The Lebesgue-Stieltjes Integral: A Practical Introduction

II

Undergraduate Texts in Mathematics Editors

S. Axler F.W. Gehring K.A. Ribet

Springer New York Berlin Heidelberg Barcelona Hong Kong London Milan Paris Singapore Tokyo

Undergraduate Texts in Mathematics Anglin: Mathematics: A Concise History and Philosophy. Readings in Mathematics. AnglinILambek: The Heritage of Thales. Readings in Mathematics. Apostol: Introduction to Analytic Number Theory. Second edition. Armstrong: Basic Topology. Armstrong: Groups and Symmetry. AxIer: Linear Algebra Done Right. Second edition. Beardon: Limits: A New Approach to Real Analysis. BakINewman: Complex Analysis. Second edition. BanchoffIWermer: Linear Algebra Through Geometry. Second edition. Berberian: A First Course in Real Analysis. Bix: Conics and Cubics: A Concrete Introduction to Algebraic Curves. Bremaud: An Introduction to Probabilistic Modeling. Bressoud: Factorization and Primality Testing. Bressoud: Second Year Calculus. Readings in Mathematics. Brickman: Mathematical Introduction to Linear Programming and Game Theory. Browder: Mathematical Analysis: An Introduction. Buskes/van Rooij: Topological Spaces: From Distance to Neighborhood. Callahan: The Geometry of Spacetime: An Introduction to Special and General Relavitity. Carter/van Brunt: The LebesgueStieltjes Integral: A Practical Introduction Cederberg: A Course in Modem Geometries. Childs: A Concrete Introduction to Higher Algebra. Second edition. Chung: Elementary Probability Theory with Stochastic Processes. Third edition. CoxfLittIe/O'Shea: Ideals, Varieties, and Algorithms. Second edition.

Croom: Basic Concepts of Algebraic Topology. Curtis: Linear Algebra: An Introductory Approach. Fourth edition. Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory. Second edition. Dixmier: General Topology. Driver: Why Math? EbbinghauslFlumlThomas: Mathematical Logic. Second edition. Edgar: Measure, Topology, and Fractal Geometry. Elaydi: An Introduction to Difference Equations. Second edition. Exner: An Accompaniment to Higher Mathematics. Exner: Inside Calculus. FinelRosenberger: The Fundamental Theory of Algebra. Fischer: Intennediate Real Analysis. FlaniganlKazdan: Calculus Two: Linear and Nonlinear Functions. Second edition. Fleming: Functions of Several Variables. Second edition. Foulds: Combinatorial Optimization for Undergraduates. Foulds: Optimization Techniques: An Introduction. Franklin: Methods of Mathematical Economics. Frazier: An Introduction to Wavelets Through Linear Algebra. Gordon: Discrete Probability. HairerlWanner: Analysis by Its History. Readings in Mathematics. Halmos: Finite-Dimensional Vector Spaces. Second edition. Halmos: Naive Set Theory. Hammerlin/Hoffmann: Numerical Mathematics. Readings in Mathematics. HarrislHirstIMossinghoff: Combinatorics and Graph Theory. Hartshorne: Geometry: Euclid and Beyond. Hijab: Introduction to Calculus and Classical Analysis. (continued after index)

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M. Carter

B. van Brunt

The LebesgueStieltjes Integral A Practioallntroduction

With 45 Illustrations

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Springer

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M. Carter B. van Brunt Institute of Fundamental Sciences Palmerston North Campus Private Bag 11222 Massey University Palmerston North 5301 New Zealand Editorial Board

S. Axler Mathematics Department San Francisco State University San Francisco, CA 94132 USA

F.W. Gehring Mathematics Department East Hall University of Michigan Ann Arbor, MI 48109 USA

K.A. Ribet

Mathematics Department University of California at Berkeley Berkeley, CA 94720-3840 USA

Mathematics Subject Classification (2000): 28-01

Library of Congress Cataloging-in-Publication Data Carter, M. (Michael), 1940The Lebesgue-Stieltjes integral; a practical introduction / M. Carter, B. van Brunt. p. cm. - (Undergraduate texts in mathematics) Includes bibliographical references and index. ISBN 0-387-95012-5 (alk. paper) 1. Lebesgue integral. I. van Brunt, B. (Bruce) II. Title. III. Series. QA312.C37 2000 515'.43-dc21 00-020065 Printed on acid-free paper.

© 2000 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Usc in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is f()rbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.

Production managed by Timothy Taylor; manufacturing supervised by Jerome Basma. Typeset by The Bartlett Press Inc., Marietta, GA. Printed and bound by R.R. Donnelley and Sons, Harrisonburg, VA. Printed in the United States of America.

9 8 7 6 543 2 1 ISBN 0-387-95012-5 Springer-Verlag New York Berlin Heidelberg

SPIN 10756530

Preface

It is safe to say that for every student of calculus the first encounter

with integration involves the idea of approximating an area by summing rectangular strips, then using some kind of limit process to obtain the exact area required. Later the details are made more precise, and the formal theory ofthe Riemann integral is introduced. The budding pure mathematician will in due course top this off with a course on measure and integration, discovering in the process that the Riemann integral, natural though it is, has been superseded by the Lebesgue integral and other more recent theories of integration. However, those whose interests lie more in the direction of applied mathematics will in all probability find themselves needing to use the Lebesgue or Lebesgue-Stieltjes integral without having the necessary theoretical background. Those who try to fill this gap by doing some reading are all too often put offby having to plough through many pages of preliminary measure theory. I t is to such readers that this book is addressed. Our aim is to introduce the Lebesgue-Stieltjes integral on the real line in a natural way as an extension of the Riemann integral. We have tried to make the treatment as practical as possible. The evaluation of Lebesgue-Stieltjes integrals is discussed in detail, as are the key theorems of integral calculus such as integration by parts and change of

v

vi

Preface

variable, as well as the standard convergence theorems. Multivariate integrals are discussed briefly, and practical results such as Fubini's theorem are highlighted. The final chapters of the book are devoted to the Lebesgue integral and its role in analysis. Specifically, function spaces based on the Lebesgue integral are discussed along with some elementary results. While we have developed the theory rigorously, we have not striven for completeness. Where a rigorous proof would require lengthy preparation, we have not hesitated to state important theorems without proof in order to keep the book reasonably brief and accessible. There are many excellent treatises on integration that provide complete treatments for those who are interested. The book could also be used as a textbook for a course on integration for nonspecialists. Indeed, it began life as a set of notes for just such a course. We have included a number of exercises that extend and illustrate the theory and provide practice in the techniques. Hints and answers to these problems are given at the end of the book. We have assumed that the reader has a reasonable knowledge of calculus techniques and some acquaintance with basic real analysis. The early chapters deal with the additional specialized concepts from analysis that we need. The later chapters discuss results from functional analysis. It is intended that these chapters be essentially self-contained; no attempt is made to be comprehensive, and numerous references are given for specific results. Michael Carter Bruce van Brunt Palmerston North, New Zealand

Contents

Preface

v

1 Real Numbers 1.1 Rational and Irrational Numbers. 1.2 The Extended Real Number System 1.3 Bounds .

1

1 6 8

2 Some Analytic Preliminaries 2.1 Monotone Sequences 2.2 Double Series . . . . 2.3 One-Sided Limits .. 2.4 Monotone Functions 2.5 Step Functi0ns . . . . 2.6 Positive and Negative Parts of a Function 2.7 Bounded Variation and Absolute Continuity

11

3 The 3.1 3.2 3.3

39

Riemann Integral Definition of the Integral Improper Integrals ... A Nonintegrable Function

11

13 16 20

24 28

29

39 44 46 VII

VIII

Contents

4 The Lebesgue-Stieltjes Integral 4.1 The Measure of an Interval. 4.2 Probability Measures .. 4.3 Simple Sets. . . . . . . . 4.4 Step Functions Revisited 4.5 Definition of the Integral 4.6 The Lebesgue Integral

49 49 52 55 56 60 67

5 Properties of the Integral 5.1 Basic Properties . . . . 5.2 Null Functions and Null Sets 5.3 Convergence Theorems. 5.4 Extensions of the Theory

71 71 75 79 81

6 Integral Calculus 6.1 Evaluation of Integrals .. . . . . . 6.2 Two Theorems of Integral Calculus 6.3 Integration and Differentiation.

87 87 97 102

7 Double and Repeated Integrals 7.1 Measure of a Rectangle . . . . . . . . . . . . . . .. 7.2 Simple Sets and Simple Functions in Two Dimensions . . . . . . . . . . . . 7.3 The Lebesgue-Stieltjes Double Integral . . 7.4 Repeated Integrals and Fubini's Theorem.

113 113

8 The 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

123

Lebesgue Spaces IY Normed Spaces .. Banach Spaces . . . Completion of Spaces The Space £1 . . . The Lebesgue IY . . Separable Spaces. . Complex IY Spaces The Hardy Spaces HP Sobolev Spaces Wk,p .

114 115 115

124 131 135 138 142 150 152 154 .

161

Contents 1· X ------------------------

9 Hilbert Spaces and L 2 9.1 Hilbert Spaces . . 9.2 Orthogonal Sets . 9.3 Classical Fourier Series 9.4 The Sturm-Liouville Problem 9.5 Other Bases for L 2 • • . . . • . 10 Epilogue 10.1 Generalizations of the Lebesgue Integral 10.2 Riemann Strikes Back . 10.3 Further Reading .

165 165 172 180 188 199 203 · 203 · 205 · 207

Appendix: Hints and Answers to Selected Exercises

209

References

221

Index

225

Real Numbers CHAPTER

The field of mathematics known as analysis, of which integration is a part, is characterized by the frequent appeal to limiting processes. The properties of real numbers playa fundamental role in analysis. Indeed, it is through a limiting process that the real number system is formally constructed. It is beyond the scope of this book to recount this construction. We shall, however, discuss some of the properties of real numbers that are of immediate importance to the material that will follow in later chapters.

1.1

Rational and Irrational Numbers

The number systems of importance in real analysis include the natural numbers (N), the integers (Z), the rational numbers (Q), and the real numbers OR). The reader is assumed to have some familiarity with these number systems. In this section we highlight some of the properties of the rational and irrational numbers that will be used later. The set of real numbers can be partitioned into the subsets of rational and irrational numbers. Recall that rational numbers are

1

2

1. Real Numbers

numbers that can be expressed in the form min, where m and n are integers with n :f= 0 (for example ~, -~(= -;5), 15(= lIS), 0(= ~)). Irrational numbers are characterized by the property that they cannot be expressed as the quotient of two integers. Numbers such as e, Jr, and ,Ji are familiar examples of irrational numbers. It follows at once from the ordinary arithmetic of fractions that if r1 and r2 are rational numbers, then so are r1 + r2, r1 - r2, r1 r2, and r1/r2 (in the last case, provided that r2 =I- 0). Using these facts we can prove the following theorem:

Ii,

Theorem 1.1.1 If r is a rational number and x is an irrational number, then (i) r + x is irrational; (ii) rx is irrational, provided that r =I- o. Proof See Exercises I-I, No. 1.

o

A fundamental property ofirrational and rational numbers is that they are both Iidense" on the real line. The precise meaning of this is given by the following theorem: Theorem 1.1.2 If a and b are real numbers with a < b, then there exist both a rational number and an irrational number between a and b. Proof Let a and b be real numbers such that a < b. Then b - a > 0, so ,Jil(b - a) > O. Let k be an integer less than a, and let n be an integer such that n > ,Jil(b - a). Then

o
0 if n = 2m + I, m > 0 so that al = 0, a2 = I, a3 = -I, a4 = 2, and so on. The process of listing the elements ofZ as a sequence can be visualized by following the arrows in Figure 1.1 starting at o. Much less obvious is the fact that the set Q is also countable. Figure 1.2 depicts a scheme for counting the rationals. Th list the rationals as a sequence we can just follow the arrowed path in Figure 1.2 starting at 0/1 = 0, and omitting any rational number that has already been listed. The set

4

1. Real Numbers

I

.

I

I

I

,I

-3/3

I

-2/3 +- -1/3 +- 0/3 +-1/3 . - - 2/3

-3/-2

! -1/2 +- 0/2 +- 1/2 2/2t ! -1/1! 0/1 -+ 1/1t 2/1t -2/1 ! -1/-1 ! .... 0/-1 -+ 1/-1 -+ 2/-1 t -2/-1 ! -2/-2.. -1/-2 -+ 0/-2 .... 1/-2

-3/-3

-2/-3

-3/2 -3/1 -3/-1

-2/2

3/3

t t

3/2 3/1

t 3/-1

t

-+ 2/-2 -+ 3/-2

-1/-3

FIGURE 1.2

0/-3

1/-3

2/-3

3/-3

Counting the rationals

Q can thus be written as

Q= {O,l,

~,-~,-1,-2,2, ~, ~,-~,-~,-~,-3,3, ... }.

The infinite sets N, Z, and Q are all countable, and one may wonder whether in fact there are any infinite sets that are not countable. The next theorem settles that question: Theorem 1.1.4 The set S ofall real numbers x such that 0 < x < 1 is not countable. Proof We use without proof here the well-known fact that any real number can be represented in decimal form. This representation is not unique, because N.nl n2n3 ... nk9999 ... and N.nl n2n3 ... (nk + 1)0000 are the same number (e.g. 2.349999 ... = 2.35); likewise N.999 and N + 1 are the same number. We can make the representation unique by choosing the second of these representations in all such cases, so that none of our decimal expressions will end with recurring 9's. We will use a proof by contradiction to establish the theorem. Suppose S is countable, so that we can list all the elements of S as a sequence:

5

1.1. Rational and Irrational Numbers

Now, each element of this sequence can be represented in decimal form, say

where for all n,j E N, Xnj is one of the digits 0, 1,2, ... ,9. The elements of 8 can thus be written in the form al a2 a3 a4

= 0.XnX12X13X14 .•• , = 0.X21X22X23X24 ••. , = 0 .X31 X32X33X34 ••. , = 0.X41X42X43X44 ••• ,

We define a real number b =

0.mlm2m3m4 .•• ,

where for eachj

E

N,

-/.. 1 , . -- I if x··1J'" mJ { 2 ifxjj=1.

Suppose, for example, that our listing of elements of 8 begins al

= 0.837124

,

a2

= 0.112563

,

a3

= 0.33.3333

,

a4

= 0.258614

,

Then: = 8 :f= 1 X22 = 1 X33 = 3 =f:. 1 X44 = 6 =f:. 1

Xn

so ml so m2 so m3 so m4

= I, = 2, = I, = I,

and so on. The decimal expansion of b therefore begins 0.1211 .... It is clear that 0 < b < I, so that bE 8, and therefore we must have b = aN for some N E N. But by definition, the decimal expansion of b differs from that of aN at the Nth decimal place, so b :f= aN and we have a contradiction. We thus conclude that our original assumption must be false, and 8 cannot be countable. 0

6

1. Real Numbers

It follows at once from this theorem that the set lR is not count-

able. In factI it is also not hard to deduce that the set of all real numbers belonging to any interval of nonzero length (however small) is not countable.

Exercises 1·1: 1. Use the method ofproofby contradiction to prove Theorem 1.1.1.

2. Give examples to show that if Xl and X2 are irrational numbers, then Xl + X2 and XIX2 may be rational or irrational. 3. Since the set of all rational numbers is countable, it follows easily that the set 8* = {x : 0 < X < 1 and X rational} is countable. Thus, if we apply the argument used in the proof of Theorem 1.1.4 to 8* instead of 8 something must go wrong with the argument. What goes wrong? 1

4. (a) Prove that the union of two countable sets is countable.

