz space. Suppose that F is an order-dense Riesz subspace of E and that T: F -> L°(S,/^) is an order-continuous Riesz 170

MAHARAM MEASURE SPACES [64 homomorphism. Then T has a unique extension to an order-continuous Kiesz homomorphism from E to L°. Proof

Just as in 62K, I use 17B. Suppose that x ^ 0 in E, and set G = {zizeF, 0 < z < x}.

? Suppose, if possible, that T[C] is not bounded above in L°. Then by 64C there is a v > 0 in L° such that kv = mp{hv ATzizeC}

V &eN.

In particular, there is a i/0 e C such that 0 A Ty0 > 0. Now, because E is Archimedean, ?/0 = sup A where A = {y:yeF+,

3 a > 0, y ^ (yo-ax)+}.

7

Asl is an order-continuous Riesz homomorphism, TyQ = sup T[A], and there is a ?/eJ. with w = vhTy > 0. Let a > 0 be such that V < (Vo-oix)+.

Now, just as in 62K, consider, for zeC and heN, zAky^xAky^xA k(yo-ax)+ — x A OLkiar^y^ — x)+ ^ a - 1 ^ + (x - a - ^ o ) ^ A otkict-^Q - x)+

So

kw = kw Akv = kw A supzeCkv A Tz kv ATZ ^ supzeCkTy A Tz

contradicting the fact that w > 0. X Since L° is Dedekind complete [64B], sup T[G] exists. By 17B, T has a unique extension to an order-continuous increasing linear map from E to L°, which by 17Ca is a Riesz homomorphism. *64E Theorem Let (X, S, /i) be a semi-finite measure space. Then it is Maharam iff the usual topology % on L°(S,/^) [63K] is complete, and in this case % is Levi. Proof By 63L, % is Hausdorff. We know it is Lebesgue and locally solid [63K], so if it is complete, L° must be Dedekind complete [24E], and (Z, lt,/i) is Maharam [64B]. 171

64]

MEASURE SPACES

Conversely, if (X, 2,/*) is Maharam, % is Levi. P Suppose that A 0 in L° be such that lev = 8vpU€Au A lev V 1ceN

[64C] Then v = sup^^fe- 3 ^ At> for every & ^ 1. Let p be a continuous Fatou pseudo-norm on L° such that y = pv > 0 [23B or 63K]. Let& ^ 1 be such that k~xA c [u:pu ^ Jy}, i.e./)(&-%) < | y for every . Then -4 fc-%

Ati) =

using the fact that {k~xu f\v:ueA}\v. X Thus ^l is bounded above. As A is arbitrary, % is Levi. Q But we know that % is Fatou [63K] and that L° is Dedekind complete [64B], so by Nakano's theorem [23K], % is complete. 64F Proposition (a) A measure space of finite magnitude is Maharam. (b) A direct sum of Maharam measure spaces is Maharam. Proof (a) follows directly from 53D, since the measure algebra of any measure space is Dedekind o*-complete [61Dd]. Now (b) follows from 53E since the measure algebra of a direct sum is the product of the measure algebras [61H]. 64G Definition (a) The last proposition shows the importance of the following concept. A measure space (X, 2, JLL) is decomposable (sometimes called strictly localizable) if it is (isomorphic to) a direct sum of measure spaces of finite magnitude. From the discussion in 61G, it is clear that (X, S, fi) is decomposable iff there is a partition (Xt}LeI of X such that: (i) XL e 2/ for every i e I; (ii) given a set E c X, E e 2 iff E n XL e 2 for every i e I; (iii) for every (b) A special case of the above [see 64Ha] is the following. A measure space ( X , 2 , / J ) is o-finite or of countable magnitude if there is a sequence {En}neIS in 2* such that X = UneNEn. (c) A weaker property is important to us because of its use in 641 below. A measure space (X, 2, /JL) is locally determined if it is semifinite and, for any set E c X,

EoFeX 172

V

MAHARAM MEASURE SPACES

[64

64H Proposition (a) A measure space of countable magnitude is decomposable. (b) A decomposable measure space is Maharam and locally determined. Proof (a) Let (X, 2, [i) be a measure space of countable magnitude. Let (En}neji be a sequence in 2^ such that X = (JneH^ for each WGN. Then (X n ) n e N is a partition of X, and Xne 2 ' for every

neN. If E ^X, then

Finally, if .EG2, ^

H X,) = li

So <Xn>weN has the required properties, and (X, 2, ju,) is decomposable. (b) By 64F, a decomposable measure space is Maharam; it is therefore semi-finite. So it is obviously locally determined. *64I Proposition Let (X,S,/*) be a locally determined measure space which is complete in the sense of 6lFf. Suppose that there is a disjoint family s/ c 2 such that = 1 = JT in the measure algebra 91 of (X, 2,/^). Then (X, 2,/*) is decomposable. Proof (a) Observe first that if FeHf, there is a countable set <s/0 c j/such that ju,(F\\Jstf0) = 0. P Since ^ is disjoint, the set must be finite for each n e N. So F)

> 0}

is countable. Consequently U^o a n ( i -® = J P\U^o b^h. belong to 2 . Now /i(H (]E) = 0 for every Hesf, so U' n fl" = 0 in 91 for every HGS/. AS sup{IT \Hesf) = 1, JST = 0, i.e. ^ = 0, as required. Q (b) Now suppose that E c X and that E nHeH for every Hes/. 173

64]

MEASURE SPACES

Then 2?e2. P Let FeU. By (a) above, there is a countable set such that /i(F\\Jsio) = O. Now E (\H ftFel, for every , so E n F n U ^ O G S - O n the other hand, 0

J/OCJ/

which is of zero measure; so E n F\\J J / O G S as (X, 2,/^) is complete. Thus E CiFelZ. But i'7 was an arbitrary member of 2'; as (X, 2,/^) is locally determined, i? e 2. Q (c) Moreover, if i ? e 2 , then ^ = ^He^ME H # ) . P (i) Suppose first that EeHf. Then there is a countable J / 0 C si such that fi(E\\J J/ 0 ) = 0. Since J / 0 is countable, iiE = p{E n U ^o) < ^He^0ME oH)^ ZH^M®

n 5).

