Measure and Integration [1st ed.] 3319290444, 978-3-319-29044-7, 978-3-319-29046-1

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Measure and Integration [1st ed.]
 3319290444, 978-3-319-29044-7, 978-3-319-29046-1

Table of contents :
Front Matter....Pages i-xi
Rings of sets....Pages 1-18
Measurability....Pages 19-42
Integrals and measures....Pages 43-73
Convergence theorems for Lebesgue integrals....Pages 75-104
Existence and uniqueness of measures....Pages 105-132
Signed measures, complex measures, and absolute continuity....Pages 133-165
Measure and topology....Pages 167-182
Product measures....Pages 183-201
The L p spaces....Pages 203-239
Fourier analysis....Pages 241-263
Standard measure spaces....Pages 265-284
Back Matter....Pages 285-300

Citation preview

Hari Bercovici · Arlen Brown · Carl Pearcy

Measure and Integration

Measure and Integration

Hari Bercovici • Arlen Brown • Carl Pearcy

Measure and Integration

123

Hari Bercovici Department of Mathematics Indiana University Bloomington, Indiana, USA

Arlen Brown Department of Mathematics Indiana University Bloomington, Indiana, USA

Carl Pearcy Department of Mathematics Texas A&M University College Station, Texas, USA

ISBN 978-3-319-29044-7 ISBN 978-3-319-29046-1 (eBook) DOI 10.1007/978-3-319-29046-1 Library of Congress Control Number: 2016933772 Mathematics Subject Classification (2010): 28-01, 42-01, 28A05, 54G05 Springer Cham Heidelberg New York Dordrecht London © Springer International Publishing Switzerland 2016 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper Springer International Publishing AG Switzerland is part of Springer Science+Business Media (www. springer.com)

To our good friend Ciprian Foia¸s

Preface

This book was written expressly to serve as a textbook for a one- or twosemester introductory graduate course in the theory of measure and Lebesgue integration, usually designated in graduate programs as real analysis. In writing this book we have naturally been concerned with the level of preparation of the prospective reader. Such a reader who has mastered the majority of our earlier book An Introduction to Analysis (which is referred to in the text as [I]) will be more than adequately prepared to profitably read this book. However, at the other end of the spectrum, any beginning graduate student who has reached a certain level of mathematical maturity, which may be taken to mean the ability to follow and construct ε-δ arguments, should have no difficulty in mastering the material in this book. We have deliberately made this book as self-contained as possible. In keeping with our pedagogical intent, we have provided in each chapter a copious supply of examples and a lengthy collection of problems. Some of the problems that appear in these problem sets are stated simply as facts (see, for example the first sentence of Problem 1D). In this case, the reader is supposed to supply a proof of the stated fact. Hints are provided for the more challenging problems. These problem sets constitute an integral part of the book and the reader should study them along with the text. Working problems is, of course, critically important in the study of mathematics because that is how mathematics is learned. In this textbook it is particularly important because many topics of interest are first introduced in the problem sets. Not infrequently, the solution of a problem depends on material in one or several previous problems, a fact that instructors should keep in mind when assigning problems to a class. While, as noted, this book is intended to serve as a textbook for a course, it is our hope that the wealth of carefully chosen examples and problems will also make it useful to the reader who wishes to study real analysis individually. This certainly has proved to be the case for our earlier books [I] and [II]. vii

viii

Preface

One novel feature of our development of integration theory in this book is that the Lebesgue integral is defined axiomatically (see Chapter 3), and only after its usual properties have been revealed is it proved that there is a one-to-one correspondence between measures and Lebesgue integrals. To our knowledge, this approach is original with the authors and it was sketched in our earlier text [II]. Our exposition has the advantage that the Lebesgue integral is developed before, and independently of, any discussion of measure spaces. This order of doing things seemed to be quite successful in classroom use. In particular, preliminary versions of this book have been used successfully in graduate courses at The University of Michigan, Indiana University, and Texas A&M University. We take this opportunity to thank the many students in these courses who pointed out inaccuracies in the text and problem sets. Any remaining mistakes are entirely the authors’ responsibility. In writing this book no systematic effort has been made to attribute results or to assign historical priorities. The association of particular names to theorems serves primarily as a memory aid. The notation and terminology used throughout the book are in essential agreement with those to be found in contemporary textbooks. In particular, the symbols N, N0 , Z, Q, R, and C always represent the systems of positive integers, nonnegative integers, integers, rational numbers, real numbers, and complex numbers, respectively. The closure of a set A in a topological space is denoted A or A− . One basic convention valid throughout the book is that all vector spaces encountered are either real or complex, and the relevant field of scalars should be clear from the context. The numbering system for examples, propositions, theorems, remarks, etc., is somewhat standard. For instance, Proposition 4.21 is the 21st fact stated in Chapter 4 and it is followed by Example 4.22. Also, Problem 3Z is the 26th problem at the end of Chapter 3 and is followed by Problem 3AA. The reader who is interested in the historical development of measure and Lebesgue integration theory would do well to consult von Neumann [vN] and Halmos [H]. Excellent comprehensive reference books on this subject are Fremlin [F] and Bogachev [B]. Useful treatments of analytic sets are to be found in Lusin [L] and Kuratowski [CK] (historical) and Kechris [AK] (contemporary). For further information on functional analysis, one might consult Brown-Pearcy [II], Dunford-Schwartz [DS], and Rudin [R]. Similarly, for Fourier analysis, Zygmund [Z], Stein [S], and Muscalu-Schlag [MS] are excellent references. Of course, of necessity, many topics in measure theory and integration are not dealt with in this book; for instance, vector measures and Haar measure. Bloomington, IN, USA Bloomington, IN, USA College Station, TX, USA February, 2016

Hari Bercovici Arlen Brown Carl Pearcy

Preface

ix

Bibliography [I] A. Brown and C. Pearcy, An introduction to analysis. Graduate Texts in Mathematics, 154. Springer Verlag, New York, 1995. [II] ———, Introduction to operator theory. I. Elements of functional analysis. Graduate Texts in Mathematics, No. 55. Springer Verlag, New York-Heidelberg, 1977. [vN] J. von Neumann, Functional Operators. I. Measures and Integrals. Annals of Mathematics Studies, no. 21. Princeton University Press, Princeton, N. J., 1950. [H] P. R. Halmos, Measure Theory. D. Van Nostrand Company, Inc., New York, N. Y., 1950. [F] D. H. Fremlin, Measure theory. Volumes 1–5. Torres Fremlin, Colchester, 2004–2006. [B] V. I. Bogachev, Measure theory. Volumes I, II. Springer-Verlag, Berlin, 2007. [L] N. Lusin, Le¸cons sur les ensembles analytiques et leurs applications. Avec une note de W. Sierpi´ nski. Preface de Henri Lebesgue. Reprint of the 1930 edition. Chelsea Publishing Co., New York, 1972. [CK] C. Kuratowski, Topologie. I et II. Part I with an appendix by A. Mostowski and R. Sikorski. Reprint of the fourth (Part I) and third (Part II) editions. ´editions Jacques Gabay, Sceaux, 1992. [AK] A. Kechris, Classical descriptive set theory. Graduate Texts in Mathematics, 156. Springer Verlag, New York, 1995. [DS] N. Dunford and J. T. Schwartz, Linear operators. Parts I–III. Wiley Classics Library. John Wiley & Sons, Inc., New York, 1988. [R] W. Rudin, Functional analysis. Second edition. International Series in Pure and Applied Mathematics. McGraw-Hill, Inc., New York, 1991. [Z] A. Zygmund, Trigonometric series. Volumes I, II. Third edition. With a foreword by Robert A. Fefferman. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 2002. [S] E.M. Stein, Harmonic analysis: real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy. Princeton Mathematical Series, 43. Monographs in Harmonic Analysis, III. Princeton University Press, Princeton, NJ, 1993 [MS] C. Muscalu and W. Schlag, Classical and multilinear harmonic analysis. Volumes I, II. Cambridge Studies in Advanced Mathematics, 137. Cambridge University Press, Cambridge, 2013.

Contents

1

Rings of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

2

Measurability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

3

Integrals and measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

4

Convergence theorems for Lebesgue integrals . . . . . . . . . .

75

5

Existence and uniqueness of measures . . . . . . . . . . . . . . . . .

105

6

Signed measures, complex measures, and absolute continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

133

7

Measure and topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

167

8

Product measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183

9

The Lp spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

203

10 Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

241

11 Standard measure spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .

265

Subject index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

285

Notation index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

299

xi

Chapter 1

Rings of sets

It is a familiar fact of elementary calculus that the integral of a function exists only if the function is continuous, or nearly so. In the theory of the Lebesgue integral, with which we are concerned in this book, continuity is replaced by a significantly less stringent requirement known as measurability. This concept, in turn, is defined in terms of a certain type of collection of sets, called a σ-algebra, and so we begin with a brief look at this and some related concepts. One says of a collection C of sets that it is closed with respect to one or another operation on sets if, whenever the operation is performed on sets belonging to C, the resulting set also belongs to C. For example, if A∪B belongs to C whenever A and B do, then C is closed with respect to (the formation of) unions. Likewise, if C is a collection of subsets of a fixed set X, and X \ A belongs to C whenever A does, then C is said to be closed with respect to (the formation of) complements, or, more simply, to be complemented. More generally, if A \ B belongs to C whenever A and B do, then C is closed with respect to (the formation of) differences. These examples bring us at once to one of the most important concepts with which we are concerned. Definition 1.1. A nonempty collection R of subsets of a set X is a ring (of sets) in X (or a ring of subsets of X) if R is closed with respect to the formation of both unions and differences. A ring S of sets in X is a σ-ring in X if it is closed with respect to the formation of countable unions, that is,  if the union C belongs to S whenever C is a countable subcollection of S. A ring S of sets in X is an algebra if it is complemented. A σ-ring S of sets in X is a σ-algebra if it is complemented. If R is a ring of sets in X, and if A, B, and C belong to R, then it is clear that A ∪ B ∪ C does so too. More generally, by mathematical induction, if A1 , . . . , An all belong to R, then so does A1 ∪· · ·∪An . This modest observation provides the proof of a key part of the following elementary proposition.

© Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1 1

1

2

1 Rings of sets

Proposition 1.2. If R is a ring [σ-ring] of subsets of a set X, then R is closed with respect to the formation of finite [countable] unions and intersections. Furthermore, the empty set ∅ belongs to R, and R is complemented if and only if X  belongs to R. Every σ-ring is a ring, and a ring R is a σ-ring if and only if n An ∈ R for every (pairwise disjoint) sequence {An }∞ n=1 of sets in R. Proof. A ring R of sets contains at least one set E, and therefore also contains ∅ = E \ E. Thus R contains X if it is complemented. Suppose now that F  is a finite collection of sets in a ring R. Then either F is empty, whereupon F = ∅, or F can be enumerated as a finite  sequence {A1 , . . . , An }, in which case A1 ∪ . . . ∪ An ∈ R as noted above. Thus F ∈ R in any case, so R is closed with respect to finite unions. Suppose next that R is a ring of sets in X, and that R has the property  in R. that n An ∈ R whenever {An }∞ n=1 is a (pairwise disjoint) sequence  A countable subcollection C of R is either finite (in which case C is already known to belong to R) or can be enumerated as an infinite sequence {An }∞ n=1 . Along with the sets An the ring R contains the pairwise disjoint sets Bn defined as:  Aj , n ∈ N, Bn = An \ j 0, show that there exists a set E ∈ S such that μ(E) < ∞ and  |f |dμ < ε for all n ∈ N. (Hint: Adapt the argument following (3.11).) X\E n



(ii) Given ε > 0, show that there exists δ > 0 such that μ(A) < δ implies A |fn |dμ < ε for every n ∈ N. (Hint: Use the Cauchy property to reduce to the case of a finite set f1 , . . . , fn , then use Problem 3N to further reduce to the case where these functions are bounded.) 3P. Let (X, S, μ) be a measure space, and let s be a measurable simple complex-valued function on X. Let us call a representation of s of the form (†)

s=

n 

αi χ E i

i=1

standard if the sets Ei , i = 1, . . . , n, are all measurable and of finite measure and are also pairwise disjoint. Show that s admits a standard representation if and only if its support has finite measure, that is, if and only if s belongs to the vector space

3 Integrals and measures

71

L0 introduced in the proof of Theorem 3.37. Show also that if s = another standard representation of s along with (†), then n 

αi μ(Ei ) =

i=1

m 

m k=1

βk χFk is

βk μ(Fk ),

k=1



n so that the number i=1 αi μ(Ei ) associated with s by means of a standard representation (†) depends only on s and not on the representation. Use this fact to prove that the functional thus defined on L0 is linear, and hence coincides with the  functional s → X s dμ introduced in the proof of Theorem 3.37. Show also that this functional is self-conjugate and that

      ≤ s dμ |s| dμ   X

X

for any s ∈ L0 .



3Q. Let (X, S, μ) be a measure space, and consider the functional f → X f dμ defined on the vector space L in the proof of Theorem 3.37. Show directly from the definition of this linear functional that it is self-conjugate and that if f ∈ L, then |f | ∈ L and

      ≤ f dμ |f | dμ.   X

X



Show  also that if f ∈ L and if E ∈ S, then f χE  ∈ L. Verify that if we write E f dμ for X f χE dμ, then the set function νf (E) = E f dμ is finitely additive on S. 3R. If (X, S, μ) is a measure space, then a measurable set E in X is said to be σ-finite with respect to μ if there exists a  countable sequence {En } of sets of finite measure with respect to μ such that E = n En . The collection of all sets in S that are σ-finite with respect to μ is a σ-ideal in S (Problem 1W). So is the collection Z of all sets of measure zero with respect to μ. (The sets of finite measure constitute an ideal in S between these two σ-ideals.) (i) Verify that if f is a function that is integrable [μ], then the support of f is σ-finite with respect to μ. (ii) Assume that E is a σ-finite set in S such that μ(F ) ≤ a for every measurable subset F of E having finite measure. Show that μ(E) ≤ a. Show by example that the assumption that E be σ-finite cannot be omitted. (iii) Let f be a complex-valued function defined and measurable on X with σ-finite support. Show that the approximating functions in Problem 2J can be taken to be integrable [μ]. 3S. Let (X, S, μ) be a measure space such that X is σ-finite. Show that there exists a finite measure ν on (X, S) such that {E ∈ S : μ(E) = 0} = {E ∈ S : ν(E) = 0}. 3T. Verify that the measures associated with the Lebesgue integrals introduced in Examples 3.4, 3.7, and 3.8 are discrete, by calculating in each case an appropriate weight for each point of X. (Prove also that these weights are uniquely determined if the σ-algebra S contains all singletons in X.) Show, in the same vein, that if the measures associated with the integrals Lγ in Problem 3D are all discrete, then the same is true of the integral L of that problem. How are matters changed if, in Example 3.8, we drop the assumption that the points of the sequence {xn } are all distinct?

72

3 Integrals and measures

3U. Let (X, S, μ) be a measure space, and let f be a nonnegative function defined and measurable on  X and having σ-finite support. Consider the supremum M of the set of integrals X s dμ of all integrable simple functions s such that 0 ≤ s ≤ f . Show that f is integrable [μ] if and only if M < +∞, and show also that, if this is the case,  then M = X f dμ. (This fact can be exploited to obtain an alternative construction of the Lebesgue integral with respect to μ. That the requirement of σ-finiteness of the support of f is really needed in this problem may be seen by considering the function f = 1 on the measure spaces in Example 3.30.) Remark 3.44. These last results show clearly that if f is a nonnegative measurable function on a measure space (X, S, μ), then the only way for f not to be integrable [μ] is for f to be too large. Indeed, if the support of f is not σ-finite, then f is too large to be integrable [μ] (Problem 3R). On the other hand, if f has σ-finite support, then there exists a monotone increasing sequence {fn } of nonnegative integrable functions that con verges pointwise tof , and either the sequence { X fn dμ} is bounded, in which case f is integrable, or limn X fn dμ = +∞. For this reason it is customary to define

 f dμ = +∞ X

whenever f is nonnegative and measurable on X but not integrable over X. It should be

noted that the custom of writing γ∈Γ tγ = +∞ when {tγ }γ∈Γ is an indexed family of nonnegative real numbers that is not summable (see Example 3.10) is a special case of this convention. 3V. Let (X, S, μ) be a finite measure space, and let f be a measurable real-valued function defined on X. (i) Suppose first that f is bounded and that the range of f is contained in the half-open interval (a, b]. Let {a = t0 < · · · < tn = b} be a partition of [a, b], and for i = 1, . . . , n, let Ei = E(ti−1 < f ≤ ti ). Let ε be a positive number and suppose that ti − ti−1 ≤ ε for all i. Show that 0≤

n 

 ti μ(Ei ) −

f dμ ≤ εμ(X). X

i=1

(ii) In the general case fix ε > 0 and let {tn }+∞ n=−∞ be a two-way infinite sequence of real numbers such that 0 ≤ tn − tn−1 ≤ ε for every integer n and such that tn → ±∞ as n → ±∞. Then f is integrable [μ] if and only if the series +∞ 

tn μ(En )

n=−∞

is absolutely convergent, and, if this is the case, then 0≤

+∞  n=−∞

 tn μ(En ) −

f dμ ≤ εμ(X). X

Remark 3.45. These observations can also be made the basis for an alternative construction of the Lebesgue integral with respect to μ when μ(X) < +∞.} 3W. Let X be a set and let S0 denote the σ-ring generated by X and the singletons in X, that is, the σ-algebra consisting of all countable subsets of X and their complements.

3 Integrals and measures

73

Let γ denote the counting measure on the measurable space (X, 2X ), and let γ0 be the restriction of γ to S0 . Show that the functions integrable [γ0 ] over X are precisely the functions integrable [γ]. (See Example 3.32. The like observation may also be made, of course, for any σ-algebra between S0 and S.) Remark 3.46. The point of this problem is that measurability is not a matter of consequence in the context of discrete measures. The integration theory for a discrete measure is completely independent of the σ-algebra S of measurable sets provided S contains all singletons. It is appropriate to recall, however, that a Lebesgue integral is a triple (X, L, L), where X is a measurable space, which presupposes, of course, that some particular σ-algebra of measurable sets has been specified. 3X. (This problem and the following one refer to Lebesgue-Borel measure in Euclidean space; see Example 3.42.) Prove that if E is a Borel set in Rd and if ε is a positive number, then there exists an open set U in Rd such that E ⊂ U and such that λd (U \ E) < ε. Use this fact to show that for any Borel set E in Rd there is a decreasing sequence {Um } of open sets containing E and having the property that limm λd (Um \ E) = 0, as well as an increasing sequence {Fm } of closed sets contained in E such that limm λd (E \ Fm ) = 0. Conclude that there are sets H and K, where H is a Gδ and K an Fσ , such that K ⊂ E ⊂ H and such that λd (H \E) = λd (E\K) = 0. (Hint: If L denotes the collection of all Borel subsets A of Rd with the property that for each positive number ε there exists an open set U such that A ⊂ U and λ(U \ A) < ε, then L is a σ-lattice (Problem 1N). Verify directly that every closed set belongs to L.) 3Y. Let {θn }∞ n=0 be a sequence of numbers such that 0 < θn < 1 for each n, and let C{θn } be the generalized Cantor set associated with this sequence ([I, Example 6P]). ∞ Show that λ1 (C{θn } ) > 0 when and only when n=0 θn < +∞. (In particular, the Lebesgue-Borel measure of the Cantor set C itself is zero.) Show too that, while μ(C{θn } ) is always strictly less than one, if r is any real number such that r < 1, then there are sequences {θn } for which r < μ(C{θn } ) < 1. (Hint: Let Fn be the nth closed set defined in the inductive construction of C{θn } . Direct calculation shows that λ1 (Fn ) = (1 − θ0 ) . . . (1 − θn−1 ), n ∈ N. Hence λ(C{θn } ) is given by the infinite product



n=0 (1

− θn ).)

3Z. Consider an algebra R of subsets of a set X, and a (finite) finitely additive measure μ : R → [0, +∞). Denote by L∞ the vector space of all bounded functions f : X → C which are measurable [R], that is, f −1 (A) ∈ R for every Borel set A ⊂ C. (i) Show that for every f ∈ L∞ there exist simple functions s1 , s2 , · · · ∈ L∞ such that |sn | ≤ |f | and |f − sn | ≤ 1/n for every n ∈ N. (ii) Show that there exists a linear functional L : L∞ → C such that L(χA ) = μ(A) for every A ∈ R and |L(f )| ≤ μ(X) supX |f | for every f ∈ L∞ . This functional is positive, that is L(f ) ≥ 0 if f ≥ 0. 3AA. Consider an algebra R of subsets of a set X, and a positive linear functional L : L∞ → C satisfying the inequality |L(f )| ≤ supX |f | for every f ∈ L∞ . Show that the formula μ(A) = L(χA ) defines a finitely additive set function on R. Remark 3.47. The preceding two problems provide an analog of the developments in this chapter for (finite) finitely additive set functions defined on a ring. The relationship  between L and μ is also indicated by the notation L(f ) = X f dμ. Note that this functional satisfies axiom (L1 ), while (L2 ) must be replaced by the condition that limn L(fn ) = L(g) if fn → g uniformly. 3BB. Let (X, S, μ) be a measure space, let A ∈ S be an infinite atom for μ, and let f : X → C be integrable [μ]. Show that f = 0 almost everywhere [μ] on A.

Chapter 4

Convergence theorems for Lebesgue integrals

Lebesgue integration is a powerful tool principally on account of several convergence theorems (Theorems 4.24, 4.29, 4.31, and 4.35), and these are the main focus of this chapter. There are, however, several other things to be established. We begin by introducing the signed and complex counterparts of a measure. We recall that finite, finitely additive set functions ϕ defined on a ring always satisfy ϕ(∅) = 0. Definition 4.1. Let (X, S) be a measurable space. Then a signed measure on (X, S) is a countably additive set function ν : S → R such that ν(∅) = 0. The value ν(E) of ν at a measurable set E is called the measure of E, and the measurable space (X, S) together with a signed measure ν on (X, S) is called a signed measure space, denoted by (X, S, ν). Similarly, a complex measure on (X, S) is a countably additive set function ϕ : S → C. The value ϕ(E) of ϕ at a measurable set E will be called the measure of E, and the measurable space (X, S) together with a complex measure ϕ on (X, S) is a complex measure space, denoted by (X, S, ϕ). Finally, when no confusion is possible, both the signed measure space (X, S, ν) and the complex measure space (X, S, ϕ) will be denoted simply by X. This terminology, while standard in the literature, is less logical than it might be. Thus a complex measure is not a measure in general, nor is a signed measure necessarily a measure. Moreover, a measure is not, in general, a complex measure, since it may assume the value +∞. It is true, however, that every measure is a signed measure, and a finite measure (or, more generally, any finite-valued signed measure) is also a complex measure. Definition 4.2. Let (X, S, μ) be a measure space, and let ϕ be a set function defined on S that is either complex-valued or extended real-valued. We say that ϕ is absolutely continuous with respect to μ or, more simply, absolutely continuous [μ] if the condition μ(E) = 0 implies ϕ(E) = 0 for every E ∈ S. We write © Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1 4

75

76

4 Convergence theorems for Lebesgue integrals

ϕμ to indicate that ϕ is absolutely continuous [μ]. In this chapter, the above definition is used primarily when ϕ is a complex measure. In this case absolute continuity takes on a stronger aspect. For the following proposition, we only consider positive measures, though the result is true for complex measures. Indeed, complex measures are linear combinations of finite measures as seen in Chapter 6. Proposition 4.3. Let (X, S, μ) be a measure space, and let ϕ : S → [0, +∞) be a measure such that ϕ  μ. Then, for every positive number ε there exists a positive number δ such that μ(E) < δ implies ϕ(E) < ε for every E ∈ S. Proof. Assume, to get a contradiction, that the conclusion of the proposition is not true for some ε0 > 0. Then there exist sets n) <  En ∈ S such that μ(E ∞ 2−n and ϕ(En ) ≥ ε0 for n ∈ N. Define FN = n≥N En and F = N =1 FN . We have ∞ ∞

μ(En ) < 2−N = 2−N +1 μ(F ) ≤ μ(FN ) ≤ n=N

n=N

for every N ∈ N, and therefore μ(F ) = 0. On the other hand, ϕ(FN ) ≥ ϕ(EN ) ≥ ε0 ,

N ∈ N.

We show that ϕ(F ) ≥ ε0 , thus contradicting absolute continuity. Indeed, the countable additivity of ϕ implies ϕ(F ) = ϕ(FN ) −



ϕ(Fn+1 \ Fn ).

n=N

∞ Since the series n=1 ϕ(Fn+1 \ Fn ) converges, we can let N → +∞ to con  clude that indeed ϕ(F ) ≥ ε0 , as claimed. Example 4.4. The notion of absolute continuity was originally applied to functions of one real variable where it is seen to be a stronger property than uniform continuity. Consider indeed μ = λ1 on the Borel sets of the real line, and let ϕ be a finite measure defined on the Borel subsets of R such that ϕ  λ1 . Define a monotone increasing function h : R → [0, +∞) by setting h(t) = ϕ((−∞, t]) for t ∈ R. Thus, we have ϕ((a, b]) = h(b) − h(a) for a, b ∈ R, a ≤ b. Proposition 4.3 implies the following statement: for every ε > 0 there exists δ > 0 suchthat, for any collection of pairwise disjoint intervals n n {(ai , bi ]}ni=1 satisfying i=1 (bi − ai ) < δ, we have i=1 |h(bi ) − h(ai )| < ε. In particular, the function h is uniformly continuous, as can be seen by looking at a single interval. More generally, a function h : R → C is said to be absolutely continuous when the preceding statement is true. This terminology is also used for functions h defined on a proper interval.

4 Convergence theorems for Lebesgue integrals

77

Definition 4.5. Let (X, S, μ) be a measure space, and let {ϕγ }γ∈Γ be an indexed family of set functions defined on S which are either complex-valued or extended real-valued.. We say that the family {ϕγ }γ∈Γ is uniformly absolutely continuous [μ] if for every positive number ε there exists a positive number δ such that μ(E) < δ implies |ϕγ (E)| < ε for every γ ∈ Γ. We need one more important property which a set function may possess. Definition 4.6. Let (X, S, μ) be a measure space, and let ϕ be a set function defined on S that is either complex-valued or extended real-valued. We say that ϕ is concentrated on sets of finite measure with respect to [μ] if for every ε > 0 there exists a set Eε ∈ S such that μ(Eε ) < +∞ and |ϕ(F )| < ε for every F ∈ S such that F ⊂ X \ Eε . (4.1) An indexed family {ϕγ }γ∈Γ is uniformly concentrated on sets of finite measure with respect to μ if the set Eε in (4.1) can be chosen independently of γ. An important kind of set function is the indefinite integral of an integrable function. Readers who recall the definition of indefinite integral given in elementary calculus may find this terminology puzzling. The connection between point functions and set functions on R is discussed later. Definition 4.7. Let (X, S, μ) be a measure space, and let f : X → C be integrable [μ]. Then the set function νf defined by setting  f dμ, E ∈ S, νf (E) = E

is called the indefinite integral of f . Let F be a family of measurable complex valued functions on X all of which are integrable [μ]. We say that the family F is uniformly integrable [μ] if the family of measures {ν|f | : f ∈ F} is uniformly absolutely continuous [μ] and uniformly concentrated on sets of finite measure with respect to μ. Indefinite integrals have appeared before, but not by name, and only in a special case. The following result summarizes their main properties. Proposition 4.8. Let (X, S, μ) be a measure space and let f be a complexvalued function that is integrable [μ] over X. Then the indefinite integral νf is a complex measure on X that is absolutely continuous and concentrated on sets of finite measure with respect to μ. Proof. If E = E1 ∪ · · · ∪ EN is any partition of a (measurable) set E into disjoint measurable sets E1 , . . . , EN , then f χE = f χE1 + · · · + f χEN and therefore

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4 Convergence theorems for Lebesgue integrals

νf (E) = νf (E1 ) + · · · + νf (EN ). Thus νf is finitely additive by virtue of the linearity of the integral. To complete the proof that νf is a complex measure it suffices to verify semicontinuity (Problem 3I). Let {Fn } be a monotone increasing sequence of measurable sets having union E. We must show that νf (Fn ) → νf (E), or what comes the same thing, that νf (Rn ) → 0, where Rn = E \ Fn , n ∈ N. Since to      Rn f dμ ≤ Rn |f | dμ, it clearly suffices to prove this latter fact for |f | in place of f , and since it also suffices to prove the other two assertions of the proposition for |f | in place of f , we assume henceforth that f itself is nonnegative. Consider the sequence {fn } of nonnegative functions defined by fn = f χFn , n ∈ N. Clearly {fn } is monotone increasing and converges pointwise to f χE . But then νf (Fn ) → νf (E) by Proposition 3.15, and the countable additivity follows. The absolute continuity νf ≺ μ is obvious. To complete the proof, we continue assuming that f ≥ 0. We introduce the sequence of measurable sets {Hn }∞ n=1 , where Hn = E(f ≥ 1/n), and the corresponding sequence {hn = f χHn }∞ n=1 of measurable functions. The sequence {Hn } is monotone increasing with union Nf , so {hn } is monotone increasing and converges pointwise to f = f χNf . Hence νf (Hn ) → νf (Nf ) = νf (X) (Proposition 3.15). Thus it suffices to show that the sets Hn are all of finite measure   with respect to μ. This follows from the fact that χHn ≤ nf . As noted in the preceding proof, the measure νf is a linear combination of finite positive measures, and therefore its absolute continuity implies the stronger property described in Proposition 4.3. Example 4.9. If ϕ is an arbitrary complex measure on a measurable space (X, S), then it is obvious that the set functions α(E) = ϕ(E)

and

β(E) = ϕ(E),

E ∈ S,

are finite-valued signed measures on (X, S). Conversely, if α and β are any two finite-valued signed measures on (X, S), then setting ϕ(E) = α(E) + iβ(E),

E ∈ S,

defines a complex measure on (X, S). If, in particular, ϕ = νf for some integrable function f on X, then α = ϕ and β = ϕ are the indefinite integrals of f and f , respectively. Example 4.10. If a real-valued function f is integrable [μ] on a measure space (X, S, μ), then the indefinite integral νf is a finite-valued signed measure on X. Let A = E(f ≥ 0) and B = E(f < 0). This partition X = A ∪ B of X into two measurable sets has the property that νf (E) ≥ 0 for every measurable subset E of A, while νf (E) ≤ 0 for every measurable subset E

4 Convergence theorems for Lebesgue integrals

79

of B. (As a matter of fact, such a partition of a signed measure space always exists, but this is a relatively deep result. Signed measures are studied in their own right in Chapter 6.) Example 4.11. Let f be an integrable real-valued function on a measure space (X, S, μ). Then the functions f + and f − are also integrable [μ] (Proposition 3.3) and it is obvious that νf = νf + − νf − setwise on S. Thus the signed measure νf is expressed as the difference of two ordinary finite measures. This construction is generalized in Chapter 6. An important element in the theory of Lebesgue integrals, and one that helps to give that theory its characteristic flavor, is the prominent role played by sets of measure zero, or null sets as they are sometimes called. Example 4.12. Let (X, S) be a measurable space, and let {wx }x∈X be a family of nonnegative weights. Then a subset Z of X is a null set for the associated discrete measure on (X, S) (Example 3.20) if and only if Z belongs to S and the weight wx of every point x of Z is zero. In particular, the only null set with respect to the counting measure on (X, S) (Example 3.21) is the empty set. Example 4.13. A singleton in R has measure zero with respect to LebesgueBorel measure λ1 (Example 3.42). Hence every countable set in R is a null set with respect to λ1 . In particular, the set Q has Lebesgue-Borel measure zero. Example 4.14. A subset Z ⊂ R satisfies λ1 (Z) = 0 if and only if for each ε > 0 there is a countable covering {(an , bn )}∞ n=1 of Z by intervals (an , bn ) such that ∞

(bn − an ) < ε n=1

(see Problem 3X). The Cantor set C ([I, Example 6O]) is covered by the system Fn of 2n intervals for each index n, and since each of the intervals in Fn has length 1/3n , the sum of the lengths of the covering intervals is (2/3)n . Thus C is a noncountable null set in R. (In this context, recall Problem 3Y). Example 4.15. Given an arbitrary measurable subset Z of a measure space (X, S, μ), there is a simple way to define a measure on X that makes Z into a null set and that leaves the measure μ unchanged on the complement E = X \Z. Indeed, if we write μE (A) = μ(A∩E) for each measurable set A in X, then μE is a measure on (X, S) with the required properties. The measure μE , which we shall call the concentration of μ on E, may also be described as the indefinite integral of χE with respect to μ whenever μ(E) < +∞. It is instructive to compare Lebesgue integration with respect to μE with Lebesgue integration on the subspace (E, S ∩ E, μ|E) (Example 3.34).