(b) Use a proof by contradiction to prove that the set of all irrational numbers is not countable.

1.2

The Extended Real Number System

It is convenient to introduce at this point a notation that is useful in

many parts of analysis; care, however should be taken not to read too much into it. The extended real number system is defined to be the set ~e consisting of all the real numbers together with the symbols 00 and -00 in which the operations ofadditionIsubtraction I¥ultiplication and division between real numbers are as in the real number system and the symbols 00 and -00 have the following properties for any l

l

1

l

l

XE ~:

(i)

(ii)

(iii) (iv)

-00

< x


o· I

00

= 00 and (-00) . x = x· (-00) = -00 for any

1.2. The Extended Real Number System

(v) 00 . x = x . 00 x < o·

7

= -00 and (-00) . x = x . (-00) = 00 for any

I

(vi) 00·00

= 00 100·(-00) = (-00)-00 = -00 1and (-00)·(-00) =

00.

The reader is warned that the new symbols 00 and -00 are defined only in terms of the above properties and cannot be used except as prescnbed by these conventions. In particular, expressions such as 00 + (-00), (L.oo) + 00, 00 ·0, 0 . 00, 0 . (-00), and (-00) . 0 are meaningless.

A number a E ~e is said to be finite if a E ~, i.e. if a is an ordinary real number. In all that follows, when we say that I is an interval with endpoints a, b we mean that a and b are elements of ~e (unless specifically restricted to finite values) with a < b, and I is one of the following subsets of ~: (i) the open interval {x

E ~ :

a < x < b}, denoted by (a, b);

(ii) the closed interval {x E ~ : a < x < b}, denoted by [a, b], where a and b must be finite;

(iii) the closed-open interval {x E [a, b), where a must be finite;

~ : a < x < b},

denoted by

(iv) the open-closed interval {x E (a, b], where b must be finite.

~ : a < x < b},

denoted by

Note that although the endpoints of an interval may not be finite the actual elements of the interval are finite. Note also that for any a E ~I the set [a a] consists of the single point a whereas the sets [a a) and (a a] are both empty. The interval (a a) is empty for all a E ~e. The only change from standard interval notation is that intervals such as (-00 1-3]1 (-00 1(0)1 (-2 1(0)1 etc. are defined. (Intervals such as [-00 13]1 [-00 100]1 (-2 100]1 etc. are not.) I

l

l

l

l

l

8

1.3

1. Real Numbers

Bounds

Let 8 be any nonempty subset of R e . A number C E ~e is called an upper bound of 8 if x < c for all x E 8. Similarly, a number d E ~e is called a lower bound of 8 if x > d for all x E 8. Evidently, 00 is an upper bound and -00 is a lower bound for any nonempty subset of~e. In general, most subsets will have many upper and lower bounds. For example, consider the set 81 = (-3, 2]Any number c E ~e such that c > 2 is an upper bound of 81 , and any number d E ~e such that d < -3 is a lower bound of 81 . Note that there is a least upper bound for 81 (namely 2) and that in fact it is also an element of 81 . Note also that there is a greatest lower bound (namely -3), which is not a member of 81 . As another example, consider the set

Here any c > 1 is an upper bound of 82, while any d < 0 is a lower bound. Note that no positive number can be a lower bound of 82 , because for any d > 0 we can always find a positive integer n sufficiently large so that lin < d, and therefore d cannot be a lower bound of 82. Thus 82 has a least upper bound 1 an~ a greatest lower bound o. As a final example, let 83 = Q. Then 00 is the only upper bound of 83 and -00 is the only lower bound. Thus 83 has a least upper bound 00 and a greatest lower bound -00. The following result (often taken as an axiom), which we state without proof, expresses a fundamental property of the extended real number system: Theorem 1.3.1 Any nonempty subset of~e has both a least upper bound and a greatest lower bound in ~e. The least upper bound of a nonempty set 8 c ~e is often called the supremum of 8 and is denoted by sup 8; the greatest lower bound of 8 is often called the infimum of 8 and denoted by inf 8. The examples given above indicate that sup 8 and inf 8 mayor may not be elements of 8; however, in the case where sup 8 or inf 8 is

1.3. Bounds

9

finite, although supS and inf S need not be in S, they must at any rate be Ilclose" to S in a sense that is made precise by the following theorem: Theorem 1.3.2 Let S c ~ be nonempty. (i) If M E ~e is finite, then M = sup S if and only if M is an upper bound of S and for each real number E > 0 (however smal~ there exists a number XES (depending on E) such that M - E < X < M. (ii) Ifm is fil}ite, then m = inf S if and only ifm is a lower bound of S and for each real number E > 0 (however small) there exists a number XES (depending on E) such that m < x < m + E.

Proof We shall prove part (i) and leave part (ii) as an exercise. Suppose M = sup S, where M is finite. Then M, being the least upper bound of S, is certainly an upper bound of S. Let E be any positive real number. Then M - E < M and so M ,- E cannot be an upper bound of S, since M is the least upper bound. Thus there must exist a number XES such that x > M - E, and since we know that M is an upper bound of S, we have M - E < X 0 there exists a number XES such that M - E < X <M. Let Kbe any finite element of~e with K < M. Then M - K > 0, so taking E = M - K we have that there exists an XES such that M - (M - K) < x < M, i.e., K < x < M. Thus K cannot be an upper bound of SI and since -00 is obviously not an upper bound of S, it follows that M must be the least upper bound I

ofS.

0

Exercises 1·3:

1. Give the least upper and greatest lower bounds of each of the following subsets of ~e, and state in each case whether or not they are elements of the set in question:

(a) (c) (e) (g)

{x: 0 < x < S} {x: x 2 > 3} {x: X is rational and x2 < 2} {x: x is rational and positive}

(b) {x: 0 < x < S} (d) {x: ~ > 2} (f) {x:x=3+~, nEN}

10

1. Real Numbers

2. IfS

c

~ehasonlyfinitelymanyelements, sayS

= {Xl,X2, ... ,xn }, then clearly S has both a greatest element and a least element, denoted by max {Xl X2, ... , xn } and min {Xl, X2 Xn }, respectively. Prove: 1

1

sup {Xl X2, inf{x1,x2, 1

::=

max{X1, X2,.· ., Xn },

,Xn }:::::

min{X1,X2, .. . ,Xn }.

1

Xn }

••• ,

3. Prove that if Sl and S2 are nonempty subsets of ~e such that Sl c S2 then sup Sl < sup S2 and inf Sl > inf S2. 4. Let S be a nonempty subset of ~e, and e a nonzero real number. Define S* by S* = {ex: XES}. (a) Prove that if e is positive, then sup S* = e(sup S) and inf(S*) = e(inf S). (b) Prove that if e is negative, then sup S* = e(inf S) and inf(S*) = e(supS). s. Prove part (ii) of Theorem 1.3.2. 1

Some Analytic Preliminaries CHAPTER

Before we can develop the theory of integration, we need to revisit the concept of a sequence and deal with a number of topics in analysis involving sequences, series, and functions.

2.1

Monotone Sequences

Convergence of a sequence on ~e can be defined in a manner analogous to the usual definition for sequences on ~. Specifically, a sequence {an} on ~e is said to converge to a finite limit if there is a finite number a E ~e having the property that given any positive real number E (however small) there is a number N E N such that Ian - al < E whenever n > N. This relationship is expressed by an ~ a as n ~ 00, or simply an ~ a. The number a is called the limit of the sequence. If for any finite number M E ~e there exists an N E N such that an > M whenever n > N, then we write an ~ 00 as n ~ 00 or simply an ~ 00, and the limit of the sequence is said to be 00; similarly, if for any finite number M E ~e there exists an N E N such

11

12

2. Some Analytic Preliminaries

that an < M whenever n > N, then we write an -+ -00 as n -+ 00 or simply an -+ -00, and the limit of the sequence is said to be -00. Let {an} be a sequence of real numbers. The sequence {an} is said to be monotone increasing if an < a n+1 for all n E N, and monotone decreasing if an > a n+1 for all n EN. For example: The sequence 1,2,3,4, ... is monotone increasing. 13'1 ' The sequence 1 '2' 4'1 ....IS monotone d ecreas1ng. The sequence I, 1,2,2,3,3, ... is monotone increasing. The sequence I, I, I, I, ... is monotone increasing and monotone decreasing. The sequence 1,0, I, 0, ... is neither monotone increasing nor monotone decreasing. If"a sequence {an} is monotone increasing with limit.e E ~e, we write an t .e (read /Ian increases to .e"). If the sequence is monotone decreasing with limit.e E ~e, we write an -J, .e (read /Ian decreases to

.e"). We shall frequently be studying sequences of functions. Let lfn} denote a sequence of functions fn : I -+ ~ defined on some interval I c ~. The sequence lfn} is said to converge on I to a function f if for each x E I the sequence lfn(x)} converges to f(x), Le., if the sequence is pointwise convergent. The notation used for sequences of functions is similar to that used for sequences of numbers: specifically, fn -+ f on I means that for each x E I, fn(x) -+ f(x). fn t f on I means that for each x E I, fn(x) t f(x) .• fn -J, f on I means that for each x E I, fn(x) -J, f(x).

The fundamental theorem concerning monotone sequences is the following: Theorem 2.1.1 Let {an} be a sequence on~. (i) If the sequence {an} is monotone increasing, then an t sup{an }. (ii) If the sequence {an} is monotone decreasing, then an -J, inf{an}.

2.2. Double Series

13

Proof We shall prove part (i) of the theorem, leaving the second part as an exercise. Let M = sup{an}. The proof of part (i) can be partitioned into two cases depending on whether or not M is finite.

Case 1: If M = 00, then for any positive real number K, we know that K cannot be an upper bound of {an}, so there exists a positive integer N such that aN > K. Since the sequence is monotone increasing, it follows that an > aN > K for all n > N, and thus an t 00(= M) by definition. Case 2: SUPl?ose M finite and let E be any positive real number. Then

by Theorem 1.3.2 there exists a positive integer N such that M-E < aN <M.

Since the sequence is monotone increasing and has M as an upper bound, it follows that

for all n > N. This implies that for all n > N, Ian

-MI
If(x)

-.el
0 such that f(x) > M whenever 0 < Ix - tl < o. A similar definition can be made for limx~tf(x) = -00. In these definitions x can be either to the left or the right of t, i.e., x is free to approach t from the left or right (or for that matter oscillate on either side of t). Often it is of use to restrict the manner in which. x approaches t, particularly if no information about f is available on one side of t, or t lies at the end of the interval under consideration. For these situations it is useful to introduce the notion of limits from the left and from the right. Such limits are referred to as one-sided limits. The limit from the left is defined as follows: limx~t- f(x) = .e if and only if for any positive real number E there exists a positive real number 0 such that t- 0< x < t

===> 1ftx) - .e I
M.

Example 2·3·1:

Let [ : ~

~ ~

be defined as if x < I, o if x = I, x/2 ifx > 1. -1

[ex)

={

Then limx~l- [ex) = -1 and depicted in Figure 2.5.

limx~l+ [ex)

= 1/2. This function is

-----------------rr,..-----.-----------.-- .

.~

iii

18

2. Some Analytic Preliminaries

fix)

1/2 - --

---------=+---

o

--------~~x

1

------1-1 6 I

FIGURE 2.5

Example 2·3·2: Let f(x) = l/(x - 1) (cf. Figure 2.6). Then limx~l+ f(x) = 00.

limx~l- f(x)

=

-00

and

The definition of a limit can be extended further to consider cases where x ~ 00 or x ~ -00. For example, let a E ~. Then limx~oo f(x) = a if and only if for any positive real number E there exists a number X such that x > X====> If(x) -

al
f(t+). Proof Suppose f is monotone increasing, and let t be any real number. Let m = inf{f(x) : t < x} and M = sup{f(x) : x < t}. Now, f(t) is finite, and since f is monotone increasing, f(t) is a lower bound of {f(x) : t < x} and an upper bound of {f(x) : X < t}. It follows that m and M are finite, and also

(2.1)

M o. By Theorem 1.3.2, there exist Xl and X2, with t < Xl and t > X2, such that m :s f(XI) < m + E and M - E < f (X2) < M. Since f is monotone increasing and m is a lower bound of {f(x) : t < x}, it follows that t < x < Xl

===> m

< f(x) < f(XI) < m

+ E = } If(x) -

ml
f(xt") < f(x 2), and it follows that if Xl, X2 E E are such that Xl < X2, then rX1 < rxz ; thus, we have associated with each X E E a distinct rational number. Since the set of all rational numbers can be listed as a sequence, it follows that the set {rx : X E E} can also be listed as a (finite or infinite) sequence. We can then list the elements of E in the same order as their associated rational numbers. Thus E (if not empty) is either finite or countably infinit~. 0

Although Theorem 2.4.3 places restrictions on the possible set of discontinuities of a monotone fu:p.ction, this set can nevertheless be quite complicated, and one must be careful not to make unjustified assumptions about it. For example, one might guess that the discontinuities of a monotone function. must be some minimum distance apart, but the following example shows that this need not be so. Example 2-4-1: Let f : JR. ---+ JR. be defined as follows: f(x)

=

0, ifx < 0, l/(n + I), if l/(n + 1) < { I, if X > 1.

X

< lin, n

= 1,2,3, ... ,

Figure 2.8 illustrates this function. Clearly, f is monotone increasing. It can be shown that f(O+) :t= (see Exercises 2-4, No.3), so f has jump discontinuities at the countably infinite set of points {I, ~, ~, and is continuous ait all other points. In fact, unlikely

°

i, ...}

.i

24

2. Some Analytic Preliminaries

f{x)

1

0

I 1 1

I I 1 1

1/2

1 1

I

I

1

1

1/3 1/4

•

0

o---e

.--

~

-I

1

1

I

1

1

1

1

1

I

1

1

I

1

1

I

0 - - -1/4 1/3 1/2

X

1

FIGURE 2.8

as it may seem, it is possible to construct a monotone increasing function that is discontinuous at every rational number! Exercises 2-4: 1. Prove part (ii) ofTheorem 2.4.1, by showing that iff is monotone decreasing, then -f is monotone increasing, and then applying part (i). 2. Prove part (ii) of Corollary 2.4.2.

3. Prove that f(O+) = 0 in Example 2-4-1.

2.5

Step Functions

Let I be any interval. A function e : I ---+ JR. is called a step function if there is a finite collection {h, h, ... ,In} of pairwise disjoint intervals such that S = hUh U ... U In C I and a set {GI, G2, ... ,Gn}

.-------------------,r1'",---------..--------.----- .

2.5. Step Functions

25

6(x) 1= [a, b) C3 - - - - - - - - - - , I

C2 - - - - CJ=C4 - - -

'9

S =II

U

=sum of hatched areas with

A((])

~'/~9/;+/' ;or/I/.r01

I2 U I3 U 14 u IS = I

appropriate signs b

/1

X

:~:

a

cs - - - - - - - - - - - . ., - - -() ~~e(

II

6 )

H

I2 I3 14

IS

FIGURE 2.9

of finite, nonzero real numbers such that (](x) = {Gj' ~f x E Ij, j = 1,2, ... , n, O,lfxEI-S.

In other words, (] is constant and nonzero on each interval Ij, and zero elsewhere in I. The set S on which (] is nonzero is called the support of (]. Note that S may be empty, So that the zero function on I is also a step function. Figure 2.9 illustrates some possible step-function configurations.