7

(ii) In general, we know that /^i? = sup {/iF: J? e 2^, F ^ E} because (X, 2,/^) is semi-finite [6lJa]. So IiE = s\ip{/iF:FeIS,

F ^ E}

^^,

e/finite}

because «s/ is disjoint. Q (d) So if we set Ho = X\ \J stf, 88 = si U {flj,} is a suitable partition of X. For by (b) above, # o e 2 , and by (c) /iH0 = 0. Now if JE/ C X and for every £Te^, then .S E 2 by (b) and n £T) = by (c). Thus (X, 2,/^) is decomposable. 64J Exercises (a) Let (X, 2,/^) be any measure space. Let / be the cr-ideal of negligible sets in 0>X [6lFd]. Show that 2' = { = {G:lE,Fe^

E c G c F,/i(F\E) = 0}

is a cr-subalgebra of SPX, and that there is a unique extension of /i to a measure/^'on 2'. Show that: (i) (X, 2', /i') is a complete measure space; (ii) (X, 2,/^) and (X, 2',/^') have isomorphic measure algebras; (iii) M(2') = {x:xeRx, 3 * / E M ( 2 ) , y = a; p.p.}; 174

MAHARAM MEASURE SPACES [64

(iv) £/>(£,/*) ~ L ° ( 2 > ' ) ; 3 ye&(/i), y = a; p.p.}; (v) S V ) = {x:xeRX, (vi) (X, 2',/^') is semi-finite or Maharam iff (X, 2,/^) is; (vii) if (X,S,/G) is decomposable, so is (X,2',/£'). (b) Let (X, 2,/^) be any measure space. Let

2' = {E:EnFeX

V.

Define/: S'-> [0,oo] by Show that: (i) (X, 2',/G') is a locally determined measure space; (ii) if 9t and S3 are the measure algebras of (X, 2,/^) and (X, S ' , / 0 respectively, then 917 £ 93^; (iii) if (X, 2,/J) is semi-finite, then / is an extension of fi; (iv) if (X, 2,/^) is Maharam, then 91 c^ 93, so that L°(S,/0 - L°(2',/0; (v) if (X,2,/*) is complete, so is ( X , 2 ' , / 0 . [Hint for (iv): if # e 2 ' , let J J e S be such that

*(c) Let (X, 2,/^) be a semi-finite measure space, with measure algebra 91. Show that these are equivalent: (i) (X, 2,/0 is of countable magnitude; (ii) 91 is of countable magnitude [53Fa]; (iii) Lc0(9l) has the countable sup property; (iv) L°(L,/i) has the countable sup property; (v) the topology % on L°(S, fi) [63K] is metrizable. (d) Let (X, 2,/^) be a measure space of countable magnitude, and v: 2 -> R a countably additive functional such that vE = 0 whenever fiE = 0. Then there is an integrable function x: X -» R such that jEx = vE for every EeH. *(e) Let (X, 2,/0 be a measure space of countable magnitude and (xn}nen a sequence in M(2). Show that neN -> 0 for the usual topology on L°(2,/0 [63K] iff every subsequence (yn)neN of weH has in turn a subsequence 0 p.p. (t). (f) A direct sum of locally determined measure spaces is locally determined. A direct sum of decomposable measure spaces is decomposable. *(g) Any Maharam algebra is the measure algebra of a decomposable measure space. [Use 53Fc, 611.] *(h) Let (X, 2,ft)be a Maharam measure space such that whenever E c X is such that EnFel, and ju,(E ()F) = 0 for every Fe2>, then . Show that (X, 2,/J) is locally determined. 175

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Notes and comments This section demonstrates the power of the abstract theory we have been studying. It shows that, for a semi-finite measure space, Dedekind completeness of the measure algebra is equivalent to a series of very important properties of the function spaces [64B]. There is an application of 64D in the next section [65A]. It is clear that 64C/D use only the Dedekind completeness of the measure algebra, not the fact that it is semi-finite. This is significant because we know that any Dedekind complete Boolean algebra can be expressed as the measure algebra of some measure space [see the note at the end of § 62]. In fact 64D is characteristic of Dedekind complete Riesz spaces satisfying 64C; these include the C^ spaces of VULIKH [chapter v, §2] or LUXEMBURG & ZAANEN R.S. [§47]. We can see from 64Ja and 64Jb that any semi-finite measure space can be converted into a complete measure space and thence into a complete locally determined space, by adding new measurable sets. In fact some important constructions for measure spaces can be adjusted so as to yield complete locally determined spaces directly [see, for example, 71A]. There do exist Maharam measure spaces which are not decomposable but which are complete and locally determined (and are therefore unaffected by the transformations in 64Ja/b), but these are all thoughly unnatural. This is why 641 is so important; it is clearly a necessary and sufficient condition for a complete locally determined measure space to be decomposable, and it is in practice nearly a necessary condition for a properly constructed measure space to be Maharam. Measures of countable magnitude, which include, for instance, all the ordinary Lebesgue-Stieltjes measures on finite-dimensional Euclidean spaces, are the simplest non-trivial decomposable measures. They have a number of special properties such as those in 64Jc-e.

65 Banach function spaces In §§ 62 and 63 we have identified two of the most important subspaces of L°. A great many other spaces have attracted interest, principally L2 and the other IP spaces. In this section, I show how an important class of normed spaces can be discussed in terms of the concepts I have introduced; specific examples are in 6XD-6XH. I begin with a powerful general result which enables us to identify the dual space Ex for a large proportion of naturally arising Riesz spaces. 176

BANAGH FUNCTION SPACES

[65

65A Theorem Let (X, S, /i) be a Maharam measure space. Let E be an order-dense solid linear subspace of L° (2,/j). Let

F = {w.weLQ^uxweL1

V ueE}.

Then F is a solid linear subspace of L°. Define a duality between E and F by writing.

(uyw} = juxw

\f ueE, weF.

Then this duality induces a Riesz space isomorphism between F and £ x . Proof (a) Because L 1 is a linear subspace of L° [63A] and x is bilinear, F is a linear subspace of L°. Because L 1 is solid [63Ea] and E is a Riesz subspace, F is solid. P Suppose that weF and that \v\ < |w|. Then for any ueE \uxv\ = \u\ x jv| < \u\ x \w\ = |^6 x w\ eLl [using 62Mc(i)]. So u x v e L1. As u is arbitrary, veF. Q (b) Clearly the duality given is bilinear, and therefore induces a linear map T: F -> JE*. Now T[F] c £x# p Suppose first that weF+ and that 0 cz J. 10 in £. Then {wxw;: us A] 10 in L 1 [621]. So inf^e^C?7^) (w) = inf we ^/^ x w = 0, because J is order-continuous on L1. Thus TweEx.

weF, Tw = Tw+-Tw~eEx. Q (c) In fact J7: F->£xisaRieszhomomorphism. T is increasing, as

Now, for any

P It is clear that

(Tw)(u) = Juxw ^ 0 V w,w ^ 0. So suppose that v A w = 0 in F, and that u e E+. Let x,y,ze M(S)+ be such that x' — u, y' — v and z' = w. Examine

in M(S). Clearly ^ ( 0 = x(t) if 2/(^) > 2(Q, and xx(t) = 0 otherwise. So x x i (2/""2;)+ = x x (2/~2;)+> while ^ x (z — y)+ = 0. Now set u± = x\ in L°. Then 0 < u± ^ ^ and % x v = %x(«;--w)+ = \xxx(y — z)+]' = [xx (y — z)+]# = wx (v — w)+ = uxv, while similarly u±xw = u±x (w — v)+ = [a^x (z-~2/)+]' = 0. 177

65]

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So

(Tv A Tw) (u) < (Tw) (ux) + (Tv) (u - u±) = J %X W + J (U — Ux) X V = 0.