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4 Convergence theorems for Lebesgue integrals

Example 4.16. It is clear from the monotonicity and countable additivity of every measure that the collection Z of all null sets in an arbitrary measure space (X, S, μ) is a σ-ideal in S. (Indeed, this fact is implicit in several of the foregoing examples.) Conversely, if Z is an arbitrary σ-ideal in the σ-algebra S of measurable sets in an arbitrary measurable space (X, S), then there are measures on (X, S) having Z as the σ-ideal of null sets; see Example 3.4. Given a property p(x) which a point x in a measure space (X, S, μ) may or may not have, we say that the property holds almost everywhere with respect to μ, or almost everywhere [μ], if there exists a null set Z with respect to μ such that p(x) holds for every x ∈ X \ Z. (If there can be no doubt as to which measure μ is intended, we may simply say that the property holds almost everywhere. When μ is a probability measure, we may also say almost surely instead of almost everywhere.) For instance, two functions f and g on X are equal almost everywhere [μ] (f = g almost everywhere [μ]) if there exists a null set Z such that {x ∈ X : f (x) = g(x)} ⊂ Z, and two subsets E and F of X are almost equal [μ] if χE = χF almost everywhere [μ]. Likewise, a sequence {fn } of complex-valued functions on (X, S, μ) converges almost everywhere to a limit g if there exists a null set Z with respect to μ such that {fn (x)} converges to g(x) for x ∈ X \ Z. It is important to note that if each of a countable sequence {pn } of properties holds almost everywhere [μ], then all of the properties pn hold simultaneously almost everywhere [μ]. This is true because a countable union of null sets is again a null set. The reason sets of measure zero are so important is that, for most purposes, they are totally negligible. The next two results illuminate this apparent paradox. Proposition 4.17. Let X be a measure space equipped with a measure μ, and let f be a measurable complex-valued function on X such that f = 0 almost everywhere [μ]. Then f is integrable [μ] and X f dμ = 0. Proof. It suffices to treat the case f ≥ 0. By hypothesis the support Nf = E(f > 0) of f has measure zero. Consider the sequence of truncations gn = and 0 ≤ gn ≤ nχN , it follows that gn is f ∧ n, n ∈ N. Since gn is measurable  integrable [μ] and that 0 ≤ X gn dμ ≤ n X χN dμ = nμ(N ) = 0 for every n. But 3.15 it follows that f is also integrable [μ] and that   then, by Proposition g dμ tends to X f dμ = 0.   X n Corollary 4.18. Let X be a measure space equipped with a measure μ and let f and g be measurable complex-valued functions on X such that f = g almost everywhere [μ].  Then fis integrable [μ] if and only if g is, and when both are integrable, X f dμ = X g dμ. Proof. Assume that f is integrable [μ]. Proposition 4.17 implies that g − f is  integrable [μ], and therefore g = f + g − f is integrable as well. Moreover, (g − f ) dμ = 0 and this yields the desired identity. The corollary follows X by symmetry.  

4 Convergence theorems for Lebesgue integrals

81

One can paraphrase these last two results in the language of quotient spaces (see [I, Chapter 3] for definitions). Let (X, S, μ) be a measure space and let L and M denote, respectively, the (complex) vector space of all complex-valued functions integrable [μ] on X, and the (complex) vector space of all measurable complex-valued functions on X. Then according to Proposition 4.17 the set M0 of those functions in M that vanish almost everywhere [μ] is a linear submanifold of L as well as of M. Let f be a function in M and let f˙ denote the coset of f in the quotient space M/M0 . Then a func tion g in M belongs to f˙ if and only if X |f − g| dμ = 0. In particular, if   f is integrable [μ] and if g belongs to f˙, then X g dμ = X f dμ. Thus it   makes sense to define X f˙ dμ = X f dμ. Indeed, once a measure μ is fixed, it is frequently advantageous to think of integration with respect to μ as a linear functional on the quotient space L/M0 rather than on L itself. In this same connection there is also a modest expansion of the concept of measurability that is sometimes useful. Definition 4.19. If (X, S, μ) is a measure space, and if f is either a complexvalued function or an extended real-valued function whose domain of definition is a measurable subset D of X such that μ(X \ D) = 0, then f will be said to be measurable with respect to [μ], or measurable [μ], on X if f is measurable on the subspace D. The point of this usage is that such a function certainly possesses measurable extensions to all of X. Corollary 4.18 indicates that it is immaterial (for purposes of integration theory) which extension is chosen, and often there is no reason to choose one at all. In the same spirit we shall say of a complex-valued function f that is measurable [μ] on X that f is integrable [μ] on X if the measurable extensions of f to X are integrable [μ] (if one is, then all are), and we declare the common integral of these extensions to be the integral  f dμ of f with respect to μ. X With this extension of the notion of integration with respect to a measure μ on a measurable space X we admit as integrable [μ] functions that are defined on various subsets of X, and such functions, of course, do not form a vector space. Thus it becomes slightly inaccurate to call integration with respect to μ a linear functional. It is clear, however, that if f and g are any two complex-valued functions that are measurable [μ] and if α and β are any complex numbers, then αf + βg, defined pointwise almost everywhere on X, is also measurable [μ], and likewise that if f and g are both integrable [μ], then    (αf + βg) dμ = α X

f dμ + β X

g dμ. X

Thus integration retains the property of linearity for all practical purposes. (Another way to view the linearity of this slightly generalized version of the integral with respect to μ is to note that each function f that is integrable [μ] “almost belongs” to a unique element f˙ of the quotient space L/M0

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4 Convergence theorems for Lebesgue integrals

  introduced above, and to identify X f˙ dμ with X f dμ.) We observe also that for the notion of integrable function thus extended it is true that if f is a function that is integrable [μ] on X, and if g is a complex-valued function that is measurable [μ] on X such that  |g|  ≤ |f | almost everywhere [μ], then g is also integrable [μ], and  X g dμ ≤ X |f | dμ. In other words, axiom (L1 ) continues to hold in a suitably generalized form, and the same is clearly true of axiom (L2 ) as well. There are still other conventions concerning the integral of a function f on a measure space (X, S, μ) that are customarily observed when the integrand f is real-valued or extended real-valued. Suppose, first, that f is an extended real-valued function that is measurable and nonnegative almost everywhere [μ] on X. If the (measurable) set E+∞ = E(f = +∞) has measure zero, then there is a nonnegative, finite real-valued and measurable function f on X equal almost everywhere to f , and it is natural in this case to call f itself integrable [μ] if the function f is integrable [μ], and to set X f dμ = X fdμ. (Once again, if one such function f is integrable [μ], then all are and all have the same integral.) If, on the other hand, f is not integrable [μ], we write  f dμ = +∞. X

 In the event that μ(E+∞ ) > 0, we also write X f dμ = +∞. (Note that we do not  say that f is “integrable [μ]” in these latter two cases.) Thus the symbol X f dμ is defined as an extended real number for any extended realvalued function f that is defined and nonnegative almost everywhere [μ] and measurable [μ] on X. (All of this could have been said somewhat differently if the support of f were σ-finite; a consideration of cases shows that for such a function f we have   f dμ < +∞ or f dμ = +∞ X

X



according as the set of integrals X g dμ of the nonnegative integrable (simple) functions g on X such that g ≤ falmost everywhere [μ] is bounded or not (in R), and also that, in any case, X f dμ is simply the supremum (in R ) of this set of integrals.) Finally, let f be an arbitrary extended real-valued function that ismeasurable [μ] on X. Then, according to the foregoing agreements, both X f + dμ and X f − dμ are defined as nonnegative extended real numbers. If the difference   f − dμ

f + dμ − X

X

(4.2)

 is defined as an extended real number (that is, if either X f + dμ or   also f − dμ is finite), then we write X f dμ for the difference in (4.2). X

4 Convergence theorems for Lebesgue integrals

83

  If both X f + dμ and X f − dμ are finite, then there is a finite real-valued  function f defined on all of X such that over X and equal   f is integrable to f almost everywhere, and we have X f dμ = X fdμ, so that in this case our last extension of the notion of integration with respect to μ does not really present us with anything new. Hence it is only the extended real valued functions f for which X f dμ = ±∞ that require further attention at this time. In this connection we note, in the first place, that assigning values ±∞ to such integrals is the exact counterpart (indeed, a generalization) of the common and useful practice of assigning ±∞ to various divergent series and indexed sums of extended real numbers. Likewise, a consideration of cases discloses that if f and g are any two extended real-valued functions   that are measurable [μ] on X and for which the symbols X f dμ and X g dμ are both defined, and if f ≤ g almost everywhere, then X f dμ ≤ X g dμ. Moreover, itis easily seen that in all  cases, given a measurable set E, if the integral X f dμ is defined then X f χE dμ is too, and we continue the practice of writing E f dμ for this latter integral. Finally, it is important to note that integrals taking values ±∞ are here introduced only for extended real-valued integrands f ; if a function f assumes any nonreal complex values, then the integral of f , if it exists, must be an ordinary complex number. The extended real integrals just introduced obey the usual rules of algebra when the operations are defined. Proposition 4.20. Let f be an extended real-valued function that is measurable [μ] on a measure space (X, S, μ) and for which the integral X f dμ is defined real number, and let a be a (finite) real number.   as an extended extended real-valued Then X af dμ = a X f dμ. Likewise, if g is another  g dμ is defined, and if function that is measurable [μ] on X and for which X  f dμ + g dμ is also defined as an extended real number, then the sum X  X   (f + g) dμ = f dμ + g dμ. Consequently, if f and g are extended X X X   both defined, real-valued functions such that X f dμ and X g dμ are  and if  a and b are real numbers for which the combination a X f dμ + b X g dμ is defined as an extended real number, then    (af + bg) dμ = a f dμ + b g dμ. X

X

X

Proof. Consider  first a single function f and a single real number a. The case in which X f dμ is finite  has already been covered, and by symmetry f dμ it suffices to treat the case X  = +∞. If a = 0, then af = 0 almost  everywhere so X af dμ = 0 = a X f dμ. If a > 0, then (af )+ = af + and − − + (af )− dμ = (af X  X (af ) = +∞ while   ) −= af , whence it follows that a X f dμ < +∞, and hence that X af dμ = +∞ = a X f dμ. Finally, if − + + a < 0, then (af )+ = (−a)f − and  (af−) = (−a)f . But+ then X (af ) dμ = −  while X (af ) dμ = (−a) X f dμ = +∞, so that (−a) X f dμ < +∞, af dμ = −∞ = a f dμ. X X

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4 Convergence theorems for Lebesgue integrals

the sum f +g of two such functions we note once again that  Next, as regards if X f dμ and X g dμ are both finite, then the desired result is already  known. Hence, bysymmetry, it suffices to treat the following two cases: (i) X f dμ = +∞ with X g dμ finite, and (ii) X f dμ = X g dμ = +∞. Moreover, in both of these cases there is a set Z1 of measure zero such that f is defined and measurable and satisfies the inequality −∞ < f ≤ +∞ on X \Z1 and likewise a set Z2 of measure zero such that g is defined and measurable and such that −∞ < g ≤ +∞ on X \ Z2 . But then f + g is defined and measurable on X \ (Z1 ∪ Z2 ), so f + g is measurable [μ]. Likewise, in both cases, we have (f + g)− ≤ f − + g − , and therefore  (f + g)− dμ < +∞. X

 + Thus in both cases everything comes down to  showing+that X (f+g) = +∞. Suppose this were not the case. Then both X (f + g) dμ and X (f + g)− dμ are finite, which implies that f + g is integrable [μ]. But then, in case (i), we find that f = (f + g) − g is also integrable [μ], contrary to hypothesis, and this contradiction proves the result in case (i). Finally, if f + g is integrable [μ], then, in case (ii), we obtain   f dμ = [(f + g) − g] dμ = −∞ X

X

by case (i), contrary to hypothesis. Thus the desired result is also valid in case (ii).  the proof, set f = af and g = bg, so that X fdμ =   Finally, to complete a X f dμ and X g dμ = b X g dμ by the first part of the proof. Since, by hypothesis, the combination a X f dμ + b X g dμ is defined as an extended   real number, we see that the same is true of the sum X fdμ + X g dμ. But then     (af + bg) dμ = (f + g) dμ = a f dμ + b g dμ. X

X

X

X

We conclude this discussion with the following partial converse of Proposition 4.17. Proposition 4.21. Let X be a measure space equipped with a measure μ, let f be an extended real-valued function  that is defined and nonnegative almost everywhere [μ] on X, and suppose X f dμ = 0. Then f = 0 almost everywhere [μ]. Proof. According to the foregoing conventions and definitions there is a subspace X0 of X such that μ(X \X0 ) = 0 and  such that f is defined, measurable, and nonnegative on X0 , and we have X0 f dμ = 0. Moreover, it is clear, as before, that we may assume without loss of generality that X = X0 . If, for some positive integer n, F is a measurable set of finite measure in X such that f (x) ≥ 1/n for every point x of F , then

4 Convergence theorems for Lebesgue integrals



85

 f dμ ≥

0= X

f dμ ≥ μ(F )/n ≥ 0, F

and therefore μ(F ) = 0 by virtue of the monotonicity of integration with respect to μ. Since the set En = E(f ≥ 1/n) is σ-finite as a subset  of the σ-finite support Nf of f , this shows that μ(En ) = 0. But Nf = n En , so μ(N ) = limn μ(En ), and the result follows.   Example 4.22. Let (R, BR , λ1 ) be the real numbers equipped with LebesgueBorel measure (Example 3.42), and let f be the function on R that is equal to +∞ at each rational  number and equal to zero at each irrational number (f = +∞χQ ). Then E f dλ1 = 0 for every Borel set E. On the other hand, the function identically equal to +∞ on R has integral +∞ over every subset of R having positive measure with respect to λ1 , and integral zero over every null set in R. Example 4.23. Let (R, BR , λ1 ) be as in Example 4.22, and let g(t) = 1/t, t = 0. Then g is measurable[μ] on R, and if E is an arbitrary Borel subset of  R+ = {t ∈ R : t > 0}, then E g dλ1 ≥ 0. In particular, (1,+∞) g dλ1 = +∞. (The function gχ(1,+∞) dominates the simple function sn = (1/2)χ(1,2) + · · · + (1/(n + 1))χ(n,n+1))   for every s dλ1 is not bounded above in R.) SimR n  n, and the sequence ilarly, (0,1) g dλ1 = +∞. If E is an arbitrary Borel subset of R that meets  either R+ or R− = {t ∈ R : t <  0} in a set of measure zero, then E g dλ1 is defined. On the other hand, (−1,+1) g dλ1 , for instance, is undefined. If h is  the function that is identically +∞ on R+ and identically −∞ on R− , then h dλ1 is defined if and only if either E ∩ R+ or E ∩ R− has measure zero. E The fact that null sets are negligible in integration theory opens up, as we have seen, the possibility of improving many of the results of Chapter 3. Thus, for example, it is obvious that if f is a function that is measurable [μ] on a measure space (X, S, μ), and if f is merely nonnegative almost everywhere [μ], then X f dμ ≥ 0. For the most part, to be sure, sprinkling “almost everywhere” into the hypotheses of theorems results in only token improvement, and we generally leave to the reader the task of reformulating (and reproving) in appropriate fashion the results of Chapter 3. In one important case, however, namely in Proposition 3.15, it is possible to effect a very significant improvement by allowing for exceptional null sets. The resulting theorem, which we now prove, is the first in the array of convergence theorems in this chapter. Theorem 4.24. (Monotone Convergence Theorem). Let (X, S, μ) be a measure space and let {fn } be a sequence of extended real-valued functions defined and nonnegative almost everywhere [μ] and measurable [μ] on X. Suppose also that the sequence {fn } is monotone increasing almost everywhere [μ]. Then

86

4 Convergence theorems for Lebesgue integrals



 lim fn dμ = lim X

n

n

fn dμ,

(4.3)

X

where limn fn is formed pointwise in R . Proof. According to the hypotheses there exists X0 ∈ S such that μ(X \ X0 ) = 0, the functions fn are all defined, nonnegative and measurable on X0 , and the sequence {fn (x)} is increasing for each x ∈ X0 . This fact implies, to begin with, that the function limn fn is measurable [μ] on X, and hence that both sides of (4.3) are defined as extended real numbers. Moreover, it clearly suffices to prove the theorem on the subspace X0 , so that, without loss of generality, we may and do assume that X = X0 . Let F denote the (measurable) set of those points x of X at which E(limn fn = +∞). limn fn (x) is finite, and let Y denote the complement  [μ], so that lim fn dμ < +∞, then If the function limn fn is integrable n X  lim f dμ = lim μ(Y ) = 0 by convention, and n n n fn dμ. But then X F   f dμ = f dμ for each index n, and the desired result follows at once n n X F from Proposition 3.15 (applied on thesubspace F ). Suppose, on the other hand, that X lim  n fn dμ = +∞. To complete the = +∞ as well. But if this proof we must show that in this event limn X fn dμ  f dμ is bounded is not true, then the increasing sequence X n  above—say by M , and another application of Proposition 3.15 shows that F limn fn dμ ≤ M < +∞. Thus Y must have positive measure. We derive a contradiction, and thus complete the proof, by showing that μ(Y ) = 0. The first step in this direction is modest but essential. The set Y is at least σ-finite with respect to μ. Indeed, Y is clearly contained in the union of the supports of the functions fn , and each fn , being integrable, has σ-finite support (see Problem 3R(i)). Hence it is enough to show that every measurable subset W of Y having finite measure actually has measure zero. But suppose W is a subset of Y such that μ(W ) = a where 0 < a < +∞. Then by Proposition 3.36 there also exist a subset G of W such that μ(G) < a/2 and an index N such that fN (x) ≥ 2M/a at every point x of W \ G. But then   fN dμ ≥ fN dμ ≥ (2M/a)μ(W \ G) > M, X

W \G

which is contrary to assumption. We conclude that μ(Y ) = 0, and the proof of the theorem is complete.   Corollary 4.25. (Theorem of Beppo-Levi) Let (X, S, μ) be a measure space, let {pn } be a sequence of extended real-valued functions defined and nonnegative almost everywhere [μ] and measurable [S] on X, and let p denote the (almost everywhere defined ) pointwise sum p(x) = n pn (x). Then 

 p dμ = pn dμ. X

n

X

4 Convergence theorems for Lebesgue integrals

87

In particular, p is integrable [μ] if and  if all of the functions pn are  only integrable [μ] and the numerical series n X pn dμ is convergent in R. Hence    p dμ < +∞ implies p (x) < +∞ at almost every point x. n n n X n Proof. Just as in the preceding proof it is easy to see that it is enough to treat the case in which the functions pn are defined and nonnegative everywhere on X. Set fn = p1 + p2 + · · · + pn , n = 1, 2, . . . , and apply the monotone convergence theorem.   Example 4.26. Let (X, S, μ) be a measure space, and let f be an extended real-valued function that is measurable [S] on X and has the property that  f dμ is defined as an extended real number. As has been noted, for any X set E in S the integral E f dμ is also defined, so that setting  νf (E) = f dμ, E ∈ S, E

defines a set function νf on S, called, as before, the indefinite integral of f with respect to μ. This notion of indefinite integral differs from the one introduced initially in that it may assume one or the other of the values ±∞ (but not both). It is an easy consequence of all that has been said up to now that νf is absolutely continuous [μ]. Example 4.27. Let (X, S, μ) be a measure space and let f be a nonnegative extended real-valued function defined almost everywhere [μ] and measurable [S] on X. Then the sequence {gn }∞ n=1 of truncates gn = f ∧ n always has the property that   f dμ = lim gn dμ. n

X

Example 4.28. Let (X, S, μ) be a measure space and {En }∞ n=1 ⊂ S. For each x ∈ X denote by N (x) the number of sets En in this sequence such that x ∈ En , with the understanding that N (x) = +∞ whenever the latter number is ℵ0 . Then  ∞

N dμ = μ(En ). R

Indeed, N is simply the sum

∞

 N dμ = R

n=1

n=1

χEn (in R ) so

∞ 

n=1

R

χEn dμ =



μ(En )

n=1

∞ by the theorem of Beppo-Levi. When n=1 μ(En ) < +∞, it follows that N is integrable [μ], so N must be finite almost everywhere [μ]. This last observation is one half of the Borel-Cantelli lemma, frequently used in probability arguments. (The other half is a converse which holds when μ is a probability measure and the sets En are statistically independent; see Problem 4II.)

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4 Convergence theorems for Lebesgue integrals

Theorem 4.29. (Fatou’s Lemma). Let (X, S, μ) be a measure space, and let {fn } be a sequence of functions defined and nonnegative almost everywhere [μ] and  integrable [μ] on X. Suppose that there exists a real number M such that f dμ ≤ M for every n, and suppose also that {fn } converges pointwise X n almost everywhere [μ] to a limit f . Then f is integrable [μ] over X and  f dμ ≤ M. X Proof. For each positive integer n the function gn = inf k≥n fk is defined and nonnegative almost everywhere. Moreover, gn is measurable [μ] and, since 0 ≤ gn ≤ fn almost everywhere, it follows that gn is integrable [μ]. Finally, the sequence {gn } is monotone increasing and convergent almost everywhere to f . The desired conclusion follows at once from the monotone convergence theorem.   Example 4.30. Let {fn }∞ n=1 be a sequence of integrable complex-valued functions on a measure space (X, S, μ) such that ∞ 

n=1

|fn − fn+1 | dμ < +∞. X

(Such a sequence might be said to be of bounded variation in the mean.) Then, by the theorem of Beppo-Levi (Corollary 4.25), the numerical series ∞

(fn (x) − fn+1 (x))

n=1

converges absolutely almost everywhere [μ]. Since this series telescopes, we see that the sequence {fn } converges almost everywhere to some limit—say f . Since {fn } is Cauchy in the mean, one sees, by applying Fatou’s lemma to the sequence {|fn − fm |}∞ m=1 , that {fn } also tends to f in the mean with respect to μ. (This final conclusion could also be obtained directly via an application of axiom (L2 ) suitably generalized, as discussed above.) The convergence theorems proved apply only to sequences of (extended) realvalued functions. We turn now to the standard convergence theorems for complex-valued functions. The sufficient conditions given in this result are also necessary; see Problems 4Q, 3N, and 3O. Theorem 4.31. Let (X, S, μ) be a measure space, let {fn } be a sequence of complex-valued functions integrable [μ] over X, and suppose that {fn } is uniformly integrable [μ] and converges almost everywhere [μ] to a limit f . Then f is integrable [μ] and {fn } converges to f in the mean. Proof. There exists a measurable subset X0 of X such that μ(X \ X0 ) = 0 and the functions fn are everywhere defined and pointwise convergent to f on X0 . The limit f is measurable on X0 and therefore measurable [μ] on X. If f is integrable [μ] over X0 , then it is also integrable [μ] over X. Similarly, by

4 Convergence theorems for Lebesgue integrals

89

 Corollary 4.18 it suffices to verify that X0 |f − fn | dμ → 0. Thus, as before, we may and do assume without loss of generality that X = X0 . For each positive integer n let νn denote the indefinite integral of |fn |. Given ε > 0, the hypothesis implies the existence of a set E0 of finite measure such that  νn (X \ E0 ) = |fn | dμ < ε/5, n ∈ N, (4.4) X\E0

and of a number δ > 0 with the property that if E is any measurable set such that μ(E) < δ, then νn (E) = E |fn | dμ < ε/5 for every n ∈ N. By Egorov’s theorem (Proposition 3.35) there exists a subset F of E0 such that μ(F ) < δ and such that {fn } converges to f uniformly on E0 \ F , and it follows that  νn (F ) = |fn | dμ < ε/5, n ∈ N. (4.5) F

Furthermore, the uniform convergence of {fn } on E0 \F implies the existence of a positive integer N such that |f (x) − fn (x)| ≤ (μ(E0 ) + 1)ε/5  for all x ∈ E0 \ F and all n ≥ N . But then E0 \F |f − fn | dμ < ε/5 for all n ≥ N . Thus f is integrable [μ] over E0 \ F . Moreover, (4.4) together with an application of Fatou’s lemma (Theorem 4.29) to the sequence {|fn |} shows that f is also integrable [μ] over X \ E0 , and that, in fact,  |f | dμ ≤ ε/5. X\E0

Similarly, by (4.5), f is also integrable over F and  |f | dμ ≤ ε/5. F

Hence f is integrable [μ] over X and we have     |f −fn | dμ ≤ (|f |+|fn |) dμ+ |f −fn | dμ+ (|f |+|fn |) dμ < ε X

E0 \F

X\E0

for all n ≥ N . Thus the sequence

 X

F

 |f − fn | dμ tends to zero.

 

Corollary 4.32. If f and the sequence {fn } satisfy the hypotheses of Theo rem 4.31, then limn X fn dμ = X f dμ in C.     Proof. Use the inequality  X f dμ − X fn dμ ≤ X |f − fn | dμ.   Remark 4.33. While the conclusion of Corollary 4.32 is certainly weaker than that of Theorem 4.31, it is nonetheless true that Theorem 4.31 can

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4 Convergence theorems for Lebesgue integrals

be derived from Corollary 4.32. Indeed, if the hypotheses of Theorem 4.31 hold for a sequence {fn }, and if it is known for any reason that the limit f is integrable, then it is easy to verify that all of the hypotheses of the theorem also hold for the sequence {|f − fn |}, which converges to zero almost everywhere Remark 4.34. Consider once again the sequence {νn } of indefinite integrals in Theorem 4.31. If {Fk } is a monotone decreasing sequence of measurable sets with limit F , then limk νn (Fk ) = νn (F ) for each index n since νn is a finite measure (Problem 3J). It is sometimes said that the sequence {νn } is equicontinuous from above at F if this convergence is uniform in n, that is, if for every monotone decreasing sequence {Fk } converging to F , and for every positive number ε there exists K such that νn (Fk ) − νn (F ) < ε for every k ≥ K and every n. It turns out that Theorem 4.31 remains valid if we assume of the sequence {νn } simply that it is equicontinuous from above at ∅. Indeed, it can be shown that this single condition actually implies that {νn } is both uniformly absolutely continuous and uniformly concentrated on sets of finite measure with respect to μ. The following two results are simple consequences of Theorem 4.31, but are of considerable importance in their own right. Theorem 4.35. (Dominated Convergence Theorem). Let (X, S, μ) be a measure space, let {fn }∞ n=1 be a sequence of complex-valued functions measurable [μ] on X, and suppose that {fn } converges almost everywhere [μ] to a limit f . Suppose also that there exists a nonnegative function g such that g is integrable [μ] on X and |fn | ≤ g almost everywhere [μ], n ∈ N. Then f is integrable [μ], and {fn } converges to f in the mean. Proof. The functions fn are clearly integrable [μ] on X. Moreover, the indefinite integrals of the functions |fn | are all dominated setwise by the indefinite integral of g, and it follows at once (Proposition 4.8) that they are uniformly absolutely continuous and uniformly concentrated on sets of finite measure with respect to μ. The theorem follows immediately from Theorem 4.31.   Theorem 4.36. (Bounded Convergence Theorem) Let (X, S, μ) be a finite measure space, and let {fn }∞ n=1 be a sequence of complex-valued functions measurable [μ] on X. Suppose that there exists a real number M such that |fn | ≤ M almost everywhere [μ] for every n and that {fn } converges almost everywhere [μ] to a limit f . Then the functions fn and f are integrable [μ] on X and {fn } converges to f in the mean. Proof. The constant function g ≡ M is integrable [μ] since μ(X) < +∞.

 

Example 4.37. (The Darboux Integral) Let f be a real-valued function defined and bounded on a real interval I = [a, b], and let P = {a = t0 < · · · < tN = b}

4 Convergence theorems for Lebesgue integrals

91

be a partition of I. For each index i = 1, . . . , n we set Mi = sup {f (t) : ti−1 ≤ t ≤ ti } ,

mi = inf {f (t) : ti−1 ≤ t ≤ ti } ,

and define DP (f ) =

n

Mi (ti − ti−1 ),

dP (f ) =

i=1

n

mi (ti − ti−1 ),

i=1

the upper and lower Darboux sums, respectively, of f based on the partition P. Concerning these sums it is evident that for any P we have m0 (b − a) ≤ dP (f ) ≤ DP (f ) ≤ M0 (b − a), where m0 and M0 denote, respectively, the infimum and supremum of f over I. More generally, if P  is another partition of I that refines P, then dP (f ) ≤ dP  (f ) ≤ DP  (f ) ≤ DP (f ).

(4.6)

Thus {dP (f )} and {DP (f )} are bounded monotone nets (increasing and decreasing, respectively) indexed by the directed set of all partitions of the interval I, and it follows that both nets converge in R ([I, Example 6J]). The two limits J(f ) = lim dP (f ) and J(f ) = lim DP (f ) P

P

are the upper and lower Darboux integrals, respectively, of f over I. If P1 and P2 are any two partitions of I, and if P+ is a common refinement of P1 and P2 , then from (4.6) we obtain dP1 (f ) ≤ dP+ (f ) ≤ DP+ (f ) ≤ DP2 (f ). This shows that any one lower Darboux sum of f is a lower bound of the entire net of all upper Darboux sums, and hence that J(f ) ≤ J(f ). (The same calculation shows that the difference J(f ) − J(f ) is given by J(f ) − J(f ) = lim(DP (f ) − dP (f )) = lim P

P

n

(Mi − mi )(ti − ti−1 ),

i=1

where Mi − mi = ω(f ; [ti−1 , ti ]) is the oscillation of f over the subinterval [ti−1 , ti ] ([I, Problem 7X]).) When J(f ) = J(f ) (and only then) f is said to be Darboux integrable over I, and the common value J(f ) = J(f ) = J(f ) is the Darboux integral of f over I. Thus far we have said little that is new. (The reader is referred to [I, Problem 2S] for basic definitions.) Our main purpose here is to relate these notions to the Lebesgue integral. As regards the Darboux sums this is easily done. Given f and the partition P as above, we simply define two auxiliary functions by setting

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4 Convergence theorems for Lebesgue integrals

gP (t) = mi

and

GP (t) = Mi ,

ti−1 < t < ti ,

for each index i = 1, . . . , n. Then gP and GP are simple Borel measurable functions defined almost everywhere on I with respect to Lebesgue-Borel measure λ1 (Example 3.42), and it is clear that 



b

dP (f ) =

gP dλ1 ,

b

DP (f ) =

a

GP dλ1 . a

To express the upper and lower Darboux integrals as Lebesgue integrals is more difficult. We need to replace the limits appearing in their definition by ordinary sequential limits, and this requires some work. Example 4.38. To begin with, let P be a partition of I, and suppose P  is another partition obtained from P be adjoining one or more partition points in some single subinterval of P—say the subinterval [ti−1 , ti ]. To fix ideas, let us consider lower sums. If we write

Σ = mj (tj − tj−1 ), j =i

then, of course, dP (f ) = Σ  + mi (ti − ti−1 ), while dP  (f ) = Σ  + Σ  , where Σ  denotes the contribution coming from those subintervals of P  obtained by subdividing [ti−1 , ti ]. But now Σ  ≤ M0 (ti − ti−1 ), and therefore dP  (f ) − dP (f ) = Σ  − mi (ti − ti−l ) ≤ Ω0 (meshP), where Ω0 = M0 − m0 is the oscillation of f over I. Thus we obtain dP  (f ) ≤ dP (f ) + Ω0 (meshP). By mathematical induction, we find that dP  (f ) ≤ dP (f ) + rΩ0 (meshP) if P  is a refinement of P obtained by further subdividing at most r of the subintervals of P. Now fix ε > 0, select a partition P0 of I such that dP0 (f ) > J(f ) − ε/2, and suppose P0 = {t0 < . . . < tN } has exactly N subintervals. Choose δ > 0 so small that N Ω0 δ < ε/2, and let P be a partition of I with meshP < δ. If P+ denotes the common refinement of P and P0 obtained by adjoining the partition points t1 , . . . tN to P, then, as we have just seen, dP+ (f ) ≤ dP (f ) + N Ω0 δ < dP (f ) + ε/2. But then, of course, dP (f ) > dP+ (f ) − ε/2 > J(f ) − ε.