"_------------~---------_._"---"" Ii

26

2. Some Analytic Preliminaries

If the support of a step function 8 has finite total length, then we associate with 8 the area A (8) between the graph of 8 and the x-axis, with the usual convention that areas below the x-axis have negative sign (we often refer to A (8) as the Ilarea under the graph" of 8). Thus 4(8) exists for the step function 8 in Figure 2.9-2, but not for that in Figure 2.9-1. If 81 , 82 , ... , 8m are step functions on the same interval I, all with supports of finite total length, and if aI, a2, ... ,am are finite real numbers, then the function 8 defined by m

8(x) =

L:~8iCX) j=d

for x E I is also a step function on I. The support of 8 has finite length, and m

A(8) = l:~A(8j). j=l

The fact that 8 is also a step function is a rather tedious and messy thing to prove in detail, but an example should be sufficient to indicate why it is true. E~ple 2-5-1:

Let 81 , 82 : [0,3) ---+ JR. be defined by _ {I, if 0 < x < 2, 81 (x) 2, if 2 < x < 3, (cf. Figure 2.10). Let 8 = 281

-

(h(x) = {-I, I,

if 0 < x < I, if 1 < x < 3

82. Then

3, ifO<x f(xz) whenever Xl < Xz (xl, Xz E I); in either case we say that f is monotone· on I. A very slight modification of the appropriate part of the prootf of Theorem 2.4.1 shows that iff is monotone onI, where I is an interval with endpoints a, b, thenf(t-) and f (t+) exist and are finite for all t such that a < t < b, and also f(a+) and f(b-:-) exist (but are n!ot necessarily finite). Furthermore, f(a+) and f(b-) are both finite iif and only if sup{f(x) : X E I} and inf{f(x) : X E I} are both finite. Lemma 2.7.1 Let I be an interval, and f : I -+ ~ a function of bounded variation on I. For any X E I, denote by Ix the ~nterval {t : t E I, t < x} C I. Then (i) 0 < V(f, Ix) < V(f, I) for all ~ E I; (ii) the function g : I -+ JR defined by g(x) = V(f, Ix) for all x E I is monotone increasing on I. Proof Part (i) follows at once trom the result proved in Exercises 1-3, No.3, and the fact that any partial subdivision of Ix is also a partial subdivision of I. To prove part (ii), let XI, Xz E I be such that Xl < Xz. Then I X1 C Ix2 , so any; partial subdivision of I X1 is also a partial subdivision of I x2 . Thus V!(f, IxJ < V(f, IxJ, which proves that g is monotone increasing on I. 0

Theorem 2.7.2 Let I be any interval. Then a funcmon f : I -+ JR has bounded variation on I if and only iff can be expressed as a difference

f = hI -

hz,

where the functions hI, hz : I -+ JR are both monotone increasing on I, and sup{h l (x) : X E I}, inf{h l (x) : X E I}, sup{hz(x) : x E I}, inf{hz(x) : x E I} are all finite. Proof We prove first that iff has,bounded variation on I, then it can be represented by the difference hI - hz, where the h k are as claimed

i i

34

2. Some Analytic Preliminaries

in the theorem. Suppose f has bounded variation on I. For each x E I, define Ix as in Lemma 2.7.1. Define hI, h z : I ~ JR by hI (x) = V(f, Ix) and hz(x) = V(f, Ix) - f(x) for each x E I. Then certainly f = hI - hz, and Lemma 2.7.1 shows that hI heas all the required properties. Also, if Xl, Xz E I are such that Xl < Xz, then hz(xz) - hZ(XI)

= V(f, Ix2 ) -

V(f, IxJ - [f(xz) - f(XI)].

Now, if S is any partial subdivision of lXI' then S* partial subdivision of I x2 , and Vs(f, IxJ

+ If(xz) -

f(XI)'1

= Vs*(f, Ix2 )
I, we have that

j,

1

= l/x. In any closed interval

c-1 dx. = [logx]~ = log e. x

Since log e -+ 00 as e -+ not exist, i.e., it diverges.

00,

the improper integral f1°O l/x dx. does

Although the definition of the Riemann integral can be extended to open or semiopen intervals, many of the results concerning the Riemann integral over a closed interval do not carry over in the extension. Example 3-2-2 indicates that continuity on a semiopen interval does not guarantee the existence of the improper integral.

46

3. The Riemann Integral

Examples 3-2-2 and 3-2-4 show that monotonicity does not imply Riemann integrability when the interval is not closed. If f : [a, b] -+ JR is Riemann integrable over [a, b], then it can be sI:own that If I is also Riemann integrable over [a, b]. For imf(x) dx. may converge proper integrals, this is no longer true, i.e., but If(x) I dx. may diverge. If f(x) dx. and If(x)1 dx. both conver~e, then the improper integral is called absolutely convergent. If fa f(x) dx. converges but f: If(x)1 dx. diverges, then the improper integral is called conditionally convergent. The integral in Example 3-2-3 is absolutely convergent. The next example requires more familiarity with improper integrals than assumed heretofore, but it provides a specific example of a conditionally convergent integral.

f:

f:

f:

f:

E~ple

3-2-5: Letf: [Jr, (0) -+ JR be defined by f(x) = (sinx)/x. Over any closed interval of the form [Jr, e], e > Jr, the functionf is Riemann integrable, and (anticipating integration by parts) we have

l

7(

c

sinx dx. = [_ COSX]c x x

7(

+l

7(

c

co~x dx.. x

Now, Icosx/x 2 1 < 1/x 2 for all x E [Jr, e], and it can be shown that limc-*oo cos x/x 2 dx. exists, since limc-*oo 1/x 2 dx. exists (the comparison test). On the other hand, it can be shown that limc-*oo Isinx/xldx. does not exist.

f:

f:

f:

The definition of the Riemann integral can thus be extended to intervals ofintegration other than closed intervals by using improper integrals. The modern approach, to be described in the next chapter, works with arbitrary intervals from the start, leading to a tidier theory, but this is a relatively minor improvement. A more fundamental weakness of Riemann's approach is revealed in the next section.

3.3

A Nonintegrable Function

Theorems 3.1.4 and 3.1.5 indicate that the class of Riemann integrable functions is a large one. In fact, it can be proved that if f : [a, b] -+ 1R is bounded on [a, b] and the set of all points of dis-

3.3. A Nonintegrable Function

47

continuity of f in [a, b] is either empty, finite, or countably infinite, thenf is Riemann integrable over [a, b]. However, if the set of points of discontinuity off is infinite but not countable, then f may not be Riemann integrable, as the following example illustrates. Example 3-3-1: Letf: [0,1] -+ JR be defined by

_ {I, if x is rational, x #- 0, I, f(x).0, if x is irrational or x = 0 or x

= 1.

Suppose that f were Riemann integrable over [0,1]. Then taking E = ~ in the definition of Riemann integrability, there must exist step functions g, G : [0,1] -+ JR such that g < f < G on [0,1] and A(G) - A(g) < ~.

Now, we have seen that any interval of nonzero length contains infinitely many rational numbers and infinitely many irrational numbers. Thus we must have g(x) < 0 and G(x) > 1 for all but a finite number (possibly zero) of points x E [0,1]. The values of g(x) and G(x) at a finite number of values ofx do not affect the values of the areas A(g) and A(G), so we must have A(g) < 0 and A(G) > 1. Thus A(G) - A(g) > I, which contradicts A(G) - A(g) < ~, and so f cannot be Riemann integrable over [0,1]. Note thatf is discontinuous at every point in the interval [0, 1]The reader might rightly ask why we should be concerned that this rather peculiar function does not have an integral in the Riemann sense. The reason is connected with the following concern: Suppose the functions fn : [a, b] -+ JR are Riemann integrable over [a, b] for all n = 1,2, ..., and fn -+ f on [a, b]. It is natural to hope that the property ofintegrability II carries over to the limit," so that we can be sure thatf is also Riemann integrable over [a, b] (and that the integral of fn tends to the integral off as n tends to 00), but this may not be the case. This concern is important, because the solutions to many problems in the calculus such as differential equations are often obtained as the limit ofa sequence of successive approximations. Unfortunately, there are sequences of functions that are Riemann integrable but that converge to a function that is not. It is necessary to find only one counterexample to destroy our hopes.

--------------_.--.-"..------------_ _--_

.

48

3. The Riemann Integral

We will now show that the nonintegrable function defined in the previous example is the limit of a sequence of Riemann integrable functions. We know that the rational numbers in (0,1) form a countably infinite set, so we can write the set of rationals in (0, 1) as {rl, r2, r3, ...}. For each j = 1,2, ... we subdivide [0,1] into three subintervals: Ijl = [0, rj), Ijz = [rj, rj], and Ij3 = (rj, 1]. We then define ej:[O,I]-+JRby ej(x) =

Now, by

{

0,

if x

I, 0,

~fx E lfx E

ej is a step function for each j =

E

Ijl' Ijz, Ij3.

I, 2, .... Define fn : [0, 1] -+

~

n

fn = Lej. j=l

Now, each fn is also a step function, and is therefore Riemann integrable over [0, 1]. Furthermore, it is evident that fn(x) = {O, I,

~x is irrational or x E {O, I, rn+l, rn+2·· .}, lfx E {rl' rz, ... , rn} .

Thus if x E [0,1] is irrational or x = 0 or x = I, thenfn(x) = 0 = f(x) for all n = 1,2, ..., and so fn(x) -+ f(x) as n -+ 00. If x E [0,1] is rational, say x = rN (N = 1,2, ...), then fn(x) = 1 = f(x) for all n > N, so again fn(x) -+ f(x) as n -+ 00. We have therefore established that the Riemann integrable sequence of functions fn converges to the nonintegrable functionf on [0, 1]. This example shows that integrability in the Riemann sense does not always carry over in the limit. It can be shown that it does under certain conditions, but these conditions are rather complicated. Because the modern theory (as we shall see) allows for a wider class of integrable functions, the conditions under which integrability carries over to the limit are much simpler and easier to use. This is one of the most important ways in which the modern theory is an improvement over the older one.

.- ---------------------"r-----------------.----.,....

I

CHAPTER

The LebesgueStieltjes Integral

We now proceed to formulate the definition of the integral that we are going to study. It results from combining the ideas of two people. The French mathematician Henri Lebesgue (1875-1941), building on earlier work by Emile Borel (1871-1956) on the measure of a set, succeeded in defining an integral (the Lebesgue integral) that applied to a wider class of functions than did the Riemann integral, and for which the convergence theorems were much simpler. The Dutch mathematician Thomas Stieltjes (1856-1894) was responsible for the notion of integrating one function with respect to another function. His ideas were originally developed as an extension of the Riemann integral, known as the Riemann-Stieltjes integral. The subsequent combination of his ideas with the measure-theoretic approach of Lebesgue has resulted in a very powerful and flexible concept of integration.

4.1

The Measure of an Interval

Let a : JR -+ JR be a monotone increasing function, and let I be an interval with endpoints a, b. We define the a-measure of I, denoted

49

50

4. The Lebesgue-Stieltjes Integral

by J.La(I), as follows: JLa([a, b])

= a(b+) -

a(a-),

JLa((a, b]) = a(b+) - a(a+), JLa([a, b)) = a(b-) - a(a-),

and if a < b,

The lI open interval" (a, a) is of course the empty set, and we define JLa((a, a)) to be zero for any a E JRe. The intervals (a, a] and [a, a) are also empty, but in those cases the fact that their a-measure is zero follows from the general definition, and need not be specified separately. It follows easily from Theorem 2.4.I(i) and Corollary 2.4.2(i) that JLa(I) > 0 for any interval I, and that if I and J are intervals with I C J, then JLa(!) < JLaU)· If a and b are finite, and a is continuous at both a and b, then we have a(a-) = a(a+) = a(a) and a(b-) = a(b+) = a(b), and so J.La(I) = a(b) - a(a) in all four cases. In particular, if a(x) = x for all x E JR, then JLa(I) = b - a is the ordinary length of the interval I. In general, the a-measure of an interval is just the change in the value of a over the interval in question; it can be thought of as a generalization of the notion of length. Example 4-1-1: Let a : JR -+ JR be defined by

a(x) =

" if x < I, - 2x + 2, if 1 <x < 2, if x = 2, if x > 2 x+2,

0, x2 3,

(cf. Figure 4.1). Then: JLa([I, 2]) = a(2+) - a(I-) = 4 - 0 = 4, JLa((l, 2]) = a(2+) - a(l+) = 4 -1 = 3, JLa([I, 2)) = a(2-) - a(l-) = 2 - 0 = 2,

------------------,r-----.----------"---"",,

I

4.1. The Measure of an Interval

a(x) 4

51

/ I I

•

3

I I I

~I r1 ? 01

2

>x

FIGURE 4.1

= a(2-) /La([2, 3]) = a(3+) /La((2,3)) = a(3-) /La([2, 2]) = a(2+) /La(( -1,3)) = a(3-) -

= 2 -1 = I, a(2-) = 5 - 2 = 3, a(2+) = 5 - 4 = I, a(2-) = 4 - 2 = 2, a( -1 +) = 5 - 0 = 5,

/La([ -8,~]) = a(~ +)

a( -8-)

/La((I,2))

-

a(1 +)

= 0 - 0 = O.

It can be seen from these examples that the a-measure of an interval

takes account of a jump in the value of a at an endpoint if and only if that endpoint is included in the interval. Note also that it is the left- and right-hand limits of a at the endpoints that determine the measure, not the value of a at the enc;lpoints. Note finally that, as the following examples illustrate, an interval that has one or both endpoints infinite may have, but does not necessarily have, infinite measure:

= a(oo-) - a(2-) = 00 - 2 = 00, /La(( -00,00)) = a(oo-) - a(( -00)+) = 00 - 0 = 00, /La(( -00,2]) = a(2+) - a(( -00)+) = 4 - 0 = 4. /La([2, 00))

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _, - - - - - - - - - - - - - - - - -r-

I

52

4. The Lebesgue-Stieltjes Integral

Exercises 4-1: 1. Let a : JR -+ JR be defined by if x < 0, if x = 0, a(x) = I, { 3 -e -x , if x > o. X,

(a) Sketch the graph of a. (b) Find /La((O, 1)), /La([O, 1]), /La(( -1,1)), /La([O, 0]), /La(( -00, I)), /La((O, (0), /La([O, (0)).

2. Let a : JR -+ JR be defined by

a(x)

=

0, I, 4, 6,

ifx < 0, if 0 < x < I, if 1 <x < 2, ifx>2.

(a) Sketch the graph of a. (b) Find /La([ -1,2)), /La ((1 , (0)), /La(( -00,4)), /La((O, 2]), /La((1/2, 3/2)), /La ((1 , 3]), /La ((1 , 3)).

4.2

Probability Measures

A particularly important type of measure arises when the function a is a probability distribution function. In this case, the variable x is referred to as a random variable, and for each real number X, the value a(X) is the probability that the random variable x has a value • no greater than X: a(X) = P(x < X).

The corresponding a-measure is then called a probability measure, and has the property that for any interval I, /La(I) = P(x

E

I).

Any probability distribution function must necessarily satisfy the conditions a(( -(0)+) = 0 and a(oo-) = I, and it follows from this

- - - - - - - - - - . - - - - - - - , , - - - - - - - - - - - - - - -..- -.... ~~

I

4.2. Probability Measures

53

a(x) 1

----+--~-----L---~x

o

B

FIGURE 4.2

that if /La is a probability measure, then /La(I) < 1 for any interval I. Example 4-2-1: The uniform distribution on the interval [A, B] (A and B finite, A < B) is the probability distribution ex defined by

ex(x) =

{

0, ~=~,

I,

if x < A, if A < x < B, if x > B

(cf. Figure 4.2). Since ex is continuous, we can say that if I is an interval with finite endpoints a, b, then /La(I) = ex(b) - ex(a), so that if A < a < b < B, then /La(!)