As u is arbitrary, Tv A Tw = 0. But this shows that T is a Biesz homomorphism [14Eb]. Q (d) It follows (because E is order-dense in L°) that T is one-to-one. P If weF and w =)= 0, then |w| > 0. So there is a weE such that 0 < w ^ |w|. Now u x |w| > 0, so \Tw\ (u) = T(\w\)(u)

= jux\w\>

0,

and Tw + 0. Q (e) Next, G = T[F] is order-dense in £ x . P Suppose that / > 0 in £ x . Let w0 > 0 in E be such that/& 0 > 0. Now since L00 is orderdense in L° [62H], we know that A = {u:ueL00, 0 < u < ^0}f ^ 0 . So there is a we A such that/w > 0. Let ux = ||w||~1w, so that/% > 0 and ||ux\ oo = 1. Consider the functional g: L00 -> R given by g(v) =f(u1xv)

V veL™.

00

Since the maps V H ^ X V I L - ^ £ and/: £ -> R are both increasing and order-continuous, ge (L°°)x. So by 63G there is a weL1 with Jwxt? = 0, so w > 0. Now suppose that u e E+. Then once again B = {V:VGL00, 0 < v ^ w}f w,

so ButifveB,

{w;xv:?;G£}t^xw. jwxv=f(uxxv)

^f(v) ^f(u),

1

since % ^ e = (^^)*. As L is an i-space, {t^xvrveJS} is bounded above in L1, i.e. wxueL1, and jwxu = suj)veBjwxv

^f(u).

1

So for any %e£, \wx u\ = wx |u\ eL , and weF. As w > 0, I 7 ^ > 0 in Ex, by (d) above. Finally, we have seen that, for any ueE+,

(Tw)(u) = jwxu ^ f(u), so 0 < Tw < / . Thus [using 15E] T[F] is order-dense in £ x . 178

BANAGH FUNCTION SPACES

[65

(f) Thus T is a Riesz space isomorphism between F and the orderdense Riesz subspace G of Ex. Define 8: G -> L° by 8 = T~\ Now 8 is a one-to-one Riesz homomorphism, and 8[G] = F is solid in L° [(a) above], so 8 is order-continuous [17E]. Therefore, S extends to a one-to-one Riesz homomorphism U: Ex -> L° [64D, using 17Cb]. Now E7[EX] c F. P Suppose t h a t / ^ 0 in E x , and consider 4 = { V : V G F + , TV < / }

Then, because G is order-dense in E x and 8 is order-continuous, ^4 f Uf in L°. But suppose that ueE. Then {|w| xv:veA}\\u\ x £7/. JH

(2T)(H)/(H)

So once again |w x Uf\ = \u\ x UfeL1. As % is arbitrary, £7/ eF. Now for a n y / e £ x , 17/ = I7/+ - Uf-eF. Q (g) But now T: F-* Ex and £7: £ x -> F are both one-to-one Riesz homomorphisms, and UT is the identity on F. It follows at once that T is an isomorphism between F and Ex. *65B Corollary The same result applies for any order-dense linear subspace E of L°, whether solid or not. Sketch of proof Just as in part (a) of the proof above, F is certainly solid. Now let G be the solid hull of E in L°. Then it is easy to see that

F = {w:weL°, uxweL1

V ueG},

so that the duality between F and G sets up an isomorphism between F and G x . Now using 17B it is easy to see that every increasing ordercontinuous functional / : E -> R extends to an increasing ordercontinuous functional on G, and therefore corresponds to an element of F. The remainder of the verification that F is isomorphic to Ex is relatively straightforward [cf. 17Gb]. 65G Definition Let (X, £,/£) be a semi-finite measure space. An extended Fatou norm on L°(L,/i) is a function p: L° -> [0,00] such that: (i) p(u + v) ^p(u)+p(v) V u,veL°; (ii) p(ocu) = \a\p(u) V usL°, a e R ; 179

65] (iii) (iv) (v) (vi)

MEASURE SPACES if \u\ ^ \v\ inL°, then p(u) ^ p(v); if 0 [see 65E below]. The conditions (v) and (vi) are designed to eliminate uninstructive complications. For examples of extended Fatou norms, see 6XE-6XG. 65D Theorem Let (X, 2,/^) be a semi-finite measure space and p an extended Fatou norm on L°(Lt/i). Then p has an associate 0, given by ^ = mv^uxw\\i:ueu>,p{u) ^ 1}. 6 is also an extended Fatou norm, and the associate of 6 is now p, i.e. p(u) = sup{||t& x t^||x: ^(t£;) ^ 1 } V ueL°. Note Here I am employing the convention of 6XE, that if v e L°\LX then lv\\x = 00. Proof I t is easy to verify, directly from the fact that || ||x is an extended Fatou norm, that 6 satisfies the conditions (i)-(iv) of 65C. If w > 0, then, because Z> is order-dense, there is a u e Z> such that 0 < u ^ w; if a = p(u), then 6{w) ^ Hwxa- 1 ^! > 0. Thus 6 satisfies condition (vi) of 65C. Before showing that 6 satisfies 65C(v), I shall show that p is the associate of 6. Of course, if p(u0) ^ 1, then 1. Consider

U ^{u'.ueU^piu)

< 1}.

Then by conditions (i)—(iii) of 65C, U is solid and convex. Moreover, if 0 c i c ( 7 and A f v in L1, then by 65C (iv) It follows from 23L (or otherwise) that U is closed for the norm topology on L1. Also, since L 1 is order-dense in L° [63F],

180

BANACH FUNCTION SPACES

[65

and there is a ux eB such that p(u^ > 1, i.e. ux £ U, Now by the HahnBanach theorem there is a continuous linear functional/: L 1 -> R such that/(%) > 1 but \fu\ ^ 1 for every ue U. Recall that (L1)' = (L1)* [260]. Consider | / | in (L1)*. As ?7 is solid,

:H^H}^l

V ueU,

while of course | / | (%) ^ |/(%)| > 1. Now the natural duality between L 1 and L00 represents L 1 as (L°°)x [63G], so it must map L00 onto an order-dense Riesz subspace of (L1)x = (L1)' [32B]. Thus if we know that {ib: we (7} f | / | in (L1)x, so by 16Db there is a we C such that >s/ \ < c We now find that 6(w) < 1. P Suppose that t^GL0 and that yo(^) ^ 1. Once again, because L 1 is order-dense in L°, {vx w:veL1, 0 ^ v ^ \u\}^ \u\xw = \uxw\, so J v x ^ : V G L 1 , 0 ^ v ^ |^|}

= sup{u)(v): vGL1, 0

As t6 is arbitrary, this shows that 6(w) < 1. Q Consequently,

11% x Hli ^ / % x ^ > 1. Thus we see that, for w0 e L°, p(u0) < 1 o ^(ii0) < 1.