4 Convergence theorems for Lebesgue integrals

93

(Note that in this calculation P is any partition of I having mesh less than δ.) Fix now one nested sequence {Pn }∞ n=1 of partitions of I with the property that limn meshPn = 0. Consider the corresponding bounded sequence of simple functions {gn = gPn }. These functions are all defined almost everywhere [μ], and the sequence is monotone increasing almost everywhere [μ] to a limit, which we denote by g. (The exceptional set is the set Z of all the partition points of the various partitions Pn , and Z is countable.) The bounded convergence theorem yields 



b

b

g dλ1 = lim n

a

a

gn dλ1 = lim dPn (f ), n

and this latter limit, as we have just seen, is precisely the lower Darboux integral of f . On the other hand, if t is a point of I \ Z, and if for each index n we denote by In the subinterval of Pn that contains t , then {In } is a nested sequence of neighborhoods of t with diamIn → 0, and it is easily seen that g(t ) = lim inf {f (t) : t ∈ In } n

coincides with lim inf t→t f (t). Thus g is equal almost everywhere [μ] to the lower envelope of f (see [I, Proposition 7.27]). These facts regarding lower Darboux integrals have valid duals, of course. The dual arguments show that J(f ) coincides with the Lebesgue integral over I of the upper envelope G of f , and that this equality is realized along any sequence {Pn } of partitions of I with mesh Pn → 0. (Alternatively, one may employ the fact that −DP (f ) = dP (−f ) for any f and P, and hence that −J(f ) = J(−f ).) We note for future reference that the above arguments show that for any ε > 0 we have both  b  b dP (f ) > g dλ1 − ε and Dp (f ) < G dλ1 + ε a

a

for any partition P of I with sufficiently small mesh. It should also be recalled that the upper and lower envelopes g and G are semicontinuous ([I, Proposition 7.27]) and therefore Borel measurable (Example 2.16). Finally we observe that 

b

J(f ) − J(f ) =

(G − g) dλ1 . a

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4 Convergence theorems for Lebesgue integrals

Thus we have proved: A bounded real-valued function f is Darboux integrable over I if and only if G = g almost everywhere [λ1 ], i.e., if and only if f is continuous almost everywhere [λ1 ] on I, and that, when this is the case, the Darboux integral J(f ) coincides with the Lebesgue integral of f . Example 4.39. (The Riemann Integral) Just as in the preceding example, let f be a real-valued function defined and bounded on an interval I = [a, b], and let P = {t0 < . . . < tn } be a partition of I. If for each index i = 1, . . . , n the real number τi belongs to the ith subinterval [ti−1 , ti ] of P, then R=

n

f (τi )(ti − ti−1 )

i=1

is a Riemann sum for f based on P and the sequence {τ1 , . . . , τn }. These sums provide the basis for still another notion of integral, namely the Riemann integral. We say that the function f is Riemann integrable over I if there exists a number R(f ) such that for any given ε > 0 there exists δ > 0 with the property that |R(f ) − R| < ε whenever R is a Riemann sum for f based on a partition P of I with meshP < δ and a sequence τi ∈ [ti−1 , ti ], i = 1, . . . , n. This number R(f ), which is clearly unique if it exists, is called the Riemann integral of f over I and is written  b f (t) dt. R(f ) = a

Since f is not assumed to be continuous, and need not assume either a greatest or a least value on any subinterval of I, the Darboux sums of f based on a partition P need not be among the Riemann sums for f based on P. On the other hand, it is clear that DP (f ) and dP (f ) are, respectively, the supremum and infimum of the various Riemann sums for f based on P. Hence if f is Riemann integrable over I, then lim(DP (f ) − dP (f )) = 0, P

and f is therefore Darboux integrable over I as well. Suppose, conversely, that f is Darboux integrable over I. Then, as was seen in Example 4.37, for given ε > 0 there exists δ > 0 such that if P is a partition of I with mesh P < δ, then J(f ) − ε < dP (f ) ≤ DP (f ) < J(f ) + ε. But then |J(f ) − R| < ε for any Riemann sum R for f based on P. Thus f is also Riemann integrable over I, and J(f ) is its Riemann integral. We have proved the following result: A bounded real-valued function f on an interval I = [a, b] is Riemann integrable over I if and only if f is continuous almost

4 Convergence theorems for Lebesgue integrals

95

everywhere [λ1 ] on I, and when this is the case, the Riemann and Lebesgue integrals of f over I coincide. Remark 4.40. The definition of the Riemann integral makes sense for complex-valued functions, and a moment’s reflection discloses that the results of Example 4.39 hold for complex-valued functions f . According to this example, the Riemann and Darboux integrals are exactly the same (for realvalued integrands). The true advantage of the Riemann integral lies in the greater flexibility in selecting approximating finite sums that is provided by its definition. Incidentally, the requirement in the definition of the Riemann integral that the function f be bounded results in no loss of generality. It is easily seen that the set of all Riemann sums for an unbounded function based on any one partition is also unbounded, no matter how that partition is chosen. Remark 4.41. Examples 4.37 and 4.39 show that Riemann integration (at least for bounded functions on bounded real intervals) is merely the restriction of Lebesgue integration with respect to λ1 to certain special functions. But that does not mean that the Riemann integral is therefore rendered superfluous. For it comes equipped with a powerful evaluation mechanism— the fundamental theorem of calculus—and as a matter of fact, for the most part, the Lebesgue integrals (with respect to λ1 ) that can actually be evaluated in the customary sense are those that are Riemann integrals, and their evaluation is effected via the fundamental theorem, just as in ordinary calculus. Example 4.42. Let f : [a, b] → R be a continuously differentiable function on some real interval [a, b] ([I, Example 3M]). Then f satisfies the hypotheses of the mean value theorem on every subinterval of [a, b]. Hence for any partition P = {a = t0 < . . . < tn = b} there is a point τi in each subinterval [ti−1 , ti ] such that f (ti ) − f (ti−1 ) = f  (τi )(ti − ti−1 ), n and the corresponding Riemann sum R = i=1 f  (τi )(ti − ti−1 ) telescopes into f (b) − f (a). Now f  is Riemann integrable, and R is accordingly arbitrarily close to the integral of f  over [a, b] provided mesh P is small enough. Thus we have the well-known formula  b f  (t) dt = f (b) − f (a). a

Example 4.43. The function g of [I, Example 7X] has Riemann integral zero over any subinterval [a, b] of (0, +∞), even though it is discontinuous at every rational number. The Riemann integral over [0, 1] of the characteristic function of the Cantor set C exists (and equals zero). On the other hand,  is a Cantor set in [0, 1] with λ1 (C)  > 0 (recall Problem 3Y), then χ  if C C

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4 Convergence theorems for Lebesgue integrals

1  is not Riemann integrable over [0, 1], while, of course, 0 χC dλ1 = μ(C). Thus integration with respect to Lebesgue-Borel measure on a real interval constitutes a proper extension of Riemann integration on that interval. Example 4.44. The function f defined by f (0) = 1 and f (t) =

sin t , t

0 < t < +∞,

is continuous and bounded on [0, +∞). Thus the Riemann and LebesgueBorel integrals of f over any interval [0, T ] exist and coincide. Define 



T

F (T ) =

T

f (t) dt = 0

f dλ1 ,

0 < T < +∞.

0

Then limT →+∞ F (T ) exists. (If for any positive number T a nonnegative inteT ger n is chosen so that nπ ≤ T < (n+1)π, then |F (T )−F (nπ)| = nπ |f (t)|dt. Since limt→+∞ f (t) = 0, this integral can be made as small as desired by taking T , and therefore n, large enough. Thus limT →+∞ ∞ F (T ) coincides with the sum of the (convergent) alternating series n=0 an , where  (n+1)π an = nπ f (t) dt, n ∈ N0 .) This limit, which is by definition the value  +∞ of the convergent improper integral  0 f (t) dt, is not the Lebesgue-Borel integral of f over [0, +∞). Indeed, [0,+∞) |f | dλ1 dominates the partial sums ∞ of the series n=0 |an |, where the numbers an are as defined above, and | ≥ 2/π(n + 1) > 1/2(n + 1), this latter series is divergent. Thus since |a n  |f | dλ = +∞. Moreover, f + and f − both have integrals with re1 [0,+∞)  spect to μ equal to +∞ over [0, +∞), so [0,+∞) f dλ1 is undefined. Thus Lebesgue-Borel integration over a ray is not an extension of (improper) Riemann integration over that ray. Remark 4.45. In order to obtain the greatest possible flexibility in their application, the main convergence theorems have been so formulated as to apply to functions that are undefined on various sets of measure zero. To maintain this degree of generality in the statement of every theorem would be both pedantic and pointless. In the following problems, and throughout most of the balance of the book, we routinely assume that we are dealing with functions that are everywhere defined, even when (indeed, especially when) it is clear that they may be allowed to be undefined on null sets.

Problems 4A. Use Proposition 4.8 to give a new proof that if a function f is integrable with respect to a measure μ, then the support of f is σ-finite with respect to μ (see Problem 3R).

4 Convergence theorems for Lebesgue integrals

97

4B. Let X be measure space with measure μ, and let p be a positive number. The collection of all those measurable scalar-valued functions f on X with the property that |f |p is integrable [μ] is called the Lebesgue space of order p on X and is denoted by Lp = Lp (X, μ). (Note that the space of all integrable functions with respect to μ, which we have regularly denoted by L, coincides in this definition with the space L1 (X, μ).) Show that L1 is a vector space (see Problem 3B). Show also that, if μ(X) < +∞, and if p ≤ q, then Lq ⊂ Lp . (The assumption that μ(X) is finite cannot be dropped. Indeed, if γ denotes the counting measure on the space N, the inclusions are all reversed; if p ≤ q, then Lp (N, γ) ⊂ Lq (N, γ).) simply of the collection of Remark 4.46. Note that the Lebesgue space Lp (N, γ)

consists all infinite sequences {ξn }∞ |ξn |p < +∞. This space is usually n=1 with the property that denoted by p . For the case p = 1, see Problem 1I. We undertake a more thorough study of the spaces Lp in Chapter 9. 4C. Let X be a measure space with measure μ, and suppose that f belongs to Lp (X, μ) for some p > 0. Show that for every positive number α the set Eα = E(|f | ≥ α) satisfies the inequality  1 μ(Eα ) ≤ p |f |p dμ. α (This generalization of Problem 3N(i) is sometimes called Markov’s inequality. The  case p = 2 is also known as Tchebysheff ’s inequality.) Show too that if |f |p dμ = 0 for some p > 0, then f = 0 almost everywhere [μ]. (In the case p = 1 this last result provides a converse to Proposition 4.17.) 4D. Let (X, S, μ) be a measure space with μ(X) > 0, and let f be a scalar-valued function that is measurable [μ] on X. Then f is said to be essentially bounded with respect to μ, or essentially bounded [μ], if there exists a real number M such that |f | ≤ M almost everywhere [μ]. Show that if f is essentially bounded, then there is a smallest real number M such that |f | ≤ M almost everywhere [μ]. This number is the essential supremum of |f | (notation: M = ess supX |f |). More generally, if E is a measurable set, then f is essentially bounded on E if f χE is essentially bounded on X, and the essential supremum of |f | on E (notation: ess supE |f |) is, by definition, the essential supremum of f χE . Show that if E is a set of finite measure and if f is measurable [μ] and essentially bounded on E, then f is integrable [μ] over E, and

 |f | dμ ≤ (ess sup E |f |)μ(E). E

4E. Let X be a measure space with measure μ, let E be any measurable subset of X, and let f be a real-valued function that is integrable over E. Suppose that α ≤ f ≤ β almost everywhere on E, where α and β denote extended real numbers. Show that

 αμ(E) ≤

f dμ ≤ βμ(E). E

(If α, β, and μ(E) are all finite, the assumption that f is integrable over E may be replaced by the assumption that f is measurable over E, and the assertion is trivial. It is precisely when one or more of the numbers α, β, and μ(E) is infinite that care must be exercised.)

98

4 Convergence theorems for Lebesgue integrals

4F. Let (X, S, μ) be a measure space, let f be a nonnegative measurable function on X,  and let E be a measurable subset of X. Let us define νf (E) = E f dμ if f is integrable [μ] over E, and νf (E) = +∞ otherwise. Show that the set function thus defined on S is a measure on (X, S). This measure is still called the indefinite integral of f , even if f is not integrable over X. It is easy to see that if f is not integrable [μ], then νf need not be concentrated on sets of finite measure with respect to μ. Show by example that it also need not satisfy the conclusion of Proposition 4.3. (It is clear, however, that νf is always absolutely  continuous with respect to μ; see the remark following Problem 2V. If the notation f dμ = +∞ is permitted for nonnegative measurable functionsf that are not integrable, then the set function νf has the simpler definition νf (E) = E f dμ for all E ∈ S.) 4G. If f and g are two scalar-valued functions, each defined on some subset of X of full measure, and if α and β are scalars, let us agree to write αf + βg for the function h(x) = αf (x) + βg(x) defined on the intersection of the domains of definition of f and g. Suppose that (X, S, μ) is a measure space and that f and g are both measurable [μ] on X. Show that h = αf + βg as just defined is again measurable [μ]. This definition does not turn the collection Mμ of all scalar-valued functions measurable ˙ μ denotes the set of equivalence [μ] into a vector space (why not?). However, if M classes f˙ of functions f in Mμ with respect to the equivalence relation f = g almost ˙ μ is a everywhere [μ], and if we write αf˙ + β g˙ = h˙ where h = αf + βg, then M vector space. Note the relation between this space and the quotient space M/M0 of the space of all measurable functions on X taken modulo the submanifold M0 of ˙ μ contains exactly functions vanishing almost everywhere [μ]. Each element f˙ of M one element of M/M0 , this element consisting of course of the functions in f˙ that are everywhere defined. 4H. Let (X, S, μ) be a measure space. Then subsets E and F of X are almost equal [μ] if and only if the symmetric difference EF is a subset of a set of measure zero. Hence, if E and F are measurable and almost equal [μ], then μ(E) = μ(F ). Clearly the relation of being almost equal [μ] is an equivalence relation on S. Let E˙ denote the equivalence class of E and let S˙ denote the set of all such equivalence classes. The ˙ μ), ˙ = μ(E), is known as the pair (S, ˙ where μ˙ is the function on S˙ defined by μ( ˙ E) ˙ F˙ ) = μ(EF ) defines measure algebra of (X, S, μ). Show that when μ(X) < ∞, ρ(E, ˙ Show also that S˙ is complete as a metric space with respect to the a metric on S. metric ρ. (Hint: Start, as one always may

in a completeness argument, with a sequence of sets {En } having the property that n ρ(E˙ n+1 , E˙ n ) < +∞ ([I, Proposition 8.6]). Show that there exists a null set Z with the property that the sequence {En \ Z} is convergent to a limit E0 (see [I, Example 1U]). Show likewise that the sequence {Em \ Z)(En \ Z)}∞ n=1 converges to (Em \ Z)E0 .) 4I. A measure space (X, S, μ) is said to be complete if every subset of a set of measure zero with respect to μ is measurable and is, therefore, also a set of measure zero. (Another way to say this is to say that the collection Z of null sets is a σ-ideal not only in S, but in the power class on X.) Show that if (X, S, μ) is a complete measure space and if f is a function that is measurable [μ], then every function g such that g = f almost everywhere [μ] is also measurable [μ]. Similarly, if (X, S, μ) is complete and if E belongs to S, then every set F that is almost equal to E[μ] also belongs to S and we have μ(E) = μ(F ). Again, if Z is a null set in a complete measure space (X, S, μ), then an arbitrary scalar-valued function defined on an arbitrary subset of X is measurable [μ] on Z.

4 Convergence theorems for Lebesgue integrals

99

4J. Let (X, S, μ) be a measure space. Consider the collection S of those subsets of X that are almost equal [μ] to some set E in S. If F belongs to S and if E1 and E2 are sets in S such that F and Ei , i = 1, 2, are almost equal [μ], then E1 and E2 are almost equal [μ] and therefore μ(E1 ) = μ(E2 ) (see Problem 4H). Thus we may and do define μ(F ) = μ(E1 ) = μ(E2 ). Show that S is a σ-algebra of subsets of X and that μ is a measure on the measurable space (X, S). Show also that (X, S, μ) is complete, and that if (X, T, ν) is any complete measure space such that S ⊂ T and μ = ν|S, then S ⊂ T and μ = ν|S. The space (X, S, μ) is the completion of (X, S, μ). Show that a property P (x) holds almost everywhere [μ] on the measure space (X, S, μ) if and only if it holds almost everywhere [μ] on the completion (X, S, μ). 4K. Let (X, S, μ) be a measure space and let (X, S, μ) be its completion. Let f be a function on X that is measurable [S] and let g be a function on X such that g = f almost everywhere [μ]. Show that g is measurable [S]. Show, conversely, that if g is a function on X that is measurable [S], then there exists a function f that is measurable [S] such that g = f almost everywhere

[μ]. (Hint: Reduce the problem to the case of simple functions by writing g = n gn , where the functions gn are simple and measurable [S].) Remark 4.47. This last result may also be phrased as follows. Consider the vector space ˙ μ of equivalence classes of functions measurable [S], where the equivalence relation is M ˙ μ that of being equal almost everywhere [μ] (see Problem 4G). Then each element g˙ of M ˙ μ formed using μ instead of μ contains exactly one element f˙ of the like space M . Moreover, the functions belonging to g˙ are precisely the various possible extensions of elements of f˙ (to arbitrary subsets of X). 4L. Show that in the monotone convergence theorem the assumption that the functions fn are nonnegative almost everywhere can be weakened. (The phrasing of the theorem in the text was chosen simply because it is more or less traditional.) Show likewise that if {fn } is a sequence of real functions on a measure space (X, S, μ) such that each fn is integrable [μ], and if {fn } is monotone decreasing almost  everywhere, then limn fn is integrable [μ] if and only if the  numerical sequence  { fn dμ} is bounded (below), and that in this case we have (limn fn ) dμ = limn fn dμ. 4M. The assumption that the functions fn in Fatou’s lemma are nonnegative almost everywhere cannot be dropped (example ?), but it can be relaxed, and the conclusion can also be strengthened slightly. Here is an alternative version of Fatou’s lemma frequently found in the literature. Let X be a measure space with measure μ, and let {fn } be a sequence of real functions on X such that each fn is integrable [μ]. Suppose that (i) there exists a function ϕ, integrable [μ] over X, such that ϕ ≤ fn almost  everywhere for all n, and (ii) lim inf n fn dμ < +∞. Then the pointwise inferior limit of {fn } is integrable [μ] and we have

 

 lim inf fn n

 dμ ≤ lim inf n

fn dμ.

(Hint: In  the proof of Fatou’s lemma as given in the text we actually have inf k≥n fk dμ.)



gn dμ ≤

4N. Use the preceding problem to obtain the following theorem. Let X be a measure space with measure μ, and let {fn } be a sequence of real functions measurable [μ] on X. Suppose that there exist functions ϕ and Φ, both integrable [μ] over X, such that

100

4 Convergence theorems for Lebesgue integrals ϕ ≤ fn ≤ Φ almost everywhere [μ] for all n. Then both lim inf n fn and lim supn fn are integrable [μ] and we have



 (lim inf fn ) dμ ≤ lim inf n

n

fn dμ



≤ lim sup

 fn dμ ≤

n

(lim sup fn ) dμ. n

Use this fact to give a new proof of the dominated convergence theorem. 4O. Let (X, S, μ) be a measure space, and let {fn } be a sequence of integrable functions on X that is Cauchy in the mean and converges almost everywhere [μ] to a limit f . Use Fatou’s lemma to show that {fn } also converges to f in the mean. 4P. (i) Give an example of a measure space (X, S, μ)and a sequence of  functions {fn } on X converging pointwise to zero such that fn dμ = 0 and |fn | dμ = 1 for every positive integer n. (Thus fn → f pointwise, together with the condition  that fn dμ → f dμ, does not imply in general that {fn } converges to f in the mean.) (ii) On the other hand, if (X, S, μ) is a measure space, and if {fn } is a sequence of realvalued integrable functions on X that converges limit  pointwise to an integrable  f in such a way that the numerical sequence { fn dμ} converges to f dμ, and if there exists a single integrable function g such that fn ≥ g almost everywhere for every n (or such that fn ≤ g almost everywhere for every n, or even such that one or the other holds for every n) then {fn } does converge to f in the mean. (Hint: Suppose fn ≥ g for all n. Show that (fn − f )− ≤ fn , and hence that  (fn − f )− dμ → 0 by the dominated convergence theorem. Then verify that the sequence { (fn − f )+ dμ} also tends to zero.) 4Q. The hypotheses of Theorem 4.31 are necessary as well as sufficient. Indeed, if {fn } is a sequence of integrable functions that is Cauchy in the mean on a measure space (X, S, μ), then the indefinite integrals of the functions |fn | are uniformly absolutely continuous and uniformly concentrated on sets of finite measure with respect to μ. 4R. Let γ denote the counting measure on the set N of positive integers, and for each m ∈ N let em = {em,n } denote the sequence defined by

em,n =

0, 1,

m = n, m = n.

The indefinite integral νm of em with respect to γ is just the unit point mass at m, and the sequence {νm } is certainly uniformly absolutely continuous  with respect to γ, but {em }∞ em dγ ≡ 1. Show m=1 converges pointwise to the sequence zero, while that the requirement that the sequence {νn } be uniformly concentrated on sets of finite measure cannot be dropped in general, even in Corollary 4.32. 4S. Let δ denote the discrete measure on N obtained by assigning to each positive integer n the weight wn = 1/2n , and let fn = 2n en , with en as in the preceding problem. Then fn is integrable [δ], when regarded as a function on N, and the  sequence {fn } tends pointwise to zero. Moreover, δ(N) =1. Nevertheless, we have fn dδ = 1 for every n. Thus the requirement that the sequence {νn } be uniformly absolutely continuous cannot be dropped either, even in Corollary 4.32.

4 Convergence theorems for Lebesgue integrals

101

4T. Let (X, S, μ) be a measure space such that μ(X) < +∞, and let {fn } be a sequence of integrable functions on X converging uniformly to a scalar-valued function f . Show that f is integrable and that {fn } converges to f in the mean. Show by example that if the requirement that μ be finite is dropped, then it is possible for the  limit f not to be integrable. Show also that even if f is integrable, thesequence { fn dμ} may fail to converge, or may converge to some limit other than f dμ. 4U. Let (X, S, μ) be a measure space, and let {fλ }λ∈Λ be a net of real-valued functions defined everywhere on X and integrable [μ]. Suppose that the net {fλ } is monotone increasing in the sense that fλ ≤ fλ almost everywhere whenever λ ≤ λ , and suppose finally that the index family Λ is countably determined ([I, Problem 4K]). Show that the pointwise limit limλ fλ is integrable if and only if the numerical net   { fλ dμ}λ∈Λ is bounded, and in this case we have (limλ fλ ) dμ = limλ fλ dμ. (Hint: See [I, Problems 1I and 4K].) Show by example that the assumption that Λ is countably determined cannot be dropped. Remark 4.48. The reason for assuming the functions fλ to be everywhere defined is to ensure that the numerical nets {fλ (x)}λ∈Λ are defined on a substantial set of points. Trivial examples show that if each fλ is permitted to be undefined on its own null set, then every point of X may fail to belong to the domain of definition of fλ for all λ in a cofinal subset of Λ (even when Λ is countably determined). In these circumstances it is somewhat questionable what “limλ fλ (x)” is to mean. What can be obtained is the following highly technical result. Suppose all of the hypotheses of the theorem in Problem 4U are satisfied, except that each fλ is only assumed to be defined almost everywhere [μ]. For each point x ∈ X let Λx denote the set of those λ such that fλ is defined at x. Suppose also that the net { fλ dμ} is bounded. Then there exists a measurable set E such that μ(X \ E) = 0 and such that for every x ∈ E the set Λx is a directed cofinal subset of Λ. Furthermore, if we define a function f by  setting f (x) =limλ∈Λx fλ (x) at each point of E, then f is integrable [μ] and we have f dμ = limλ∈Λ fλ dμ. 4V. Show that the theorem of Beppo-Levi (Corollary 4.25) holds for indexed sums of integrable functions provided that the index family is countable. (Simple examples show that it does not hold in general for uncountable indexed sums.) 4W. Let (X, S, μ) be a measure space, and let Λ be a countably determined directed set of indices. Suppose given a net of functions {fλ }λ∈Λ converging pointwise to a limit f , where each fλ is everywhere defined on X and integrable [μ]. For each index λ let νλ denote the indefinite integral of fλ , and suppose the family {νλ }λ∈Λ is uniformly absolutely continuous and uniformly concentrated on sets of finite measure with respect to μ. Show that f is integrable [μ] and that the net {fλ } converges to f in the mean. State and prove a dominated convergence theorem for nets in place of sequences. 4X. Let (X, S, μ) be a measure space, let f be a scalar-valued function on X that is measurable [μ], and let {fn } be a sequence of functions each of which is measurable [μ]. Write Eε,n = E(|f − fn | ≥ ε) for ε > 0 and every n ∈ N. Then {fn } is said to converge to f in measure if limn μ(Eε,n ) = 0 for every ε > 0. If the functions f and fn are all everywhere defined, and if the sequence {fn } converges uniformly to f , then it also converges to f in measure. Show that if μ(X) < +∞ and if {fn } merely converges to f almost everywhere, then {fn } also converges to f in measure. (Convergence in measure is studied in more detail in Chapter 9.)

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4 Convergence theorems for Lebesgue integrals

4Y. Let (X, S, μ) be a measure space. Show that if the functions f and fn of the preceding problem are all integrable [μ], and if the sequence {fn } converges to f in the mean, then it also converges to f in measure. Show, in addition, that Theorems 4.29, 4.31, and 4.35 are all valid with convergence almost everywhere replaced by convergence in measure. Show, finally, that if {fn } converges to f in measure, then a subsequence {fnk } converges to f almost everywhere [μ]. (Hint: Choose fnk so μ(E(|f − fnk | ≥ 1/k)) < 1/2k .) 4Z. Consider a function f : I → R, where I = [a, b] is a finite interval. Show that the following conditions are equivalent. (i) f is Riemann integrable. (ii) Given ε > 0, there exist continuous functions g, h : I → R such that g ≤ f ≤ h and  (h − g) dλ1 < ε. I (iii) Given ε > 0, there exist polynomials g, h with real coefficients such that g ≤ f ≤ h on I and I (h − g) dλ1 < ε. 4AA. This problem provides an interesting use of the Riemann in proving a

∞ integral n theorem in analysis. Consider a power series f (x) = n=0 an x with an ≥ 0 for n ∈ N0 and with the property that the series converges in R for x ∈ (0, 1) and satisfies lim(1 − x)f (x) = 1. x↑1





(i) Show that limx↑1 (1 − x) n=0 an xn p(xn ) = [0,1] p dλ1 for every polynomial p. (ii) Show that the result in (i) also holds for any Riemann integrable function p defined on [0, 1]. (Hint: Use Problem 4Z(iii).) (iii) Apply (ii), with p the characteristic function of an interval, to conclude that limN N −1 N n=0 an = 1. The proof suggested above is due to Karamata. Part (iii) is an example of a Tauberian theorem, which result. If {an } ⊂ C

is a partial converse to the following

∞ satisfies limN N −1 N a = 1, then lim (1 − x) a xn = 1. n n x↑1 n=0 n=0 4BB. Consider a compact Hausdroff topological space X, and a subalgebra A of the algebra C(X) of all continuous complex-valued functions on X. Assume that A contains the constant functions, and for every two distinct points x, y ∈ X there exists a real-valued function u ∈ A such that u(x) = u(y). Stone’s extension of the Weierstrass approximation theorem states that, under these assumptions, every function f ∈ C(X) can be approximated uniformly on X by functions in A. This problem sketches a proof discovered by V. Machado. Assume, to get a contradiction, that there exists a function f ∈ C(X) such that the number d = inf sup |f (x) − g(x)| g∈A x∈X

is not zero. Denote by F the collection of those closed nonempty subsets C ⊂ X with the property that inf sup |f (x) − g(x)| = d.

g∈A x∈C

We have F = ∅ since X ∈ F .

4 Convergence theorems for Lebesgue integrals

103

(i) Let  C ⊂ F be a collection which is totally ordered by inclusion. Show that C∈C C ∈ F . (ii) Consider a minimal element C0 ∈ F , that is, C0 has no proper closed subset which belongs to F . Assume that C0 contains at least two elements, and select a realvalued function u ∈ A such that supx∈C0 u(x) = 1 and inf x∈C0 u(x) = 0. Define closed sets C1 = {x ∈ C0 : u(x) ≤ 2/3} and C2 = {x ∈ C0 : u(x) ≥ 1/3} so that C1 and C2 are proper closed subsets of C0 such that C1 ∪ C2 = C0 . Choose functions gj ∈ A with the property that supx∈Cj |f (x) − gj (x)| < d for j = 1, 2, and define n

n

g = [1 − (1 − un )2 ]g1 + (1 − un )2 g2 ∈ A. Show that supx∈C0 |f (x) − g(x)| < d if n is sufficiently large. (iii) Derive a contradiction to the assumption that d > 0. 4CC. A trigonometric polynomial (with period 1) is a function of the form q(x) =

N 

cn e2πinx ,

x ∈ R,

n=−N

where N ∈ N and cn ∈ C for |n| ≤ N . Let f : [0, 1] → C be a continuous function such that f (0) = f (1). Prove the Weierstrass approximation theorem for trigonometric polynomials: for every ε > 0 there exists a trigonometric polynomial q such that |f (x) − q(x)| < ε for all x ∈ [0, 1]. If f is real-valued then q can be chosen to be real-valued as well. 4DD. In Problem 4Z, assume that I = [0, 1] and show that conditions (i–iii) are also equivalent to the following condition. (iv) Given ε > 0, there  exist real-valued trigonometric polynomials g, h such that g ≤ f ≤ h on I and I (h − g) dλ1 < ε. 4EE. Consider a number ξ ∈ R \ Q, and denote by ξn = nξ − [nξ] ∈ [0, 1] the fractional part of nξ. Use Problem 4DD to show that lim

n→∞

f (ξ1 ) + f (ξ2 ) + · · · + f (ξn ) = n



1

f (t) dt 0

for every Riemann integrable function f : [0, 1] → R. In particular, lim

n→∞

card{j : ξj ∈ [a, b], j = 1, 2, . . . , n} =b−a n

for every interval [a, b] ⊂ [0, 1]. (Sequences {ξn } which satisfy this last condition are said to be equidistributed in [0, 1]. The result in this problem and its proof are due to H. Weyl.) 4FF. Consider real numbers ξ1 , ξ2 , . . . , ξd which are linearly independent over the rational field Q, and let A = [a1 , b1 ] × · · · × [ad , bd ] be a cell contained in [0, 1]d . Show that the limit lim

n→∞

card{j ∈ {1, . . . , n} : jξi − [jξi ] ∈ [ai , bi ] for all i = 1, . . . d} n

exists and equals λd (A).

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4 Convergence theorems for Lebesgue integrals

4GG. Let f : R → R be an absolutely continuous function. Show that f maps every λ1 -null set to a λ1 -null set. 4HH. Let (X, S, μ) be a measure space, let F be a family of complex-valued functions integrable [μ] over X. Show that F is uniformly integrable [μ] if and only if the following three conditions are satisfied.