=

a- A

b- a

length of I B - A - B - A == B - A = length of [A, Bf b- A

Since the only changes in the value of ex occur within the interval [A, B], it follows that for any interval I, /La(I)

=

length of I n [A, B] length of [A, B] .

In this case /La(I) can be interpreted as the probability that a random number generator, programmed to select a random number in the interval [A, B], will in fact select a number in I. Example 4-2-2: A discrete distribution is a probability distribution that is constant except for jump discontinuities at a finite or countably infinite

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _--,..

•

.

......

I

54

4. The Lebesgue-Stieltjes Integral

f*)

1

!

11/2

o

-1

• I

i

>x

1

FIGURE 4.3

number of points. An example is the function ex defined by if x < -I, ~,if -1 < x < I, { I, ifx>1 0,

ex(x)

=

(cf. Figure 4.3). In this case we have 0, if I contains neither 1 nor -I, /La (x) = ~ , if I contains 1 or -1 but not both, { I, if I contains both 1 and -1. This corresponds to a random variable x such that P(x

1

= -1) = P(x = 1) = -. 2

For example, x might be the outcome of tossing a coin if IIheads" is scored as 1 and Itails" as -1. Exercises 4-2:

1. If x is a random variable that can take only one value A (with probability 1), what is the corresponding probability distribution function ex? 2. If x is a random variable that can take exactly n values AI, A2, ... ,An (where Al < A2 < ... < An), each with a probability I In, what is the corresponding probability distribution function (¥?

--------------------r----------------.----..--..- . . .

I

4.3. Simple Sets

4.3

55

Simple Sets

A simple set is a subset of JR that can be expressed as the union of a finite collection of disjoint intervals. If S is a simple set, say S = U~lIj where II, h, ... ,Im are disjoint intervals, and if a : JR -+ JR is a monotone increasing function, then the a-measure of S is defined by m

/La(S) =

L /La(Ij). j=l

A given simple set can, of course, be subdivided into disjoint intervals in many different ways, but the value of its a-measure is independent of the way in which it is subdivided. Note also that (i) /La(S) > 0 for any simple set S;

(ii) if Sand T are simple sets such that S

C

T, then /La (S) < /La(T).

Some other elementary properties of simple sets are explored in the exercises. Note finally that a simple set is said to be a-finite if it has finite a-measure. Exercises 4-3: 1. It is true (though rather tedious and unenlightening to prove in general) that if Sand T are simple sets, so are S U T, S nT, and S - T = {x : XES and x E T}. Verify this for each of the following cases: (a) S

= [1,3) U (4,8), T = (2,5] U (6, 7];

(b) S

= (2, 3) U [5, 7], T = [1,4] U [6, 8);

(c) S

= (1,2] U [5, 6), T = [2,4] U (5,7).

2. Prove that if Sand T are disjoint simple sets, then /La (S U T) = /La(S) + /La(T) for any monotone increasing function a : JR -+ JR. Give examples to show that if Sand T are not disjoint, then /La(S U T) mayor may not equal /La(S) + /La(T). 3. Use what was proved in the preceding exercise to show that if Sand T are simple sets such that T C Sand T is a-finite, then /La(S - T) = /La(S) - /La(T) for any monotone increasing function a : JR -+ JR. (Note that T is required to be a-finite in order to avoid

56

4. The Lebesgue-Stie1tjes Integral

having the meaningless expression side.)

00 -

00

on the right-hand

4. Give examples to show that if Tis not a subset of S, then JLa(S-T) mayor may not equal JLa(S) - JLa(T).

4.4

Step Functions Revisited

Let a : JR -+ JR be a monotone increasing function. Let I be any interval, and let 0 : I -+ JR be a step function. It is clear from the relevant definitions that the support of 0 is a simple set. We say that 0 is a-summable if the support of 0 is a-finite. In that case we associate with 0 a real number Aa(O) defined by n

Aa(B) = LCjJLa(lj) j=l

using the notation introduced in Section 2-5. If a(x) = x for all x E JR, so that JLa(Ij) is just the ordinary length of the interval I j , then Aa(O) is just the area A(O) under the graph of 0, as defined in Section 2.5. In general, Aa(O) can be thought of as a generalized area" for which IIlengths" along the x-axis are measured by a-measure rather than by ordinary length. Note that if the endpoints of I are both finite, then any step function 0 : I -+ JR is a-summable for all monotone increasing functions a : JR -+ JR. lI

Example 4-4-1: Let a be defined as in Example 4-1-1, and let 01 defined by

:-

[0,3] -+ JR be

0 (x) = {-I, if 0 < x < I, 1 2, if 1 <x < 3

(cf. Figure 4.4). Then

= a(I-) /La ([1 , 3]) = a(3+) -

JLa([O, 1))

=0a(l-) = 5 -

a(O-)

= 0, 0 = 5,

0

------------------..."...--------_._-----"-----

.

4.4. Step Functions Revisited

(h(x) 2

.,

.,

I I I I I I I

I I I I I I

1

X

1I

0

57

2

I I

3

-1

FIGURE 4.4

(h(x) 2

.,

9

I I I I I I

1

X

1I

0

2

I

3

-1

FIGURE 4.5

and so A a (81 )

= (-1)0 + 2(5) = 10.

Suppose we modify the definition of 81 very slightly, to give

8 (x) z

= {-I, 2,

if 0 < x < I, if 1 < x < 3

(cf. Figure 4.5). In this case

= a(1+) JLa((I,3]) = a(3+) JLa([O, 1])

= 1 - 0 = I, a(1 +) = 5 - 1 = 4,

a(O-)

and so A a (8z)

= (-1)1 + 2(4) = 7.

________________r - - - - - - - - - - - - -..- ---.. . .

58

4. The Lebesgue-Stieltjes Integral

8(x)

20--------

-------1 -----+--------~x

o

FIGURE 4.6

Note that while the area A(8) under the graph is the same for these two functions, the values of A a(81 ) and Aa(~) are different. This is because at the single point where 81 and 82 have different values, a has a discontinuity, and so the interval consisting of that single point has positive a-measure. Clearly, discontinuities in a complicate matters! E~ple

4-4-2: Let a be the discrete distribution function defined in Example 4-2-2, and let 8 : JR -+ JR be defined by if x < 0, 2, if x > 0

8(x) = { I,

(cf. Figure 4.6). Then /La(( -00, OD /La((O, (0))

= a(O+) = a(oo-) -

a(( -(0)+) a(O+)

1

=2-

=1-

1 2

0

1

= 2'

1

= 2'

and so

We conclude this section by listing a number of basic properties that are straightforward to prove and intuitively reasonable, so we will omit the proofs. (i) If8 is a nonnegative a-summable step function, then A a (8) > 0; also, Aa(O) = O.

-----------'--------".---------------"._-'"

4.4. Step Functions Revisited

59

a(x)

FIGURE 4.7

(ii) If 81 and 82 are a-summable step functions on the same interval I such that 81 < 82 on I, then A a (8l ) < A a (82 ).

(iii) If8 is an a-summable step function, then so are 181,8+, and 8-, and we have A a (8) = A a (8+) + A a (8-) and Aa (181) = A a (8+) A a (8-). (iv) If 81 , 82, ... ,8m are a-summable step functions on the same interval I, aI, a2, ... , am are finite real numbers, and 8 : I -+ JR is defined by m

8(x) = L~8j(x) j=l

for all x E I (i.e., 8 = L:;l aj8j), then 8 is also an a-summable step function on I, and m

A a (8)

=L

ajAa (8j).

j=l

Exercises 4-4: Let a : JR -+ JR be defined by a(x) = {

~x,

I,

if x < 0, if x > 0

(cf. Figure 4.7). For each of the following step functions 8: (a) sketch the graph of 8; (b) determine whether or not 8 is a-summable, and if it is, find A a (8).

--------------.,,----------_._--_

......

60

4. The Lebesgue-Stieltjes Integral

1. 8: (-2,1) -+ JR defined by 8(x) = { 3, -I,

if -2 < x < -1 , if-l <x < 1.

2. 8: [-1,1] -+ JR defined by 8(x) = { -2,

I,

if -1 < - x < 0, ifO<x 3.

4. 8: (-00,0) -+ JR defined by 8(x)

={

if x < -I, I, if -1 < x
O.

Definition of the Integral

We are now in a position to set up the necessary machinery for defining the Lebesgue-Stieltjes integral. Throughout this section we take I to be a given interval with endpoints a, b, and ex : JR -+ JR to be a given monotone increasing function. Let f : I -+ JR be a function that is nonnegative on I. A sequence 81,82 ,83, ... is said to be admissible for f if it satisfies all the following conditions: (a) 8j is an ex-summable step function on I, for eachj = 1,2,3, ...; (b) 8j(x) > 0 for each x

E

I and eachj

(c) 0 0. Since La(lf - fnl) -+ 0, there exists a positive integer N( E) such that n > N(E)

=> La(lf - fn!)

I

Ii

E

< 2'

(5.2)

5.1. Basic Properties

73

Also, from equation (5.1), we have that 2 n > E

1

=> -

E

< n 2

=> La(lfn -

E

en I) < -. 2

(5.3)

By Theorem 4.5.4 we have that for each n = I, 2, ... ,

Thus, ifn > max{N(E), 2/ E} then equations (5.2) and (5.3) imply that E

La(lf - en I) < 2

+

E

2 =

E,

and so La(lf - en I) -+ 0, and f is integrable over I with respect to (X by definition. The rest of the theorem follows from Theorem 5.1.1, Theorem 4.5.5(iii), and the definition of the integral (see Exercises 5-1, No.1). 0 Theorem 5.1.3 Iff: I -+ lR is integrable over I with respect to (x, then

Proof By definition,

thus,

[f da by Theorem 5.1.1.


If (x) I (since x E BN ), and so we can say that If (x) I < L~=l fn(x) for all x E 8. On the other hand, if x E 1-8, then x e Bn for each n = 1,2, ... , and so L~=lfn(x) = o = f(x). Thus L~=l fn converges on I, and If I < L~=l fn on I. By

78

5. Properties of the Integral

Theorems 4.5.2(ii), 4.5.3, and 4.5.4 we then have 00

o 2

90

6. Integral Calculus

a(x)

•

1

--~O-+---+1---

°

6.1. Evaluation of Integrals

91

a(x) 3

-------

2

1 X

0 FIGURE 6.2

Then,

l

eX da =

[0,(0)

1

eX da

+ [,

[0,0]

C da (by Theorem 6.1.1)

(0,00)

= eO(a(O+) -

a(O-)) +

(

C d(3 - e- 2x )

1(0,00)

(by Theorems 6.1.6 and 6.1.8). Applying Theorem 6.1.7 to the second integral gives

l

eX da

= 1(2 -

[0,(0)

0)

1 +1

+ (

C(2e- 2x )

ax

1(0,00)

= 2+

2e-

X

ax

(by Theorem 6.1.3(i))

[0,(0) 00

=2

= 2 + lim

c~oo

2e -x

ax

(an improper Riemann integral)

[- 2e- X]C

°

= 2 + (0 - (-2)) = 4.

Note again that although a(x) = 3 -e- 2x for all x E [0, (0), we cannot conclude that ~Oloo) e!' da = ~O,oo) e!' d(3 - e- 2x ). Again, the discontinuity in a at the endpoint of the interval is crucial; the condition in Theorem 6.1.8 that I must be open cannot be disregarded.

92

6. Integral Calculus

---o---~x

FIGURE 6.3

2 I I

1

t 1

---('J-----~x

2

FIGURE 6.4

E~ple

6-1-3: Let a : lR ~ lR be defined by a(x)

(Figure 6.3). Let h ,f2 : lR

~

= f 1 (X)

={

x,

x+

if x < 0, I, if x >

°

lR be defined by

{x,I, ifif xx=j:.= 2,2,

.r:: ( ) }2 X

= {x,

x

°

if =j:. 0, I, if x =

(Figure 6.4). Now, the set {2} has zero a-measure, and Theorem 5.2.3 implies that ~1,3] h 00 = ~1,3] X 00. Thus,

1

[1,3]

f1 00 =

1

xd(x + 1)

(by Theorems 6.1.3 and 6.1.8)

(1,3)

,I

6.1. Evaluation of Integrals

=

i 1

93

x(1)dx (by Theorem 6.1.7)

(1,3) 3

=

xdx (a Riemann integral) by Theorem 6.1.3

=[~J:=~-~=4. However, we cannot say that ~-I,I]h 00 = ~-1,1] X 00, because the set to} does not have zero a-measure. Instead, we proceed thus:

i

[-1,1]

r; 00 =

i

h 00 + { h 00 + ( h 00

[-1,0)

1[0,0]

1(0,1]

(by Theorem 6.1.1)

=

1°

xdx + h(O)(a(O+) - a(O-))

+ {I xdx 10

-1

(by Theorem 6.1.6) (using reasoning similar to that used in the first part to deal with the first and third integrals). Therefore,

i

hOO =

[-1,1]

2 0) + [X Jl = 2 °

2 X [ O + 1(1 J 2 -1

1.

Note that in contrast, ~ -1,1] x 00 = 0, the calculation being the same as the preceding one except that h(O) is replaced by 0, the value of the integrand at O. Example 6·1·4: Let a : lR --* lR be defined by

ifx x -2x+2, if 1 if x 3, if x x+2, 0,

2

a(x) =

< I,

<x < 2, = 2, > 2

(Figure 6.5). Then,

{ ~oo= { ~oo+ ( ~oo+i

1[0,3)

1[0,1)

1[1,1]

+ { x2 00 + 1[2,2]

i

(2,3)

x2 00

(1,2)

x2 00 (by Theorem 6.1.1)

94

6. Integral Calculus

a(x)

/

4

I I

•

3

I I I

2

~

1

0

X

2

1

FIGURE 6.5

x 2 da + 12 (a(1 +) - a(1-))

= ( 1(0,1)

+ 22 (a(2+) -

a(2-)) +

+

1

x 2 d(x 2

-

2x + 2)

(1,2)

1

x 2 d(x + 2)

(2,3)

(by Theorems 6.1.3 (i), 6.1.6, and 6.1.8) = 0 + 1 (1 - 0)

1 +1 +1

+

1

x 2 (2x - 2)

(2,3)

(2x

3

[1,2]

-

~) ax + 8 +

2

= 9

(zx3 -

= 9+

2)

x2 1 ax (by Theorems 6.1.5 and 6.1.7)

+ = 1

ax + 4(4 -

(1,2)

[~ -

2; [;J:

109

6

.i

3

1 Jl- ax

[2,3].

(by Theorem 6.1.3)

x2 ax (Riemann integrals)

6.1. Evaluation of Integrals

95

a(x)

I

3 2

1

1

0

X

FIGURE 6.6

Exercises 6·1: 1. Let a : JR --* JR be defined by if x < 0, ifO<x1

e3x

a(x)= {

2,' 2x+l,

(cf. Figure 6.6). Let [ : JR --* JR be defined by

[(x) = {ex, (cf. Figure 6.7). Evaluate I: (i) (-1,0) (iv) (-1,1]

2X ,

if x < I, if x > 1

h[ da for each ofthe following intervals (ii) [-1,0] (v) [1,3]

(iii) (-1,1) (vi) (-00,0)

2. Let [x] denote the integer part of x, i.e., the largest integer n such that n < x; for example, [2.71] = 2, [3] = 3, [-1.82] = -2. (a) Sketch the graph of the function [x] : JR --* JR.