Since both /> and <j) satisfy the positive-homogeneity condition 65C(ii), p = 0 in L°, §5(t;) = yo(v) > 0; so there is a weL° such that #(w) < 1 and ||v x w\\t > 0. Then 0 < v A |^| < v and #(# A |^|) < oo, so 6 satisfies the condition (v) of 65C, and is an extended Fatou norm. 65E Proposition Let (X, 2,/^) be a Maharam measure space and p an extended Fatou norm on L°(L,ju,). Then p, restricted to LP, is a Fatou norm, and the topology it induced on LP is Levi and complete. 181

65]

MEASURE SPACES

Proof The conditions (i)-(iv) and (vi) of 65C make it clear that p is a Fatou norm on LP. If A is a non-empty set in (LP)+ which is bounded and directed upwards, then it must be bounded above in L°. P ? For otherwise, by 64C, there is a v > 0 in L° such that

{uAkv:ueA}fkv

V &eN.

Now it follows that, for any & e N, Jcp(v) = p(kv) < $wpueAp{u

A ho) ^ mpUEAp(u)

< oo,

so p(v) — 0; which contradicts 65C(vi). X Q As L° is Dedekind complete [64B] it follows that w0 = sup A exists in L°. Now A f w0, so p(w0) < &wpueAp(u) < oo, and WOELP. Thus A is bounded above in LP. AS A is arbitrary, the norm topology on LP is Levi. It follows by Nakano's theorem [23K], or otherwise [see 65la], that LP is complete. 65F Proposition Let (X, 2,/*) be a Maharam measure space and p an extended Fatou norm on L°(S, /i). Let 6 be the associate of p [65D]. Then Ld = {w:weLQ, wxueL1 VUELP}, so Le may be identified with (LP)X. Proof (a) HUELP and we Ld, let a, = p(u).Ifot = 0, then w = 0 and u x w = 0 G L1. If a > 0, then oo > #(w) ^ ||a"~

HJ

l l l l j

Thus uxweL1 and ||ttx w^ < p(u)6(w). (b) Conversely, suppose that weL° and that wxueL1 for every UELP. Then |w| x |w| = |wx w| el^for every UELP. Set JS = {v:veL^, 0 ^ v < |^|}. For every VEB, VE(LP)', where v(u) = juxv V UELP, ||$|| = sup{|t(u)\ :p(u) < 1} = sup{||^x vWi'.pfa) ^ 1}

and

using the fact that {u: p(u) < 1} is solid. Morever, for any u e LP, *VPveB\Hu)\ < SUVvesj \u\ X V ^ j \u\ X \w\ < 00.

So by the uniform boundedness theorem (since LP is a Banach space, 65E)

182

'

oo > sup vei? ||$|| = mVveBd(v)

= 0(\w\) = 0(w),

BANACH FUNCTION SPACES

[65

since Bf \w\ because U is order-dense in L°. Thus weLd. As w is arbitrary, this shows that

Le = {w.wxueL1

V us LP},

as required. (c) Now by 65A above the natural duality between LP and L$, writing (u, v) = J u x v, represents Le as (LP)X. 65G Corollary If (X, 2, ju,) is a Maharam measure space and p an extended Fatou norm on L°(2, /^), then LP is perfect in the sense of 33A. Proof For if d is the associate of/?, then p is the associate of 6 [65D], so (Z>)xx = (L*)x = £P. 65H Proposition Let (X, S,/^) be a Maharam measure space and p an extended Fatou norm on L°(L,JLL). Then (L^)' = (LP)~, and the following are equivalent: (i) the norm topology on LP is Lebesgue; (ii) whenever 0 c: A\ 0 in Z>, infweJ,yo(it) = 0; (iii) (Lpy = (L/>)X; (iv) whenever (un}nex is a disjoint sequence in (L°)+ and then neH -> 0. Proof (a) (L^)' = (Z>)~ by 25G, because by 65E I> is a Banach lattice. (b) Of course (i) and (ii) are equivalent by definition, and (i) and (iii) are equivalent by 25M. Because the norm topology on LP is Levi [65E], we see that, for a disjoint sequence (un)neJS in (Z>)+, sup neH />(Si< n ^) < oo iff

{un:neN} is bounded above in LP. SO (iv) is equivalent to 'whenever (un}n€}X is a disjoint sequence in (LP)+ which is bounded above, then which by 24J is equivalent to (i). 651 Exercises (a) Let (X, £, /i) be any semi-finite measure space, and p an extended Fatou norm on L°(£, /i). Show that (i) p is a Fatou 183

65]

MEASURE SPACES

norm on 2>; (ii) if (un}neV is a bounded increasing sequence in Z>, then supneHwn exists in L° and belongs to Z> [use 62 J]; (iii) Z> is complete [use 25Na]; (iv) if 0 is the associate of p, Ld = {W.WXUEL1

V UELP},

just as in 65F; (v) the whole of 65H is still true [use (ii) and (iii) above instead of 65E]. *(b) Let (X, 2, /£) be a semi-finite measure space, and/? an extended Fatou norm on L°(L,/i). Give L° its usual topology [63K]. Then the embedding L? £ L° is continuous. Notes and comments The theorem 65A/B gives a general method of describing order-continuous linear functionals on most of the usual function and sequence spaces [see 6XD]; it includes, for instance, the identifications (I00)* = l\ (P) x = I00, and (co)x = I1 in 2XA-2XC, as well as (L°°)x = L 1 [63G] and (L1)* = L00 [64B]. Moreover, it can be shown that if E is any Riesz space such that Ex separates the points of E, then E can be represented as an order-dense Riesz subspace of L°(£,/£)forsomeMaharammeasurespace(X,2,/£) [FREMLIN A.K.S. II]; so that 65B gives a picture of the duality between E and Ex which is in some sense complete. The examples 6XE-6XH give some idea of the kind of spaces that can be generated by extended Fatou norms. They include most of the solid normed function spaces that have attracted interest. In fact, using the representation theorem mentioned above and the method of 6XH, it is not hard to show that if E is any Dedekind complete Riesz space with a Fatou norm inducing a Levi topology, and if Ex separates the points of JE, then E is an Z> space. When the underlying measure space is Maharam, then 65E-65H give a description of these function spaces which in its own terms is fairly complete. They are all perfect and their topologies are Fatou and Levi. The question of when they are Lebesgue is tackled in 65H. For the sake of simplicity, the underlying measure space is assumed Maharam all through this work. But in 651a I list those results which can be extended to arbitrary semi-finite measure spaces. They give a good idea of the way in which, for metric spaces, it is often enough to be able to handle sequences. In 6X11 give a example of a class of spaces which is very similar to the function spaces of this section except that the defining functions are not norms and the topologies are consequently not locally convex. 184