(i) supf ∈F X |f | dμ 0. Then for each n it is a simple matter to construct a cell Z"n containing Zn in its interior and such that |Z"n | < |Zn | + ε/2n . (Replace each edge of Zd by an interval having the same center as that edge but t times longer for some t > 1. The cell Z"n thus obtained satisfies |Z"n | = td |Zn |. Then take t to be sufficiently close to 1.) The set Z is covered by the open interiors (Z"n )◦ , and consequently [a1 + ε, b1 ] × · · · × [ad + ε, bd ] ⊂ (Z"1 )◦ ∪ · · · ∪ (Z"N )◦ for some suitably large N , by the Heine-Borel theorem ([I, Corollary 8.33]). Thus, by (D), d #

(bj − aj − ε) ≤ |Z1 | + · · · + |ZN | + ε ≤



|Zn | + ε.

n=1

j=1

The countable subadditivity of | · | on Z follows by letting ε tend to zero. Hence for every cell Z in Rd we have λ∗d (Z) = |Z|. Lebesgue outer λ∗d measure is translation invariant on Rd , that is λ∗d (E) = λ∗d (E + x),

E ⊂ Rd , x ∈ Rd ,

where E + x = {y + x : y ∈ E}. Indeed, this equality is easily verified when E is a cell, and its extension to arbitrary sets is obtained directly from the definition of λ∗d . Example 5.21. Let f be a monotone increasing, real-valued function defined on R, and suppose f is right continuous on R (that is, f (t) = lims↓t f (s) for every t ∈ R). Let γf be the gauge associated with f as in Example 5.12. The proof of the countable subadditivity of γf is quite like that for the gauge | · | of Example 5.20. Let {(an , bn ]}∞ n=1 be a sequence of intervals in H covering an interval (a, b], where we assume a < b and an < bn , n ∈ N, and fix ε > 0. For each n, choose bn > bn such that γf ((an , bn ]) < γf ((an , bn ]) + ε/2n+1 (using the right continuity of f at bn ). Similarly, choose a ∈ (a, b) such that γf ((a , b]) > γf ((a, b]) − ε/2 (using right continuity of f at a). The sequence {(an , bn )} of open intervals covers the closed interval [a , b], and it follows by the Heine-Borel theorem ([I, Corollary 8.33]) that some finite number (a1 , b1 ), . . . , (aN , bN ) of these intervals covers [a , b]. But then the half-open intervals (a1 , b1 ], . . . , (aN , bN ] also cover the half-open interval (a , b]. As noted in Example 5.12, the gauge γf extends to a finitely additive set function νf on the ring Hr , and therefore γf ((a , b]) ≤ γf ((a1 , b1 ]) + · · · + γf ((aN , bN ]).

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5 Existence and uniqueness of measures

From this it follows that γf ((a, b]) ≤



γf ((an , bn ]) + ε,

n=1

and letting ε → 0 we conclude that γf is countably subadditive. Hence the outer measure μ∗f generated by γf extends γf . By Proposition Example 5.22. Let f be the monotone increasing function on R obtained by setting

0 if t ≤ 0, f (t) = 1 if t > 0, and let γf be the gauge associated with f as in Example 5.12. The sequence of intervals {(−n, 0]}∞ n=1 covers the ray (−∞, 0], and γf ((−n, 0]) = 0 for every index n. Similarly, the sequence of intervals {(1/n, n]}n=1 provides a covering of the ray (0, +∞) with γf ((1/n, n]) = 0 for every n. Thus, applying (5.4) one sees that the outer measure generated by γf is identically zero. It follows from Proposition 5.19 that γf is not countably subadditive. (Note that f is left continuous rather than right continuous.) If it is assumed that the outer measure μ∗ generated by a gauge γ extends γ, then it is obvious that the measure μ defined by μ∗ also extends γ if and only if the sets belonging to the domain C of γ are all measurable [μ∗ ]. (Note that in this case it follows that the entire σ-algebra generated by C is contained in the σ-algebra M of sets measurable [μ∗ ].) In this connection the following result provides an important beginning. Proposition 5.23. Let γ be a gauge defined on C ⊂ 2X , and let μ∗ be the outer measure on X generated by γ. Then a set A ⊂ X is measurable [μ∗ ] if and only if it splits all of the sets in C additively, that is, if and only if μ∗ (C) = μ∗ (C ∩ A) + μ∗ (C \ A),

C ∈ C.

Proof. The necessity of the condition is obvious. To prove its sufficiency, let A have the stated property, let B be an arbitrary subset of X, and fix ε > 0. It is immediate that A splits B additively if μ∗ (B) = +∞, so we assume that μ∗ (B) is finite. Then there exists a covering {Cn } of B consisting of sets in C such that n γ(Cn ) ≤ μ∗ (B) + ε. For each n set Cn = Cn ∩ A and Cn = Cn \ A. Then   Cn and B \ A ⊂ Cn , B∩A⊂ n

so μ∗ (B ∩ A) ≤

n

μ∗ (Cn )

n

and

μ∗ (B \ A) ≤

n

μ∗ (Cn ).

5 Existence and uniqueness of measures

115

Hence μ∗ (B ∩ A) + μ∗ (B \ A) ≤

[μ∗ (Cn ) + μ∗ (Cn )] =

n

μ∗ (Cn ).

n

But μ∗ (Cn ) ≤ γ(Cn ) for each n as was noted above. Thus

γ(Cn ) ≤ μ∗ (B) + ε, μ∗ (B ∩ A) + μ∗ (B \ A) ≤ n

and since ε is arbitrary, this shows that A also splits B additively, and hence   that A is measurable [μ∗ ]. Example 5.24. The gauge in Example 5.20 satisfies the hypothesis of Proposition 5.23, and the σ-ring generated by Z is BRd . The measure λd generated by this gauge is Lebesgue measure on Rd , or d-dimensional Lebesgue measure. The restriction of λd to BRd is d-dimensional Lebesgue-Borel measure. It is convenient to use the notation λd for both of these measures. It will be clear from the context whether we work with Lebesgue measure or with its restriction to BRd . It was noted in Example 5.20 that Lebesgue outer measure is translation invariant. It follows that a set E ⊂ Rd is Lebesgue measurable if and only if E + x is Lebesgue measurable for every x ∈ Rd , in which case λd (E) = λd (E + x),

x ∈ Rd .

Example 5.25. The measure μf that arises from the gauge γf in Example 5.22 is called the Lebesgue-Stieltjes measure determined by the  function f . The integral of a function h relative to μf is usually written as R h(t) df (t). More generally, suppose that I ⊂ R is an interval and f : I → R is a monotone increasing, right-continuous function. Then one can use the procedure of Example 5.22 to construct the Lebesgue-Stieltjes measure μf on I. Example 5.26. Let I = [a, b], a < b, be a compact interval and let f : I → R be a monotone increasing, right-continuous function. For every point t ∈ I, μf ({t}) equals the jump of f at the point t, that is, the difference between f (t) and the left-hand limit of f at t. It follows that f is continuous on I if and only if μf does not assign positive measure to any singleton. We can similarly characterize those functions f with the property that μf |BI is absolutely continuous relative to the Lebesgue-Borel measure λ1 |BI . This happens precisely when f is absolutely continuous in the following sense: for every ε > 0 there exists δ > 0 such that for every  finite collection {(ai , bi ]}ni=1 n of npairwise disjoint subintervals of I satisfying i=1 (bi − ai ) < δ we have i=1 (f (bi ) − f (ai )) < ε. The equivalence of these notions of absolute continuity, as defined above, follows easily from Proposition 4.3. The notion of an absolutely continuous function can be extended to arbitrary functions f : I → C,where I ⊂ R is an arbitrary interval. We say that such an f is absolutely continuous on I if for every ε > 0 there exists δ > 0

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5 Existence and uniqueness of measures

n such that for every finite disjoint subinn collection {(ai , bi ]}i=1 ofpairwise n tervals of I satisfying i=1 (bi − ai ) < δ we have i=1 |f (bi ) − f (ai )| < ε. It is obvious that f is absolutely continuous on I if and only if its real and imaginary parts are.

Example 5.27. Lebesgue measure on Rd is a proper extension of LebesgueBorel measure. Indeed, the σ-algebra BRd is generated by countably many open sets, and it follows easily that the cardinality of BRd equals 2ℵ0 . On the other hand, Problem 3Y shows that the ternary Cantor set C satisfies λ1 (C) = 0. Since Lebesgue measure is complete, it follows that every subset of C is measurable [λ1 ]. Thus the cardinality of the σ-algebra of Lebesgue ℵ0 ℵ0 measurable sets is at least 2card C = 22 . Of course, 22 is the cardinality of 2R . The Cantor-Bernstein theorem [I, Theorem 4.1] implies that there are ℵ0 exactly 22 Lebesgue measurable sets in R. One may wonder whether all subsets of R are Lebesgue measurable. Using the axiom of choice, one can construct sets which are not Lebesgue measurable. This is done as follows. Given two real numbers x, y ∈ R, write x ≡Q y if y − x ∈ Q. Then ≡Q is an equivalence relation, and the equivalence classes are of the form Q + x with x ∈ R. Since each such set is dense in R, the axiom of choice implies the existence of a set E ⊂ (0, 1) which contains exactly one element in each equivalence class. Observe now that  (E + x), R= x∈Q

so that +∞ = λ1 (R) ≤

λ∗1 (E + x).

x∈Q

All the terms in the sum above are equal to λ∗1 (E), and thus λ∗1 (E) > 0. On the other hand, the sets E +x, x ∈ Q, are pairwise disjoint, and E +x ⊂ (0, 2) if x ∈ (0, 1). If E were measurable [λ1 ] we could conclude that

λ∗1 (E + x) ≤ λ1 ((0, 2)) = 2, x∈Q∩(0,1)

and therefore λ∗1 (E) = 0. This contradiction shows that E is not measurable [λ1 ]. The set E constructed above shows, more generally, that there does not exist a translation-invariant measure μ on 2R such that μ((0, 1)) = 1. Banach proved that there do exist translation invariant, finitely additive set functions μ : 2R → [0, +∞] such that μ((0, 1)) = 1. One easy consequence of Proposition 5.23 is the following somewhat special result.

5 Existence and uniqueness of measures

117

Proposition 5.28. Let R be a ring of subsets of a set X and let γ be a finitely additive gauge defined on R. Then the sets in S(R) are all measurable with respect to the outer measure μ∗ generated by γ. Proof. Let P and Q be sets in R. According to Proposition 5.23 it suffices to show that P splits Q additively with respect to μ∗ , and it suffices to consider fix ε > 0 and let {Rn } be a countable the case μ∗ (Q) < +∞. To this end  ∗ (Q) + ε. The sets Rn ∩ P covering of Q by sets in R such that n γ(Rn ) ≤ μ ∗ belong to R  and cover Q ∩ P , so μ (Q ∩ P ) ≤ n γ(Rn ∩ P ). Similarly, μ∗ (Q \ P ) ≤ n γ(Rn \ P ), and therefore

[γ(Rn ∩ P ) + γ(Rn \ P )] = γ(Rn ) ≤ μ∗ (Q) + ε μ∗ (Q ∩ P ) + μ∗ (Q \ P ) ≤ n

n

since γ(Rn ∩ P ) + γ(Rn \ P ) = γ(Rn ) for each n by the finite additivity of γ. The result follows.   Proposition 5.28 answers a question that may well not be worth asking— for instance, when the outer measure μ∗ is itself without interest (recall Example 5.22). When Proposition 5.28 is combined with Proposition 5.19, however, a useful result, which needs no further proof, is obtained. Corollary 5.29. Let R be a ring of subsets of a set X and let γ be a countably subadditive and finitely additive gauge defined on R. Then the outer measure μ∗ generated by γ and the measure μ defined by μ∗ both extend γ. In particular, every set in S(R) is measurable [μ]. Remark 5.30. The hypotheses of Corollary 5.29 were selected so as to make as clear as possible how this result follows from those that precede it. It should be observed that a gauge γ (defined on a ring of sets R) that is both finitely additive and countably subadditive on R is, in fact, countably additive on R, and that the converse is also true. That is, a countably additive gauge defined on a ring of sets is also finitely additive and countably subadditive. Thus Corollary 5.29 might equally well have been stated for a countably additive gauge on R. Such a set function is usually called a ring measure on R. Definition 5.31. A gauge γ defined on a collection C ⊂ 2X is said to be extendible if there exists a measure μ on a σ-algebra S in X such that C ⊂ S and μ|C = γ. Equivalently, γ is extendible if there exists an outer measure μ∗ on X such that μ∗ |C = γ and every set E ∈ C is measurable [μ∗ ]. Proposition 5.32. Assume that C ⊂ 2X and γ : C → [0, +∞] is an extendible gauge. Denote by μ∗ the outer measure defined by γ. Then: (1) For every set A ⊂ X there exists a set B ⊂ X which is measurable [μ∗ ], B ⊃ A, and μ∗ (B) = μ∗ (A). (2) For every increasing sequence {An } ⊂ 2X with union A, we have μ∗ (A) = limn μ∗ (An ).

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5 Existence and uniqueness of measures

Proof. To prove (1), assume first that μ∗ (A) < +∞. It follows from the given m ∈ N, there exists }∞ definition of μ∗ that, n=1 ⊂ C  a sequence {Cm,n  ∗ for all n ∈ N and γ(C ) ≤ μ (A) + 1/m. such that A ⊂ n Cm,n m,n n  We then set B = m n Cm,n . This set is measurable [μ∗ ] because γ was assumed extendible. If μ∗ (A) = +∞ we can choose B = X. Consider next an increasing sequence {An } ⊂ 2X , and use (1) to find sets ∗ ∗ ∗ B n ⊃ An which are measurable [μ ] and μ (Bn ) = μ (An ). Replacing Bn by k≥n Bk , we may assume that the sequence {Bn } is increasing as well. We have then    ∗ ∗ μ (A) ≤ μ Bn = lim μ∗ (Bn ) = lim μ∗ (An ), n

n

n

where we used Proposition 3.28 in the first equality. The inequality μ∗ (A) ≥   limn μ∗ (An ) is true because A ⊃ An for all n. The reader will recall from Definition 1.24 in Chapter 1 the notion of an analytic set in a metric space. Analytic sets play an important role in certain areas of mathematics like descriptive set theory, fractals, etc., and therefore at times it is useful to know that such analytic sets are measurable. For measures on metric spaces defined on the Borel subsets of the space, one can prove in some generality that analytic subsets are measurable as well (relative to the corresponding outer measure). Proposition 5.33. Consider a separable, complete metric space X, an extendible gauge γ defined on the collection G of open subsets of X, and the outer measure μ∗ determined by γ. Then every analytic set A ⊂ X with finite outer measure is measurable [μ∗ ]. Proof. Assume that the set A is the result of the Suslin operation (Definition 1.24) applied to the system {Fn1 ,...,nk } of closed subsets of X. We may assume in addition that each set Fn1 ,...,nk is nonempty, has diameter less than 1/k , and contains Fn1 ,...,nk ,nk+1 for all nk+1 ∈ N (Problem 1Z(i)). Fix now ∞ ε > 0, and use the notation Fn = k=1 Fn1 ,...,nk for n = (n1 , n2 , . . . ) ∈ NN . Proposition 5.32 implies that ⎡ ⎤  Fn ⎦ = μ∗ (A), lim μ∗ ⎣ m→∞

n1 ≤m

and thus there exists m1 ∈ N such that ⎡ ⎤  μ∗ ⎣ Fn ⎦ > μ∗ (A) − ε. n1 ≤m1

5 Existence and uniqueness of measures

119

Proceeding inductively, we find a sequence m ∈ NN such that ⎡ ⎤  μ∗ ⎣ Fn ⎦ > μ∗ (A) − ε − ε2 · · · − εk > μ∗ (A) − n1 ≤m1 ,...,nk ≤mk

Define now a subset of A by setting Kε = Kε =

∞ 

 nj ≤mj ,j∈N



ε . (5.6) 1−ε

Fn . We claim that

Fn1 ,...,nk .

(5.7)

k=1 n1 ≤m1 ,...,nk ≤mk

The inclusion Fn ⊂ Kε is obvious since nj ≤ mj for all j. Conversely, fix a point x in the set on the right-hand side of (5.7), so that for each k ∈ N we have x ∈ Fn1 (k),...,nk (k) for some numbers n1 (k) ≤ m1 , . . . , nk (k) ≤ mk . A simple diagonal argument shows that there exists n ∈ NN such that, for each j, the equality nj (k) = nj is true for infinitely many values of k ≥ j. Then clearly x ∈ Fn . The equality (5.7) shows that Kε is closed and totally bounded, and therefore compact because X is complete. Let G be an arbitrary open set containing Kε . The compactness of Kε implies that {x ∈ X : dist(x, Kε ) ≤ 1/k} ⊂ G for some integer k, and therefore   Fn ⊂ Fn1 ,...,nk ⊂ G. n1 ≤m1 ,...,nk ≤mk

n1 ≤m1 ,...,nk ≤mk

By (5.6), μ∗ (G) ≥ μ∗ (A) − ε/(1 − ε), and thus μ∗ (Kε ) ≥ μ∗ (A) − ε/(1 − ε). Repeating this  construction for a sequence εn → 0, we obtain a measurable subset C = n Kεn ⊂ A such that μ∗ (C) = μ∗ (A). We know from Proposition 5.32 that μ∗ (A) = μ∗ (B) for some measurable B ⊃ A, and we conclude that μ∗ (B \ A) ≤ μ∗ (B \ C) = μ∗ (B) − μ∗ (C) = 0. The conclusion follows be cause sets of outer measure zero are measurable [μ∗ ] by Proposition 5.8.  One way to prove that a gauge is extendible is to prove that it is countably additive on C, but the direct verification of that fact can be tedious. In this connection the following somewhat technical result is sometimes useful. In it we assume that the collection of sets on which the given gauge is defined is a lattice of sets (Problem 1K). Proposition 5.34. Let L be a lattice of subsets of a set X and let γ be a countably subadditive gauge on L satisfying the condition (S) γ(L) + γ(M ) ≤ γ(L ∪ M ) + γ(L ∩ M ) for every pair of sets L, M in L. Suppose in addition that the following condition is satisfied for some set L0 belonging to L: (H) For every ε > 0 there exists a subset Q ⊂ L0 such that L \ Q ∈ L for every L ∈ L and γ(L0 \ Q) < ε.

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5 Existence and uniqueness of measures

Then L0 is measurable [μ∗ ], where μ∗ is the outer measure on X generated by γ. Proof. Proposition 5.19 implies that μ∗ extends γ. By Proposition 5.23 it suffices to show that L0 splits the sets of L additively. Fix L ∈ L such that μ∗ (L) < +∞, and ε > 0. Select Q ⊂ L0 satisfying the conditions in (H), and set M = L \ Q, L = L ∩ L0 , L = L \ L0 . Then L ⊂ M and L ∪ M = L. Moreover, L ∩ M ⊂ L0 \ Q so μ∗ (L ∩ M ) < ε. An application of (S) yields μ∗ (L ) + μ∗ (M ) ≤ μ∗ (L) + ε and therefore μ∗ (L ) + μ∗ (L ) ≤ μ∗ (L) + ε. Since ε is arbitrary, this implies μ∗ (L ) + μ∗ (L ) ≤ μ∗ (L), and the result follows.   Corollary 5.35. Let L, γ and μ∗ be as in the foregoing proposition, and let L0 denote the collection of all those sets L0 in L that satisfy (H). Then: (1) All of the sets in the σ-ring S(L0 ) are measurable [μ∗ ]. (2) If L0 contains all sets L0 ∈ L satisfying γ(L0 ) < +∞, all the sets in the σ-ring S(L) are measurable [μ∗ ]. Proof. Part (1) follows immediately from Proposition 5.19. To prove part (2) we need a slight modification of the preceding proof. Fix L ∈ L such that μ∗ (L) < ∞, and ε > 0. Under the hypothesis of (2), L = L ∩ L0 belongs to L0 , and therefore we can select Q ⊂ L such that μ∗ (L \ Q) < ε. Apply now (S) to L and L \ Q to obtain μ∗ (L ) + μ∗ (L \ Q) ≤ μ∗ (L) + μ∗ ((L ∩ L0 ) \ Q) < μ∗ (L) + ε, where we used the obvious identity L ∩ (L \ Q) = (L ∩ L0 ) \ Q. The desired  conclusion follows again by using μ∗ (L ) ≤ μ∗ (L \ Q) and letting ε → 0.  Definition 5.36. A complex- or extended real-valued set function λ defined on a lattice L of subsets of a set X is said to be modular, or to satisfy the modular law if (M) λ(L) + λ(M ) = λ(L ∪ M ) + λ(L ∩ M ),

L, M ∈ L.

Any modular set function on L satisfies (S) (for “semimodularity”) of Proposition 5.34 (Clearly too, any finitely additive set function ν on a ring R of subsets of X must be modular, and, in fact, Corollary 5.35 can be used to give a new proof of Proposition 5.34 Indeed, to see that every set R0 in R also satisfies (H) simply set Q = R0 .) We observe that if ρ is a gauge on a lattice L and if L, M ∈ L are disjoint, then (S) reduces to γ(L) + γ(M ) ≤ γ(L ∪ M ). Thus condition (S) together with countable subadditivity implies countable additivity. (Of course, if γ is to be extendible, it must be countably

5 Existence and uniqueness of measures

121

additive.) Condition (S) was chosen for the formulation of Proposition 5.34 principally because even for certain modular gauges it is easier to verify than (M). Example 5.37. Let X be a locally compact topological space, and denote by Cc (X, R) the vector space of continuous functions f : X → R with the property that the support Nf = E(f = 0) is relatively compact in X. Consider a positive linear functional ϕ : Cc (X, R) → R. For each open set U in X we define γϕ (U ) = sup{ϕ(f ) : 0 ≤ f ≤ χU , f ∈ Cc (X, R)}. Then γϕ is a gauge defined on the collection G of all open subsets of X, and γϕ (U ) < +∞ if U is relatively compact. Indeed, if U is relatively compact, by Urysohn’s lemma (see [I, Chapter 9]), there exists f ∈ Cc (X, R) such that χU ≤ f ≤ 1, and the positivity assumption implies that γϕ (U ) ≤ ϕ(f ). We show that γϕ is extendible. In order to verify the countable subadditivity of γϕ , let V be open in X, let {Un }n∈N be an arbitrary open covering of V , let f ∈ Cc (X, R) satisfy 0 ≤ f ≤ χV , and let ε > 0. We set K = {x ∈ X : f (x) ≥ ε}. If fε = (f −ε)∨0, then 0 ≤ fε ≤ χK and f ≤ fε + ε. Since K is compact there exists a positive integer N such that K ⊂ U1 ∪ · · · ∪ UN , and there exist functions g1 , . . . , gN ∈ Cc (X, R) such that 0 ≤ gi ≤ χUi , i = 1, . . . , N , and such that if g = g1 + · · · + gN , then 0 ≤ g ≤ 1 on X, while g(x) = 1 for all x ∈ K. (This is just a partition of unity subordinate to {U1 , . . . , UN }; see [I, Problem 9I].) Thus fε = fε g = fε g1 + · · · + fε gN , whence it follows that ϕ(fε ) ≤ γϕ (U1 ) + · · · + γϕ (UN ). Since f ≤ fε + ε, we conclude that ϕ(f ) ≤ γϕ (U1 ) + · · · + γϕ (UN ) + ε ≤



γϕ (Un ) + ε.

n=1

∞ By the definition of γϕ this shows that γϕ (V ) ≤ n=1 γϕ (Un ) because ε is arbitrary. Next, if U ⊂ X satisfies γϕ (U ) < +∞ and ε > 0, there exists a function f ∈ Cc (X, R) such that 0 ≤ f ≤ χU and such that ϕ(f ) > γϕ (U ) − ε. Fix ε > 0, and set K = {x ∈ X : f (x) ≥ ε/2}. Then K is closed, so U \ K is open along with U . If g ∈ Cc (X, R) and 0 ≤ g ≤ χU \K , then it is readily verified that f + g ≤ χU + ε/2. Hence, if we set h = (f + g − ε/2) ∨ 0, so 0 ≤ h ≤ χU and f + g ≤ h + ε/2, then ϕ(f ) + ϕ(g) ≤ ϕ(h) + ε/2 ≤ γϕ (U ) + ε/2. But then ϕ(g) ≤ γϕ (U ) − ϕ(f ) + ε/2 < ε, whence it follows that γϕ (U \ K) ≤ ε, and thus U satisfies condition (H) of Proposition 5.34.

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5 Existence and uniqueness of measures

Finally, to show that γϕ satisfies condition (S) of Proposition 5.34, let U and V be two open subsets of X, and let f and g be any two continuous functions on X such that 0 ≤ f ≤ χU and 0 ≤ g ≤ χV . If we set h = (f + g) ∧ 1

and

k = f + g − h,

then it is easy to see that 0 ≤ h ≤ χU ∪V and 0 ≤ k ≤ χU ∩V . Since f + g = h + k, we have ϕ(f ) + ϕ(g) = ϕ(h) + ϕ(k) ≤ γϕ (U ∪ V ) + γϕ (U ∩ V ), and (S) follows. Thus ρϕ is indeed extendible by Corollary 5.35(2). So far in this chapter we have investigated at some length the question of the existence of various measures. It is desirable to complement these results with a uniqueness theorem. The following result settles the matter quite effectively. Theorem 5.38. Let μ and ν be two measures on the same space X (each measure defined, in general, on its own σ-algebra of measurable sets). Suppose that μ and ν agree and are finite-valued on a collection D of subsets of X that is closed with respect to the formation of (finite) intersections. Then μ and ν agree on the σ-algebra S(D). Proof. Let D0 be a fixed set belonging to D, and let D0 denote the trace of D on D0 . Since D is closed with respect to the formation of intersections, D0 has the same property, consisting as it does of all those sets in D that are subsets of D0 . Next let Q0 denote the collection of all those subsets of D0 on which μ and ν agree. Since μ and ν are measures, and since μ(D0 ) = ν(D0 ) < +∞, it is clear that Q0 is a σ-quasiring (see Chapter 1 for definitions). Since D0 ⊂ Q0 we have S0 ⊂ Q0 , where S0 = S(D0 ), by Theorem 1.23. But S0 is just the trace of S(D) on D0 . Thus μ and ν agree on every set in S(D) that is a subset of D0 . Since D0 was an arbitrary set in D, this shows that μ(E) = ν(E) for every E in S(D) for which there exists D ∈ D such that E ⊂ D. To complete the proof, let E be an arbitrary set in S(D). There exists a sequence {Dn } in D that covers E (Problem 1C(iii)). Let {En } be the disjointification of the sequence {E ∩ Dn }. Then En ⊂ Dn , and therepairwise disjoint fore μ(En ) = ν(En ) for every index n. But {En } is a   sequence of sets with union E, and it follows that μ(E) = n μ(En ) = n ν(En ) = ν(E).   Example 5.39. The collection of closed cells in Rd is closed with respect to the formation of intersections, so Theorem 5.38 applies and says that Lebesgue-Borel measure λd is uniquely determined by the fact that λd (Z) = |Z| for all such cells. Example 5.40. Let X = R and consider the counting measure γ on X along with the measure 2γ. These two measures agree at every set of the ring H of finite unions of half-open intervals [a, b), a < b, since both are identically

5 Existence and uniqueness of measures

123

equal to +∞ on H except at the empty set ∅. But γ and 2γ do not agree, of course, on the σ-ring generated by H , since this ring is precisely BR and contains all finite sets. Thus the requirement that μ and ν be finite-valued on the collection D in Theorem 5.38 is indispensable for the validity of the theorem. (The requirement that D be closed with respect to the formation of intersections also cannot be dropped from the hypotheses of Theorem 5.38, as may be learned by experimenting with a three point space.) To this point we have only used outer measures as a source of abstract measures via Proposition 5.23. As we have seen before, however, some things are much more interesting in the context of a metric space. We conclude this chapter with a brief sketch of Carath´eodory’s theory of outer measures on metric spaces. Definition 5.41. Let X be a metric space and let μ∗ be an outer measure on X. Then μ∗ is a Carath´eodory outer measure if every Borel set in X is measurable [μ∗ ]. The following simple proposition is useful. Proposition 5.42. An outer measure μ∗ on a metric space X is a Carath´eodory outer measure if all the open sets in X are measurable [μ∗ ], or if all the closed sets are. If X is separable, and if all of the sets in some topological base for X are measurable [μ∗ ], then μ∗ is a Carath´eodory outer measure. Proof. Each of the stated collections of sets generates the σ-algebra BX .

 

Example 5.43. Lebesgue outer measure on Rd is a Carath´eodory outer measure. Similarly, the outer measures νf∗ associated with the various monotone increasing real-valued functions f on R are all Carath´eodory outer measures. If μ∗ is a Carath´eodory outer measure on a metric space (X, ρ), and if E and F are disjoint closed sets in X, then μ∗ (E ∪ F ) = μ∗ (E) + μ∗ (F ), since E ∪ F is split additively by the open set X \ E. Similarly, if A and B are arbitrary subsets of X such that d(A, B) = inf{ρ(x, y) : x ∈ A, y ∈ B} > 0, then μ∗ (A ∪ B) = μ∗ (A) + μ∗ (B), since A ∪ B is split additively by X \ A− . This last named fact has a useful converse. To prove it we need a lemma. Lemma 5.44. Let μ∗ be an outer measure on a metric space X satisfying the condition μ∗ (A ∪ B) = μ∗ (A) + μ∗ (B) whenever d(A, B) > 0,

(5.8)

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5 Existence and uniqueness of measures

and let {An }∞ n=1 be a sequence of subsets of X such that d(Am , An ) > 0 whenever m = n and neither Am nor An is empty. Then ∞  ∞ 

∗ An = μ∗ (An ). μ n=1

n=1

Proof. Since d(A1 ∪ · · · ∪ An , An+1 ) > 0 whenever A1 ∪ · · · ∪ An and An+1 are both nonempty, it is clear (mathematical induction) that n

μ∗ (A1 ∪ · · · ∪ An ) =

μ∗ (Ak )

k=1

 ∞  μ∗ (Ak ) ≤ μ∗ k=1 Ak for every n, so ∞  ∞

 ∗ ∗ μ (Ak ) ≤ μ Ak ,

for each index n. But then

n

k=1

k=1

k=1

and the lemma follows. Proposition 5.45. If μ∗ is an outer measure on a metric space (X, ρ) and if condition (5.8) holds, then μ∗ is a Carath´eodory outer measure. Proof. We show that if A is a subset of X and U is an open subset of X, then U splits A additively with respect to μ∗ , whence the statement follows by Proposition 5.42. In this endeavor we may and do assume without loss of generality that ∅ = U = X (for otherwise U doesn’t split A at all) and that μ∗ (A) < +∞ (since any splitting of a set of infinite outer measure is additive). Let Fn = {x ∈ X : d(x, X \ U ) ≥ 1/n} for each n ∈ N, so the sequence {Fn } increases and has union U , and define D1 = F1 and Dn+1 = Fn+1 \ Fn ,

n ∈ N.

The sequence {Dn }∞ n=1 need not satisfy the hypothesis of Lemma 5.44, but if m and n are positive integers such that n ≥ m + 2, and if neither Dm nor Dn is empty, then d(Dm , Dn ) > 0. (If x ∈ Fm and y ∈ / Fm+1 then ρ(x, y) ≥ (1/m) − (1/(m + 1)) = 1/m(m + 1).) Hence the lemma does apply ∞ to the sequence {A ∩ D2n }∞ n=1 as well as the sequence {A ∩ D2n−1 }n=1 , and we conclude that   ∞ ∞

 ∗ ∗ μ (A ∩ D2n ) = μ A ∩ D2n ≤ μ∗ (A) < +∞ n=1

n=1

and also that ∞

n=1

 ∗

μ (A ∩ D2n−1 ) = μ



A∩

∞  n=1

 D2n−1

≤ μ∗ (A) < +∞.

5 Existence and uniqueness of measures

125

∞ But then n=1 μ∗ (A ∩ Dn ) < +∞ too, and it follows (Problem 5N) that the ∞ sum of the series n=1 μ∗ (A ∩ Dn ), that is, the limit of the sequence {μ∗ (A ∩ (D1 ∪ · · · ∪ Dn ) = μ∗ (A ∩ Fn )},   ∞ coincides with μ∗ A ∩ n=1 Fn = μ∗ (A ∩ U ). From this it follows that the sequence {μ∗ (A ∩ Fn ) + μ∗ (A \ U )} tends upward to μ∗ (A ∩ U ) + μ∗ (A \ U ). But d(Fn , X \ U ) > 0 for every n (for which Fn = ∅), so μ∗ (A ∩ Fn ) + μ∗ (A \ U ) ≤ μ∗ (A) for every n, and we finally obtain the desired result μ∗ (A ∩ U ) + μ∗ (A \ U ) = μ∗ (A).