(b) Evaluate the following integrals:

, i

96

6. Integral Calculus

fix)

---+--------~x

o

1

FIGURE 6.7

(i) ~O,S](X2 + 1) d[x]

(ii) ~O,S] ex d(x + [x])

(iii) ~1/4,S/4][X] d[2x]

(iv) ~1/4,S/4][2X] d[x]

3. Let (X : lR --* lR be a probability distribution function correspond-

ing to a random variable x. We define the mean of x (also called the expectation or the expected value) to be E(x) =

1

xda.

(-00,00)

(a) Calculate the mean of the uniformly distnbuted random variable defined in Example 4-2-1. •

(b) Calculate the means of the random variables defined in Example 4-2-2 and Exercises 4-2, NO.2. (c) If x is a random variable that can take exactly n values AI, A2, ... ,An (where Al < A2 < ... < An) with probabilities PI, P2, ... ,Pn, respectively (where L:.f=I Pi = 1), find the corresponding probability distribution function (x, and the mean of this random variable.

6.2. Two Theorems of Integral Calculus

6.2

97

Two Theorems of Integral Calculus

In this section we will look at the form taken in the LebesgueStieltjes theory by two theorems that are familiar to you in the context of elementary integral calculus. An important aid in evaluating elementary integrals is the I/change of variable" theorem

l

b

f[u(t)]u' (t) dt =

lU(b)

a

f(x) dx,

u(a)

where we have made the substitution x = u(t) in order to simplify the integral on the left. The same thing can be done within the Lebesgue-Stieltjes theory. First we need a definition. We say that a function u : lR ~ lR is strictly increasing on an interval I if u(Xl) < u(X2) for all Xl, X2 E I such that Xl < X2. Suppose now that u : lR --* lR is continuous and strictly increasing on I, and write u(I) = {u(x) : x

E

I}.

Then it is easy to see that u(I) is an interval. For example, if u(x) = x 3 + 1 for all X E lR, then u(( -2,1)) = (-7,2), u([3, (0)) = [28, (0), u(( -00, -1)) = (-00,0), and so on. We can now state the "change of variable" theorem for the Lebesgue-Stieltjes integral.

Theorem 6.2.1 (Change of Variable) Let I be any interval, and u : lR --* lR be a function that is continuous and strictly increasing on the interval I. Then { (f 0 u) du =

11

1

f dx,

u(I)

where f 0 u denotes the composition off and u, defined by (f 0 u)(x) = f[u(x)] for all X E I. If, in addition, u is differentiable on I, then this result can be written in the form

( if 0 u)u' dx =

11

1

u(I)

f dx.

98

6. Integral Calculus

Finally, if a : lR --* lR is monotone increasing, then

l

(fou)d(aou) =

1

fda.

u(I)

I

All three results hold in the sense that if one side exists, then so does the other, and the two are equal.

The condition that u should be strictly increasing on I is not really a restriction in practice. If u is not strictly increasing, the interval of integration can usually be split up into subintervals on which u is either strictly increasing or strictly decreasing, or constant, and each of these can be dealt with separately (note that if u is strictly decreasing then -u is strictly increasing, so the theorem can still be used with the obvious modifications). When evaluating integrals in practice, therefore, one is usually safe, provided that u is continuous (and for the most part u is also differentiable). The second matter we will consider here is integration by parts. Recall that for the Riemann integral, the technique of integration by parts centers on the formula

{fg' dx+ {f'gdx =

[fg]~.

For the Lebesgue-Stieltjes integral, this result takes the basic form

Ifdg+ 19df =

!.Lfg(I),

where f and g can be allowed to be functions of bounded variation, using the approach outlined in Chapter 5. However, a correction term is needed in order to take account of cases where f and g have • points of discontinuity in common. In order to understand the need for a correction term, it is sufficient to consider the simplest case, where I is the closed interval [a, a] = {a} consisting of a single point. We then have (

f dg

= f(a)[g(a+) -

g(a-)]

= f(a)J.Lg({a}),

J{a}

( g df = g(a)[f(a+) - f(a-)] = g(a)J.Lf({a}), J{a}

J.Lfg({a})

= (fg)(a+) -

(fg)(a-)

= f(a+)g(a+) -

f(a-)g(a-).

6.2. Two Theorems of Integral Calculus

99

Thus,

(1.

{a}

f dg

+

1.

{a}

g df ) - J.Lfg({a})

+ g(a)J.Lf({a}) f(a)J.Lg({a}) + g(a)J.Lf({a})

[t(a+)g(a+) - f(a-)g(a-)]

- [t(a+)g(a+) - f(a+)g(a-)

+ f(a+)g(a-) -

= f(a)J.Lg({a}) =

f(a-)g(a-)]

+ g(a)J.Lf({a}) - [t(a+)J.Lg({a}) - g(a-)J.Lf({a})] (f(a) ~ f(a+))J.Lg({a}) + (g(a) - g(a-))J.Lf({a}) = A(a), say.

= f(a)J.Lg({a}) =

Then also A(a) = [tea) - f(a-)

+ f(a-) -

+[g(a) - g(a+)

f(a+)]J.Lg({a})

+ g(a+) -

g(a-)]J.Lf({a})

= [tea) - f(a-)] J.Lg ({a}) - J.Lf({a})J.Lg({a}) +[g(a) - g(a+)]J.Lf({a})

= [tea) - f(a-)] J.Lg ({a})

+ J.Lg({a})J.Lf({a})

+ [g(a) -

g(a+)]J.Lf({a}).

Adding gives 2A (a) = [2f (a) - f (a+) - f (a-)] J.Lg({a})+ [2g(a) - g(a -) - g(a+)] J.Lf ({a}),

and so finally,

We then have

1.

{a}

f dg

+

1.

{a}

gdf = J.Lfg({a})

+ A(a).

Note first that iff is continuous at a, thenf(a) = ~(f(a+)+f(a-)) and J.Lf({a}) = 0, and so A(a) = 0; similarly if g is continuous at a. Thus it is possible for A to be nonzero only if both f and g are discontinuous at a. Three important special cases are: (i) If f(a) = ~(f(a+) A(a) =

o.

+ f(a-))

and g(a) = ~(g(a+)

+ g(a-)), then

100

6. Integral Calculus

(ii) Iff and g are both continuous on the right at a, so that f (a+) = f(a) and g(a+) = g(a), then 1

A(a) = -(f(a+) - f(a-))J.Lg({a})

2

1

+ -(g(a+) 2

g(a-))J.Lf({a})

= J.Lf({a})J.Lg({a}).

(iii) Iff and g are both continuous on the left at a, so that f( a-) = f(a) and g(a-) = g(a), then 1

A(a) = z(f(a-) - f(a+))J.Lg({a})

1

+ Z(g(a-) -

g(a+))J.Lf({a})

= -J.Lf({a})J.Lg({a}). It is now easy to see why the general theorem for integration by

parts takes the form given below. Note that iff and g are functions of bounded variation, then by Theorems 2.7.2 and 2.4.3 the set ofpoints of discontinuity off and g is either empty or countably infinite.

Theorem 6.2.2 (Integration by Parts) Let f, g : I ~ lR be functions of bounded variation, and let S denote the set ofpoints at which f and g are both discontinuous. Then [ f dg

+ [gdf =

I

J.Lfg(I)

+ LA(a),

I

aeS

where

In particular, (i) IfS is empty, oriff(a) = ~(f(a+)+f(a-)) andg(a) = ~(g(a+)+ g(a-)) for all a E S, then

[f dg + [gdf =

/Lfg(I).

(ii) Iff and g are continuous on the right at all points of S, then [ f dg I

+

[g

df = J.Lfg(I)

I

+L aeS

.i

J.Lf({a})J.Lg({a}).

6.2. Two Theorems of Integral Calculus

101

(iii) Iff and g are continuous on the left at all points of 8, then

I 1 fdg

+

I

9df = JJ-fg(I) -

I

L JJ-f({a})JJ-g({a}). aES

Example 6-2-1: Consider the integral ~OI3) i2 da that was discussed in Example 6-1-4. Using integration by parts, bearing in mind that x2 has no points of discontinuity, we have {

~[OI3)

X2~da + (

~[OI3)

a d(x 2 ) = JJ-xla([O, 3))

= (x 2a)(3-) = 9a(3-) - 0

(x 2a)(0-)

= 9(5) = 45.

Thus, {

~[O,3)

~ da =

45 - (

a d(i2)

= 45 - (

ad(~)

= 45 -

a (2x)dx by Theorem 6.1.7

~[OI3) ~(O,3) {

~(O,3)

by Theorem 6.1.3

= 45 -

!o3 2xa dx by Theorem 6.1.3 (a Riemann integral)

= 45 -

[[2 2x(x'! - 2x +2)dx +12x(x +2)dxJ 3

= 45 -

[

::..4 - -4x3 2 3

[2

+ 2i2 J2 - ~3+ 2i2 J3 1

3

[8- +8- G-~ +2)J -[18+18+8)J

=45-

3:

c:

109 6 '

as before.

\i

2

102

6. Integral Calculus

I ~

JA,-

.......

I

--+----1 a t L-

I b

.....1

---1\,-

---y-It

V Jt ·

FIGURE 6.8

Exercises 6·2: 1. Use integration by parts to evaluate the integrals in Example 6-1-1 and Exercises 6-1, No. 2(b)(i). 2. (a) Investigate what happens when integration by parts is tried as a method for evaluating the integral in Example 6-1-2. Which hypothesis of Theorem 6.2.2 does not hold in this case? (b) Use integration by parts to evaluate ~OIOO) e-x 00, where ex is as defined in Example 6-1-2. Check your answer by evaluating the integral directly.

6.3

Integration and Differentiation

We now examine some connections between differentiation and integration. Here we restrict ourselves to the Lebesgue rather than the Lebesgue-Stieltjes integral; that is to say, integration is with respect to x throughout. For the first theorem, we need some notation. Let I be any interval with endpoints a, b and let t be any real number such that a < t < b. We denote by It, It the intervals defined by It = {x : x E I, x < t}, It = {x: x E I, t < x} (cf. Figure 6.8). Theorem 6.3.1 (Fundamental Theorem of Calculus) (i) If f : I --* lR is integrable over I, then both F(t) = ht fax and G(t) = 1;t fax are absolutely continuous on I and differentiable a.e. on I, and P' (t) = f(t), G' (t) = -f(t) a.e. on I. If, in addition, f is continuous on I, then Ha.e.H can be replaced by HeverywhereH in the preceding statement.

6.3. Integration and Differentiation

103

(ii) IfF: I --+ IR is absolutely continuous on I, then it is differentiable a.e. on I, F' is integrable over I, and F(t) = F' ax + C for all t t E I, where C is constant on I.

h

E~ple

6-3-1: The error function erf(t) is defined as an ordinary Riemann integral erfCt)

J;r l'

=

e-x' ax

for all t > 0. It is important in statistics (in connection with the normal distribution), and it also arises in the context of certain partial differential equations connected with heat flow. The complementary error function erfc(t) is defined by erfc(t) =

2 ,.;n

joo et

x2

ax

for all t > 0. It follows at once from Theorem 6.3.1(a) that for all t > 0,

d[

dt erf(t)

]=

t 2 ,.;ne-

2

d[ erfc(t)] and dt

=-

2 t2 ,.;ne.

E~ple

6-3-2: Let a : lR --+ lR be a probability distribution function. If there exists a function f : lR --+ lR such that f > on lR and aCt) = -oo,t] f ax for all t E lR, then f is call a density of a. From Theorem 6.3.1, we know that this happens if and o:p.ly if a is absolutely continuous, and that in this case a' = f a.e. onlR. Clearly, discrete distnbutions, as defined in Section 4-2, do not have densities, since their distnbution functions are discontinuous. If the functions f and g are both densities of a, then f = g a.e., so in this sense we can say that the density of a, if it exists, is unique. As an example, take the case of the uniform distnbution defined in Example 4-2-1:

°

a(x)

=

0, ~=~, { I,

Ie

if x B

(Figure 6.9). Here we can take the density f of a to be the derivative of a where it exists (which is everwhere except at A and B), and

,i

104

6. Integral Calculus

a(x) 1

_ _ _ _+-_-JLo A

I I I I I I LI- - ~ x

B

FIGURE 6.9

r---r----r f{x)

l/(B-A)

FIGURE 6.10

define f (arbitrarily) to be zero at A and B, giving

~X
B, If A < x < B

(Figure 6.10). The previous theorem dealt with differentiation of an integral that has a variable interval of integration. It is also important to be able to differentiate functions of the form get)

= [f(t,x)dx,

where the variable t appears in the integrand, not in the interval of integration. It is natural to ask whether we can find g' (t) by interchanging the order of the differentiation and integration operations. Thus g'(t)

= ~[[f(t,X) dx] = [ ~f(t,x)dx,

6.3. Integration and Differentiation

105

where in the right-hand integral, x is held constant while differentiation is carried out with respect to t. Simple examples suggest that this is correct. Consider, for example, get)

= (I sin(2t + 3x) dx = [_~ cos(2t + 3X)]X=1 k 3 x~ 1 1 = - 3 cos(2t + 3) + 3 cos2t.

By direct differentiation, g/ (t) hand, ~

1 1

= ~ sin(2t + 3) -

-a sin(2t + 3x) dx =

o at

=

1

~ sin 2t. On the other

1

2 cos(2t + 3x)dx

0

]X=l

2 - sin(2t + 3x) [ 3 x=o

2

= - sin(2t + 3) 3 = g/(t)

2 - sin 2t, 3

as expected. The general theorem, which tells us that this process, called IIdifferentiation under the integral:' is legitimate, is as follows: Theorem 6.3.2 (Differentiation Under the Integral) Let 1 and J be any intervals. Let the real-valued function f(t, x) be such that (i) f(t, x) is defined for all t E J, X E I; (ii) f(t, x) is integrable with respect to x over I; (iii) For each t E J, :/(t, x) exists a.e. on I; (iv) For each closed subinterval J* C J, there exists a function A : I --+ IR such that A is integrable over I, and I ~f(t, x)1 < A(X) for all x E I and t E J*. Then for each t E J, a/atf(t, x) is integrable with respect to x over I, and

~ [ [f(t, x) ax] = [:/Ct, x) ax. Example 6-3-3: Consider the function get) = JoT( In(l + t cos x) dx, where -1 < t < 1. Note that since I cosxl < 1 for all x, we have that It cosxl < It I < 1

106

6. Integral Calculus

for all t E (-1, 1). Thus 1 + t cos x > 0 for all x and all t and so both In(1 + t cos x) and

a

-a In(1 + t cos x)

+ tcosxl

(-1, 1),

cosx

= 1 + t cosx

t are continuous for all x and all t E (-1,1). Take any closed subinterval [a, b] C max{la/, lb/}. We have that for all x and all t and so 11

E

(-1, I), and let k = E [a, b], It cosxl < k < I,

> l-Itcosxl > 1 - k >

o.

Thus cosx 1 1. By differentiating both sides of this equation with respect to t, evaluate

~

{Jr

Jo

1

(t - cos X)2 dx.

4. Given thatg(t) = J~ sin(x-t)dx, findg'(t) by using Leibniz's rule. Check by evaluating get) directly and then differentiating. 5. Find g/(t) if get)

=

j

t 21

-

x

t

sin(tx) dx,

where t > 1. 6. Assuming that g is continuous on the interval [0, (0), show that the function yet)

1

1

(t

= k C2 sin(kt) + CI cos(kt) + k Jo

g(x) sin {k(t - x)} dx,

for t > 0, satisfies the differential equation

d2y dt 2

+ ~y =

get),

where k > 0, together with the initial conditions yeO) = CI, y/(O) = C2. 7. Assuming that g is continuous on the interval [0, (0), show that the function yet) = -1

n!