EXAMPLES FOR CHAPTER 6

[6X

6X Examples for Chapter 6 I begin with simple examples of measure spaces; the most important one is of course Lebesgue measure [6XAb, 6XB]. Because Lebesgue measure is diffuse [6XBa], the associated L° space is not locally convex [6XBb]. The next paragraph [6XC] is an examination of a natural inverse-measure-preserving function, showing how the ideas of § 54 are linked to Fubini's theorem. The rest of the section is a discussion of some simple function spaces, based on the ideas of § 65, showing how the results of Chapters 2 and 3 can be applied. 6XA Measure spaces Define /i: 2 -> [0, oo] by fiE — n

(a) Let X be any set, and set 2 = SPX.

if E has n members, where neN;

= oo otherwise. Then (X, 2,/^) is a complete decomposable measure space. As pE = 0 iff E=

0,

its measure algebra 21 is isomorphic to 2, and L°(2,/^) ~ M(2) = R x . L°°(«) - 2>°(2) = t»(X) [4XCa], a n d L 1 ^ ) s £ V ) = P(X) [cf. 5XB]. The usual topology on L° [63K] is precisely the product topology on R* [cf. 1XD]. (b) Lebesgue measure For the construction of Lebesgue measure on R, I refer you to WIDOM, chapter 1, or BARTLE, p. 96, chapter 9; an alternative methodis suggestedin 7XA. Its basic property is that /i[<x,fi] =fi—0Lwhenever a < fi in R. It is complete and of countable magnitude. (c) For further examples, see MTJNROE [chapter in] and Chapter 7 below; also HALMOS M.T., chapter xi, §§57-60, for a description of Haar measure. 6XB Lebesgue measure on [0,1] Now let us take X to be the unit interval [0,1], and 2 the class of Lebesgue measurable subsets of X; let [i be the restriction of Lebesgue measure to 2. Then (X, 2, /i) is a complete measure space and fiX — 1. (a) Suppose that 2?e2 and that [iE > 0. Let n ^ 1 be such that a = rr1 < fiE. Since [0,1] = \Ji R be given by gx = f(x') for every xeM(2,). Let x0 > 0 in M(S) be such that gx0 > 0. For each n e N , observe that i, 2-^(i+ 1)]), n

so there must be an i < 2 such that where Now let

yn = xo x X(i2~~% 2"-(i+1)]). w,

2~n. If we set

so that g(zn) = 2»and/*{r- zn(y) * E == {y: sup ne

Y) = oo},

we see that

.t(y) 4= 0},

Egz

so that Let Then z' ^ zn in L°, so

fiE z

2-< = 0.

= sup^z

for every n e N, which is impossible. X Q *(c) Consequently, if % is the usual topology on L° [63K], then the dual of L° for X is {0} [22D]. In particular, % is not locally convex, though it is complete and metrizable [64E, 64Jc]. For further striking properties of X, see PRYCE U.S.

*6XC An inverse-measure-preserving function Let / be the unit interval [0,1], and let X = I2 c R2; let ju, be the restriction of Lebesgue planar measure to subsets of X, and 2 the domain of ft. Let (Y,T,v) be Lebesgue linear measure on / , as in 6XB above. Define / : X -> Y by writing f(oc,fi) = a for all ot,/3el. Then / is inversemeasure-preserving. Let 21 be the measure algebra of (X, 2,/^) and 95 that of (Y, T, v). Then the measure-preserving ring homomorphism n: S3 -> % of 6IE is given by n = {f_1[E]).

186

=

(Exiy

EXAMPLES FOR CHAPTER 6

[6X

So n: L°°(95) -» L°°(3l) is given by n(x') = (a® 1)' V aJGL°°(T), where 1 = # / and # (g) 2/ is given by (x ® y) (a, /?) = s(a) */(/?) V a, /? G J. Similarly, rr: L^SB) -> L^Sl) [54D] is given by 7r(aO = (s® 1)" V ze&(v). On the other hand, Pn: L\%) -> L^SB) is given by (Pnu, v) = O , TTV) V ^ G ^(31),

t; G L°°(93)

[54F], i.e. [using the representation of the duality between L1(3l) and L°°(9l) given in 63P]. Now Fubini's theorem [MUNROE, §28, or WILLIAMSON, p. 63, § 4.2] tells us that, given xe z(ot) = (x(a,ft)v{dfi) exists and that

^-p.p. (a),

J zdv = J #eZ/£.

So, applying this to x x (y ® 1), (P7Tx',y') = jxx(y®

l)d/i = jy(a)z(a)v(da) =

As this is true for every 2/GL G0 (T), Pnx' = z\ Thus Pn ' averages' members of L1(3l) over vertical lines, in a fashion perfectly analogous to 5XC. 6XD Sequence spaces Suppose, in 6XAa, that X = N. Then M(S), which is isomorphic to L°(S, /i), becomes RN, and we may call its subspaces 'sequence spaces'. A Riesz subspace E of RN is easily seen to be order-dense iff it includes s0, the space of sequences with only finitely many non-zero terms. In this case we may use 65B to identify Ex with {y.xxyel^N)

V xeE},

identifying I1 with S 1 - L1. Observe that (RN)X = s0 and (s0)* = RN. This result also includes the identifications of (/1)x, (/°°)x and (co)x given in 2XA-2XC. 187

6X]

MEASURE SPACES

6XE Extended Fatou norms For any semi-finite measure space (X, 2,/^), we may extend || ||x and || H^ to the whole of L°(2,/£) by writing ||^||1==oo for UEL°\L\ IKHoo = oo for ueL°\L 1 in R. Set q = (p—l^p, so that p ^ + q-1 = 1. Let (X,2,/£) be any semi-finite measure space. If x, y e M(2), x =y ox = y p.p. => \x\* = \y\P p.p. o (\x\*Y = (\y\*>)\ Thus we may define \x'\& to be equal to (|#| p )'. Now we write

M , = (II Mix)1'* for each u e L0(S, /i), where || || x is permitted to take the value oo, as in 6XE. Then || \\p is an extended Fatou norm and its associate is || || r Proof (i) Holder's inequality If a, J3 e R, then p-iaP + q-1/?*^ a/3 [HARDY, LITTLEWOOD & P6LYA,

§2.5]. It follows that if u,v ^ OinL0, |^|a ^ uxv.