Thus far we have dealt only with the generation of outer measures from gauges on abstract sets. On a metric space there is a different way of generating measures called the Carath´eodory procedure. In outlining this construction it will be convenient to employ the following notation. If C is a collection of subsets of a metric space X, and if r is an arbitrary positive number, then Cr will denote the subcollection of consisting of those sets C in C such that diam(C) ≤ r. Definition 5.46. Let X be a metric space and let γ be a gauge defined on a collection C ⊂ 2X . Then for each positive number r the restriction of γ to Cr is a gauge on X and generates an outer measure μ∗r . Moreover, if 0 < r < s, then Cr ⊂ Cs , so μ∗r (A) ≥ μ∗s (A) for every A ⊂ X. Thus the net {μ∗r }r>0 is monotone increasing, and therefore converges to an outer measure μ∗0 . (See Problem 5E; the directed set employed here is the ray (0, +∞) directed downward.) The outer measure μ∗0 will be said to be generated by γ via the Carath´eodory procedure, and the measure μ0 defined by μ∗0 will be referred to as the measure generated by γ via the Carath´eodory procedure. d Example 5.47. Consider the  collection Z of closed cells in R and the gauge | · |. In any covering A ⊂ n Zn of a set A by closed cells, we can always replace each cell Zn in the covering by any cellular partition of itself without changing the value of the sum of the volumes of the cells. Thus each of the approximating outer measures μ∗r , r > 0, in this case coincides with Lebesgue outer measure, and so therefore does the outer measure obtained by the Carath´eodory procedure.

Example 5.48. In a metric space X define a gauge γ : 2X → [0, +∞] by γ(A) = diam A, A ∈ 2X . The outer measure ξ1∗ generated by this gauge by the Carath´eodory procedure is known as (one-dimensional ) Hausdorff outer measure on X. In the special case X = Rd , consider the restriction of the gauge γ to the col" denotes lection of closed convex subsets of Rd . If C is a subset of Rd , and if C

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5 Existence and uniqueness of measures

" = diam C (since every closed ball the closed convex hull of C, then diam C " whence it follows that the Hausdorff outer that contains C also contains C), measure ξ1∗ is also the outer measure generated by this restricted gauge. But if σ is a line segment in Rd , and {Cn } is a countable covering of σ by closed convex sets, then {Cn ∩ σ} is a countable covering of σ by subsegments of σ, and it follows that ξ1∗ (σ) is just the length of σ. In particular, ξ1∗ coincides with Lebesgue outer measure on R. In this connection see Problems 5S and 5T. The Carath´eodory procedure is more complicated than the method described in Proposition 5.15, but it frequently yields a more useful end product. In particular, outer measures generated by the Carath´eodory procedure always define a measure on Borel sets. Proposition 5.49. If γ is a gauge defined on a collection C of subsets of a metric space X, and if μ∗0 is the outer measure on X generated by γ by the Carath´eodory procedure, then μ∗0 is a Carath´eodory outer measure. Proof. By Proposition 5.45 it suffices to verify condition (5.8) for μ∗0 . Let A, B ⊂ X satisfy d(A, B) = d0 > 0 and let r ∈ (0, d0 ). It suffices to prove {Cn } be a countable covering (5.8) when μ∗0 (A∪B) < +∞. Fix ε > 0, and let of A ∪ B by sets belonging to Cr such that n γ(Cn ) is no greater than μ∗r (A ∪ B) + ε. Denote by {Cn } the collection of those sets in the covering {Cn } that meet A, and by {Cn } the collection of those that meet B. Then {Cn } and {Cn } have no sets in common, since a set of diameter less than d0 cannot meet both A and B. Thus {Cn } is a countable covering of A, {Cn } is a countable covering of B, and they have no sets in common. Moreover,

γ(Cn ) and μ∗r (B) ≤ γ(Cn ). μ∗r (A) ≤ n

n

But then μ∗r (A) + μ∗r (B) ≤

γ(Cn ) ≤ μ∗r (A ∪ B) + ε ≤ μ∗0 (A ∪ B) + ε.

n

Letting first r and then ε tend to zero, the result follows.

 

As has already been suggested, a central difficulty in dealing with an outer measure generated by the Carath´eodory procedure lies in relating the values of that outer measure to those of the gauge that generates it. In this connection we offer the following elementary result. Proposition 5.50. Let γ be a gauge defined on a collection C of subsets of a metric space X, and let μ∗0 denote the outer measure generated by γ via the Carath´eodory procedure. If μ∗ is the outer measure generated by γ in the elementary manner of Proposition 5.15, then μ∗ ≤ μ∗0 . Hence if γ is countably subadditive, then γ ≤ μ∗0 setwise on C.

5 Existence and uniqueness of measures

127

Proof. The outer measure μ∗ is dominated setwise by each of the preliminary outer measures μ∗r used in the construction of μ∗0 . If γ is countably subaddi  tive, then μ∗ |C = γ (Proposition 5.19). So far we have focused on constructing measures from ‘raw data’ such as gauges. There is a basic construction which yields a new measure from a given measure and a measurable map. Assume that (X, S, μ) is a measure space, (Y, T) is a measurable space, and ϕ : X → Y is a measurable map. Define ν : T → [0, +∞] by setting ν(E) = μ(ϕ−1 (E)) = μ({x ∈ X : ϕ(x) ∈ E}),

E ∈ T.

Then ν is a measure. Indeed, if {En }∞ n=1 ⊂ T is a sequence of pairwise disjoint sets with union E, then the sets {ϕ−1 (En )}∞ n=1 ⊂ S are pairwise disjoint as well with union ϕ−1 (E). Countable additivity follows immediately from this observation. Definition 5.51. The measure ν constructed above is called the image or push-forward of μ via ϕ, and it is denoted μ ◦ ϕ−1 or μϕ . If (Y, T) = (R, BR ), the measure μϕ is called the distribution of ϕ. More generally, if ϕ = (ϕ1 , ϕ2 , . . . , ϕd ) : X → Rd is measurable [S, BRd ], the measure μϕ is called the joint distribution of ϕ1 , ϕ2 , . . . , ϕd . Distributions of random variables are frequently used in probability. A particularly important case is the distribution μ|f | , where f : X → C is a measurable function. The numbers μ|f | ((t, +∞]) = μ({x ∈ X : |f (x)| > t}) can be used to determine whether the function f is integrable [μ]; see Example 8.13. Example 5.52. Assume that (X, S, μ) is a measure space and f : X → R is a constant function—say f (x) = α. Then μf is a point mass μ(X)δα at α:

μ(X), α ∈ E, μf (E) = 0, α∈ / E. More generally, if f = partition of X, we have

n j=1

αj χσj , where the sets {σj }nj=1 ⊂ S form a

μf =

n

μ(σj )δαj .

j=1

Here, of course, δαj is the unit point mass at αj , also known as the Dirac measure at αj . Proposition 5.53. Let (X, S, μ) be a measure space, let (Y, T) be a measurable space, and let ϕ : X → Y be a measurable map. A measurable function f : Y → C is integrable [μ ◦ ϕ−1 ] if and only if f ◦ ϕ is integrable [μ], and

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5 Existence and uniqueness of measures



f d(μ ◦ ϕ−1 ) = Y

 f ◦ ϕ dμ X

when either of these functions is integrable. Proof. It suffices to prove the result for positive functions f . Then the monotone convergence theorem shows that we can restrict ourselves to simple functions, and linearity allows us to further restrict ourselves to characteristic functions. When f = χE with E ∈ T, we have f ◦ ϕ = χϕ−1 (E) , and the result amounts to the definition of μ ◦ ϕ−1 in this special case. Proposition 5.54. Let (X, S, μ) be a probability space, and let f : X → R be a measurable function. Endow the segment (0, 1) with the restriction of Lebesgue measure λ1 . There exists a unique monotone decreasing, right continuous function g : (0, 1) → R such that μg = μf . Proof. Simply define g(t) = inf{β ∈ R : μf ((β, +∞)) > t},

t ∈ (0, 1).

The assertions of the proposition are now routine verifications.

 

The function g constructed above is called the decreasing rearrangement of f . It can be defined when μ(X) = +∞ provided that μ({x ∈ X : |f (x)| > t}) < +∞ for every t > 0, in which case g is monotone decreasing and right-continuous on (0, +∞).

Problems 5A. Let μ∗ be an outer measure on a space X. Show that if A, B ⊂ X, and if either A or B is measurable, then μ∗ (A ∪ B) + μ∗ (A ∩ B) = μ∗ (A) + μ∗ (B). 5B. Let μ∗ be an outer measure on a space X, and let A be a subset of X. Then the restriction of μ∗ to the power class on A is an outer measure on A. (We denote this outer measure by μ∗ |A.) Show that if E is measurable [μ∗ ], then E ∩ A is measurable [μ∗ |A]. In the event that A is itself a measurable set, the measure defined by μ∗ |A coincides with the contraction of μ to A, where μ denotes the measure defined by μ∗ . 5C. Let μ∗ be an outer measure on a space X, and let A be a subset of X. Define ν  (B) = μ∗ (A ∩ B) for every subset B of X. Show that ν  is an outer measure on X. (This outer measure is called the concentration of μ∗ on A.) Show that if E is measurable [μ∗ ], then E, E ∩ A, and E \ A are all measurable [ν  ]. How is ν  related to the contraction μ∗ |A of μ∗ to A as defined in the preceding problem?

5 Existence and uniqueness of measures

129

5D. Let μ∗ be an outer measure on a space X, and let {En } be a disjoint sequence of [μ∗ ]-measurable subsets of X with E. Then for any subset A ⊂

union

X, measurable ∗ (E ∩ A) and μ∗ (E \ A) = ∗ μ or not, we have μ∗ (E ∩ A) = n n n μ (En \ A). Similarly, if {Fn } is an increasing sequence of measurable sets, then limn μ∗ (Fn ∩A) = μ∗ (limn Fn ∩ A) for arbitrary A. (The same goes for decreasing sequences provided some μ∗ (Fn ∩ A) is finite.) 5E. If μ∗1 and μ∗2 are two outer measures on a space X, then μ∗1 + μ∗2 is also an outer measure on X. Show likewise that if {μ∗λ }λ∈Λ is a monotone increasing net of outer measures on X, then the limit limλ μ∗λ is an outer measure. In particular, if {μ∗α }α∈A is indexed family of outer measures on X, then the setwise indexed sum

an arbitrary ∗ α∈A μα is an outer measure on X. (Hint: The finite subsets F of A form a directed set under the inclusion ordering.) 5F. Let μ∗ be an outer measure on a space X, and let {An }∞ n=1 be an infinitesequence n of subsets of X. Define B = A , n ∈ N , and set B = limn Bn = ∞ n i i=1 n=1 An .

∞ ∗ (A ) < +∞, then lim μ∗ (B ) = μ∗ (B). (Hint: Show that Prove that if μ n n n n=1 μ∗ (B \ Bn ) → 0.) 5G. The following are all examples of outer measures. Show in each case that the only measurable sets are the sets of outer measure zero and their complements. (i) Let X be an arbitrary set. Set μ∗ (∅) = 0 and let μ∗ (A) = 1 for every nonempty subset A ⊂ X. (ii) Let γ denote the counting measure on N, and for each set A ⊂ N, let δ ∗ (A) = lim supn γ(A ∩ {1, . . . , n})/n. (iii) Let X be a set whose cardinal number is ℵα where α > 0 (see the paragraph following [I, Theorem 5.8]). If A ⊂ X is infinite, then card A = ℵη for some 0 ≤ η ≤ α. For such a set, define n, η=n 0 for every positive integer n such that An = ∅ and An+1 = X. Let μ∗ be a Carath´ eodory outer measure on X, let {A } be a metrically nested sequence of subsets of X, and n  let A = n An . Prove that μ∗ (A) = limn μ∗ (An ). 5O. What are the outer measures generated by the gauges in Problems 5H(ii) and 5H(iii) by means of the Carath´ eodory procedure? 5P. Let X be a separable metric space and let C denote the collection of all bounded subsets of X. For each positive real number p, we define γp (∅) = 0 and γp (C) = (diam C)p for all C ∈ C, C = ∅. The outer measure ξp∗ generated by γp via the Carath´ eodory procedure is known as the Hausdorff outer measure of dimension p on X, and the measure ξp defined by ξp∗ is the Hausdorff measure of dimension p on X. Show that the same outer measure is obtained if the gauge is restricted to the closed [open] sets in C.

5 Existence and uniqueness of measures

131

Remark 5.55. If λ∗d denotes Lebesgue outer measure on Rd then λ∗d and the Hausdorff outer measure ξd∗ on Rd are not identical, except when d = 1. It can be shown, however, that they are constant multiples of one another. Thus it is not uncommon to find in the literature a slightly different definition of the Hausdorff measure of dimension greater than γd = v(d) · γd , where v(d) is a proportionality constant chosen so as one based on a gauge  to make Hausdorff measure of dimension d coincide with Lebesgue measure in Euclidean space of dimension d. Incidentally, a 0-dimensional version of Hausdorff measure on an arbitrary separable metric space X is obtained by defining γ0 (∅) = 0 and γ0 (C) = 1 for all nonempty sets C ⊂ X. It is easily seen, however, that the outer measure generated by this gauge by means of the Carath´ eodory procedure is simply the counting measure on X. 5Q. Suppose that X is a separable metric space. Let B denote the σ-algebra of Borel sets in X, let γ be a countably subadditive gauge defined on B, and let μ∗0 denote the outer measure generated by γ via the Carath´ eodory procedure. Let E be a Borel set in X, and let {Pn }n∈N be a sequence of partitions of E into countably many Borel subsets such that (i) all the elements of Pn have diameter less than 1/n, and (ii) Pn+1 is a refinement of Pn in the sense that every element of Pn+1 is a subset of an element of Pn for all n ∈ N. Show that



μ∗0 (E) = lim n

γ(F ).

F ∈Pn

5R. (i) Let X be a metric space and let f : X → Y be a mapping of X into

an arbitrary set Y . If E is a subset of X and if P is a partition of E ⊂ X, then g = F ∈P χf (F ) has for its value at each point y of Y the number (either a positive integer or +∞) of those sets F belonging to P such that F ∩ f −1 ({y}) = ∅. Let {Pn } be a sequence of partitions of E that satisfies

properties (i) and (ii) in the preceding problem. Show that the functions gn = F ∈Pn χf (F ) form a monotone increasing sequence which converges pointwise on Y to N (y) = γ(E ∩ f −1 ({y})), where γ denotes the counting measure on X. (ii) Let X be a separable metric space, let (Y, S, μ) be a measure space, and let f : X → Y have the property that f (E) belongs to S for every Borel set E ⊂ X. Define γ(E) = μ(f (E)) for every Borel set E ⊂ X, and let μ∗0 denote the outer measure generated by the gauge γ via the Carath´ eodory procedure. Show that if E ⊂ X is a Borel set, then μ∗0 (E) < +∞ if and only if the function N (of part (i)) is integrable [μ], in which case we have μ∗0 (E) =

 N dμ.

(5.9)

Y



(If we employ the notation Y g dμ = +∞ when g is a nonnegative function on Y that is measurable [S] but not integrable [μ], then (5.9) holds without exception. It should be noted that the requirement that f map Borel sets onto measurable sets is not as restrictive as it might seem. Indeed, if both X and Y are separable, complete metric spaces, μ is σ-finite, and S ⊃ BY , then every continuous mapping f meets the requirement. See Problems 1Z and 1AA and Proposition 5.33.) 5S. Assume that μ is a measure on (Rd , BRd ) such that μ([0, 1]d ) = 1 and μ is translation invariant, that is μ(A + x) = μ(A) for all A ∈ BRd and x ∈ Rd . Prove that μ(A) = λd (A) for every A ∈ BRd .

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5 Existence and uniqueness of measures

5T. Denote by L the lattice of elementary figures (that is, finite unions of cells) in Rd . We are interested in functions γ : L → C which are modular (see Definition 5.36), translation invariant (as in Problem 5S), invariant under permutations of the coordinates, and such that γ([a1 , b1 ]× · · · ×[ad , bd ]) is a continuous function of the variables a1 , . . . , ad and b1 , . . . , bd . Denote by V the vector space consisting of all functions γ satisfying these properties. For instance, the restriction λd |L belongs to V . (i) Assume that γ ∈ V and show that there exist scalars λj ∈ C, j = 0, 1, . . . , d, such that

× [0, xd ]) = λ0 + dk=1 λk sk , x1 , . . . , xd ∈ R, (†) γ([0, x1 ] × · · ·

where sk = i1 1, and for each m < n we have either Gn ⊂ Gm or Gn ∩ Gm = ∅. These sets can be constructed by induction. The existence of G1 satisfying μ(G1 ) ≥ 1 follows from the fact that X ∈ C. Assume that the sets G1 , G2 , . . . , Gn−1 have been constructed, and partition the set X into 2n−1 subsets (some of them empty), each of the form F = A1 ∩ A2 ∩ · · · ∩ An−1 , where Aj is either Gj or X \ Gj , j = 1, 2, . . . , n − 1. As noted above, there is one such set F which belongs to C, and therefore we can choose Gn ∈ S such that Gn ⊂ F and μ(Gn ) ≥ 1 + μ(Gn−1 ). Observe that there cannot exist an infinite sequence n1 < n2 < · · · such that the sets {Gnj }j are pairwise disjoint because in that case ⎛ ⎞ ∞ ∞ 

+∞ > μ(X) = μ ⎝X \ Gnj ⎠ + μ(Gnj ) j=1

⎛ = μ ⎝X \

∞ 



j=1

Gnj ⎠ + ∞ = +∞.

j=1

We conclude that there exists an infinite sequence n1 < n2 < · · · such that Gnj ⊃ Gnj+1 for all j ∈ N. We have then μ(Gnj \Gnj+1 ) ≤ −1 and considering the partition ⎞ ⎛ ⎛ ⎞ ∞ ∞   Gnj ⎠ ∪ ⎝ (Gnj \ Gnj+1 )⎠ Gn1 = ⎝ j=1

we obtain

j=1

⎛ 1 ≤ μ(Gn1 ) = μ ⎝ ⎛ = μ⎝

∞  j=1 ∞ 

⎞ Gnj ⎠ + ⎞



μ(Gnj \ Gnj+1 )

j=1

Gnj ⎠ − ∞ = −∞,

j=1

a contradiction.

 

6 Signed measures, complex measures, and absolute continuity

135

If μ is any complex measure on a measurable space (X, S), the set functions E → μ(E) and E → μ(E), E ∈ S, are clearly finite signed measures. Corollary 6.2. If μ is a complex measure on the measurable space (X, S) then there exists a constant c ≥ 0 such that |μ(E1 )| + |μ(E2 )| + · · · + |μ(En )| ≤ c for every finite collection {Ej }nj=1 ⊂ S of pairwise disjoint sets. The smallest constant c with this property is called the total variation of μ. Proof. It suffices to prove the corollary when μ is real-valued because |μ(E)| ≤ n |μ(E)|  + |μ(E)| for E ∈ S. Let  {Ej }j=1 ⊂ S be pairwise disjoint, and set F1 = {Ej : μ(Ej ) ≥ 0}, F2 = {Ej : μ(Ej ) < 0}. We have then n

|μ(Ej )| = μ(F1 )−μ(F2 ) ≤ sup{μ(E) : E ∈ S}−inf{μ(E) : E ∈ S} < +∞

j=1

 

by Proposition 6.1.

One way to build a signed measure on a measurable space (X, S) is to start with two measures μ and ν on (X, S), one of which is finite, and form the difference μ − ν. It is of interest to know that this simple construction produces the most general signed measure. Theorem 6.3. (Hahn Decomposition) Given a signed measure μ on a measurable space (X, S), there exists a set A ∈ S such that, for every measurable set E, we have μ(E) ≥ 0 when E ⊂ A and μ(E) ≤ 0 when E ⊂ X \ A. If A ∈ S is another set with the same property, we have μ(E) = 0 for every measurable set E ⊂ AA . The partition X = A ∪ (X \ A) is called a Hahn decomposition of X with respect to μ. Proof. The last assertion is immediate. Indeed, every subset A ⊂ AA can be written as E = E1 ∪E2 , with E1 ⊂ A\A and E2 ⊂ A \A. Then μ(E1 ) ≥ 0 because E1 ⊂ A and μ(E1 ) ≤ 0 because E1 ⊂ X \ A . Similarly, μ(E2 ) = 0. In proving existence we assume without loss of generality that μ(X) < +∞, and thus 0 ≤ M = sup{μ(E) : E ∈ S} < +∞ by Proposition 6.1. It suffices to produce a set A ∈ S such that μ(A) = M . Indeed, if A is such a set and E ⊂ A is measurable, we have M = μ(A) = μ(E) + μ(A \ E) ≤ μ(E) + M by the definition of M , and thus μ(E) ≥ 0. On the other hand, if E ⊂ X \ A is measurable, then M ≥ μ(A ∪ E) = M + μ(E), so μ(E) ≤ 0.

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6 Signed measures, complex measures, and absolute continuity

To produce a set A with the desired properties, choose sets An ∈ S such that μ(An ) > M − 1/2n for n ∈ N. We conclude the proof by showing that μ(A) = M , where ∞  ∞  Am . A = lim sup An = n

n=1 m=n

Given a measurable set E ⊂ An+1 , we have M−

1 ≤ μ(An+1 ) = μ(An+1 \ E) + μ(E) ≤ M + μ(E) 2n

so μ(E) ≥ −1/2n+1 . Fix now n ∈ N and define Bm = Am+1 \ (An ∪ An+1 ∪ · · · ∪ Am ) for m ≥ n, so μ(Bm ) ≥ −1/2m+1 by what has just been proved. Then     ∞ ∞ ∞  

Am = μ An ∪ Bm = μ(An ) + μ(Bm ), μ m=n

m=n

m=n



and thus the sets Cn = m≥n Am satisfy M ≥ μ(Cn ) ≥ M − 1/2n−1 . The sequence {Cn } is decreasing with intersection A, and the calculation μ(C1 ) = μ(A) +



μ(Cn \ Cn+1 ) = μ(A) + lim

n=1

N →∞

N

μ(Cn \ Cn+1 )

n=1

= μ(A) + lim [μ(C1 ) − μ(CN +1 )] = μ(A) + μ(C1 ) − M. N →∞

shows that μ(A) = M , as claimed.

 

The uniqueness assertion in the preceding result allows us to give the following definition. Definition 6.4. Let μ be signed measure on a measurable space (X, S), and let X = A ∪ (X \ A) be a Hahn decomposition of X with respect to μ. The measures defined (unambiguously) for E ∈ S by μ+ (E) = μ(E ∩ A), μ− (E) = −μ(E \ A), and |μ|(E) = μ+ (E) + μ− (E), are called the positive variation, negative variation, and variation of μ, respectively. The following result is essentially a summary of what has been proved about signed measures. Theorem 6.5. The positive and negative variations μ+ and μ− of a signed measure μ on a measurable space (X, S) are measures on (X, S), at least one of which is finite, and μ = μ+ − μ− setwise on S. Moreover, μ+ and μ− are minimal in this respect in the sense that if ν1 and ν2 are any two measures on (X, S) such that μ = ν1 −ν2 , then there exists a finite measure δ on (X, S)

6 Signed measures, complex measures, and absolute continuity

137

such that ν1 = μ+ + δ and ν2 = μ− + δ. If μ is finite or σ-finite, then so are μ+ and μ− . Proof. Assume that μ = ν1 − ν2 , where ν1 and ν2 are measures on (X, S), and let X = A ∪ (X \ A) be a Hahn decomposition of X with respect to μ. Then the measure δ(E) = ν2 (E ∩ A) + ν1 (E \ A)

E ∈ S,

satisfies the required conditions. The remaining assertions of the theorem are immediate.   The representation of a signed measure μ as the difference of its positive and negative variations is called the Jordan decomposition of μ. Example 6.6. Let f : R → R be a right-continuous, monotone increasing function. We have seen in Example 5.21 that there exists a Borel measure μ on R satisfying the relation μ((a, b]) = f (b) − f (a),

a, b ∈ R, a ≤ b.

(6.1)

Two functions define the same Borel measure if and only if they differ by a constant. Indeed, every (positive) Borel measure μ which assigns finite measure to compact subsets of R is obtained in this way. Thus we can recover the function f (normalized by the condition f (0) = 0) by the formula

μ((0, t]), t ≥ 0, f (t) = (6.2) −μ((t, 0]), t < 0. Assume now that μ is a signed Borel measure on R, which we assume finite for simplicity, and define a function f : R → R by (6.2). The function f is still right-continuous, but it is not increasing if μ takes some negative values. Instead, f has bounded variation in the sense that there exists a constant c ≥ 0 such that n

|f (bj ) − f (aj )| ≤ c j=1

for any finite collection {[aj , bj )}nj=1 of pairwise disjoint intervals. (The smallest constant c in this inequality is called the total variation of the function f .) Indeed, the sum above is nothing but n

|μ((aj , bj ])| ≤ |μ|(R).

j=1

The Jordan decomposition of μ shows that there exist bounded, right continuous, monotone increasing functions f1 , f2 : R → R with the property that f (t) = f+ (t) − f− (t). Indeed, this fact can be proved using solely the fact

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6 Signed measures, complex measures, and absolute continuity

that f has bounded variation (Problem 6A). Therefore any right continuous function f of bounded variation is seen to determine a Borel signed measure μ satisfying (6.1). This discussion can be extended to signed Borel measures μ which assign finite measure to every bounded interval, in which case the corresponding function f has finite variation on every bounded interval but its total variation may be infinite. The fact that a signed measure has a Jordan decomposition enables us to define the integral of a function with respect to it. Definition 6.7. Let μ be a signed measure on a measurable space (X, S) A measurable function f : X → C is said to be integrable with respect to μ (or integrable [μ]) if it is integrable [|μ|] or, equivalently, it is integrable both [μ+ ] and [μ− ]. If f is integrable [μ], then the integral of f with respect to μ is defined to be    f dμ = f dμ+ − f dμ− . X

X

X

Similarly, if ζ : S → C is a complex measure on (X, S) written as ζ = μ + iν in terms of its real and imaginary parts, a measurable function f : X → C is said to be integrable [ζ] if it is integrable with respect to both μ and ν. If f is integrable [ζ], then the integral of f with respect to ζ is defined to be    f dζ = f dμ + i f dν. X

X

X

Just as in the case of signed measures, integrability [ζ] is equivalent to integrability relative to a (positive) measure |ζ| called the variation of ζ. Observe that for a signed measure μ and a measurable set E we have ( ) |μ|(E) = sup |μ(F )| : F ∈ F , where the supremum is taken over the collection of all finite families F of pairwise disjoint measurable subsets of E. Thus the following definition does extend Definition 6.4. Definition 6.8. Consider a measurable space (X, S) and a complex measure ζ : S → C. The variation of ζ is the set function |ζ| : S → [0, +∞) defined by ( ) |ζ|(E) = sup |ζ(F )| : F ∈ F , E ∈ S, where the supremum is taken over the collection of all finite families F of pairwise disjoint measurable subsets of E. The number |ζ|(X) is precisely the total variation of ζ defined earlier. Proposition 6.9. Let ζ be a complex measure on the measurable space (X, S), and set μ = ζ, ν = ζ. Then:

6 Signed measures, complex measures, and absolute continuity

(1) (2) (3) (4)

139

The function |ζ| is a finite (positive) measure on (X, S) . We have |μ|(E) ∨ |ν|(E) ≤ |ζ|(E) ≤ |μ|(E) + |ν|(E) for E ∈ S. A function f : X → C is integrable [ζ] if and only if it is integrable [|ζ|]. The map f → X f dζ, defined on the linear space L of functions that are integrable [|ζ|], is a linear functional satisfying       f dζ  ≤ |f | d|ζ|, f ∈ L.   X

X

Proof. To prove (1) we verify that |ζ| is countably additive. Let {En }∞ n=1 ⊂ S  be a collection of pairwise disjoint sets, set E = n En , and let Fn be a finite collection of pairwise disjoint subsets of En , n ∈ N. Fix N ∈ N, and observe that F1 ∪ · · · ∪ FN is a finite collection of disjoint subsets of E, and therefore N

|ζ|(E) ≥ |ζ(F )|. n=1 F ∈Fn

Take now the supremum over all collections {F1 , . . . , FN } to conclude that  N |ζ|(E) ≥ n |ζ|(En ). n=1 |ζ|(En ), and let N → ∞ to obtain |ζ|(E) ≥ For the opposite inequality, let F be a finite collection of pairwise disjoint measurable subsets of E, and denote Fn = {F ∩ En : F ∈ F}. The countable additivity of ζ yields

|ζ(F )| ≤

F ∈F

=



|ζ(F ∩ En )|

F ∈F n=1 ∞

n=1 F ∈Fn

|ζ(F )| ≤



|ζ|(En ).

n=1

Taking the supremum over all such collections F we obtain the countable additivity of |ζ|.  Parts (2), (3), and the nlinearity of the map f → X f dζ follow easily from the definitions. If s = j=1 αj χEj is such that the sets Ej ∈ S are pairwise disjoint, then        n n

   n  ≤  s dζ  =  α ζ(E ) |α ||ζ(E )| ≤ |α ||ζ|(E ) = |s| d|ζ|. j j j j j j     X X  j=1  j=1 j=1 The inequality is extended to an arbitrary function f ∈ L by writing f as the limit of simple functions sn satisfying sn ≤ |f | and applying the dominated convergence theorem (Theorem 4.35).   The Jordan decomposition of a measure has an analog for complex measures. In order to describe the proper setting of this analog, we recast this decomposition as follows. Let μ be a signed measure on (X, S), and let

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6 Signed measures, complex measures, and absolute continuity

X = A ∪ (X \ A) be a Hahn decomposition of X with respect to μ. Then the measure μ is equal to the indefinite integral of the function f = χA − χX\A with respect to |μ|. That is, we have  μ(E) = f d|μ|, E ∈ S. E

The extension to the complex case requires the remarkable theorem of Radon and Nikod´ ym which we prove next. We recall that the notation ν  μ for two measures indicates that ν is absolutely continuous with respect to μ, that is, μ(E) = 0 implies ν(E) = 0 for every measurable set E (see Definition 4.2). Theorem 6.10. (Radon-Nikod´ ym Theorem). Consider two measures μ, ν on a measurable space (X, S) such that μ is σ-finite and ν  μ. There exists a measurable function f : X → [0, +∞] such that  ν(E) = f dμ, E ∈ S. (6.3) E

In other words, ν = νf is the indefinite integral of f with respect to μ. The function f , which is uniquely determined almost everywhere [μ], is called the Radon-Nikod´ ym derivative of ν with respect to μ, and is denoted by dν/dμ. A function g : X → C is integrable [ν] if and only if gf is integrable [μ], and      dν dμ (6.4) g dν = gf dμ = g X X X dμ when either of these integrals is defined. Proof. The uniqueness statement is easily verified. Indeed, let f and f0 be two Radon-Nikod´ ym derivatives of ν with respect to μ, and let ε > 0. Consider the measurable set A = {x : f (x) > f0 (x) + ε}, and let B ⊂ A have finite measure with respect to μ. Then we have   εμ(B) ≤ f dμ − f0 dμ = μ(B) − μ(B) = 0, B

B

and we conclude that μ(A) = 0 since μ is σ-finite. In other words, f ≤ f0 + ε almost everywhere [μ], and therefore f ≤ f0 almost everywhere [μ] because ε is arbitrary. By symmetry, f = f0 almost everywhere [μ]. Next, we reduce the proof to the case of a finite measure μ. Let {En }∞ n=1 be a measurable partition of X into sets of finite measure with respect to μ, and assume the theorem valid for each of the subspaces En . Denote by ym derivative of ν|En with respect to μ|En , and define fn the Radon-Nikod´ f : X → [0, +∞] by setting f (x) = fn (x),

x ∈ En , n ∈ N.