I

t

(t - x)ng(x) dx

0

satisfies the differential equation dn+ly dt n+l = get) together with the initial conditions yeO) = 0, yCI)(O) = 0, ... ,ycn)(O)

= O.

Double and Repeated Integrals

CHAPTER

Lebesgue-Stieltjes integrals of functions of more than one variable can be defined using the same approach as was used in Section 4.5 for functions of one variable. For the sake of simplicity we will discuss only functions of two variables. The process for functions of more than two variables is completely analogous.

7.1

Measure of a Rectangle

We define a rectangle to be a set ofthe form II x 12 C JR2, where II and h are intervals. For monotone increasing functions aI, a2 : JR --+ JR we define the al x a2-measure of II x h, denoted by fJ-al xa2 (II X 12 ), by fJ-al xa2 (II

X h)

= fJ-al (II) X fJ-a2 (h).

For example, if al and a2 are the functions defined in Exercises 4-1, problems 1 and 2, respectively, then fJ-al

((0, 1))

fJ-a2 ((0,

= 1-

1)) = 0,

e- I ,

=3fJ-a2 ([0, 1)) = I,

fJ-al ([0,

1))

e- I ,

113

114

7. Double and Repeated Integrals

and therefore JL CX I XCX 2((O, 1) x [0,1)) = 1 - e- l , JLcxIXCX2([O, 1) x [0,1)) = 3 - e- l .

JL CX I XCX 2((O, 1) x (0,1)) = 0,

7.2

Simple Sets and Simple Functions in Two Dimensions

A simple set in]R2 is a subset of]R2 that can be expressed as the union of a finite collection of disjoint rectangles. Just as for simple sets in ]R, we can define the measure of a simple set in ]R2. If aI, a2 : ]R --+ ]R are monotone increasing functions and S is a simple set of the form m

S = U(Ilj

X

hj),

j=l

where h,l x 12,1, h,2 X h,2, ... , 11,m X 12,m are pairwise disjoint rectangles, then the al x a2-measure of S is defined by m

JLcxIXCX2(S) = LJLcxl xcx2(I1j x h,j)' j=l

The properties of simple sets in ]R2 and their measures are the same as those described in Section 5-3 for simple sets in ]R. We can now define simple functions of two variables by analogy with step functions of one variable (see Sections 2-5 and 4-4). We could continue to use the term "step functions:' but customary usage restricts this term to functions of one variable. • A function 8 : ]R2 --+ ]R is a simple function if there is a simple set n

S = U(11 ,}0 x h ,}0) j=l

8(x,y) = {

0,

if (x, y)

E

]R2 - S.

7.4. Repeated Integrals and Fubini's Theorem

115

The set S is called the support of8. The properties of step functions given in Section 2-5 carry over without difficulty to simple functions. If aI, a2 : lR -+ lR are monotone increasing functions, we define the generalized "volume" A CX1 xCXz (8) in a way exactly analogous to the definition of A cx (8) in Section 4-4.

7.3

The Lebesgue-Stieltjes Double Integral

Let S be a subset of lR2 and let f : S -+ lR be a function. We extend the definition of f to lR2 by defining f(x, y) to be zero if (x, y) E lR2 - S. Let aI, a2 : lR -+ lR be monotone increasing functions. The Lebesgue-Stieltjes double integral of f, denoted by

L2 f f de

IJlj

x 1Jl2),

is defined by a process that is almost word-for-word the same as that used for the single-variable integral in Section 4-5. The only change is that a-summable step functions on I are replaced by al x a2summable simple functions on ]R2. All the elementary properties analogous to those proved in Sections 5-1 and 5-2 carry over and are proved in the same way, and the same goes for the convergence theorems of Section 5-3 and the definitions of measurable functions and measurable sets given in Section 5-4. In practice, the evaluation of double integrals is invariably done, as in elementary calculus, by converting them to repeated integrals.

7.4

Repeated Integrals and Fubini's Theorem

Let f : lR2 -+ lR be a function. For any y E lR we define the singlevariable function f(·, y) : x -+ f(x, y), and for any x E lR we likewise define the function f(x,·) : y -+ f(x, y). Let aI, a2 be monotone

116

7. Double and Repeated Integrals

increasing functions. If for eachy E lR, fIRf("y)001 exists, then this defines a function 12 : y --+ f IR f(·,y)OO 1. If fIR 12 002 exists, we call this a repeated integral of f and write it as fIR fIR f 001 da 2. If for each x E lR, fIRf(x,') 002 exists, we define fI : x --+ fIRf(x,') 002, and if fIR fI 001 exists, it gives us the repeated integral of f with the opposite order of integration, written fIR fIR f 002 001. In most cases calculation shows that fIR fIR f 001 002 and fIR fIR f 002 001 have the same value, but this is not always the case. Consider, for example, the improper Riemann repeated integrals 1 1 1 11 ---.-;:.X - Y dx dy and X - Y 3 dy dx. 0 0 (x + y) 1o 0 (x + y)3

11

We have

1 11 1 o

0

Y (x + y)3 dxdy X -

111(1 1 = + 0

-

[1 1 + (1 1 + 1

+

Y

JX=l dy

(x + y? x=o

(x

y)

o

1

Y + +--Y (1 + y? y y

1

-1

0

=

(x

0

2Y ) y? - (x + y)3 dxdy

1

1

=

o (1 1

+ y?

J1

1 1)

~

dy 1

= [ 1+yo=-2'

However, 1 11 ---:X - Y dydx = 1 o 0 (x + y)3

11 + 11 1 + 1

0

1

0

Y - x dxdy (interchanging x andy)

(x 1

= -

o

0

y)3

•

x-y dxdy (x y)3

1

2'

and so the two repeated integrals have different values. It turns out, though we shall not prove it, that this cannot happen if either repeated integral is absolutely convergent. We can easily verify that this condition does not hold in our example. We

7.4. Repeated Integrals and Fubini's Theorem

117

y 1 r----."

y>x y<x --f---~~X

o

1

FIGURE 7.1

investigate

1 11 1 + o

0

Ix -YI ---dxdy (x y)3

by splitting the region ofintegration into two parts, one where y < x and one where y > x (cf. Figure 7.1). Then,

118

7. Double and Repeated Integrals

y

y=x 2 ----\--

FIGURE 7.2

and so this integral (and likewise J; Jo (~.;~13 dy dx) does not converge. The fundamental theorem that relates double and repeated integrals using the absolute convergence condition is called Fubini's theorem: l

Theorem 7.4.1 (Fubini's Theorem) If aI, a2 : ]R -+ ]R are monotone increasing functions and f : ]R2 -+ al x a2-measurable, then the existence ofanyone ofthe integrals

LJ Ifl

dCetl

x

et2), LL Ifl det det2, LL Ifl det2det l

]R is

l,

implies the existence and equality ofthe integrals

LJfdCetl x et2), LLf detldet2 ' LLf det2detl . In practice, the functions that arise are almost always measurable, so Fubini's theorem justifies the use ofrepeated integrals to evaluate double integrals, provided that one of the repeated integrals is absolutely convergent. The details can be very messy, so we will confine ourselves to one example. Example 7-4-1: Let 8 = {(x, y) : < X sup If'(x) - g'(x) I = xE[a,b]

11f' - g'llco.

Example 8-3-3: Let H(D(c; r)) denote the set of all functions holomorphic (analytic) in the closed disk D(c; r) = {z E C : Iz - cl < r}, r > O. This set forms a complex vector space, and the function II . lie defined by

IIflle = sup If(z) I zED(e;r)

provides a norm for the space. In fact, it can be shown that (H(D(c;r)),II·lIe)isaBanachspace.LetT¢,bbetheoperatormap ping H(D(a; r)) to H(D(a - b; r)) defined by T¢,bf = ei¢f(z - b),

where ¢ E lR is a constant. The operator T¢,b is a one-to-one and onto mapping from (H(D(a; r)), II . lIa) to (H(D(a - b; r)), II . II a-b), and since sup leitPf(z - b) - eitPg(z - b)1 zED(a-b;r) sup If(z - b) - g(z - b)1 ZED(a-b;r) = sup If(z) - g(z) I = IIf - gila, ZED(a;r)

IIT¢,bf - T¢,bglla-b =

the operator is also an isometry. The two normed spaces are thus isometric. Given an incomplete normed vector space (X, 1\.1\), it is natural to enquire whether the space can be made complete by enlarging the vector space and extending the definition of the norm to cope with the new elements. The paradigm for this process is the completion of the rational number system Q to form the real number system R. This example has two features, which, loosely speaking, are as follows:

8.3. Completion of Spaces

137

(i) the completion does not change the value of the norm where it was originally defined, Le., IIrll e = Ilrll e , for any rational number r; (ii) the set Q is dense in the set lR (cf. Section 1-1). The first feature is obviously desirable: We wish to preserve as much as possible the original normed vector space, and any extended definition of the norm should not change the value of the norm at points in the original space. The second feature expresses the fact that the extension of tne set Q to the set lR is a IIminimal" one: Every element added to Q is required for the completion. We could have Ilcompleted" Q by including all the complex numbers to form the complex plane C, which is complete, but this is overkill. The completion ofthe rational numbers serves as a model for the general completion process. Feature (i) can be framed for general normed vector spaces in terms of isometries. In order to discuss feature (ii) in a general context, however, we need to introduce a general definition of density. Let (X, II . II) be a normed vector space and W ex. The set W is dense in X if every element of X is the limit of some sequence in W. Density is an important property from a practical viewpoint. If W is dense in X, then any element in X can be approximated by a sequence in W to any degree of accuracy. For example, the sequence {an} of Example 8-2-1 consists purely of rational numbers and can be used to approximate ,J2 to within any given (nonzero) error. A fundamental result in the theory of normed vector spaces is that any normed vector space can be completed. Specifically, we have the following result: Theorem 8.3.1 Given a normed vector space (X, II . Ilx), there exists a Banach space (Y, II . II y) containing a subspace (W, II .II y) with the following properties: (i) (W, II . II y) is isometric with (X, II . Ilx); (ii) W is dense in Y. The space (Y, II . II y) is unique except for isometries. In other words if (Y, II . II y) is also a Banach space with a subspace (W, II . II y) haVing properties (i) and (ii), then (Y, II . II y) is isometric with (Y, II . II y).

138

8. The Lebesgue Spaces Y

The space (Y, 1I·lIy) is called the completion of the space (X, 1I·lIx). The proof of this result would lead us too far astray from our main subject, integration. We refer the reader to [25] for the details. Exercises 8-3: 1.

(a) Suppose that Z is dense in W, and W is dense in Y. Prove that Z is dense in Y.

(b) Suppose that the completion of (X, 1I·lIx) is (Y, 1I'lIy) and that P is dense in X. Prove that (Y, II . II y) is also the completion of (P, II . IIx). 2. Let P[a, b] denote the set of polynomials on the interval [a, b], and let PQ[a, b] denote the set of polynomials on [a, b] with rational coefficients. Prove that PQ[a, b] is dense in P[a, b]. 3. Weierstrass's theorem asserts that any function in C[a, b] can

be approximated uniformly by a sequence of polynomials, Le., P[a, b] is dense in C[a, b] with repect to the II . 1100 norm. Use Exercises 8-3, No.2, to deduce that any function in C[a, b] can be approximated uniformly by a sequence in PQ[a, b].

8.4

The Space L 1

Having made our brief foray into functional analysis, we are now ready to return to matters directly involved with integration. Example 8-2-2 shows that the normed vector space (C[a, b], II . IIR) is not complete. We know, however, that this space can l]e completed, but it is not clear exactly what kinds of functions are required to complete it. In this regard, the norm itself can be used as a rough guide. Clearly, a function f need not be in C[a, b] for the Riemann integral of IfI to be defined. This observation suggests that perhaps the appropriate vector space would be R[a, b], the set of all functions f : [a, b] ~ lR such that IfI is Riemann integrable. This Ilexpansion" of C[a, b] to R[a, b] solves the immediate problem, since the sequence {fn} in Example 8-2-2 would converge to a function f E R[a, b], but it opens the floodgates to sequences such as that defined in Section

i i

8.4. The Space L1

139

3-3-1 that do not converge to functions in R[a, b]. Although II . IIR is not a norm on R[a, b] (Exercises 8-2, No. 2(b)), this problem can be overcome. Any hopes of using R[a, b] to complete the space, however, are dashed by Example 3-3-1, because this example indicates that (R[a, b], II . IIR) is not complete. Recall that Example 4-3-1 motivated us initially to seek a more

general integral to accommodate functions such as f(x) = •

{I,0,

I, if x IS IrratIonal or x = 0, 1. ~fx ~s ~atio.nal, x#- 0,

Eventually, we arrived at the Lebesgue integral. The function f plays a role in the completion of (R[a, b], II . IIR) analogous to that played by the number ,J2 in the completion of (Q, II . lie). The Lebesgue integral essentially leads us to the appropriate space and isometry for the completion of (R[a, b], II . IIR) (and (C[a, b], II . IIR)). Let A l [a, b] denote the space of all functions f : [a, b] ~ lR that are (Lebesgue) integrable on the interval [a, b] and let II . III be the function defined by IIfll1

= [

If (x) Idx.

J[a,b]

The set Al[a, b] forms a vector space, but II . III is not a norm on it because there are nonzero functions g in A1[a, b] such that IlglII = 0, Le., ifg = a.e. then IIglII = 0. Functions that fail to be norms solely because they cannot satisfy this condition are called seminorm8, and the resulting spaces are called seminonned vector spaces. Notions such as convergence and Cauchy sequences are defined for seminormed vector spaces in the same way they are defined for normed vector spaces. The problem with the seminorm on A 1 [a, b] is not insurmountable. The essence of the problem is that IlglI = IIfll whenever f = g a.e. (Theorem 5.2.3 (iii)). The set Al[a, b], however, can be partitioned into equivalence classes based on equality a.e. Let L 1 [a, b] denote the set of equivalence classes of Al[a, b]. An element F of L 1 [a, b] is thus a set of functions such that iff1,fz E F, tq~nfi = fz a.e. Since any element f of F can be used to represent the equivalence

°

140

8. The Lebesgue Spaces Y

class, we use the notation F = [f].l Addition is defined as

[f] + [g] = [f + g], and scalar multiplication as

= [af].

a[f]

The set L 1[a, b] forms a vector space, and if II . 111 is defined by

1I[f]1I =

{

ira,b]

If(x) Idx,

then (Ll[a, b], II . 111) is a normed vector space. The candidate for the completion of the space (C[a, b], II·IIR) (and the space (R[a, b], II . IIR)) is the space (Ll[a, b], II . III). In the notation of the previous section, we have X = C[a, b], II . IIx = II . IIR, Y = Ll[a, b], and II . lIy = II . III. Let W = {[f] E Ll[a, b] : [f] contains a function in C[a, bU, and let T be the operator that maps a function f E C[a, b] to the element [f] E W. Now, every element of C[a, b] has a corresponding element in W, and no equivalence class in W contains two distinct functions in C[a, b]; therefore, T is a one-to-one, onto operator from C[a, b] to W. Moreover, Theorem 4.6.1 implies that

IITfl1I

= 11[f]11I = { If(x) Idx ira,b]

.t

If(x)ldx=

IlfllR,

so that T is an isometry. The space W is thus isometric with C[a, b]. To establish that (Ll[a, b], II· III) is the completion of (C[a, b], II ·IIR) it remains to show that W is dense in Ll[a, b] and that GLl[a, b], II . III) is a Banach space. We will not prove that W is dense in Ll[a, b]. The reader is referred to [37] for this result. We will, however, sketch a proof that (Ll[a, b], II . III) is complete. Theorem 8.4.1 The normed vector space (Ll[a, b],

II . III) is a Banach space.