Consequently ||t^ x v||x ^ 1 whenever \\u\\p < 1 and ||t>||ff ^ 1 (since p~1 + q~1= 1). So |[^x v||x < H^llpll^llg whenever H^H^ and \\v\\q are finite, for

\\au\\p = |a| \\u\\p V a e R ,

ueL°.

(ii) It follows that

IKIlp = TOp{lk for every ueL°.

xv

lli : llvll« < 0

P We have just seen that \\u\\p ^ suplWuxvliilvWe < 1}

for every u such that \\u\\p < oo, and therefore for every ueL°. If H^IIP = 0, then |^|^ = Osoi^ = 0, and the equality is certainly satisfied. 188

EXAMPLES FOR CHAPTER 6 [6X If 0 < \\u\\p < oo, set v = \u\*!*. Then

II^IU = (II l^i^I[x)a/^ = (II |^l^IU) a/ « = (1KIU)^« = ^ say. So \\fiv\\Q = 1, and using the fact that 1 -f (p/q) = p, so that p — (p/q) = 1. So in this case also the equality is satisfied. Finally, if H ^ = oo, we use the fact that (X, 2,/*) is semi-finite to see that oo = \\u\\p = a u p { \ \ w \ \ p : w e S ( W ) , O^w^ \u\} , 0 ^ w < \ u \ , \\v\\q < 1} (iii) It follows at once that || ||p satisfies the conditions (i)-(iv) of 65C, and it satisfies (v) and (vi) because || \\t does. Similarly || ||g is an extended Fatou norm, and is of course the associate of || ||p. (b) We normally write Lp for the Banach space

{u:\\u\\p)' = (LP)- = (Z>) x may be identified with L«, a t least when (X, S, fi) is Maharam. Thus, in this case, IP is reflexive. * Actually, Lp always has the countable sup property (because L1 does), so is necessarily Dedekind complete, even when (X, £ , / 0 is not Maharam. We can see that the norm topology of Z> is Levi because the norm topology of L 1 is Levi; thus IP is always complete [see 25Na]. The argument of 65A shows that Lq can be thought of as an order-dense Riesz subspace of (Lp)x, which b y 17F is solid. Now, because LQ is a Banach space and the norm on 2> is that induced b y the duality [part (ii) of the proof of (a) above], a %8{Lp, L 5 )-bounded set is || Unbounded; so Z> is Levi for S£S(L^,L«). I t follows from 33B that L* can be identified with (L«) x . Similarly, L« = (L^) x = (2>)', so again Z> is reflexive. 6XG L 2 spaces If in 6XF we set p = q = 2, we get the outstanding special case f Til ro T(\ 5 ^ L2 = {u:ueL°, uxueL1}. The identification between L2 and (L2)' corresponds of course to an inner product, writing j x v V u, veL2. 189

6X]

MEASURE SPACES

We know that L2 is complete; therefore it is a real Hilbert space. In view of its great importance, it is worth noting that many of the arguments of 6XF can be simplified in this special case. *6XH Orlicz spaces Let us consider the properties that the unit ball U of a Banach function space Z> defined by an extended Fatou norm p must have. It is a subset of an L° space which is solid and convex [65C(i)-(iii)], and JTJ £ U [65C(iv)]. If u > 0 in L°, there is a v G U such that 0 < v ^ u [65C(v)], and if cm e U for every a e R, then u = 0 [65C(vi)]. Clearly these properties are necessary and sufficient for U to be {u: p(u) < 1} for some extended Fatou norm p. We may regard 6XF as defining || \\p by Now the function a\-+ap: R+ -> R + is a continuous monotonic function which, for p ^ 1, is convex, i.e. (yot + (l-y)/3)*> ^

yav

+ (1 - y) ft*

whenever a, JS ^ 0 and 0 < y ^ 1. This is clearly the essential property to ensure that {u: J \u\p < 1} is convex. So now suppose that [0, oo] is any convex monotonic nondecreasing function satisfying the following: (i) O is continuous on the left, and O(0) = 0; (ii) there is an a > 0 such that O(a) < oo; (iii) (\x(t)\)dt < 1}. (Because O is permitted to take the value oo, we must be careful to choose x such that O(|a;(£)|) < oo for all t; if this is impossible, then u$U. Because O is monotonic, the function t h-» O( \x(t) |) will always be measurable, using the criterion of 62D.) It is immediate from the definition, because O is convex and monotonic, that U is solid and convex. It is a little harder to see that J U c JJ. But, for any u ^ 0 in U, we may define Qu e L1 by where x' = u. Now if 0 X a sublattice containing 0 and closed under countable intersections. Let A: JT -> R+ be an increasing function such that: (i) A0 = 0 ;

(ii) ifH, KeJfa,ndH(]K= 0, A(H[)K) ^ XH + XK; (iii)

IfF^eJfandH^F, , K X -> [0, oo] by A^^L

for every i c j , Set and let ju, be the restriction of A* to S. The whole of the rest of the proof consists of a demonstration t h a t fi is the required measure. (a) Let us begin with two simple observations. If A, B c: X and A n J5 = 0 , then A^-4 U B) > \*A + A^S. P AHc^ + AH8B ^ sup{A(#[) < A*(il U B), using condition (ii). Q (b)

; K ^ A) sup {\+(K (because A* is obviously increasing) by (a) above. Q (c) Suppose that E,Fe"L. Then X\E and JE? u -F e S. IfEnF = 0 , then A*{E u F) = A*E + \*F. P It is immediate from the defimtion of 2 that if EeH then . NOW let KG JT. Then

194

SEQUENTIALLY SMOOTH FUNCTIONALS

[71

(because .Fe 2) = A*(K nF) + A*((K\F) (]E) + A*((K\F)\E) (because 2Je2, using (b) above with A = K\F) = A*(K n (E u F)) + A*(K\(E u F)) (because FeZ, using (b) above with A = K n (E U F)). As K is arbitrary, 2?u.Fe2. Finally, if E(\F = 0, then, again using (b), . Q (d) So we see that S is a subalgebra of ^ X If KeJf, ^iT = AiT, because A is increasing. So if H,

then

n F) + sup{AiT: As H is arbitrary, J ' G S . Thus X c S and /^ = A^ = A on «yf. (e) Now suppose that {En}neV is an increasing sequence in S and that i£ = UneH^n- T h e n E el> and fiE = ]imn_^ao/iEn. P I shall rely heavily on the fact that (by (c) above) 12 = {F:Fe'L, A*F < 00} is a subring of £PX and A^: U -> R is additive. Let HeX* and e > 0. For each neN, choose an Hnetf such that Hn c j f f ^ and AJ3^ ^ \*{H\En) - 2~ne. Now we know that, for each neN,Hn\jHn+1c:H\En;8o X{Hn U Hn+1) ^ and

A*(Hn+1\Hn)

= A ( ^ U Hn+1) - XHn ^ 2 - e .