6 Signed measures, complex measures, and absolute continuity

141

Then clearly f is a Radon-Nikod´ ym derivative of ν with respect to μ. It remains to prove the theorem under the additional assumption that μ is a finite measure. Under this assumption, use the Hahn decomposition (Theorem 6.3) for the signed measure rμ−ν to find, for each positive rational number r, a set Ar ∈ S such that ν(E) ≤ rμ(E) for every measurable set E ⊂ Ar , and ν(E) ≥ rμ(E) for every measurable set E ⊂ X \ Ar . These measurable sets are not necessarily nested. However, if s, r ∈ Q+ and s > r, we have sμ(Ar \ As ) ≤ ν(Ar \ As ) ≤ rμ(Ar \ As ), and thus μ(Ar \ As ) = ν(Ar \ As ) = 0. Using this fact it is not difficult to verify that the set  Bt = Ar , t ∈ (0, +∞), 0 0}, {x : g(x) < 0} must also have positive measure [μ], and a contradiction is obtained by integrating g over such a set.   An important consequence of the Radon-Nikod´ ym theorem is the existence of conditional expectations, which we now define. Definition 6.13. Consider a measure space (X, S, μ), a σ-algebra T ⊂ S, and a [μ]-integrable function f : X → C. A [μ]-integrable function g : X → C is called a conditional expectation of f relative to T if (1) g is measurable [T], and  (2) E f dμ = E g dμ for every E ∈ T. We use the notation g = E[f |T] when a conditional expectation exists. One can of course modify E[f |T] on a [μ]-null set in T and obtain another conditional expectation. This becomes an issue when we deal with uncountably many functions. Conditional expectations are most commonly used in probability spaces where they always exist. The following statement is somewhat more general. Corollary 6.14. Let (X, S, μ) be a measure space, and let T ⊂ S be a σ-algebra such that μ|T is σ-finite. Then every [μ]-integrable function f : X → C has a conditional expectation relative to T, and the following equalities hold almost everywhere [μ] E[(f + g)|T] = E[f |T] + E[g|T], E[λf |T] = λE[f |T], E[hf |T] = hE[f |T], provided that f, g are integrable [μ], λ ∈ C, and h : X → C is bounded and measurable [T]. Proof. The existence and uniqueness almost everywhere [μ] follow from Corollary 6.12 applied with μ|T in place of μ and ν(E) = E f dμ, E ∈ T, since ν is clearly absolutely continuous relative to μ|T. The first two equalities follow immediately from the uniqueness of conditional expectations. It suffices, via Proposition 2.36, to verify the third one when h = χF for some F ∈ T, and in this case it follows again from the uniqueness of conditional expectations. Indeed,     χF E[f |T] dμ = E[f |T] dμ = f dμ = χF f dμ E

for every E ∈ T.

E∩F

E∩F

E

 

144

6 Signed measures, complex measures, and absolute continuity

Definition 6.15. Two measures μ and ν on a measurable space (X, S) are said to be equivalent (notation: μ ≡ ν) if μ  ν and ν  μ. Two measures on (X, S) are equivalent if and only if they possess exactly the same null sets, and this is an equivalence relation on the collection of all measures on (X, S). Proposition 6.16. Let λ, μ and ν be σ-finite measures on a measurable space (X, S) and assume that λ  μ  ν. Then    dλ dλ dμ = dν dμ dν almost everywhere with respect to all three measures. In particular, if μ and ν are equivalent, then 1 dμ = dν dν/dμ almost everywhere with respect to both μ and ν. Proof. For every E ∈ S,   λ(E) = E

dλ dμ

 

 dμ =

E

dλ dμ



dμ dν

 dν.

This proves the desired identity by the uniqueness assertion in Theorem 6.10.   The antithesis of the concept of equivalence for measures is that of singularity. Definition 6.17. Consider two measures μ, ν on a measurable space (X, S). We say that μ and ν are (mutually) singular, or that μ is singular with respect to ν (notation: μ ⊥ ν or ν ⊥ μ), if there exists a set A ∈ S such that μ(A) = ν(X \ A) = 0. The proof of the following result is sketched in Problem 6CC. Proposition 6.18. (Lebesgue decomposition) Consider two σ-finite measures μ, ν on a measurable space (X, S). There exist unique measures ν1 , ν2 on (X, S) such that ν = ν1 + ν2 , ν1  μ, and ν2 ⊥ μ. The expression ν = ν1 + ν2 is called the Lebesgue decomposition of ν relative to μ. We return now to the complex analog of the Jordan decomposition. Proposition 6.19. Let ζ be a complex measure on a measurable space (X, S). Then there exists a measurable function ϕ : X → C such that |ϕ| = 1 al most everywhere [|ζ|] and ζ(E) = E ϕ d|ζ| for all E ∈ S. The function ϕ is uniquely determined almost everywhere [|ζ|].

6 Signed measures, complex measures, and absolute continuity

145

Proof. The conclusion of the proposition, except for the statement that |ϕ| = 1 almost everywhere [|ζ|], follows from Corollary 6.12. To prove that |ϕ| = 1 almost everywhere [|ζ|], write the set {z ∈ C : |z| = 1} as a countable union of balls of the form B = {z : |z − reit | ≤ ε} where r = 1, t ∈ R, and ε < |r − 1|/2. It suffices to show that |ζ|(ϕ−1 (B)) = 0 for any such ball B. For a point x ∈ ϕ−1 (B) we have (e−it ϕ(x)) ≥ 1 + ε [|ϕ(x)| ≤ 1 − ε] if r > 1 [r < 1]. For every measurable set F ⊂ ϕ−1 (B) we obtain  −it (e ζ(F )) = (e−it ϕ)d|ζ| ≥ (1 + ε)|ζ|(F ) [|ζ(F )| ≤ (1 − ε)|ζ|(F )] F

for r > 1 [r < 1]. Thus, if F is a finite collection of pairwise disjoint measurable subsets of ϕ−1 (B),

|ζ(F )| ≥ (e−it ζ(F )) ≥ (1 + ε) |ζ|(F ) F ∈F

F ∈F

*

F ∈F

|ζ(F )| ≤ (1 − ε)

F ∈F

+ |ζ|(F )

F ∈F

for r > 1 [r < 1]. Taking the supremum over all such collections F yields |ζ|(ϕ−1 (B)) ≥ (1 + ε)|ζ|(ϕ−1 (B)) [|ζ|(ϕ−1 (B)) ≤ (1 − ε)|ζ|(ϕ−1 (B))], and   this only happens when |ζ|(ϕ−1 (B)) = 0. We have seen in Example 6.11 that Radon-Nikod´ ym derivatives are related to the derivatives encountered in differential calculus. We investigate now to what extent this relation holds for Borel measures on R. We first introduce some notation related to limits of functions ϕ : Ix → R, where Ix is the collection of all finite open intervals in R containing a point x ∈ R. Given such a function and α ∈ R, we write lim ϕ(I) = α

I→x

if for every ε > 0 there exists δ > 0 such that |ϕ(I) − α| < ε for every interval I ∈ Ix such that λ1 (I) < δ. The relations lim ϕ(I) = ±∞

I→x

are defined analogously. Note that Ix is a directed set, directed downwards by inclusion, and to say of a function ϕ : Ix → R that limI→x ϕ(I) = α is equivalent to saying that ϕ(I) converges to α along the net Ix . The limit limI→x ϕ(I) exists if and only if lim sup ϕ(I) = inf sup{ϕ(I) : I ∈ Ix , λ1 (I) < δ} I→x

δ>0

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6 Signed measures, complex measures, and absolute continuity

and lim inf ϕ(I) = sup inf{ϕ(I) : I ∈ Ix , λ1 (I) < δ} I→x

δ>0

are equal, in which case limI→x ϕ(I) is the common value of these two extended real numbers. Expressions of the form limI→x ϕ(I) can also be defined when ϕ is defined on sufficiently small intervals containing x. The fraction μ(I)/ν(I) in the following statement is defined to be 0 if μ(I) = ν(I) = 0 and +∞ if μ(I) = 0 < ν(I). Theorem 6.20. Consider two Borel measures μ, ν on R, both of which assign finite measure to each bounded interval, and let dν = h dμ + dρ be the Lebesgue decomposition of ν relative to μ; that is, h is a Borel nonnegative function and ρ ⊥ μ. Then the limit lim

I→x

ν(I) μ(I)

exists and equals h(x) for [μ]-almost every x ∈ R . This limit exists and equals +∞ for [ρ]-almost every x ∈ R. The proof requires a covering lemma which we discuss first. Assume that A ⊂ R and C is a collection of open intervals. We say that C is a Vitali covering of the set A if, for every x ∈ A and every ε > 0, there exists I ∈ C such that x ∈ I and λ1 (I) < ε. Proposition 6.21. Consider a Borel measure on R which assigns finite measure to each bounded interval, a Borel set A ⊂ R with μ(A) < +∞, and a Vitali covering C of A. Then for every ε > 0 there exists a finite collection {Ij }nj=1 of pairwise disjoint intervals in C such that ⎞ ⎛ n  Ij ⎠ < ε. μ ⎝A \ j=1

Proof. With ε > 0 given, set δ = ε/10. We have μ(A) = lim μ(A ∩ [−N, N ]), N →∞

and thus there exists N ∈ N such that μ(A ∩ [−N, N ]) > μ(A) − δ. An application of Proposition 7.10(2) (with X = [−N, N ]) allows us to choose a compact set B ⊂ A ∩ [−N, N ] satisfying μ(B) > μ(A) − δ. Note  μ({x}) ≤ μ(B) for every finite set S ⊂ B, and therefore first that x∈S  μ({x}) ≤ μ(B) < +∞. We can thus choose a finite set S ⊂ A such that x∈A

x∈B\S

μ({x}) < δ.

6 Signed measures, complex measures, and absolute continuity

147

It is  possible to find pairwise disjoint intervals J1 , J2 , . . . , JM in C such that M S ⊂ m=1 Jm . Taking into account that some endpoints of the intervals Jm may be atoms, we see that     M M   Jm ≤ μ B \ Jm + δ. μ B\ m=1

Choose a compact set K ⊂ B \

m=1

M m=1

* μ

M 

B\

Jm such that +



Jm \ K

< δ,

m=1

and denote by C1 the collection of those intervals I ∈ C with the property that  M   I∩ Jm = ∅. m=1

Then C1 is a Vitali covering of K, and we see next that it suffices to prove the proposition with K, C1 , and ε − 3δ in place of A, C, and ε, respectively. Indeed, if {Ii }ni=1 ⊂ C1 satisfies the conclusion of the proposin tion after these replacements, then the collection {Jm }M m=1 ∪ {Ii }i=1 satisfies the original  conclusion of the proposition. Note for later use that  μ({x}) ≤ x∈K x∈B\S μ({x}) < δ. Set ⎛ η = inf μ ⎝K \

n 

⎞ Ij ⎠ ,

j=1

where the infimum is taken over all finite collections {Ij }nj=1 of pairwise disjoint intervals from C1 . We argue by contradiction, assuming that η > ε − 3δ. Choose a pairwise disjoint collection {Ii }ni=1 ⊂ C1 such that ⎛ ⎞ n  μ ⎝K \ Ij ⎠ < η + δ, j=1

and then choose a compact set K1 ⊂ K \ ⎛⎡ μ ⎝⎣ K \





n

j=1 Ij

such that



Ij ⎦ \ K1 ⎠ < δ.

j=1

The n collection C2 consisting of those intervals in C1 which are disjoint from j=1 Ij form a Vitali covering of K1 . By compactness of K1 , there exist intervals L1 , L2 , . . . , Lm ∈ C2 such that

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6 Signed measures, complex measures, and absolute continuity

K1 ⊂

m 

Lk .

k=1

We may assume without loss of generality that m is the smallest integer for which such a cover exists, and thus it follows that no three of the intervals {Lk }m k=1 have a common point. (Indeed, if three intervals (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) have a point in common, their union is (min{a1 , a2 , a3 }, max{b1 , b2 , b3 }) = (ap , bq ) = (ap , bp ) ∪ (aq , bq ) for some p, q ∈ {1, 2, 3}.) Therefore {1, 2, . . . , m} = B1 ∪ B2 , where B1 and and so are the B2 are disjoint, the intervals {Lk } k∈B1 are pairwise disjoint,  intervals {Lk }k∈B2 . The sets K1 \ k∈B1 Lk and K1 \ k∈B2 Lk are disjoint, so one of them has measure at most μ(K1 )/2. Assume for definiteness that    μ(K1 ) . Lk ≤ μ K1 \ 2 k∈B1

We consider next the collection of pairwise disjoint intervals F = {Ij }nj=1 ∪ {Lk }k∈B1 for which we have  μ K\



⎛⎡

 I

≤ μ ⎝⎣ K \

I∈F

 ≤ μ K1 \

n 

⎤ Ij⎦ \

j=1



 Lk



⎞ Lk ⎠ + δ

k∈B1

+ 2δ

k∈B1



η+δ μ(K1 ) + 2δ < + 2δ < η. 2 2

The first inequality is true because the total measure of the points in K1 which are endpoints of the intervals Ij is less than δ. The last inequality above is true if δ < ε/8, and it contradicts the definition of η, thus concluding the proof.   We proceed now with the proof of Theorem 6.20. Proof. Given a positive rational number α, we define Borel sets , , ν(I) ν(I) < α , Fα = x ∈ R : lim sup >α . Eα = x ∈ R : lim inf I→x μ(I) I→x μ(I) We show next that every compact set K ⊂ Eα satisfies the inequality ν(K) ≤ αμ(K). Indeed, fix such a set K, let G ⊃ K be an open set, and note that the

6 Signed measures, complex measures, and absolute continuity

149

collection C of intervals I ⊂ G satisfying ν(I) ≤ αμ(I) forms a Vitali covering of K. Thus, given ε > 0, we can find pairwise disjoint intervals {Ij }nj=1 in C such that ⎞ ⎛ ν ⎝K \

n 

Ij ⎠ < ε.

j=1

We deduce that ν(K) < ε +

n

ν(Ij ) ≤ ε + α

j=1

n

μ(Ij ) ≤ ε + αμ(G).

j=1

We can now let ε → 0 and μ(G) → μ(K) to conclude that ν(K) ≤ αμ(K). Switching the roles of μ and ν and replacing α by 1/α, we deduce that every compact set K ⊂ Fα satisfies the inequality ν(K) ≥ αμ(K). With these observations out of the way, we are ready to show that lim sup I→x

ν(I) ν(I) = lim inf I→x μ(I) μ(I)

for [μ]-almost and [ν]-almost every x ∈ R. For this purpose, it suffices to show that, given positive rational numbers α < β, the set Eα ∩ Fβ is null [μ] and [ν]. Indeed, for a compact set K ⊂ Eα ∩ Fβ we have μ(K) ≤ ν(K)/β ≤ (α/β)μ(K), which is only possible when μ(K) = ν(K) = 0. Next we show that the limit in the statement of the theorem is infinite [ρ]-almost everywhere. It suffices to show that the set Eα is null [ρ] for every rational α > 0. The inequality ν(K) ≤ αμ(K) for compact sets K ⊂ Eα shows that the concentration of ν on such a set K is absolutely continuous [μ]. Thus ρ(Eα ) = 0, as was to be proved. Observe also for further use that the inequality h(x) ≤ α must also be satisfied [μ]-almost everywhere on Eα . Indeed, if K ⊂ Eα is a compact set such that μ(K) > 0 and h(x) > α for x ∈ K, we conclude that  ν(K) ≥ h dμ > αμ(K), K1

contrary to what we just proved. Thus we have h ≤ α almost everywhere [μ] on Eα . Since α is arbitrary, we conclude that h(x) ≤ lim

I→x

ν(I) μ(I)

almost everywhere [μ].

Finally, given a positive rational number β, we claim that h ≥ β almost everywhere [μ] on Fβ . Indeed, otherwise there would exist a compact set

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6 Signed measures, complex measures, and absolute continuity

K ⊂ Fβ such that h < β on K, μ(K) > 0, and ρ(K) = 0. Thus  h dμ < βμ(K), ν(K) = K

contrary to the inequality ν(K) ≥ βμ(K), which concludes the proof.

 

The special case μ = λ1 of Theorem 6.20 is due to Lebesgue. It can be viewed as an extension of the fundamental theorem of calculus using Lebesgue integration in place of Riemann integration. Theorem 6.22.(1) Let u : R → C be integrable [λ1 ]. Then the function t t → −∞ u dλ1 is differentiable almost everywhere [λ1 ], and d dt



t −∞

u dλ1 = u(t)

almost everywhere [λ1 ].

(2) Let f : R → R be a monotone increasing function. Then f is differentiable almost everywhere [λ1 ], the derivative f  is integrable on every finite interval, and  b f  dλ1 f (b) − f (a) ≥ a

for all a, b ∈ R such that a < b. Equality holds in this equation precisely when f is absolutely continuous on [a, b]. Proof. It suffices to prove (1) when u ≥ 0, in which case the measure ν defined by dν = u dλ1 satisfies  t u dλ1 = ν((−∞, t]), t ∈ R. −∞

Theorem 6.20 implies the following assertion for [λ1 ]-almost every t ∈ R: given sequences of real numbers {an } and {bn } such that an < t < bn and limn→∞ (bn − an ) = 0, we have  bn lim

n→∞

an

u dλ1

bn − a n

= u(t).

Fix t0 ∈ R for which this assertion is true, and let {bn } be a sequence of real numbers such that t0 < bn and limn→∞ bn = t0 . We choose for each n ∈ N a number an < t0 such that t0 − an < 1/n and    bn  bn   u dλ 1 1   an u dλ1 − t0 < .   bn − a n bn − t 0  n

6 Signed measures, complex measures, and absolute continuity

We see that

151

 bn

u dλ1 = u(t) bn − t t and this verifies that the right derivative of t → −∞ u dλ1 at t0 equals u(t0 ). A similar argument for the left derivative concludes the proof of (1). For (2) we observe that the function g defined by t

lim

n→∞

g(t) = lim f (t + h), h↓0

t ∈ R,

is right-continuous, monotone increasing, and differs from f only at the (countably many) points of discontinuity of f . There exists a Borel measure ν on R such that ν((a, b]) = g(b) − g(a) for a < b (see Example 5.21). Writing the Lebesgue decomposition dν = u dλ1 + ρ with ρ ⊥ λ1 , we see that 

b

g(b) − g(a) = ρ([a, b)) +

u dλ1 ,

a, b ∈ R, a < b.

a

The equality g  = u almost everywhere [λ1 ] follows as in the proof of part (1), and it is clear that f  = g  at all points of continuity of f where g  exists. The equality  b u dλ1 f (b) − f (a) = a

holds when ρ((a, b)) = 0, f is right-continuous at a and left-continuous at b. This implies the final assertion of part (2).   Example 6.23. Borel measures on R which assign finite measure to bounded sets can be further decomposed into a discrete and continuous component. A Borel measure μ on R is said  to be continuous if μ({x}) = 0 for every x ∈ R, and discrete if μ(E) = x∈E μ({x}) for every E ∈ BR . If μ((a, b]) = f (b)−f (a) for a, b ∈ R, a ≤ b, for some right-continuous, monotone increasing function f : R → R, it is clear that μ is continuous if and only if f is a continuous function. Indeed, μ({x}) = lim(f (x) − f (y)), y↑x

x ∈ R,

is precisely the size of the jump discontinuity of f at x. Given an arbitrary Borel measure μ on R, one defines

μ({x}), E ∈ BR , μd (E) = x∈E

and μc = μ − μd , which represents μ as the sum of a discrete measure μd and a continuous one μc . It is an interesting fact, however, that there exist measures which are continuous but singular [λ1 ]. To construct such a measure it suffices to exhibit a continuous, monotone increasing function f : R → R

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6 Signed measures, complex measures, and absolute continuity

which is not constant, but f  = 0 almost everywhere [λ1 ]. Such a function can be constructed using the Cantor ternary set C ⊂ [0, 1] which satisfies λ1 (C) = 0, as follows We recall that [0, 1] \ C is the union of open intervals {Ii,j : i, j ∈ N, j ≤ 2i−1 } such that λ1 (Ii,j ) = 3−i , and Ii,j is on the left of Ii,j+1 for j < 2i−1 . Define a function f on R \ C by setting ⎧ ⎪ for t < 0, ⎨0 f (t) = 1 for t > 1, ⎪ ⎩ 2j−1 for t ∈ Iij , i ∈ N, j = 1, 2, . . . , 2i−1 . 2i The function thus defined is monotone increasing on R \ C, and its range is dense in [0, 1]. It follows that its right and left limits at each point t ∈ C must be equal, and therefore f extends to a continuous, monotone increasing function on R, sometimes called the Cantor function. Clearly f  (t) = 0 on R \ C, and thus f determines a continuous singular Borel measure on R. Note also that this function satisfies  1 f  (t) dt = 0 1 = f (1) − f (0) > 0

so by Theorem 6.22, f is not absolutely continuous on [0, 1]. Example 6.24. Let J ⊂ R be an arbitrary interval, and let f : J → R be a function. We say that f is a convex function on J if the inequality f (tx + (1 − t)y) ≤ tf (x) + (1 − t)f (y)

(6.6)

holds for all x, y ∈ J and t ∈ (0, 1). The function is strictly convex if the inequality is strict whenever x = y. If f is convex on J and x < y, the preceding inequality can be rewritten as f (tx + (1 − t)y) − f (x) f (y) − f (x) f (y) − f (tx + (1 − t)y) ≤ ≤ . [tx + (1 − t)y] − x y−x y − [tx + (1 − t)y] Since tx + (1 − t)y is an arbitrary point in (x, y), we conclude that f (y) − f (x) f (y1 ) − f (x1 ) ≤ y−x y1 − x 1

(6.7)

whenever x < y ≤ x1 < y1 . We deduce that the difference quotients (f (y) − f (x))/(y − x) are bounded above and below when x and y are allowed to vary in a compact interval I ⊂ J. In particular, |f (y) − f (x)| ≤ cI |y − x|,

x, y ∈ I,

for some constant cI ≥ 0. Obviously, this implies that f is absolutely continuous on the interval I.

6 Signed measures, complex measures, and absolute continuity

153

Corollary 6.25. Let J ⊂ R be an open (possibly unbounded ) interval, and let f : J → R be a convex function. Then: (1) There exists a monotone increasing function u : J → R such that 

b

f (b) − f (a) =

u(t) dt,

a, b ∈ J, a < b.

a

(2) The derivative f  (t) exists and equals u(t) for all but countably many points t ∈ J. (3) The second derivative f  (t) exists and satisfies f  (t) ≥ 0 almost everywhere [λ1 ] on J. (4) The function f is strictly convex on J if and only if the function u is strictly increasing on J. Conversely, if f : J → R is twice differentiable and satisfies f  ≥ 0 on J, then f is convex. Proof. The existence of u and v and the equality f  = u almost everywhere [λ1 ], follow from the absolute continuity of f on all compact intervals I ⊂ J. Relation (6.7) implies that f  (a) ≤ f  (b) if a < b and these two derivatives exist. It follows that the function u can be assumed, after alteration on a [λ1 ]-null set, to be monotone increasing. With this assumption, we see that f  (a) exists for all points of continuity for u, and u only has countably many points of discontinuity. Thus (1) and (2) are proved, and (3) follows from Theorem 6.22. To verify (4), consider x, y ∈ J and t ∈ (0, 1) such that x < y, and set z = tx + (1 − t)y. We have f (z) − tf (x) − (1 − t)f (y) = t[f (z) − f (x)] + (1 − t)[f (z) − f (y)]  z  y =t u dλ1 − (1 − t) u dλ1 , x



and

z

z

u dλ1 ≤ tu(z)λ1 ([x, z]) = t(1 − t)(y − x)u(z).

t x

An analogous calculation shows that  y (1 − t) u dλ1 ≥ t(1 − t)(y − x)u(z). z

Thus (6.6) fails to be strict precisely when u is constant on the interval (x, y). This also proves the last statement since, if f  exists and is nonnegative, it follows that f  is monotone increasing, and the calculation above (with u = f  ) proves the convexity of f .   The following extension of (6.6) is due to Jensen.

154

6 Signed measures, complex measures, and absolute continuity

Theorem 6.26. Consider an arbitrary (possibly unbounded ) interval J ⊂ R and a continuous convex function f : J → R. Let (X, S, μ) be a probability space and let h : X → J be an integrable function. Then X h dμ ∈ J,  f (h(x)) dμ(x) exists, and X    h dμ ≤ f (h(x)) dμ(x). f X

X

Proof. If J  left endpoint α, then h(x) ≥ α for all x ∈ X, and  has a finite therefore X h dμ ≥ X α dμ = α. An analogous argument for the right endpoint shows that t0 = X h dμ does indeed belong to J. With the notation of Corollary 6.25, we have  t f (t) − f (t0 ) = u(t) dt ≥ u(t0 )(t − t0 ), t ∈ J, t0

and therefore f (u(x)) − f (t0 ) ≥ u(t0 )(u(x) − t0 ),

x ∈ X.

The conclusion of the theorem follows by integrating this inequality.

 

Example 6.27. Fix p ∈ [1, +∞). The function f (t) = tp satisfies f  (t) = p(p−1)tp−2 > 0 for t > 0 and is therefore convex. Applying (6.6) with t = 1/2 yields the inequality (x + y)p ≤ 2p−1 (xp + y p ),

x, y ≥ 0, p ≥ 1.

(6.8)

On the other hand, if p ∈ (0, 1), the derivative f  (t) = ptp−1 is decreasing on (0, +∞), and therefore −f is convex. We deduce that (x + y)p ≥ 2p−1 (xp + y p ),

x, y ≥ 0, p ∈ (0, 1).

For p ∈ (0, 1) we also have 



y

(x + y)p − xp =

y

p(x + t)p−1 dt ≤ 0

ptp−1 dt = y p , 0

so (x + y)p ≤ xp + y p ,

x, y ≥ 0, p ∈ (0, 1).

(6.9)

Example 6.28. The function f (t) = et is convex on R since f  = f > 0. Thus etx+(1−t)y ≤ tex + (1 − t)ey , x, y ∈ R, t ∈ (0, 1). Setting a = etx , b = e(1−t)y , this can be written as ab ≤ ta1/t + (1 − t)b1/(1−t) ,

a, b ≥ 0, t ∈ (0, 1).

(6.10)

6 Signed measures, complex measures, and absolute continuity

155

There is a higher dimensional analog of the Lebesgue differentiation theorem 6.22(1). For the proof we require a different covering lemma, originally due to Vitali. A collection C of open balls in Rd is a Vitali covering of a set K ⊂ Rd if every element of K is contained in balls in C of arbitrarily small radius. Lemma 6.29. Consider a Vitali covering C of a bounded set K ⊂ Rd . There exists a sequence of pairwise disjoint balls {Bn } ⊂ C such that    Bn = 0. λd K \ n

Proof. We may assume that all the balls in C intersect K. Denote by α1 the least upper bound of the radii of the balls in C. If α1 = +∞ then a single ball in C covers K. Otherwise, we construct the sequence Bn inductively. First choose a ball B1 = B(x1 , r1 ) ∈ C such that r1 > α1 /2. Assume that balls for j = 1, 2,. . . , n. If there are no balls Bj = B(xj , rj ) have been constructed n n in C which are disjoint from j=1 Bj , then K ⊂ j=1 Bj and ⎛ λ d ⎝K \

n 

⎞ Bj ⎠ = 0,

j=1

so the lemma is proved. Otherwise, denote byαn+1 the supremum of the n radii of those balls in C which are disjoint from j=1 Bj , and choose Bn+1 = n B(xn+1 , rn+1 ) ∈ C such that Bn+1 ∩ j=1 Bj = ∅ and rn+1 > αn+1 /2. Since the disjoint balls {Bn }∞ n=1 are contained in a set of finite measure, we must have lim αn = 0. n→∞

We now show that

⎛ K \⎝

n 

⎞ Bj ⎠ ⊂

j=1

Given x ∈ K \ and



n j=1

∞ 

B(xj , 5rj ).

j=n+1

 Bj , we can choose B = B(y, r) ∈ C such that x ∈ B ⎛ B∩⎝

n 

⎞ Bj ⎠ = ∅.

j=1

Observe now that B ∩ Bj = ∅ for some j ≥ n + 1. Indeed the hypothesis that B ∩ Bj = ∅ for all j implies r < αn for all n, and thus r = 0. Let N be the first integer such that B ∩ BN = ∅. Then r ≤ αN < 2rN , and this implies that x ∈ B ⊂ B(xN , 5rN ). We conclude that

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6 Signed measures, complex measures, and absolute continuity

⎛ λ d ⎝K \







Bj ⎠ ≤ λd ⎝K \

j

n 

⎞ Bj ⎠ ≤ 5d

j=1



λd (B(xj , rj )),

j=n+1

and the desired conclusion is obtained as n → +∞.

 

We record for further use a version of Vitali’s lemma for ordinary covers by open balls. Lemma 6.30. Let K ⊂ Rd be a compact set, and let C a collection of open balls whose union contains K. There exists a finite collection {B(xj , rj )}N j=1 of pairwise disjoint balls in C such that K⊂

N 

B(xj , 3rj ).

j=1

In particular, λd (K) ≤ 3d

N j=1

λj (B(xj , rj )).

Proof. Since K is compact, we may assume that C is finite. Choose the balls B(xj , rj ) inductively so B(x1 , r1 ) is the largestball in C, and B(xn+1 , rn+1 ) n is the largest ball in C which is disjoint from j=1 B(xj , rj ). The inductive process stops  at some integer N such that there are no balls in C which are N disjoint from j=1 B(xj , rj ). It suffices to show now that any ball B(x, r) ∈ C N is contained in j=1 B(xj , rj ). Indeed, let n be the smallest integer such that B(x, r) ∩ B(xn , rn ) = ∅, and note that r ≤ rn , and hence B(x, r) ⊂   B(xn , 3rn ). Given x ∈ Rd , denote by Bx the collection of open balls in Rd that contain x. Given a function ϕ : Bx → R and α ∈ R, we write lim ϕ(B) = α

B→x

if for every ε > 0 there exists δ > 0 such that |ϕ(B) − α| < ε for every ball B ∈ Bx with radius less than δ. Infinite limits are defined analogously. Setting lim sup ϕ(B) = inf sup{ϕ(B) : B ∈ Bx with radius < δ} B→x

δ>0

and lim inf ϕ(B) = sup inf{ϕ(B) : B ∈ Bx with radius < δ}, B→x

δ>0

it is clear that the limit limB→x ϕ(B) exists precisely when these two (extended real) numbers coincide. As in the case of functions defined on intervals, limB→x ϕ(B) can be viewed as limBx ϕ(B), where the limit is taken along the net Bx .

6 Signed measures, complex measures, and absolute continuity

157

Theorem 6.31. Let f : Rd → C be integrable [λd ]. Then the limit  1 lim f dλd B→x λd (B) B exists and equals f (x) for [λd ]-almost every x ∈ Rd . Proof. As usual, it suffices to consider the case of positive functions f . As in the proof of Theorem 6.22, we define for each α > 0 we define Borel sets ,  1 d Eα = x ∈ R : lim inf f dλd < α , B→x λd (B) B ,  1 d f dλd > α . Fα = x ∈ R : lim sup B→x λd (B) B We show that f (x) ≤ α for [λd ]-almost every x ∈ Eα . Assume to the contrary, via regularity, that there exists a compact set K ⊂ Eα such that f (x) > α on K and λd (K) > 0. For any open set G ⊃ K, the collection C consisting of those balls in Rd contained in G and satisfying the inequality  1 f dλd < α λd (B) B is a Vitali covering of K, and therefore a pairwise disjoint subcollection {Bn } of C covers K except for a null set [λd ]. We have then ∞  1 λd (K) ≤ λd (Bn ) ≤ f dλd α n=1 Bn n=1   1 1 ≤ f dλd + f dλd . α K α G\K ∞

Since G is an arbitrary open set containing K, the last integral above can be made arbitrarily small, and thus  1 f dλd < λd (K), λd (K) ≤ α K which is not possible. An analogous argument shows that f (x) ≥ α for [λd ]almost every x ∈ Fα . Thus the intersection Eα ∩ Fβ is λd -null when α < β, and this establishes the existence of the limit in the statement for [λd ]-almost every x. The fact that the limit is f (x) follows by noting that, in the contrary case, there would exist a compact set K of positive measure for which one of the following is true for some rational α > 0: (a) K ⊂ Eα and f (x) ≥ α for x ∈ K; (b) K ⊂ Fα and f (x) ≤ α for x ∈ K. But we have seen that such sets do not exist.  

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6 Signed measures, complex measures, and absolute continuity

The conclusion of Theorem 6.31 is still valid if f is only assumed to be integrable on every bounded subset of Rd . This is seen by applying the theorem to the functions χB(0,N ) f with N ∈ N. Example 6.32. If E ⊂ Rd is a measurable set, we have lim

B→x

λd (B ∩ E) =1 λd (B)

(6.11)

for almost every x ∈ E [λd ], as can be seen by applying Theorem 6.31 to f = χE . Points x for which (6.11) holds are called density points of E. Definition 6.33. Let f : Rd → C be a function which is integrable [λd ]. A point x ∈ Rd is called a Lebesgue point for f if  1 lim |f − f (x)| dλd = 0. B→x λd (B) B Corollary 6.34. Let f : Rd → C be a function which is integrable [λd ]. Then [λd ]-almost every point in Rd is a Lebesgue point for f . Proof. For each complex number z with rational real and imaginary parts there exists a λd -null set Ez such that  1 |f − z| dλd = |f (x) − z|, x ∈ Rd \ Ez . lim B→x λd (B) B Denote by E the union of these sets Ez , so E is also λd -null. Every point x∈ / E is a Lebesgue point for f . Indeed,   Proof. lim sup B→x

1 λd (B)



1 B→x λd (B)



|f − f (x)| dλd ≤ |f (x) − z| + lim B

|f − z| dλd B

= 2|f (x) − z| for every z ∈ C with rational real and imaginary parts, and the right-hand side can be made arbitrarily small.