Although this is standard notation, there is some danger of confusion with the notation used for the span that takes sets as arguments

1

i i

8.4. The Space L1

141

Proof We prove that the seminormed space CAl[a, b], II . III) is complete. The completeness of (Ll[a, b], II . III) then follows upon identification of the functions with their equivalence classes in Ll[a, b]. Let lfn} be a Cauchy sequence in (A l[a, b], II . III). Given any E > 0 there is thus an integer N such that IIfn - fm II < E whenever m > Nand n > N. In particular, there is a subsequence lfnk} of lfn} with the property that

Let m

gm

=L

Ifnk+l - fnk!,

k=l

and let g = limm~oo gm denote the pointwise limit function. Note thatg(x) need not be finite for all x E [a, b]. Let [a, b] = h UIz, where II denotes the set of all points such thatg(x) < 00. We will show that Iz must be a null set. Now, gm E Al [a, b] and

IIgmllJ

=

1

[a,b]

Igm(x)1 ax
0 there is an integer N such that IIfn - fmllI

=

1

Ifn(x) - fm(x) I

[a,b]

ax < E

for any m > N, n > N. Let k be sufficiently large so that nk > Nand let m = nk. Then for n > N,

o
1 because g E Lq[a, b] for any q. Thus, V[a, b] C LI[a, b] for all P > 1. The next result indicates that if 1 such that 11 (f) I < c IIf II for all f EX. The norm of a bounded functional 1 is defined as the smallest number c such that 11 (f) I < c IIf II for all f E X and denoted by 111 II. Since ex = is a legitimate choice for a scalar, we see that

°

1(0 . fI)

= 1(0) = 01(fl) = 0,

Le., 1(0) = 0. The norm of 1 is thus given by

11111

= sup 11(f) I [EX

(any choice of c >

°satisfies 101

f#O


O. Now, f has a support of finite length, and hence there is some number fJ > 0 such that f = 0 outside the interval [-fJ, fJ]. Certainly f (restricted to the interval [-fJ, fJ]) is in the space C[ -fJ, fJ], and thus there is a polynomial PE such that E

sup IPE(X) - f(x) I < (2fJ)l/p

xE[-P,Pl

(8.1)

Moreover, we know from Exercises 8-3-2 that the set PQ [ -fJ, fJ] of polynomials with rational coefficients is dense in P[ - fJ, fJ], and hence we can find a polynomial with rational coefficients satisfying inequality (8.1). In fact, we can find a polynomial qE with rational coefficients such that inequality (8.1) is satisfied and qE(-fJ) = qE(fJ) = O. Now the polynomial qE canbe extended to a functiongE = qE' X[-P,Pl

152

8. The Lebesgue Spaces £P

defined on ~. Here X[-P,P] is the characteristic function defined by

X_

ex) =

[ P,P]

{I,

if x E [-,8, ,8], 0, otherwise.

Thus, IIgE - flip = { ( IgE(X) - f(x)I Pdx}l/P =

JJR


nl be an orthonormal set in the Hilbert space X, and let f E X. The numbers if, 4>n} are called the Fourier coefficients of f with respect to M, and the series L~=l if, 4>n}4>n is called the Fourier series of f (with respect to M). Now, for any mEN, m

o < IIf -

m

L if, 4>n}4>n 11 n=l

2

= if -

m

m

L if, 4>n}4>n, f - L if, 4>n}4>n} n=l n=l m

= if,f} - if, Lif, 4>n}4>n} - (Lif, 4>n)4>n,f}

n=l

n=l

m

m

+ (L if, 4>n}4>n, L n=l

if, 4>n)4>n}

n=l m

= IIfll

2

-

m

+ II L

2 L (f, 4>n) if, 4>n} n=l

2

n=l

m

=

if, 4>n}4>n 11

m

IIfll 2 - 2 L Iif, 4>n}1 2 + II Lif, 4>n}4>nIl 2 , n=l

n=l

and Pythagoras's equation implies that m

m

II Lif,4>n}4>nIl = L 1Iif,4>n}4>nIl 2 2

n=l

n=l m

= L

Iif, 4>n} 12 114>n 11 2

n=l m

= L Iif, 4>n}1 n=l

2

and therefore we have m

o < IIf -

L if, 4>n}4>nIl n=l

2

m

=

IIfll

2

-

L 1(f,4>n}1 n=l

2

.

,

9.2. Orthogonal Sets

177

Hence, m

L /(f, rPn)/2 < I/fl/ 2,

n=l

and since the sequence {am} = {L::=1 I (f, rPn) 12 } is a monotonic increasing sequence bounded above, we have that {am} converges to some a E JR and Bessel's inequality holds: 00

L I(f, rPn)/2 < I/fl/ 2. n=l

Bessel's inequality indicates that the Fourier series of f with respect to M is absolutely convergent (in the norm). This means that the Fourier series is convergent (in the norm) and that its sum is independent of the order in which terms are added. Thus, the series L~=l (f, rPn)rPn converges to some functionPMf EX. Now, iff E [M]..L (= M..L), all the Fourier coefficients off are zero, and thus P Mf = O. Alternatively, iff E [M], then it can be shown thatPMf = f. For the general f EX, the set [M] is a closed subspace in the Hilbert space X, and the projection theorem indicates that f = PMf + h, where PMf E [M] and h~M]..L. The function PMf is thus the Ilclosest" approximation in [M] to f. An orthonormal set M = {rPn} in the separable Hilbert space X is called an orthonormal basis of X if for every f EX, 00

f

= L(f, rPn)rPn. n=l

If M is an orthonormal basis, then m

lim "(f, rPn) rPn £...J

m-+oo

= f,

n=l

and using the derivation ofBessel's inequality (and continuity of the norm function) we have that

178

9. Hilbert Spaces and £2

We thus arrive at Parseval's formula 00

2

IIfll =

L I(f, rPn) 1 2

•

n=l

Suppose now that f E M.1. Since M is an orthonormal basis of X, f = L~l (f, rPn}rPn and (f, rPn) = for all rPn E M; consequently, by Parseval's formula we have that IIfll = 0, and the definition of a norm implies that f = O. In this manner we see that if M is an orthonormal basis, then M.1 = 0, i.e., M is a total orthonormal set. Yet another implication of M being an orthonormal basis is that [M] = X. This follows from the projection theorem, since [M].1 c M.1 = 0, and so

°

X

--

--.1

--

--

= [M] E9 [M] = [M] E9 0 = [M].

It is interesting that the above implications actually work the

other way as well. For example, if M is an orthonormal set in X such that Parseval's formula is satisfied for all f EX, then it can be shown that M is an orthonormal basis. In summary we have the following result:

Theorem 9.2.3 Let X be a separable Hilbert space and suppose that M = {rPnl is an orthonormal set in X. Then the following conditions are equivalent: (i) M is an orthonormal basis; (ii) IIf11 2 = L~l 1(f, rPn) 12 for all f E X; (iii) M is a total set; (iv) [M] = x. It can be shown that every separable Hilbert space has an orthonormal basis. Formally, we have the following result: Theorem 9.2.4 Let X be a separable Hilbert space. Then there exists an orthonormal basis for X. The proof of this result is based on a Gram-Schmidt process analogous to that used in linear algebra to derive orthonormal bases. Parseval's formula is remarkable in this context because it essentially identifies all separable Hilbert spaces with the space e2 . Now, e2 is a separable Hilbert space, and given an orthonormal basis M = {rPnl

i i

9.2. Orthogonal Sets

179

on a general separable Hilbert space X there is a mapping T : X -+ e2 defined by the Fourier coefficients if, cPn}. In other words, T maps f E X to the sequence {if, cPn}} E e2 . Parseval's formula shows that T is an isometry from X to e2 . On the other hand, given any sequence {an} E e2 , the function defined by L~I ancPn is in X, since X is a Hilbert space and the series is convergent. We thus have that any separable Hilbert space is isometric to the Hilbert space e2 . In fact, an even stronger result is available, viz., all separable complex Hilbert spaces are isomorphic to e2 . This means that at an Iialgebraic level" a complex separable Hilbert space is indistinguishable from the space e2 , i.e., at this level there is only one distinct space. A similar statement is true for real Hilbert spaces, where the space e2 is replaced by its real analogue. In passing we note that any Hilbert space X possesses an Ilor_ thogonal basis," though it may not be countable. The cardinality of the set is called the Hilbert dimension of the space. If X and Xare two Hilbert spaces both real or both complex with the same Hilbert dimension, then it can be shown that X and X are isomorphic. This is a generalization ofthe situation in finite-dimensional spaces, where, for example, all n-dimensional real Hilbert spaces are isomorphic to JRn. Exercises 9-2:

1. Show that for any integers m, n

1

cos(mx)cos(nx)dx =

[-Jr,Jr]

{](o, ,

ifm =1= n, if m = n,

and

1

sin(mx) cos(nx) ax = 0.

[-Jr,Jr]

2. The first three Legendre polYnomials are defined by Po(x) = I, PI (x) = x, and P2(X) = ~(3x2 - 1). (a) Show that the set P onL 2 [-I,I].

=

{Po, PI, P2} forms an orthogonal set

(b) Construct an orthonormal set M from the set p' and find the Fourier coefficients for the function f(x) = ex.

180

9. Hilbert Spaces and £2

3. The Rademacher functions are defined by rn(x) = sgn(sin(Zn.rrx)) for n

= I, Z, ... , where sgn denotes the signum function sgn C) x

={

-I, I,

ifx < 0, if x > 0.

(a) Show that the set M = {rnl is an orthogonal set on £2[0,1].

(b) Iff is definedbYf(x) = cos(ZJl'X), showthatf J.. rnforalln = I, Z, ... and deduce that M cannot form a total orthogonal set. 4. Let X be an inner product space and let M = {rPnl be an orthonormal basis of X. Show that for any f, g EX, 00

if, g}

= L if, rPk} (g, rPk). k=1

9.3

Classical Fourier Series

We saw in Section 9-1 that £2 is a Hilbert space, and we know from Theorem 8.6.1 that all IY spaces are separable. Theorem 9.2.4 thus implies that the space £2 must have an orthonormal basis. It turns out that the classical Fourier series (i.e., trigonometric series) lead to an orthonormal basis for £2[a, b]. In this section we study classical Fourier series and present some basic results with little detail. There are many specialized texts on the subject ofFourierseries such as [15] and [45], and we refer the reader to these works for most the details. A particularly lively account of the theory, history, and applications of Fourier series can be found in [Z4]. For convenience we focus primarily on the space £2[ -Jr, Jr] and note here that the results can be extended mutatis mutandis to the general closed interval. A classical Fourier series is a series of the form 1

Zao +

L (an cos(nx) + bnsin(nx)), 00

n=l

(9.1)

9.3. Classical Fourier Series

181

where the an's and bn's are constants. We know from Example 9-2-1 that for any nonzero integers m, n,

1

sin(mx) sin(nx) ax = {O,

[-Jr,Jr]

]"(,

~ffm =1= n,

1

m - n.

We also know from Exercises 9-2, No. I, that for all integers m, n,

1

cos(mx) cos(nx)

[-Jr,Jr]

ax =

{' 0

if m =1= n,

]"(,

if m

= n,

and

1

sin(mx) cos(nx) ax

= o.

[-Jr,Jr]

In addition, it is evident that for any integer n,

1

sin(nx) ax

=

[-Jr,Jr]

Let

1

cos(nx) ax

= o.

[-Jr,Jr]

= {rPn} and \II = {'lfrn} denote the sets of functions defined by rPn(x)

=

cos(nx) ~

,

'lfrn(x)

=

sin(nx) ~

,

for n = I, 2, .... Then the above relationships indicate that the set S = {11.J2]r} U U \II forms an orthonormal set in £2[-]"(,]"(]. The Fourier coefficients of a function [ E £2[_]"(,]"(] with respect to S are given by

1) .J2]r 11 (, .J2]r f, -

= -

(f, rPn) = (f, 'lfrn) =

[-Jr,Jr]

1 ~1

~

'\/ ]"(

[-Jr,Jr]

'\/ ]"(

[-Jr,Jr]

[(x)

ax ,

[(x) cos(nx) ax, [(x) sin(nx)

ax,

and the Fourier series is (9.2)

\

,

182

9. Hilbert Spaces and £2

which is equivalent to expression (9.1) with the familiar coefficient relations ao

.!.1 .!.1 = .!.1 =

an = bn

f(x) dx,

re

[-Jr,3l']

re

[-Jr,Jr]

re

[-Jr,Jr]

f(x) cos(nx) dx, f(x) sin(nx) dx.

Fourier series can be expressed in atidier, more symmetric, form using the relation einx = cos(nx) + i sin(nx). Using this relation, the series (9.1) can be written in the form 00

inx

'L...J " cn e

(9.3)

,

-00

where the

Cn

are complex numbers defined by Cn

The set of functions B

= -1

1

2re [-Jr,Jr]

= {f3n},

.

f(x)e znx dx.

where

f3n(x) =

einx

$'

forms an orthonormal set for L~[ -re, re], and for any f E L~[ -re, re] (and consequently for any f E L 2 [ -re, reD the corresponding Fourier series is 00

PBf =

L(f, f3n)f3n.

(9.4)

-00

Now, for any f E L 2 [-re, re], Bessel's inequality guarantees that the Fourier series (9.2) (and (9.3)) converges in the \I . 112 norm to some function Psf E L 2 [ -re, re]. The central question here is whether the set S forms an orthonormal basis for L 2 [-re, re], so that Psf = f a.e. In fact, it can be shown that S forms an orthonormal basis for L 2 [ -re, re]. If we combine this fact with Theorem 9.2.3, we have the following result:

9.3. Classical Fourier Series

183

Theorem 9.3.1 Let f E L 2 [-re, re] and let

k=n

=

L

if, 13k}13k·

k=-n

Then: • (i) IIsn - fll2 -+ 0 as n -+ 00; (ii) Parseval's relation is satisfied:

Ilfll~ = 1if, 1}1 2+

00

L (I if, rPn}1

2

+ 1if, 1/Ik}1 2 )

n=l

re

00

2 2 2 = _a 20 + re "Ca L....Jn + bn)

n=l 00

00

= L 1if, f3n}1 = 2re L Icn l2 . 2

-00

-00

From the last section we know that all separable Hilbert spaces are isomorphic to the Hilbert space £2 and Parseval's relation is a manifestation of this relationship. This observation leads to the following result: I

Theorem 9.3.2 CRiesz-Fischer) Let {c n } E £2. Then there is a unique function f E L 2 [-re , re] such that the Cn are the Fourier coefficients for f. Note that Ilunique" in the above theorem means that the sequence {c n } determines an equivalence class of functions modulo equality a.e. An immediate consequence ofTheorem 9.3.1 and the projection theorem (Theorem 9.2.2 ) is the following result: Theorem 9.3.3 Let f E L 2 [-re, re]. Then for any E > 0 there exists a positive integer n and a trigonometric polynomial ofdegree n, say On = L~-n qkf3k such that liOn - fl12 < E. Moreover; among the trigonometric polynomials of

184

9. Hilbert Spaces and £2

degree n, the closest approximation to f in the II . 112 norm is that for which the qk correspond to the Fourier coefficients. In other words, trigonometric polynomials can be used to approximate any function in L 2 [-re, re], and the Fourier coefficients provide the best approximation among such polynomials in the II . 112 norm. Theorem 9.3.1 guarantees that the Fourier series will converge in the II ·112 norm, but this is not the same as pointwise convergence, and an immediate question is whether or not a Fourier series for a function f E L 2 [ -re, re] converges pointwise to f. More explicitly, for a fixed x E [-re, re] does the sequence of numbers {sn(x)} converge, and if so does sn(x) -+ f(x) as n -+ oo? The answer to this question is complicated, and much of the research on Fourier series revolved around pointwise convergence. Some simple observations can be made. First, the orthonormal basis defining the Fourier series is manifestly periodic, so that if sn(x) converges for x = -re, it also converges for x = re, and s( -re) = sere). The existence of a Fourier series does not require thatf( -re) = f(re), so pointwise convergence will fail at an endpoint unless f satisfies this condition. More generally, we see that any two integrable functions f and g such that f = g a.e. produce the same Fourier coefficients, so that for a specific function f, we expect that the best generic situation would be that sn(x) -+ f(x) a.e. We shall not Ilplumb the depths" of the vast results concerning pointwise convergence of Fourier series; however, we will discuss two results of interest in their own right that make crucial use of the Lebesgue integral. Prima facie, it is not obvious that the coefficients an and bn in the Fourier series have limit zero as n -+ 00. Given that arguments such as x = 0 lead to series such as L~=o an, this is clearly a concern. An elegant result called the Riemann-Lebesgue theorem resolves this • concern and is of interest in its own right. We state the result here in a form more general than is required for the question at hand.