So, if we set Kn = C[i^nHi for each n e N , 2e, and 1

=

\Hn+1-A*{Hn+1\Kn)

Let Z = RneH-^n ^ ^ W - W

ek n 0 W t h a t

A*(H\E) > l i n v ^ , A^Zf n En) + AK. To compute AK, consider \+(H\K). If i^G JT and F c Z ? ^ , then T I 6 N I ^ a n d neN -> 0 ; t n i s i s where we use the 7-2

195

71] REPRESENTATION OF LINEAR FUNGTIONALS condition (iv) of the theorem, and this is why the decreasing sequence had to be found. Now we see that AF = limn^A(F n As F is arbitrary, A*(H\K) ^ limn^A*(H\Kn), and XK > \imn^AKn > limn^ So

As e is arbitrary, \E

AH;

as H is arbitrary, Eel,. Moreover, if, above, H £ E, then K = 0; so by equation (*), < 3e- A s e i s arbitrary, lim,,^^ A*(H\En) = 0, so

Now, because H is arbitrary, as required. Q (f) Thus (X, S, fi) is a measure space, and clearly it has the property (a).If2?£2and/£i? > 0, there is a KeX such that K c jJandAif > 0; now fiK = AK < oo. Thus (X, 2,/0 is semi-finite. If 2? c X and # n iTeS for every KeJf, then A^^

n E) + A*

for every Ke2^,80 Eel,; thus S has the property (/?), and as X c £/, it follows at once that (X, 2,/j) is locally determined. HE ^ Fell and / ^ = 0, then for any KeJf, A*{K [)F) = AK,

so Eel. Thus (X, l,/i) is complete. 71B Definition Let X be a set and E a Riesz subspace of Rx. An increasing linear functional / : E -> R is sequentially smooth if (fxn}nen -> 0 whenever (xn)neji is a sequence in E such that 196

SEQUENTIALLY SMOOTH FUNGTIONALS

[71

Remarks In §72 I shall introduce 'smooth' functional. Note that the property of being sequentially smooth is not intrinsic to the functional/and the space E; it depends on the embedding of E in R x . A sequentially order-continuous increasing linear functional on E must be sequentially smooth, but the converse is false. For instance, it is not hard to show that there is no non-trivial sequentially order-continuous increasing linear functional on C([0,1]), but every increasing linear functional on C([0,1]) is smooth. Consequently we must, at several points below, distinguish between bounds in E and bounds in Rx. I shall use the phrase ' inf A = z in R x ' to mean that z(t) = infxeAx(t) for every teX; while 'supA = z in E' would mean that z was the least member of E which was an upper bound for A. 71C Definition Let X be a set and E a Riesz subspace of R x . I shall say that a functional / : E -> R is an integral if there is a measure /i on X such that J xd/i exists and is equal to fx for every xe E. In this case, I shall call/ 'the integral with respect to /i9. An integral is always linear, increasing and sequentially smooth [Lebesgue's theorem, 63Md]. To obtain a converse result, we need an extra condition. 71D Definition Let X be a set and E a Riesz subspace of R x . I shall call E truncated if x A ;yX e E for every x e E. 71E Lemma Let X be a set and E a truncated Riesz subspace of Rx. Let x e E. Then (a) x A a#X e E for every a > 0 (b) if H = {t: x(t) > 1}, then there is a sequence (%n)nejs in E such that (xn}nelj( | xH in R x . Proof

(a) For x A a#X = a(arxx A #X). (b) Set n

xn =

2n(xAXX-xA*nxX),

where otn = 1 - 2~ . 71F Lemma Let (X, £,/£) be a measure space and E a truncated Riesz subspace of Rx such that, for every xe E, Hx = {t:x{t) 197

71]

REPRESENTATION OF LINEAR FUNCTIONALS

Suppose that / is a sequentially smooth increasing linear functional on E such that, for each x e E, y>

XHX}.

Then / is the integral with respect to ji. Proof

(a) Let x e E+, and for each n, i e N set

Set

ynr = 2i 0. Then »B = a ; A 2 ^ I - a ; A 2 ^ I e E

V neN,

x

and (xn}nEl[ f x in R , i.e. <x — xn)nelSi j 0 in R x . So there is an n e N such thsbtfx < fxn + e, because/is sequentially smooth. Now set H = {t:x(t) ^ 2 - ^ } G S , n

and set y = 2 XH, so that xn ^ y. Let z e E be such that z ^ ^J? and

fz^fiH + 2~ne. Then /a: ^ c + X = e+f(2"z)-f(2*z-xn) ^ e + 2nfz-f{2nz-xn)d/i (applying (a) above to 2nz — xn) 2nju,H - f yd/i + j xndju, = 2e+jxnd/i ^ 2e+ (xd/i. 198

SEQUENTIALLY SMOOTH FUNGTIONALS

[71

As e is arbitrary, fx^j xd/i. Thus fx = J xdju, for every x e £ + . Clearly nowfx = j xdfi for every x e E, i.e. / is the integral with respect to fi. 71G Theorem Let X be a set and E a truncated Riesz subspace of Kx. Let/: £ -> R be a sequentially smooth increasing linear functional. Then there is a complete locally determined measure /ionX such that / is the integral with respect to [i. Proof (a) Let $f be the family of all those sets K £ X such that there exists a sequence (xn)nelx in E with xK = mfnelssxn in R x . Define A: J T - > R + b y AZ = i n f { f x : x e E , x ^

x K)

v

KeJlT.