Problems 6A. Let f : R → R be a function of bounded variation. (i) Show that f has left and right limits at every point in R. Moreover, if J(x) = lim |f (x + h) − f (x − h)| h↓0

6 Signed measures, complex measures, and absolute continuity

159



denotes the jump of J at x, then x∈R J(x) < +∞. In particular, f has at most countably many points of discontinuity. (ii) Assume that f is right-continuous and define g : R → [0, +∞) by the requirement that g(t) is the variation of f on the interval (−∞, t] for t ∈ R, that is,

g(t) = sup

⎧ n ⎨ ⎩

⎫ ⎬

|f (bj ) − f (aj )|

j=1



,

where the supremum is taken over all finite collections of pairwise disjoint intervals (aj , bj ] ⊂ (∞, t]. Show that g is monotone increasing, right continuous, limt→−∞ g(t) = 0, and limt→∞ g(t) equals the total variation of f . In addition, show that f − g is monotone increasing as well. Derive a new proof of Theorem 6.3 for bounded Borel signed measures on R. (iii) Show that the derivative f (x) exists almost everywhere [λ1 ]. Moreover, the function f  is integrable [λ1 ], and R |f  | dλ1 is at most the total variation of f . a. (iv)Are there analogues of the preceding results for complex valued functions with bounded variation? 6B. (F. Riesz) Consider a continuous function h : [0, 1] → R, and define an open subset G ⊂ (0, 1) to consist of those points x ∈ (0, 1) for which there exists y ∈ (0, x) such that h(y) < h(x). Show that each connected component (a, b) of the set G satisfies h(a) ≥ (b). (In fact h(a) = h(b) unless a = 0 or b = 1.) 6C. (F. Riesz) Consider a continuous, monotone increasing function f : [0, 1] → R. (i) Apply Problem 6B to the function h(x) = f (x) − βx for a fixed β > 0 to show that the set  f (x) − f (y) B = x ∈ (0, 1) : lim sup >β x−y y↑x satisfies λ1 (B) ≤ (f (1) − f (0))/β. Conclude that the set consisting of those points x ∈ (0, 1) at which f has an infinite left derivative is null [λ1 ]. (ii) Fix positive numbers α < β, define a Borel set by

 C=

x ∈ (0, 1) : lim sup y↑x

f (x) − f (y) f (x) − f (y) > β, lim inf t}) dt.

0

Consider the Borel measure ν on (0, +∞) defined by dν(t) = dt/t, and define h(t) = tμ({x : |f (x)| > t})1/p ,

t > 0.

The above equation can then be written as f p = p1/p hp , where hp is calculated relative to the measure ν. This observation leads naturally to a refinement of the spaces Lp . Namely, fix p ∈ (0, +∞), define the function h as above, and set f p,q = q 1/p hq ,

q ∈ (0, +∞),

f p,∞ = h∞ . The Lorentz space L (μ) consists of those classes [f ] ∈ L0 (μ) with the property that f p,q < +∞. Note that the map [f ] → f p,q is not generally a norm on Lp,q (μ). Clearly, f p = f p,p so Lp,p (μ) = Lp (μ). The space Lp,∞ (μ) is usually called weak Lp (μ) because it is a ‘slightly’ larger space. To see this, we recall the Tchebysheff inequality (Problem 4C):  μ({x : |f (x)| > t}) = 1 dμ p,q



{x:|f (x)|>t}

≤ μ({x:|f (x)|>t})

f pp |f |p dμ ≤ p , p t t

t > 0.

This yields immediately that tμ({x : |f (x)| > t})1/p ≤ f p , that is, f p,∞ ≤ f p . For some values of p and q, a closely related expression does define a norm on Lp,q (μ). The weak L1 space appears in many estimates, for instance in the study of maximal functions which we now introduce. Definition 9.38. Given d ∈ N and a function f ∈ L0 (λd ), the maximal function M f : Rd → [0, +∞] is defined by  1 (M f )(x) = sup |f | dλd , x ∈ Rd , {B:x∈B} λd (B) B where the supremum is taken over all open balls B in Rd containing x. There is an obvious connection between the differentiation Theorem 6.31 and the maximal function since       1 1   f dλd  ≤ |f | dλd ≤ (M f )(x)  λd (B) λd (B) B B

9 The Lp spaces

227

when x ∈ B. More generally, many convergence results can be obtained by showing that the relevant limit is dominated by the maximal function. Note that {x ∈ Rd : (M f )(x) > α} is an open set for every α > 0, and therefore M f is measurable [λd ]. It is clear that M (f + g) ≤ M f + M g,

M (tf ) = tM (f ),

f, g ∈ L0 (λd ), t ≥ 0,

that is, M is subadditive and positive homogeneous on L0 (λd ). If f is essentially bounded, then M f is bounded and M f ∞ ≤ f ∞ . Example 9.39. Consider the function f = χ(−1,1) on R. The maximal function of f can be calculated explicitly: ⎧ ⎪ x ∈ (−1, 1), ⎨1, (M f )(x) = 2/(1 + x), x > 1, ⎪ ⎩ 2/(1 − x), x < −1. This maximal function is not integrable [λ1 ], but M f 1,∞ = 4. The following result was first proved by Hardy and Littlewood when d = 1. Theorem 9.40. For every d ∈ N and every f ∈ L1 (λd ) we have [M f ] ∈ L1,∞ (λd ), and M f 1,∞ ≤ 3d f 1 . Proof. Fix t > 0 and let K ⊂ {x ∈ Rd : (M f )(x) > t} be a compact set. The collection of those open balls B with the property that  1 |f | dλd > t λd (B) B covers K. Let B1 , B2 , . . . , BN be the balls constructed in Lemma 6.30. We have N  N

3n 3n f 1 , λd (K) ≤ 3n λd (Bj ) ≤ |f | dλd ≤ t j=1 Bj t j=1 and therefore tλd ({x : (M f )(x) > t}) ≤ 3n f 1 . These inequalities for t > 0 yield the desired conclusion.   The estimate just proved shows that the maximal operator is of weak type (1, 1) in the sense of the following definition. Definition 9.41. Let (X, S, μ) and (Y, T, ν) be measure spaces, let D ⊂ L0 (μ) be a linear manifold, and let T : D → L0 (ν) be an arbitrary map. We say that T is of (strong) type (p, q), p, q ∈ (0, +∞], if there exists a constant

9 The Lp spaces

228

c ≥ 0 such that T f q ≤ cf p for every f ∈ D. We say that T is of weak type (p, q), p ∈ (0, +∞], q ∈ (0, +∞), if there exists a constant c ≥ 0 such that T f q,∞ ≤ f p for every f ∈ D. In order to make some statements uniform, it is convenient to agree that weak type (p, ∞) is the same as strong type (p, ∞), p ∈ (0, +∞]. The Marcinkiewicz interpolation theorem produces strong type inequalities at intermediate indices from weak type estimates at the endpoints of an interval. For the statement, we require the notation

f (x), |f (x)| > α, (α) f (x) = 0, otherwise, and f(α) = f − f (α) for the truncations of a function f ∈ L0 (μ) at some level α > 0. These functions satisfy the inequalities p

p

p

p

f (α) p0 ≤ α1− p0 f pp0 and

f(α) p1 ≤ α1− p1 f pp1 provided that 0 < p0 < p < p1 < +∞. These inequalities follow immediately because |f (x)|p0 = |f (x)|p |f (x)|p0 −p ≤ αp0 −p |f (x)|p if |f (x)| > α and

|f(α) (x)|p1 ≤ αp1 −p |f (x)|p

if |f (x)| ≤ α. Theorem 9.42. (Marcinkiewicz) Let (X, S, μ) and (Y, T, ν) be measure spaces, let D ⊂ L0 (μ) be a linear manifold, and let T : D → L0 (ν) be an arbitrary map. Assume that f (α) and f(α) belong to D for every f ∈ D and every α > 0, and that T is subadditive, that is, |T (f + g)| ≤ |T (f )| + |T (g)| almost everywhere [μ],

f, g ∈ D.

If T is of weak type (p0 , q0 ) and of weak type (p1 , q1 ), where 0 ≤ p0 ≤ q0 ≤ +∞ , 0 ≤ p1 ≤ q1 ≤ +∞ and q0 = q1 , then T is of strong type (pτ , qτ ) for each τ ∈ (0, 1), where 1 1−τ τ 1 1−τ τ = + , = + , pτ p0 p1 q τ q0 q1

τ ∈ (0, 1).

Proof. We give complete details under the additional assumptions that p0 = p1 , and q0 = +∞ = q1 . The special cases when p0 = p1 or one of the numbers q0 , q1 is infinite are sketched in the problems. Fix τ ∈ (0, 1) and write p = pτ , q = qτ . We assume without loss of generality that p0 < p1 and

9 The Lp spaces

229

T (f )q0 ,∞ ≤ c0 f p0 , T (f )q1 ,∞ ≤ c1 f p1 ,∞ ,

f ∈ D.

In other words, q

q

ν({y : |(T f )(y)| > t}) ≤ cj j

f pjj , tj

f ∈ D, j = 0, 1.

We need to prove that there exists c > 0 such that T (f )q ≤ cf p ,

f ∈ D.

(9.12)

Start with a function f ∈ D ∩ Lp (μ), and recall that  T (f )qq

+∞

=q

tq−1 ν({y : |(T f )(y)| > t}) dt.

(9.13)

0

Given s > 0, we have |T (f )| ≤ |T (f (s) )| + |T (f(s) )|, so t t }) + ν({y : |T (f(s) )| > }) 2 2 ≤ (2c0 )q0 t−q0 f (s) qp00 + (2c1 )q1 t−q1 f(s) qp11 .

ν({y : |(T f )(y)| > t}) ≤ ν({y : |T (f (s) )| >

It is useful to use different cutoff levels for different values of t. More specifically, we use the value s = s(t) = s1 tγ , where γ=

p1 (q − q1 ) p0 (q − q0 ) = q1 (p − p1 ) q0 (p − p0 )

(9.14)

and s1 is a positive constant. The rationale for these choices is made clear later. We have  +∞ tq−q0 −1 f (s(t)) qp00 dt (9.15) T (f )qq ≤ (2c0 )q0 q 0  +∞ tq−q0 −1 f(s(t)) qp11 dt, +(2c1 )q1 q 0

and we need to estimate the two integrals in terms of f p . We consider first the case in which q0 < q1 , when we must have γ > 0, s(t) is monotone increasing, and q0 < q < q1 . The integral  +∞ tq−q0 −1 f (s(t)) qp00 dt 0

can be written as g0 rr00 = g0 (t) = f (s(t)) pp00 ,

 +∞ 0

g0 (t)r0 dρ0 (t), where

dρ0 (t) = tq−q0 −1 dt,

r0 = q0 /p0 > 1.

9 The Lp spaces

230

By Lemma 9.16, we have  g0 r0 =

+∞

tq−q0 −1 h(t)g0 (t) dt

0

for some Borel measurable function h : (0, +∞) → R with the property that 

r

hr0 = 0

+∞



h(t)r0 dρ0 (t) ≤ 1,

0

where r0 = q0 /(q0 − p0 ) is the conjugate exponent of r0 . Note that s−1 (t) = (t/s1 )1/γ , so the Fubini-Tonelli Theorem 8.4 yields *  +∞

g0 r0 = 

tq−q0 −1 h(t)

0

{x:|f (x)|>s(t)}

0

+∞

|f (x)|p0

=

+

X

0

|f (x)|p0 dμ(x) dt

1 χ(0,s−1/γ |f (x)|1/γ ) h(t) dρ(t) dμ(x). 1

We apply the H¨older inequality applied to the inner integral to obtain  +∞ χ(0,s−1/γ |f (x)|1/γ ) h(t) dρ(t) ≤ χ(0,s−1/γ |f (x)|1/γ ) r0 hr0 1

0

1



(q −q)/γr0 |f (x)|(q−q0 )/γr0 s1 0 (q − q0 )1/r0

,

and therefore (q −q)/γ

g0 r0

s 0 ≤ 1 q − q0



(q −q)/γ

|f (x)|p0 +(q−q0 )/γr0 dμ(x) = X

s1 0 f pp q − q0

by the choice (9.14) of γ. Thus the first term on the right side of (9.15) is 

+∞

(2c0 )q0 q 0

tq−q0 −1 f (s(t)) qp00 dt = (2c0 )q0 qgrr00 (q −q)/γ

≤ (2c0 )q0 q

s1 0 f rp0 p . q − q0

Similarly, to estimate the second term we write  +∞  +∞ tq−q1 −1 f(s(t)) qp11 dt = g1 (t)r1 dρ1 (t) = g1 rr11 , 0

0

9 The Lp spaces

231

where g1 (t) = f(s(t)) pp11 ,

dρ1 (t) = tq−q1 −1 dt,

r1 = q1 /p1 > 1.

The quantity g1 r1 is evaluated the same way, with χ(s−1/γ |f (x)|1/γ ,+∞) in 1 place of χ(0,s−1/γ |f (x)|1/γ ) , and this yields 1



+∞

q0

(2c0 ) q 0

tq−q1 −1 f(s(t)) qp11 dt = (2c1 )q1 qg1 rr11 (q −q)/γ

≤ (2c1 )q1 q

s1 1 f rp1 p , q1 − q

and thus (q −q)/γ

T (f )qq ≤ (2c0 )q0 q

(q −q)/γ

s1 0 s 1 f rp1 p . f rp0 p + (2c1 )q1 q 1 q − q0 q1 − q

(9.16)

We can now set s1 = f ηp to obtain the desired inequality (9.12) provided that (q0 − q)η (q1 − q)η + r0 p = + r1 p = q. γ γ A simple calculation shows that η = γp

r1 − r0 q1 − q0

satisfies these requirements. It remains to consider the case in which q0 > q1 . In this case we have γ < 0 so the function s(t) is decreasing, and thus one must interchange the intervals −1/γ −1/γ (0, s1 |f (x)|1/γ ) and [s1 |f (x)|1/γ , +∞) in the preceding argument. One obtains the same inequality as (9.16) with q0 − q and q − q1 in place of q − q0 and q1 − q, respectively. Strong type (p, q) is obtained with the same choice   of the constant s1 . Corollary 9.43. Let d be a positive integer and let p ∈ (1, +∞). There exists a constant c = c(d, p) such that M (f )p ≤ cf p ,

f ∈ L0 (λd ),

where M (f ) denotes, as usual, the maximal function. Proof. Denote by D the linear manifold in L0 (λd ) consisting of measurable functions defined on the entire Rd . Then M : D → D is subadditive, it has weak type (1, 1) by Theorem 9.40, and clearly M (f )∞ ≤ f ∞ ,

f ∈ D.

The desired result follows immediately by interpolation.

 

9 The Lp spaces

232

The inequalities p0 ≤ q0 and p1 ≤ q1 were necessary in the proof of Theorem 9.42 because of the use of H¨ older’s inequality for the exponents r0 = q0 /p0 and r1 = q1 /p1 . These restrictions can be removed under the assumption that T is linear and of strong type (p0 , q0 ) and (p1 , q1 ), although we need p0 , p1 ≥ 1. We require a lemma from complex analysis which we present without proof (see Problem 9EE). Lemma 9.44. Set Ω = {x + iy : 0 < x < 1, y ∈ R}, and let u : Ω → C be a bounded continuous function such that the restriction u|Ω is analytic. If |u(iy)| ≤ 1 and |u(1 + iy)| ≤ 1 for every y ∈ R then |u(z)| ≤ 1 for every z ∈ Ω. Theorem 9.45. (M. Riesz and G. O. Thorin) Let (X, S, μ) and (Y, T, ν) be measure spaces, let D ⊂ L0 (μ) be a linear manifold containing all simple integrable functions on X, and let T : D → L0 (ν) be a linear map. Assume that f (α) and f(α) belong to D for every f ∈ D and every α > 0. If T is of strong type (p0 , q0 ) and of strong type (p1 , q1 ), where 1 ≤ p0 , q0 ≤ +∞, 1 ≤ p1 , q1 ≤ +∞, and q0 = q1 , then T is of strong type (pτ , qτ ) for each τ ∈ (0, 1), where 1−τ τ 1 1−τ τ 1 = + , = + , pτ p0 p1 q τ q0 q1

τ ∈ (0, 1).

More precisely, if T (f )qj ≤ cj f pj ,

f ∈ D, j = 0, 1,

(9.17)

then T (f )qτ ≤ c1−τ cτ1 f pτ , 0

f ∈ D, τ ∈ (0, 1).

(9.18)

Proof. We argue first that it suffices to prove the inequality (9.18) when f is a simple function. Given an arbitrary function f ∈ D ∩ Lpτ , we find integrable simple functions (sn )∞ n=1 such that limn→∞ f − sn pτ = 0 and limn→∞ T (sn ) = T (f ) almost everywhere [ν]. Then the Fatou Lemma (Theorem 4.29) implies T (f )qτ ≤ lim sup T (sn )qτ n→∞

and (9.18) follows from the corresponding inequality with sn in place of f . The construction of sn is routine and the details are sketched in Problem 9FF. We also assume that the exponents p0 , p1 , q0 , q1 are all finite and leave to the reader the easy adaptation of the argument to the case in which one or more of these equals +∞. Consider simple functions f=

M

k=1

λk χEk ∈ L0 (μ),

g=

N

=1

ρ χF ∈ L0 (ν),

9 The Lp spaces

233

with nonzero complex coefficients λk , ρ , pairwise disjoint collections {Ek }N k=1 in S, {F }N =1 in T, and M

|λk |μ(Ek ) =

k=1

N

|ρ |ν(F ) = 1.

=1

Define αζ , βζ ∈ C by αζ =

1−ζ ζ + , po p1

βζ =

1−ζ ζ + , q0 q1

ζ ∈ C,

and observe that the functions fζ =

M

λk |λk |αζ χEk , |λk |

k=1

gζ =

N

ρ |ρ |βζ χF , |ρ |

ζ ∈ C,

=1

belong to D by hypothesis. The function u : C → C defined by  1 u(ζ) = 1−ζ ζ gζ T (fζ ) dν = k0 k1 Y  N M

1 = 1−ζ ζ |λk |αζ |ρ |βζ χF T (χEk ) dν, ζ ∈ C, k0 k1 k=1 =1 Y is clearly analytic. If ζ = τ + iσ for some τ ∈ [0, 1] we have |fζ | =

M

|λk |αζ χEk =

k=1

M

|λk |1/pτ χEk ,

k=1

and thus fζ pτ = 1. Analogously, gζ qτ = 1, ζ = τ + iσ, τ ∈ [0, 1]. The hypothesis implies that |u(ζ)| ≤ 1 if ζ ∈ {0, 1}, and Lemma 9.44 yields the inequality |u(ζ)| ≤ 1 when ζ ∈ (0, 1). In particular, if ζ = τ ∈ (0, 1) we obtain      gτ T (fτ ) dν  ≤ k 1−τ k1τ . (9.19) 0   Y

Every simple function h ∈ Lpτ (μ) with hpτ = 1 can be written as h = fτ for some simple function f satisfying f 1 = 1, and an analogous observation  about Lqτ (ν) shows that (9.19) implies that      gT (f ) dν  ≤ k 1−τ k1τ , 0   Y

when f ∈ L0 (μ) and g ∈ L0 (ν) are simple and satisfy f pτ = gqτ = 1. Replacing f by f /f pτ for an arbitrary simple function f ,

9 The Lp spaces

234

     gT (f ) dν  ≤ k 1−τ k1τ f p τ 0   Y

when f ∈ L0 (μ) and g ∈ L0 (ν) are simple and satisfy gqτ = 1. Lemma 9.16 implies now the desired inequality T (f )qτ ≤ k01−τ k1τ f pτ for simple functions f ∈ L0 (μ).

 

Problems 9A. Let I be an uncountable set, and endow V = CI with the family of seminorms {qi }i∈I defined by qi ((λj )j∈I ) = |λi | for i ∈ I. Show that 0 ∈ V does not have a countable neighborhood base in the topology defined by these seminorms. In particular V is not metrizable. Determine the dual space V ∗ of V. 0 9B. Find a sequence of functions {fn }∞ n=1 ⊂ L (R, BR , λ1 ) which converges in measure to zero, but has the property that the sequence {fn (t)}∞ n=1 does not converge for any t ∈ R. 9C. Prove an analog of Theorem 9.15 where L0 is replaced by Lp with p ∈ (0, +∞) and the measure μ is not assumed to be σ-finite. 9D. Denote by c0 the vector subspace of ∞ consisting of those sequences in ∞ which converge to zero. (i) Show that c0 , endowed with the ∞-norm, is a Banach space. 1 ∗ (ii) Given x = (ξn )∞ n=1 ∈  , define a functional ϕx ∈ (c0 ) by setting ϕx ((ζn )∞ n=1 ) =

∞ 

ξn ζn ,

(ζn )∞ n=1 ∈ c0 .

n=1

Show that ϕx is indeed continuous, and ϕx  = x1 . Also show that every functional ϕ ∈ (c0 )∗ is of the form ϕ = ϕx for some x ∈ 1 . Thus 1 can be identified with the dual of a Banach space. (In fact 1 can be identified with the dual of many quite distinct Banach spaces.) ∞ p ∗ (iii) Fix p ∈ (0, 1). Given x = (ξn )∞ n=1 ∈  , define a functional ϕx ∈ ( ) by setting ϕx ((ζn )∞ n=1 ) =

∞ 

ξn ζn ,

p (ζn )∞ n=1 ∈  .

n=1

Verify that ϕx is continuous and show that every continuous linear functional on p is of this form. 9E. (Day) Consider the restriction μ of λ1 to the interval [0, 1], and fix p ∈ (0, 1). Show that there is no nonzero continuous linear functional on the F -space Lp (μ). Prove the same statement for the space L0(μ) with the topology of convergence in measure given by the quasinorm q([f ]) = [0,1] min{1, |f |} dμ, f ∈ L0 (μ). (Hint: If ϕ is a continuous linear functional, then the set {u ∈ Lp (μ) : |ϕ(u)| < 1} is convex and it must contain Vε = {u ∈ Lp (μ) : up < ε} for some ε > 0. Every element in Lp (μ) can be written as (u1 + u2 + · · · + un )/n with u1 , . . . , un ∈ Vε for sufficiently large n.)

9 The Lp spaces

235

9F. Let (X, S, μ) be a measure space. (i) Let ϕ be a continuous linear functional on L∞ (μ). Define a function ν : S → C by setting ν(E) = ϕ([χE ]) for E ∈ S. Show that ν is finitely additive, that μ(E) = 0 implies ν(E) = 0, and that ν has finite total variation. More precisely, n 

|ν(Ej )| ≤ ϕ

j=1

for every finite collection (Ej )n j=1 of pairwise disjoint sets in S. (ii) Conversely, let ν : S → C be a finitely additive set function of finite variation c such that μ(E) = 0 implies ν(E) = 0. Show that there exists ϕ ∈ L∞ (μ)∗ such that ϕ([χE ]) = ν(E) for all E ∈ S and ϕ = c. (iii) Consider the case in which X = N and ν is the counting measure. Let F be a family of subsets of N which is closed under the formation of finite intersections, does not contain any finite set, and is maximal subject to these conditions. Define ν : 2N → [0, +∞) by setting ν(E) = 1 for E ∈ F and ν(E) = 0 for E ∈ / F . Show that there exists a functional ϕ ∈ (∞ )∗ of norm one such that ϕ(χE ) = ν(E) for all E ∈ 2N . This functional ϕ is not represented by any sequence in 1 . (Hint: Show that for every set A ⊂ N that does not belong to F there exists a set B ∈ F such that A ∩ B = ∅.) 9G. Consider two σ-finite measure spaces (X, S, μ), (Y, T, ν), and let f : X × Y → [0, +∞) be measurable S × T. Prove the following extension of Minkowski’s inequality:

! %

&p f (x, y) dν(y)

X

"1/p dμ(x)

 !

"1/p



Y

f (x, y)p dμ(x) Y

dν(y).

X

(Hint: Use Lemma 9.16. The usual Minkowski inequality corresponds to the case where ν is the counting measure on a set with two elements.) 9H. Show that Lp (λ1 ) ⊂ Lq (λ1 ) if p, q ∈ (0, +∞] and p = q. 9I. Let (X, S, μ) and (Y, T, ν) be two σ-finite measure spaces. Given functions f ∈ L0 (μ) and g ∈ L0 (ν), define a function f ⊗ g ∈ L0 (μ × ν) by setting (f ⊗ g)(x, y) = f (x)g(y),

x ∈ X, y ∈ Y.

(i) Show that f ⊗ gp = f p gp for all f ∈ L0 (μ), g ∈ L0 (ν), and for every p ∈ (0, +∞]. Here we employ the usual conventions that 0 · (+∞) = 0 and (+∞) · (+∞) = +∞. (ii) Show that the linear manifold generated by {[f ⊗ g] : f ∈ Lp (μ), g ∈ Lp (ν)} is dense in Lp (μ × ν) if p ∈ (0, +∞). For p = +∞, this linear manifold is dense in the weak* topology on L∞ (μ × ν) given by the identification of L∞ (μ × ν) with L1 (μ × ν)∗ . (iii) Let {fα }α∈A and {gβ }β∈B be orthonormal bases in L2 (μ) and L2 (ν), respectively. Prove that {fα ⊗ gβ }(α,β)∈A×B is an orthonormal basis in L2 (μ × ν). 9J. Let V be a complex Hilbert space, and let M ⊂ V be a closed subspace. Denote by M⊥ = {f ∈ L2 (μ) : f, g = 0, g ∈ M} the orthogonal complement of M. Show that M ∩ M⊥ = {0} and that M + M⊥ = V . (Hint: Choose an orthonormal basis {fα }α∈A for M and find by maximality an orthonormal set {gβ }β∈B such that {fα }α∈A ∪ {gβ }β∈B is an orthonormal basis for V. Then {gβ }β∈B is an orthonormal basis for M⊥ .)

9 The Lp spaces

236

9K. Let (X, S) be a measurable space, and let μ and ν be two equivalent, σ-finite measures on S (Definition 6.15). Denote by ϕ = dν/dμ the Radon-Nikod´ ym derivative of ν with respect to μ. Show that the map [f ] → [f ϕ1/p ] is a linear isometry from Lp (ν) onto Lp (μ) for each p ∈ (0, +∞). (When p ∈ (0, 1), the isometry is relative to the metric determined by the quasinorm f → f pp . For p = +∞, we have L∞ (μ) = L∞ (ν).) 9L. Let (X, S, μ) be a measure space, let T ⊂ S be a σ-algebra such that μ|T is σ-finite, and fix p ∈ [1, +∞). (i) Let f, g : X → C be integrable [μ] and such that f p < +∞, gp < +∞, and g is measurable [T]. Show that E[gf |T] = g E[f |T] almost everywhere [μ]. (ii) Let h : X → C be integrable [μ]. Show that E[h|T]p ≤ hp for all p ∈ [1, +∞]. (Hint: Use (i) and Lemma 9.16.) 9M. Let p, p1 , p2 , . . . , pn ∈ (0, +∞] satisfy the identity

 1 1 , = p pj n

j=1

where we use the usual convention 1/ + ∞ = 0. Consider a measure space (X, S, μ) and functions fj ∈ Lpj (μ) for j = 1, 2, . . . , n. Show that the product f = f1 f2 · · · fn belongs to Lp (μ) and f p ≤ f1 p1 f2 p2 · · · fn pn . 9N. Let (X, S, μ) be a σ-finite measure space, and let p, q ∈ (0, +∞] satisfy p < q. (i) Show that Lp (μ) ∩ Lq (μ) ⊂ Lr (μ) ⊂ Lp (μ) + Lq (μ) for every r ∈ (p, q). (ii) If f ∈ Lp (μ) ∩ Lq (μ), the function r → f r is continuous on the interval [p, q]. (For q = +∞, continuity at q means that limr→+∞ f r = f ∞ .) 9O. Given d ∈ N, a function f ∈ L0 (λd ), and a vector x ∈ Rd denote, as usual, by fx the translate of f by x, that is, fx (y) = f (y − x), y ∈ Rd . Fix p ∈ (0, +∞) and two functions f, g ∈ Lp (λd ). Prove that lim

x Rd →+∞

fx + gpp = f pp + gpp .

9P. Define a function h : R → R by setting

⎧ ⎪ ⎨1, h(t) =

−1,

⎪ ⎩0,

0 ≤ t < 12 , 1 ≤ t < 1, 2

t ∈ R \ [0, 1).

Define the functions hj,k (t) = 2j/2 h(2j (t − k)) for j, k ∈ Z and t ∈ R. Show that the vectors {[hj,k ]}j,k∈Z form an orthonormal basis for L2 (λ1 ). Moreover, those functions hj,k which are supported in [0, 1] provide an orthonormal basis for the 1 space {[f ] ∈ L2 (λ1 |[0, 1]) : 0 f dλ1 = 0}.

9Q. Exhibit a finite measure space (X, S, μ) and a function f ∈ L∞ (μ) with the prop erty that | X f h dμ| < f ∞ for every h ∈ L1 (μ) such that h1 ≤ 1.

9 The Lp spaces

237

9R. Consider a measure space (X, S, μ) and a sequence {[gn ]}n∈N ⊂ L∞ (μ) which converges to zero in measure.



(i) Suppose in addition that supn∈N gn ∞ < +∞. Show that limn→∞ X f gn dμ = 0 for every f ∈ L1 (μ). (In other words, {[gn ]}n tends to zero in the weak* topology of L∞ (μ) when this space can be identified with the dual of L1 (μ).) (ii) Show that the conclusion of (i) need not hold if the norms gn ∞ are not bounded. 9S. Let {en }n∈N be an orthonormal sequence in a complex Hilbert space V. Show that 0 belongs to the closure of the set A = {en +nem : n, m ∈ N} in the weak topology. Show, moreover, that there is no sequence {xn }n∈N ⊂ A which converges to zero weakly. (This proves that the weak topology is not metrizable on V. Hint: Observe that {en } converges weakly to 0. Use uniform boundedness.) 9T. Consider the space 1 of absolutely convergent series of complex numbers. (i) Show that 0 belongs to the closure of the set {x ∈ 1 : x1 = 1} in the weak topology. 1 (ii) (Schur) Let {xn }∞ n=1 ⊂  be a sequence which converges weakly to 0. Show that limn→∞ xn 1 = 0. (Thus the unit ball of 1 fails to be metrizable in the weak topology of 1 .) 9U. Let (X, S, μ) be a σ-finite measure space, and fix p ∈ (0, +∞). (i) Show that Lp (μ) is a separable space if and only if the following condition is satisfied: there exists a sequence {En }n∈N ⊂ S such that for every F ∈ S we have μ(EF ) = 0 for some set in the σ-algebra generated by the collection {En }n∈N . If Lp (μ) is separable, show that L0 (μ) is separable when given the topology of convergence in measure. (ii) Under what conditions is L∞ (μ) separable? 9V. Let V be a real vector space and let C ⊂ V be a nonempty convex set. A nonempty convex set F ⊂ C is called a face of C if C \ F is convex. A point x ∈ C is said to be an extreme point of C if {x} is a face of C. (i) Let F be a face of C, and let F  be a face of F . Show that F  is a face of C. (ii) Let {F γ }γ∈Γ be a family of faces of C that is totally ordered by inclusion. Show that γ∈Γ Fγ is either empty or a face of C. (iii) Let ϕ : V → R be a linear functional such that ϕ|C has a maximum value α. Show that the set {x ∈ C : ϕ(x) = α} is a face of C. 9W. (Kre˘ın-Mil’man) Let V be a normed vector space, and let K ⊂ V ∗ be a nonempty convex set that is compact in the weak* topology. (i) Use Problem 9V(ii) and Zorn’s lemma to show that the set of weak*-closed faces of K has minimal elements relative to inclusion. (ii) Show that a minimal closed face of K consists of exactly one point. Thus K has at least one extreme point. (Hint: Let F be a weak*-closed face of K that contains two points ϕ = ψ. Choose x ∈ V such that ϕ(x) = ψ(x) and apply Problem 9V(iii) to the functional ϕ → ϕ(x) to show that F has a face F   F .) 9X. Show that the closed unit balls of c0 and L1 (λ1 ) possess no extreme point. Deduce that there is no Banach space V whose dual is linearly isometric to either c0 or L1 (λ1 ).