Theorem. 9.3.4 (Riemann-Lebesgue) Let I C JR be some interval and f E L 1 (f). If {An} is a sequence of real numbers such that An -+ 00 as n -+ 00, then

1

{(x) caS(AnX) ax ---> 0 and

as n -+

00.

1

{(x) sin(AnX) ax ---> 0

9.3. Classical Fourier Series

185

Proof The proof of this result is of particular interest because the convergence theorems of Chapter 5 are of little help, and we must return to the definition ofthe integral itself. We sketch here the proof for the cosine integral. Suppose first that I = [a, b] and that f : I ~ 1R is bounded on I. If M denotes an upper bound for IfI, then [

f(x) COS(AnX) dx < M

J[a,b]

cos(Anx)dx

[ J[a,b]

=M

l

b cOS(AnX)d
0, is given by 00

PMf(x)

= LCneirurxlA, -00

where en

= ~ i>(x)e-inm 0 for all x E [a, b]. Then: (i) the spectrum for the Sturm-Liouville problem is an infinite but countable set; (ii) the eigenvalues are all real; (iii) to each eigenvalue there c(J)rresponds precisely one eigenfunction (up to a'constant factor), i.e., the eigenvalues are simple; (iv) the spectrum contains no finite accumulation points. Thus, under the conditions of the above theorem, the SturmLiouville problem always has an infinite (but countable) number of eigenfunctions. The spectrum is a countable set, so we can regard it as a sequence V.n} and impose the condition IAn I < IAn+ll for all n. Since there are no finite accumlll1ation points, we see that IAnI ~ 00 as n ~ 00. Suppose that the conditions of Theorem 9.4.1 are satisfied, and that Am and An are distinct eigenvalues with corresponding eigenfunctions Ym, Yn, respectively. Then, £Ym = -AmP(X)Ym, £Yn

= -r-AnP(x)Yn,

and therefore Ym£Yn - Yn£Ym = (Am - An)P(X)YmYn.

The above equation indicates th~t

1

(Ym(X)£Yn(X) - Yn(X)£Ym(X)) dx

[a,b]

= (Ym, £Yn) - (Yn, £Ym)

= (Am -

An)

1

P(X)Ym(X)Yn(X) dx.

[alb]

Now, the eigenfunctions are real, so that (Ym, £Yn) since £ is self-adjoint, (Am - An)

1

P(X)Ym(X)Yn(x}dx

[a,b]

i i

= (£Yn, Ym) -

= (£Yn,Ym), and (Yn, £Ym)

= O.

192

9. Hilbert Spaces and L 2

By hypothesis, the eigenvalues are distinct (Am =f. An), and the above calculation shows that [

ira,b]

P(X)Ym (X)Yn (X) dx =

o.

(9.10)

If P = I, the above equation im(plies that Ym ..L Yn for m =f. n, and thus the eigenfunctions are orthgonal. By hypothesis we have that p(x) > 0, so that in any event the set of functions {.JPYn} is orthogonal. As the eigenfunctions are by definition nontrivial solutions, we know that IIPYn 112 =f. 0, so that this set of functions can always be normalized to form an orthonormal set in L 2 [a, b]. The Sturm-Liouville problem thus produces eigenfundtions from which orthonormal sets can be derived. Given a continuous function p positive on the interval [a, b], it is always possible to define anothet norm for L 2 [a, b], viz., pllYII2 = { [

ira,b]

P(X)y2(X)dx}1/2

(9.11)

Since P is positive and continuous on [a, b], there exist numbers Pm and PM such that 0 < Pm < p(x) tS PM for all x E [a, b]. This implies that PmllyII2 < pllYII2 < PMIIYII2,

so that the p 1\·112 norm is equivalent to the 1\·112 norm. Consequently, the vector space L 2 [a, b] equippe

0, then the general solution is y(x) = A cos( vlIx) + B sine,..fix),

where A and B are constants. 'the condition yeO) = 0 implies that A = 0, and the condition yen) ::::::l 0 implies that B sine Jin) =

o.

(9.14)

Equation (9.14) is satisfied for BI =f. 0 only if A = n 2 for some integer n, and in this case equation (9.12) has the nontrivial solution Yn(x) :;::: sin(nx).

i I

(9.15)

194

9. Hilbert Spaces and L 2

The set {n 2 } corresponds to the spectrum. Hence from Theorem 9.4.2 we know that the set {Yn/IIYnIl21 =l:: {Yn.J2hr} forms an orthonormal basis for L2 [0, n'). Example 9-4-2: Mathieu Functions Consider the differential equation Y"

+ (A -

2B c'os(2x))y

= 0,

(9.16)

along with the boundary conditions (9.13). Equation (9.16) is called Mathieu's equation, and B is some fixed number. Here, rex) = p(x) = 1 and q(x) = -2B cos(2x). Note ~hat when B = 0 equation (9.16) reduces to equation (9.12). Now, unlike the previous example, we cannot solve equation (9.16) in clbsed form, and it is clear that the eigenvalues will depend on the parameter B. Nonetheless, the above results indicate that for any B, there is a set {An (B)} of eigenvalues with corresponding eigenfunction;s that when normalized will yield an orthonormal basis for L 2 [O, Jr]. The solutions to equation (9.16) are well-known special functions call~d (appropriately enough) Mathieu functions. The intricate details cdncerning these functions can be found in [28] or [41]. Suffice it h~re to say that corresponding to the spectrum {An(B)}, Mathieu's eqllationhas eigenfunctions sen (x, B) that are periodic with period 2Jr artd reduce to sine functions 3 when B = o. The Mathieu functions {s~n(x, B)} thus form an orthogonal basis for L 2 [O, Jr]. The Sturm-Liouville problem; can be posed under more general conditions. These generalizatlions lead to bases for L2 that are widely used in applied mathematics and numerical analysis. The generalizations commonly made Gorrespond to either: (i) relaxing the conditions on p and r at the endpoints so that these functions may vanish (or even be discontinuous) at x = a or x = b (or both); or (ii) posing the problem on an unbounded interval. The general solutions to the differential equations with these modifications are usually unbounded on the interval, and the homo3The notation sen comes from Whit!taker and Watson [41] and denotes "sine-elliptic." There are "cosine-elliptic" functions cen with analogous properties.

i i

9.4~

The Sturm-Liouville Problem

195

geneous boundary conditions (9.6) are often replaced by conditions that ensure that the solution i~ bounded, or that limit the rate of growth of the function as x appr~aches a boundary point of the interval. These generalized versions of the Sturm-Liouville problem are called singular Sturm-Liouville problems. The theory underlying singular Sturm-Liouville problems and the corresponding results are more complicated than those fd>r the regular Sturm-Liouville problems. For example, the spectruIill may consist of isolated points or a continuum, and not every poirtt in the spectrum need correspond to an eigenvalue. The singular Sturm-Liouville problem is studied in some depth in [9] and [40]. More general references such as [5] and [11] give less detailed but clear, succinct accorunts of the basic theory. We content ourselves here with a few exanp.ples that lead to well-known bases for L 2 . The special functions atising in these examples have been studied in great detail by numer 1. (Why is this always possible?) Then choose a2 E 8 such that a2 > max{2, all, a3 E 8 such that a3 > max{3, a2}, and so on. Use a similar approach to choose a sequence bn such that b n ,J, m. Exercises 2-4 3. Hint: GivenE > O,choosensuchthat1/(n+1) < E.Itfollowsfrom the definition off that 0 < x < lin =} 0 < f(x) < 1/(n + 1) < E.

Exercises 2-6 1. Hint: Prove that If+(x) - g+ (x) I < If(x) - g(x)1 for all x E I by considering the four cases f(x) > 0 and g(x) > 0, f(x) > 0 and g(x) < 0, f(x) < 0 and g(x) > 0, f(x) < 0 and g(x) < O.

2. Similar to 1.

3. Hint: For each x E I, If (x) I = If(x) - g(x) + g(x) I < If(x) - g(x) I+ Ig(x) I, therefore If(x)-g(x)1 > If(x)I-lg(x)l. Interchangingf and g gives If(x) - g(x) I > Ig(x) I - If(x)l· Since IIf (x) I - Ig(x) II must equal either If (x) I - Ig(x) I or Ig(x) I - If(x)I, the result follows.

Appendix: Hints and Answers to Selected Exercises

211

Exercises 2·7 1. Hint: Suppose f has bounded variation on I. Choose a point a E I,

and let x be any point in I. Denote by Ix the closed interval with endpoints a and x. Use the fact that {Ix} is a partial subdivision of I, together with the definition of bounded variation, to obtain the required result. 2. Hint: For fg, use n

n

If(bj)g(bj) - f(tlj)g(tlj) I = L

L j==l

•

If(bj)g(bj) - f(bj)g(aj)

j==l

+f(bj)g(tlj) - f(aj)g(aj) I n

If(bj)llg(bj) - g(tlj) I

inf{f(x) : X Elk} and sup{f(x) : X Elk'} < sup{f(x) : X Elk}. 2. Since the partition Q = p U pI is a refinement of both P and pI, Lemma 3.1.2 implies that Sp(f) > SQ(f) and§.o.(f) > §,p,(f). Since SQ (f) > SQ (f) for any partition Q, we have that

Sp(f) > SQ(f) > and the lemma thus follows.

~(f)

2: Sp,(f),

Appendix: Hints and Answers to Selected Exercises

213

Exercises 4-1 1. (b) Jla((O, 1)) = l-e-l, Jla([O,I]) = 3 -e- 1 , Jla((-I,I))4 - e-l, Jla([O,OD = 2, Jla(( -00,1)) = 00, Jla((O,oo)) = I, Jla([O,oo)) = 3. 2. (b) Jla([-I,2)) = 4, Jla((I,oo)) = 2, Jla((-oo,4)) Jla((O,2]) = 5, Jla((~, ~)) = 3, Jla([I, 3]) = 5, Jla((1, 3))

= = 2.

6,

Exercises 4-2 ifx < A, if x > A.

1. a(x) = {

0.,

2. a(x) =

0, Un,

I,

{ I,

if x < AI, if Ai < x < Ai+l (i = 1,2, ... , n - I), if x > An.

Exercises 4-3 1. (a) S U T = [1,8). S n T = (2,3) U (4,5] U (6, 7]. S - T = [1,2] U (5,6] U (7,8).

(b) S U T

= [1,4] U [5,8).

SnT

= (2,3) U [6, 7].

S- T

= [5,6).

(c) S U T = (1,4] U [5, 7). S n T = [2,2] U (5,6). S - T = (1,2) U [5,5]. Exercises 4-4 1. Aa(O)

= O.

2. Aa(O) = O. 3. Aa(O) = 3. 4. Aa(O) = ~.

5. 0 is not a-summable. Exercises 4-5

2. Hint: The difficulty with this one is that it is too easy! Since La*(lfl) = L a *(/) = 0, you can just take On to be the zero function on [0,1], for each n = 1,2, ....

214

Appendix: Hints and Answers to Selected Exercises

Exercise 4-6 (a) Let n be the integer part of c, i.e., the largest integer not exceeding c. Then ifn O. Since f is continuous on [a, b] this means there is some interval [a ,,8] C [a, b] containing c such that If(x) I > 0 for all

216

Appendix: Hints and Answers to Selected Exercises

I:

It

x E [a, ,8]. Now IIfllR = Ifex)1 ax > If(x)1 ax > 0, which contradicts the hypothesis that IIfllR = O. Therefore, II . IIR satisfies property (ii).

(b) Hint: Let c be any number in the interval [a, b] and let f be the function defined by f(x) = 0 if x i= c and f( c) = 1. What is the norm of this function? 3. Hint: Properties (i) and (ii) follow from the inequalities IIfllI,oo > IIflloo and IIfll1,l > IIfIIR. Property (iv) follows from the inequality If(x) + g(x) I < sUPxE[a,bj If(x) I + sUPxE[a,b] Ig(x)l. 4. Hint: ISn+1 - Snl < IlIOn, and ifn > m, then ISn - Sml = I(SnSn-1) + (Sn-1 - Sn-2) + ... + (Sm+1 - Sm)l. sUPxE[a,bj

5. Since IIfllb < ,8l1flla we can choose y = 1/,8. Similarly, we can choose 8 = Ila.

Exercises 8-2 1. Suppose that an ~ a as n ~ 00. Then, for any E > 0 there is an N such that lIam - all < E/2 whenever m > N. Now, lIam - all = lI(am - an) + (an - a) II > lIam - an II - lIan - aiL and thus

lIam - an II - lIan - a II < lIam

E

-

a II < 2'

so that ifn > N, then

lIam - an II -

E

E

2 < 2'

Thus, for any E > 0 there is an N such that lIam - anll < whenever n, m > N.

E

2. Let n be any positive integer. Note that SUPXE[-l,lj Ifn(x) fn+1(X)1 is achieved at x = I/2 n+1, where fn+1CX) = 0 and fn(x) = 1 - 2m- Cm+1) = ~. Thus IIfn(x) - fn+l(X) II 00 = ~ for all n, and {in} cannot be a Cauchy sequence.

Exercises 8-3 1. (a) Let y be any element in Y and choose any E > O. Since W is dense in Y, there is a il; E W such that 1Iil; - YII < E/2. Similarly, since Z is dense in W, there is a z E Z such that

Appendix: Hints and Answers to Selected Exercises

217

liz - wll < E/2. Consequently, for any Y E Y and any E > 0 there is a z E Z such that liz - YII < E, i.e., Z is dense in Y. (b) Use part (a) and the definition of completion. 2. Hint: Use the fact that the set of rational numbers is dense in the

set of real numbers.

Exercises 8-5 1. (a) Hint: To show the triangle inequality rllf+gll p < rllfllp+rllgllp apply the Minkowski inequality with F = rl/Pf and G = rllpg.

(b) See the discussion after equation (9.11). 3. Since f E £2[0, 1], we have that f E £1[0, 1]; since k is bounded, there is an M < 00 such that Ik(x, Y)I < M for all (x, y) E [0, 1] X [0, 1]. Hence,

r

I(KD(x) I