I propose to prove that 3T and A satisfy the conditions of 71 A. (b) Obviously 0 eX*, setting xn = 0 for each WGN. If £T, I G JT, let nGN and (yn)nejsi be sequences in £ such that and #J5L = infneN2/w in R x ; then so H \J Ke Ctif. If (Kn}neJX is any sequence in JT, then choose for each neN a sequence 0. But x ^ (x A inf ij

^ V

= J1 n n»erf: (e-1+1) (*»-*) (0 > i}But { ^ ( e ^ + l ) ^ - « ) ( * ) > 1}GJT for every nsN, by 7lEb. So, because Cft* is closed under countable intersections [(b) above], . NOW suppose that z is any member of £ such that z

x ^ xF >

so

As 2J is arbitrary, XF ^ ( < (1 + e) (e + *H) + Bap{AK:KeJr,

K c

As e is arbitrary, AJ7 < AH + swp{AK:KeX',

K c

and 7lA(iii) is satisfied. (e) Finally, suppose that (Kn}n€n ^ 0 in Jf*. For each WGN, choose a sequence (xni)ieIS in £ such that ^i?^ = inf^eNx^ in R x . Set Then (yn)ne1x | and inf neN ^ = infneN^J5Tn = 0 in R^. So (fyn)nex -> 0. But yw > XKn, so neN -> 0. (f) Thus all the conditions of 71A are satisfied, and there is a complete locally determined measure /i on X extending A. But now we find [using 7lEb again] that ju, satisfies the conditions of 7IF, as all the sets Hx actually belong to X\ So/is the integral with respect to fi. 71H Exercises (a) Show that under the conditions of 71 A, there is only one extension of A to a complete locally determined measure with the property (a). (b) Suppose that X is a set and J f a sublattice of SPX containing 0 . Let A: JT -> R+ be an increasing functional satisfying the conditions (i)-(iii) only of 71 A. Then A has an extension to an additive functional on the subring of ^ X generated by CtiT. 200

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[71

(c) Let (X, T, v) be any measure space. Let $T = 17 and let A be the restriction of v to T*. Show that CrfT and A satisfy the conditions of 71 A, and that the measure ju, produced by 71A is precisely the complete measure on X obtained by applying 64Ja and 64Jb to (X,T, v). (d) Let X be a set, 91 a subring of 0*X, and /i: 91 -> R an increasing additive functional. Show that fi has an extension to a measure on Xiff, whenever {An)neV is a decreasing sequence in 91 such that Oneit^n — & > then (ju>An}nex -> 0. [Hint: either show t h a t

is sequentially smooth, using the method of 42Jc, and apply 71G, or apply 71A directly to a suitable J f ^ 91.] *(e) Let X be a set and E a Riesz subspace of Rx containing %X, Let S o be the smallest c-subalgebra of SPX such that E £ M(S0), i.e. the cr-subalgebra of 0*X generated by the sets {t: x(t) > a} as x runs through E and a through R. Let / : E -> R be a sequentially smooth increasing linear functional. Show that there is a unique measure /£0 with domain 2 0 such that / is the integral with respect to /£0. [Hint: define $T, S and fi as in 71G. Show that JJL and ju,0 agree on Jf, and therefore on So.] Notes and comments I t is a slightly surprising fact that, in 71G, we cannot dispense with the condition that E should be truncated; there is a counterexample in 7XB. Otherwise, it is clear that this theorem goes about as far as it can, since if we have any measure space (X, T, v), we can set E = S1(^), the space of y-integrable functions on X, and that the theorem will then generate the ' complete locally determined extension' of v described in 64Ja-b [cf. 71Hc above]. Readers who have encountered the Caratheodory ' outer measure' construction for measure spaces will recognize that 71A is based on a similar' inner measure' method. Up to a point they are interchangeable (thus 7lHd can be used to tackle product measures, while an outer measure method will give 71G), but outer measures are probably more important for general measure theory. However, I shall not have occasion to use them in this book. The advantage of 71A is that it directly generates a locally determined measure space, which outer measures often fail to do. 71G can also be thought of as a way of extending a sequentially smooth functional to a whole Sx-space; the point about an S1-space being that it is closed under countable suprema and infima, at least of sets which are order-bounded in itself. Indeed there exist proofs 201

71]

REPRESENTATION OF LINEAR FUNCTIONALS

which set out to make this extension directly [see LOOMIS, § 12, pp. 29 et seq.]; this approach has the advantage, when E is not truncated, that equivalents of S 1 and L1 can be constructed even though they may not correspond to any measure. I t is a remarkable fact that the range space R in this result can be replaced by a Dedekind cr-complete Riesz space F iff F is weakly cr-distributive [WEIGHT E.T.]. 72 Smooth functionals: quasi-Radon measure spaces A natural adaption of 7IB leads us to the concept of 'smooth' functional, in which the decreasing sequence of 7 IB is replaced by a directed set. In order to give a description of smooth functionals which will correspond to 71G, we need to consider a special type of measure space. Because its important applications are to functionals defined on spaces of continuous functions, I express the theory in terms of measures on topological spaces, the 'quasi-Radon' measures. 72A Definition A quasi-Radon measure space is a quadruple (X, %, 2, fi), where (X, 2, fi) is a measure space and % is a topology on X such that: (i) (X, 2, fi) is complete and locally determined; (ii) % £ 2 (i.e. every open set is measurable); (iii) if E e 2 and /iE > 0, there is a G e % such that JLLG < oo and fi(E ()O)>0; (iv) fiE = sup{jiF:F ^E,F closed} for every Ee2; (v) if 0 c ^ f in S, then

Remark This definition probably strikes you as undesirably long and technical. I hope that by the end of this section it will be apparent that there are good reasons for each feature. The following theorem is a reason for many of them. 72B Theorem composable.

A quasi-Radon measure space (X,£,2,/£) is de-

Proof I shall apply the criterion of 641. Let us call a set F e 2 supporting if fi(GftF) > 0 whenever G is an open set meeting F. Let iTbe the collection of all disjoint families of 202

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[72

supporting sets of finite measure. By Zorn's lemma, 9£ has a maximal member s/. 1 Now suppose, if possible, that in the measure algebra 51 of (X, 2 , / J ) . Then there must be a non-zero ae% such that a n F' = 0 for every F es/; i.e. there is an i ? e 2 such that fiE > 0 but ju,(E nF) = 0 for every F e stf. (i) Let Goe% be such that /*(?0 < oo and /*((?0 n E) > 0 [72A(iii)]. Every Fes/ is supporting, so

^ = {i^e^^nG 0 * 0} = {F:Fes/,ji(FnG0)>0} which is countable, because /^(?0 < oo and s/ is disjoint, so that

{F:Fes/9/i{F(\G0)>2-^} cannot have more than 2n/iG0 members. So also, because /i(E (]F) = 0 for every Fes/0, fi{E n U^o) = °? a n ( i x = /i(O0 f]E) > 0. Now E1()F = 0 for every JP GS/. (ii) Let ^ = {G:(?G2, ffc^^nffjsO}. Then 0 there is a 6r e & such that /^G ^ /^J? — e. Now n ^ ) ^ /6(G n #i) +MH\G) ^ e. As e is arbitrary, /^(£T n E±) = 0. Set Fo = E±\H; then ^ = [iEx > 0, J i n J1 = 0 for every FGS/, FO C