9 The Lp spaces

238

9Y. (Jordan-von Neumann) Let (V,  · ) be a Banach space. Show that V is a Hilbert space if and only if it satisfies the parallelogram identity x + y2 + x − y2 = 2x2 + 2y2 ,

x, y ∈ V.

(Hint: Use (9.8) to define the scalar product.) 9Z. Let n ≥ 3 be an integer and let ω ∈ C be a primitive root of order n of unity, that is, ω n = 1 = ω k , k = 1, . . . , n − 1. Prove the identity x, y =

n−1 1  k ω x + ω k y2 n k=0

for any two vectors x, y in a complex Hilbert space. 9AA. Let (X, S, μ) be a measure space. Show that the inequality f + g1,∞ ≤ f 1,∞ + g1,∞ is not generally true for f, g ∈ L0 (μ), but that the space L1,∞ (μ) is a topological vector space if a set G ⊂ L1,∞ (μ) is declared to be open precisely when, for every u ∈ G we have {v ∈ L1,∞ (μ) : u − v1,∞ < ε} ⊂ G for some ε > 0 (depending on u). 9BB. Let (X, S, μ) be a measure space, and suppose that 0 < q1 < q ≤ q2 ≤ +∞. Show that Lq1 ,∞ ∩ Lq2 ,∞ ⊂ Lq . Deduce a proof of the case p1 = p2 of Theorem 9.42. Alternatively, verify that the proof of Theorem 9.42 applies with minor changes when p1 = p2 . 9CC. Maintaining the notation in the proof of Theorem 9.42, suppose that p0 ≤ p1 and either q0 or q1 is infinite. (i) Calculate the (limiting) values of the exponents γ and η. (ii) Show that for an appropriate value of b > 0, and for a = bf ηp , we have ν({y : |T (f (s(t)) )(y)| > t/2}) = 0 for every t. (iii) Use (ii) to replace the right-hand side of (9.15) by a single integral. Estimate this integral to prove that the conclusion of Theorem 9.42 holds. 9DD. In the proof of Theorem 9.42, use the constant a = cf ηp for an appropriate constant c > 0 to show that

! T (f )qq ≤ 2q q

"

1 1 + c1−τ cτ1 f qp , q − q0 q1 − q 0

f ∈ Lp (μ), q0 < q1 .

9EE. Let u be a function satisfying the hypothesis of Lemma 9.44. Apply the maximum 2 modulus principle in a large rectangle to the function u(z)eε(z −1) , ε > 0, to obtain the conclusion of the lemma. 9FF. Suppose that a linear map T satisfies the hypotheses of Theorem 9.45, and let f ∈ D. a. Suppose that f ∈ Lp0 (μ)∩Lp1 (μ), and let (sn )∞ n=1 be a sequence of simple functions such that |sn | ≤ |sn+1 | ≤ |f | and limn→∞ sn (x) − f (x)pj = 0 for j = 1, 2. Show that the sequence (T (sn ))∞ n=1 has a subsequence that converges to T (f ) almost everywhere [ν]. b. Suppose that f ∈ Lpτ (μ) for some τ ∈ (0, 1). Use the truncations f (1) and f(1) to prove the existence of a sequence (sn )∞ n=1 of simple functions such that limn→∞ sn − f pτ = 0 and (T (sn ))∞ n=1 converges to T (f ) almost everywhere [ν]. 9GG. Prove Theorem 9.45 when one or more of the exponents p0 , p1 , q0 , q1 is infinite.

9 The Lp spaces

239

9HH. Suppose that (X, S, μ) is a measure space, 0 < p0 < p1 < +∞, t ∈ (0, 1), p = t (1−t)p0 +tp1 , and f ∈ Lp0 ∩Lp1 . Theorem 9.45 implies that f p ≤ f 1−t p0 f p1 . Give a direct proof of this inequality using the following outline. a. Reduce the inequality to the case in which f p0 = f p1 = 1 via replacing f and μ by af and bμ, respectively, for some positive constants a, b. b. Suppose that f p0 = f p1 = 1. Use f (1) and f(1) to show that f p ≤ 2. c. Apply the argument in (b) to the function fn : X n → C defined by fn (x1 , . . . , xn ) = f (x1 ) · · · f (xn ),

x1 , . . . , xn ∈ X,

and to the product of n copies of μ to deduce that f n p ≤ 2. Let n → ∞ to reach the desired conclusion. 9II. Let (X, S, μ) be a probability space, and let {fn }n∈N be a sequence in L∞ (μ) with the property that supn f ∞ < +∞. Show that limn→∞ fn 1 = 0 if and only if limn→∞ fn 2 = 0. (Of course, 1 and 2 can be replaced by any two positive numbers.)

Chapter 10

Fourier analysis

Periodic phenomena such as sunrises and seasons have fascinated people since prehistoric times. The astronomers of antiquity tried to describe the various observed periodic motions of the planets in terms of uniform circular motions or sums of such motions (cycles and epicycles). A mathematical form of this idea was studied sporadically in the 17th century when it was realized that some functions f : R → C that satisfy f (x + 1) = f (x) for x ∈ R can be written as ∞

cn e2πinx , x ∈ R, (10.1) f (x) = n=−∞

where the coefficients are calculated as  1 cn = e−2πint f (t) dt,

n ∈ Z.

0

For instance, the Taylor series for log(1 − x) yields such a representation for a discontinuous function, namely

∞ 1

e2πinx

(1 − x), x ∈ (0, 1), sin(πnx) = = 2 2πin πn 0, x = 0. n=1 n∈Z\{0} 1 (One can proceed then to write 0 (1 x)2 dx = 1/3 in terms of the coef− ∞ ficients 1/(πn), and thereby obtain n=1 n−2 = π 2 /6. Euler used different techniques to calculate this sum.) Fourier found this kind of development to be crucial in describing the solutions of the heat equation, and audaciously declared that every periodic function can be written in the form (10.1). (This unproved claim delayed the publication of Fourier’s work.) There were several failed attempts to prove the claim as well as a number of positive results and

© Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1 10

241

242

10 Fourier analysis

interesting examples. A large part of modern analysis evolved from this study. It eventually became clear that the functions f to be so represented must satisfy some requirements, and that the very meaning of (10.1) may need to be modified in order to allow for sufficient generality. For instance, we have seen in Chapter 9 that functions f : (0, 1) → C that are square integrable [λ1 ] have a Fourier expansion of the form (10.1) in the sense that 2 2 N 2 2

2 2 cn en 2 = 0, lim 2f − N →∞ 2 2 n=−N

2

where en (t) = e2πint for t ∈ (0, 1) and n ∈ Z. In this chapter we discuss in greater detail the representation of functions as Fourier series and Fourier integrals, focusing on functions of one variable mainly to simplify notation. We present a fairly limited outline of the subject and we only consider very briefly issues of pointwise convergence. In the discussion of Fourier series it is often natural to replace the interval (0, 1) by the unit circle T = {ζ ∈ C : |ζ| = 1} = {e2πit : t ∈ [0, 1)} and to endow T with the measure m obtained by pushing forward λ1 |[0, 1) via the map t → e2πit . This measure satisfies m({e2πit : t ∈ [α, β)}) = β − α,

0 ≤ α ≤ β ≤ 1,

and is called normalized arclength measure on T. The measurable sets [m] are the sets of the form {e2πit : t ∈ E}, where E ⊂ R is measurable [λ1 ]. The translation invariance of λ1 corresponds to rotation invariance for m. More precisely, m(ωE) = m(E) for every ω ∈ T and every set E ⊂ T that is measurable [m]. We also have m(E −1 ) = m(E) if E is measurable [m], where E −1 = {1/ζ : ζ ∈ E}. This follows from the invariance of λ1 under the change of variable t → −t. Fourier coefficients were defined for functions in L2 (m) = L2 (T, m), but their definition makes sense for functions in L1 (m), and even for complex Borel measures on T; see Problem 10A. Definition 10.1. Given f ∈ L1 (m), the Fourier coefficients of f are the numbers  f"(n) = ζ −n f (ζ) dm(ζ), n ∈ Z. T

10 Fourier analysis

243

The series

f"(n)ζ n

n∈Z

is called the Fourier series of f . The following result is due to Riemann and Lebesgue. Lemma 10.2. For every f ∈ L1 (m) we have |f"(n)| ≤ f 1 ,

n ∈ Z,

and lim|n|→∞ |f"(n)| = 0. Proof. The first assertion is immediate because   |f"(n)| ≤ |ζ −n f (ζ)| dm(ζ) = |f (ζ)| dm(ζ) = f 1 , T

T

n ∈ Z.

Proposition 9.6, with the maps f → f"(n) in place of Tn , allows us to reduce the proof of the last assertion to elements f in a set that spans a dense linear manifold in L1 (T). Thus it suffices to verify that assertion when f (ζ) = ζ k for some k ∈ Z, and the assertion is immediate in this case because f"(n) = 0 for n = k.   Corollary 10.3. (The Hausdorff-Young inequalities) Suppose that p ∈ [1, 2] and f ∈ Lp (m). Then we have * +1/p

 p |f"(n)| ≤ f p , n∈Z

where p = p/(p − 1) denotes the conjugate exponent of p. Proof. The Riesz-Thorin interpolation theorem (Theorem 9.45) applied to the map f → {f"(n)}n∈Z with μ = m and ν the counting measure on Z shows that it suffices to prove the corollary for p = 1 and p = 2. These two cases follow from the preceding lemma and from the Bessel inequality (9.10), respectively.   Example 10.4. Equality is achieved in Corollary 10.3 when f (ζ) = ζ n for some n ∈ Z. Unlike the case of functions in L2 (m), the partial sums (SN f )(ζ) =

N

f"(n)ζ n ,

N ∈ N, ζ ∈ T,

n=−N

do not generally converge to f in L1 (m). To understand this phenomenon it is useful to consider the convolution of functions.

244

10 Fourier analysis

Definition 10.5. Given two functions f, g : T → C that are measurable [m] and a point ζ ∈ T, the convolution f ∗ g is defined at ζ if  |f (ω)g(ζ/ω)| dm(ω) < +∞, T

and we define

 (f ∗ g)(ζ) =

f (ω)g(ζ/ω) dm(ω). T

It is easy to see that the set of points where f ∗ g is defined is measurable [m] and f ∗ g is measurable [m] as well. A simple change of variables shows that f ∗ g = g ∗ f . Indeed, the measure m is invariant under the transformation ω → ζ/ω, so setting w = ζ/ω yields   |f (ω)g(ζ/ω)| dm(ω) = |f (ζ/w)g(w)| dm(w). T

T

The same calculation, without the absolute values, yields (g∗f )(ζ) = (f ∗g)(ζ) at points ζ where (f ∗g)(ζ) is defined. In many cases of interest, f ∗g is defined everywhere or almost everywhere [m]. Example 10.6. The sequence of functions N

DN (ζ) =

ζ n,

N ∈ N, ζ ∈ T,

n=−N

is called the Dirichlet kernel. For any f ∈ L1 (m) and N ∈ N we have  (SN f )(ζ) =

f (ω) T

N

(ζ/ω)n dm(ω) = (f ∗ DN )(ζ),

ζ ∈ T.

n=−N

Example 10.7. Fix p ∈ [1, +∞], denote by p the conjugate exponent, and  suppose that f ∈ Lp (m) and g ∈ Lp (m). Then f ∗g is defined and continuous on T. Indeed, the invariance of m under the transformation ω → ζ/ω and H¨older’s inequality (9.3) yield 0

 T

|f (ω)g(ζ/ω)| dm(ω) ≤

11/p 0 |f | dm

p

p

T

= f p gp ,

T

11/p

|g(ζ/ω)| dm(ω)

ζ ∈ T.

Thus f ∗ g is defined everywhere on T and f ∗ g∞ ≤ f p gp . The continuity of f ∗ g follows when p < +∞ from the fact that the map  ω → gω ∈ Lp (T), where

10 Fourier analysis

245

gω (ζ) = g(ζ/ω),

ζ, ω ∈ T,



is continuous from T to Lp (m). This fact is verified as in Proposition 9.36, noting first that it is obvious when g is a continuous function. Example 10.8. Suppose now that f, g ∈ L1 (m). The Fubini-Tonelli theorem (Theorem 8.4) yields 0 1    |f (ω)g(ζ/ω)| dm(ω) dm(ζ) = |f (ω)| |g(ζ/ω)| dm(ζ) dm(ω) T T T T = |f (ω)|g1 dm(ω) = f 1 g1 , T

where we also used the rotation invariance of m to evaluate   |g(ζ/ω)| dm(ζ) = |g(ζ)| dm(ζ) = g1 , ω ∈ T. T

T



We conclude that T |f (ω)g(ζ/ω)| dm(ω) < +∞ for [m]-almost every ζ. Therefore f ∗ g is defined almost everywhere [m] and    |(f ∗ g)(ζ)| dm(ζ) ≤ |f (ω)g(ζ/ω)| dm(ω) dm(ζ). T

T

T

We conclude that f ∗ g ∈ L1 (m) and f ∗ g1 ≤ f 1 g1 . We note for further reference the identity f ∗ g(n) = f"(n)" g (n),

f, g ∈ L1 (T), n ∈ Z.

This is verified using Theorem 8.4 and the rotation invariance of m:   ζ −n f (ω)g(ζ/ω) dm(ω)dm(ζ) f ∗ g(n) = T 0 1 T −n = ω f (ω) g(ζ/ω)(ζ/ω)−n dm(ζ) dm(ω) T T −n = ω f (ω)" g (n) dm(ω) = f"(n)" g (n). T

One can interpolate between the observations in the last two examples to obtain an inequality due to Young. Proposition 10.9. Suppose that the numbers p, q, r ∈ [1, +∞] satisfy 1 1 1 + = + 1. p q r

246

10 Fourier analysis

For every f ∈ Lp (m) and g ∈ Lq (m) we have f ∗ g ∈ Lr (m) and f ∗ gr ≤ f p gq . Proof. Fix first f ∈ L1 (m) and consider the linear map T : L1 (m) → L1 (m) defined by T g = f ∗ g, g ∈ L1 (m). We have the inequalities T g1 ≤ f 1 g1 ,

g ∈ L1 (m),

and T g∞ ≤ f 1 g∞ ,

g ∈ L∞ (m),

and Theorem 9.45 yields T gq ≤ f 1 gq ,

g ∈ Lq (m), q ∈ [1, +∞).

To conclude the proof, we fix now g ∈ Lq (m) and apply Theorem 9.45 to the inequalities f ∗ g∞ ≤ f p gq ,



f ∈ Lq (m),

and f ∗ gq ≤ f 1 gq ,

f ∈ L1 (m),  

to obtain the desired result.

Example 10.10. If p, q, and r are as in Proposition 10.9 and f (ζ) = g(ζ) = ζ n for some n ∈ Z, then f ∗ gr = f p gq . The result is also sharp when p = q = r = 1 in the following sense. Lemma 10.11. For any g ∈ L1 (μ) we have sup{f ∗ g1 : f ∈ L1 (m), f 1 ≤ 1} = g1 . Proof. Set In = {e2πit : t ∈ (0, 1/n)} and fn = nχIn so fn 1 = 1 for n ∈ N. It suffices to show that limn→+∞ g − fn ∗ g1 = 0. Indeed, by Theorem 8.4,        ng(ζ/ω) dm(ω) dm(ζ) g − fn ∗ g1 = g(ζ) − T In        = [g(ζ) − g(ζ/ω)] dm(ω) dm(ζ) n In T ≤ ng − gω 1 dm(ω) ≤ sup g − gω 1 , In

ω∈In

where gω (ζ) = g(ζ/ω). The result follows from the continuity, noted earlier,   of the map ω → gω . Example 10.12. The preceding lemma yields, in the special case g = SN , sup{SN f 1 : f ∈ L1 (m), f 1 ≤ 1} = DN 1 ,

N ∈ N.

10 Fourier analysis

247

The numbers LN = DN 1 ,

N ∈ N,

are called the Lebesgue constants and can be approximated fairly precisely. Writing ζ 1/2 = eπit for t ∈ (−1/2, 1/2), we have 2N

DN (ζ) = ζ −N

ζn =

n=0

=

ζ

−(N + 12 ) 1 ζ− 2

so



−ζ

1/2 0

lim t↓0

1

− ζN+ 2

LN = 2 We have

ζ −N − ζ N +1 1−ζ

1 2

sin((2N + 1)πt) , sin(πt)

=

| sin((2N + 1)πt)| dt. sin(πt)

0 1 1 1 1 π − = , t sin(πt) πt 3!

so the difference 1 1 − sin(πt) πt is bounded for t ∈ [0, 1/2). Therefore the difference between LN and LN =

2 π



1/2 0

| sin((2N + 1)πt)| dt t

is bounded. A change of variable yields LN

2 = π



(2N +1)π 0

2 | sin s| ds = s π



π 0

* 2N

1 sin s t + nπ n=0

+ ds.

We observe next that 0≤

1 2N 0 2N ∞

1 1 t 1 − ≤ < +∞. = 2 nπ s + nπ nπ(s + nπ) πn n=1 n=1 n=1

 2N 2N 2N Approximating n=1 1/n by 1 (1/t) dt = log(2N ), we see that n=1 1/(s+ nπ) differs by a bounded quantity from π −1 log N , and this shows that LN differs by a bounded quantity from  π 2 2 log N sin s ds = 2 log N. π2 π 0 In particular, limN →∞ LN = +∞.

248

10 Fourier analysis

Corollary 10.13. There exists a function f ∈ L1 (m) such that sup{SN f 1 : N ∈ N} = +∞. For such functions f , the sequence {SN f }N ∈N does not have a limit in L1 (m). Proof. The result follows immediately from Theorem 9.5 applied to the sequence of linear operators SN : L1 (T) → L1 (T). Since sup{SN  : N ∈ N} = sup{LN : N ∈ N} = +∞, the result follows immediately from Theorem 9.5 applied to the sequence of linear operators SN : L1 (T) → L1 (T).   We know, of course, that limN →∞ f − SN f 2 = 0 if f ∈ L2 (m), and thus limN →∞ f − SN f 1 = 0 as well for such functions. For general functions f ∈ L1 (m), the behavior of the sequence {SN }N ∈N can be improved by averaging it in various ways. Definition 10.14. Let {an }n∈Z be a bounded family of complex numbers. ∞ The series n=−∞ an is said to be Abel summable to the complex number s if ∞

lim r|n| an = s. r↑1

n=−∞

The boundedness assumption implies that the above series converges absolutely for r ∈ [0, 1). Proposition 10.15. Let {an }n∈Z be a bounded family of complex numbers N ∞ such that the limit s = limN →∞ n=−N an exists. Then n=−∞ an is Abel summable to s. Proof. Replacing a0 by a0 − s allows us to assume that s = 0. Setting sN =  N n=−N an for n ∈ N and s−1 = 0, we have ∞

n=−∞

r|n| an =



n=0

rn (sn − sn−1 ) =



(rn − rn+1 )sn .

n=0

For fixed N ∈ N and r ∈ [0, 1), we have the estimate     ∞ N ∞    

    n n+1 n n+1 (r − r )sn  ≤  (r − r )sn  + sup |sn | (rn − rn+1 ).      n>N n=0

n=0

n=N +1

The last sum above equals rN +1 < 1. We conclude that   ∞     |n| r an  ≤ sup |sn |, lim sup    n>N r↑1 n=−∞ and the proposition follows by letting N → ∞ in this inequality.

 

10 Fourier analysis

249

Example 10.16. Let us examine the Abel summability of the Fourier series of a function f ∈ L1 (m). The series ∞

fr (ζ) =

r|n| f"(n)ζ n ,

ζ ∈ T,

n=−∞

converges uniformly for any r ∈ [0, 1), and fr is therefore a continuous function of ζ. Corollary 4.25 allows us to interchange summation and integration to obtain * ∞ + 

|n| n fr (ζ) = f (ω) r (ζ/ω) dm(ω) = (Pr ∗ f )(ζ), T

n=−∞

where Pr (ζ) =



n=−∞

r|n| ζ n =

1 1 1 − r2 + − 1 = , 1 − rζ 1 − rζ −1 |r − ζ|2

r ∈ [0, 1), ζ ∈ T.

The family of functions {Pr }r∈[0,1) is called the Poisson kernel. Lemma 10.17. The Poisson kernel has the following properties: (1) (2) (3) (4)

P  r (ζ) ≥ 0 and Pr (ζ) = Pr (1/ζ) for r ∈ [0, 1) and ζ ∈ T. P (ζ) dm(ζ) = 1 for r ∈ [0, 1). T r The function t → Pr (e2πit ) is decreasing for t ∈ [0, 1/2). For every δ ∈ (0, 1/2], lim max{Pr (e2πit ) : t ∈ [δ, 1/2]} = 0. r↑1

Proof. Property (1) is immediate from the closed form of Pr , and (2) is obtained if we integrate the series defining Pr . Property (3) is seen by observing that |r − e2πit |2 = 1 + r2 − 2r cos(2πt) and that cos(2πt) is decreasing on [0, 1/2). Finally, the maximum in (4) is

1+

r2

1 − r2 , − 2r cos(2πδ)

and the denominator has limit 2 − 2 cos(2πδ) = 0 as r ↑ 1. The following result is due to H. A. Schwarz. Proposition 10.18. For every f ∈ C(T) we have lim f − Pr ∗ f ∞ = 0. r↑1

 

250

10 Fourier analysis

Proof. Lemma 10.17(2) allows us to write  f (ζ) − (Pr ∗ f )(ζ) = [f (ζ) − fω (ζ)]Pr (ω) dm(ω), T

ζ ∈ T, r ∈ [0, 1),

where fω (z) = f (ζ/ω). For fixed δ ∈ (0, 1/2] we have  |f (ζ) − (Pr ∗ f )(ζ)| ≤ |f (ζ) − fω (ζ)|Pr (ω) dm(ω) |ω−1| 1, that is, show that sup{f ∗ gq : f ∈ L1 (m), f 1 ≤ 1} = gq for every g ∈ Lq (m). 10E. Given f ∈ L1 (m) and ζ ∈ T, define the rotation fζ of f by fζ (z) = f (z/ζ), z ∈ T. Show that f ζ (n) = ζ −n f (n) for n ∈ Z. 10F. Let f : T → C be a function of finite variation in the sense that the function t → f (e2πit ) has finite variation on [0, 1). The total variation v of f is defined as the least upper bound of all sums of the form |f (ζ1 ) − f (ζ2 )| + · · · + |f (ζn−1 ) − f (ζn )| + |f (ζn ) − f (ζ0 )|, where n is a natural number and ζj = e2πitj with 0 ≤ t1 ≤ · · · ≤ tn ≤ 1. (i) Show that |f (n)| ≤ v/n for n ∈ Z \ {0}. (ii) Set ζn = e2πi/n and show that nf − fζn pp =

n−1  k=0

fζ k − fζ k+1 pp ≤ vf − fζn p−1 ∞ n

n

for all p ∈ (1, +∞). (iii) Show that f is continuous on T if and only if lim nf − fζn 22 = 0.

n→∞

(iv) Show that the condition in (iii) is equivalent to lim n

n→∞

 k∈Z

k 2 |1 − ζn | |f (k)|2 = 0.

260

10 Fourier analysis (v) Use the inequality k2 k 2 ≤ |1 − ζn | = 4 sin2 n2

%

kπ 2n

& ≤

π 2 k2 , n2

1≤k≤

n , 2

to show that the condition in (iv) is equivalent to

n

k2 |f (k)|2

k=−n

lim

n

n→∞

= 0.

(vi) Use (i) and Problem 9II to further simplify this condition to

n k=−n

lim

k|f (k)|

n

n→∞

= 0.

(This condition is satisfied if lim|k|→∞ k|f (k)| = 0.) 10G. Let μ be a complex Borel measure on T such that μ(T) = 0, and define a function f : T → C by f (e2πit ) = μ({e2πis : s ∈ [0, t)}),

t ∈ [0, 1).

(i) Show that f has total variation equal to |μ|(T). (n)/(2πin) for n ∈ Z \ {0}. (ii) Verify that f (n) = μ (iii) (Wiener) Show that μ has no point masses if and only if

n k=−n

lim

n

n→∞

μ (k)

= 0.

(iv) Use a measure of the form μ − cm to show that (iii) holds without the assumption that μ(T) = 0.



10H. Use the fact that sup{LN : N ∈ N} = +∞ and the formula (SN f )(1) = T f DN dm to show that there exist functions f ∈ C(T) for which the sequence {(SN f )(1)}N ∈N does not converge. 10I. Use Proposition 10.18 to provide another proof of the Weierstrass approximation theorem for trigonometric polynomials. 10J. Prove a version of Lemma 10.22 without the assumption that the function g is continuous. Show that the measure μ is uniquely determined by g. 10K. Verify that the function

x →

sin x , x

1,

x = 0 x = 0,

does not belong to L1 (λ1 ). Deduce that there are functions f ∈ L1 (λ1 ) such that dT ∗ f ∈ / L1 (λ1 ) for any positive integer T . (The function dT is defined in Example 10.31.) 10L. Characterize those functions f ∈ L1 (λ1 ) that satisfy f 1 = sup{|f (t)| : t ∈ R}. 10M. Let μ be a complex Borel measure on R. Define the Fourier transform μ :R→C  of μ by μ (t) = R e−2πixt dμ(x). Show that |μ | is bounded by the total variation |μ|(R) and that μ is a uniformly continuous function on R.

10 Fourier analysis

261

10N. Suppose that f ∈ L1 (R) and on R and

f 

 R

|xf (x)| dλ1 (x) < +∞. Show that f is differentiable

= g , where g(x) = −2πixf (x), x ∈ R. 2

10O. Define f : R → R by f (x) = e−πx , x ∈ R. a. Show that f  (x) = 2πxf (x) and f  (t) = 2πtf (t), x, t ∈ R. b. Show that f = f . c. More generally, suppose that p is a complex polynomial of degree n. Show that the Fourier transform of the function p(x)f (x) is of the form q(t)f (t), where q is also a complex polynomial of degree n. (Hint: calculate the Fourier transforms of higher order derivatives dn f (x)/dxn .) 10P. Suppose that f, g ∈ L4/3 (λ1 ), so f ∗ g ∈ L2 (λ1 ). a. Show that f ∗ g = f g almost everywhere [λ1 ]. (The function f is defined because L4/3 (λ1 ) ⊂ L1 (λ1 ) + L2 (λ1 ).) b. Prove the inequality f ∗ g22 ≤ f ∗ f 2 g ∗ g2 . 10Q. Let {sn }n≥0 be a sequence of complex numbers, and set σn =

s0 + s1 + · · · + sn−1 , n

n ∈ N.

Assume that the that limn→∞ σn =

sequence {sn }n≥0 converges to s ∈ C and show

∞ s. (When sn = n aro k=−n ak and limn→∞ σn = s, one says that n=−∞ an is Ces` summable to s.) 10R. Let f ∈ L1 (m), and let n = 0, 1, . . . , and σn (ζ) =



n=−∞

an ζ n be its Fourier series. Set sn (ζ) =

s0 (ζ) + s1 (ζ) + · · · + sn−1 (ζ) , n

n

k=−n

ak ζ k ,

n ∈ N , ζ ∈ T.

a. Show that σn = f ∗ Kn , where Kn (e2πit ) =

sin2 (nπt) , n sin2 (πt)

t ∈ R, n ∈ N,

is the Fej´ er kernel. b. Prove an analog of Lemma 10.17 for the sequence {Kn }n∈N . c. Show that limn→∞ σn (ζ) = f (ζ) for every point ζ ∈ C at which f is continuous, and the limit is uniform on any compact set of continuity for f . 10S. The function Kn (e2πit ) is not monotone decreasing for t ∈ [0, 1/2]. Show however that there exist probability measures μn , n ∈ N, on [0, 1/2] and a constant c > 0 such that  Kn (ζ) ≤ c

hs (ζ) dμ(s),

ζ ∈ T, n ∈ N .

[0,1/2]

Deduce that, for every f ∈ L1 (m), we have limn→∞ f ∗ Kn = f almost everywhere [m]. 10T. Fix p ∈ (1, 2) and let p be the conjugate exponent of p. Let g ∈ Lp (λ1 ) be such that gp = 0, let n ∈ N, and define f (t) =

n  k=1

where a1 , . . . , an , b1 , . . . , bn ∈ R.

e2πiak t f (t − bk ),

262

10 Fourier analysis (i) Show that f pp is arbitrarily close to ngpp if the numbers |bk − b |, k = , are sufficiently large. (Hint: Use Problem 9O.)   (ii) Show that f pp is arbitrarily close to n g pp if the numbers |ak − a | are sufficiently large. (iii) Deduce that the ratio f p /f p can be made arbitrarily small for appropriate choices of n and a1 , . . . , an , b1 , . . . , bn . 2 (iv) Calculate f p /f p when f (t) = e−πt , t ∈ R.

10U. Suppose that f ∈ L2 (λ1 ) is supported in an interval of length L. Show that f is infinitely differentiable and its nth derivative gn satisfies gn 2 ≤ (πL)n f 2 , n ∈ N. 10V. Suppose that f : R → C is integrable [m]. (i) Show that the series g(e2πit ) =

∞ 

f (t + k),

t ∈ R,

k=−∞

converges almost everywhere [λ1 ] and the function g is integrable [m] over T. (ii) Show that the Fourier coefficients of g are f (n), n ∈ Z. (iii) Suppose, in addition, that g is everywhere defined and continuous on T, and

∞ that n=−∞ |f (n)| < +∞. Prove the Poisson summation formula ∞ 

f (n) =

n=−∞

∞ 

f (n).

n=−∞

(iv) Define ψ : (0, +∞) → R by ψ(x) = identity due to Jacobi:

∞ n=1

1 + 2ψ(x) 1 = √ , 1 + 2ψ(1/x) x

e−πn

2

x,

x > 0. Prove the following

x > 0. 2

(Hint: Apply the Poisson summation formula to f (t) = e−πxt .) 10W. Fix an integer n ≥ 2, and let G = Z/nZ denote the additive group of integers modulo n. We denote by gk ∈ G the equivalence class of an integer k ∈ Z, so G = {g1 , . . . , gn } and g0 = gn is the zero element. Denote by μ the counting measure on G. (i) Set ω = e2πi/n and define functions ek : G → C by ek (gj ) = ω kj ,

j, k = 1, . . . , n.

2 Show that {n−1/2 ek }n k=1 is an orthonormal basis in L (μ). (ii) Given an arbitrary function f : G → C, define the Fourier transform f : G → C by the formula

f (gk ) = f, ek  =

n 

f (gj )ek (gj ),

k = 1, . . . , n.

j=1

Show that f ∞ ≤ f 1 and f 2 = n1/2 f 2 for every f : G → C. Deduce an analog of the Hausdorff-Young inequalities.

10 Fourier analysis

263

(iii) Verify the following version of the Fourier inversion formula: f (gk ) =

n 1 f (gj )ek (gj ), n

k = 1, . . . , n,

j=1

for every function f : G → C. (iv) Define an analog of convolution for functions in L1 (μ) and verify the identity f ∗ g = f g. (v) Prove that the Young inequalities (Proposition 10.9) hold for functions defined on G. Remark 10.43. The functions ek in the preceding problem are precisely the characters of the group G, that is, the homomorphisms ϕ from the (additive) group G to the multiplicative group T. Similarly, the functions ζ → ζ n , n ∈ Z, are the continuous characters of the group T, and the maps x → e2πitx , t ∈ R, are the continuous characters of the additive group R. Many of the results in this chapter extend to functions defined on an arbitrary Hausdorff, locally compact, Abelian group. The analogs of characters for noncommutative groups are irreducible unitary representations.

Chapter 11

Standard measure spaces

There are many examples of measure spaces that present various pathologies and on which some of the deeper theorems of measure theory fail. Elements of the class of standard measure spaces, to be defined shortly, do not display any of these pathologies and, in addition, can be classified up to a natural notion of isomorphism. The results we present here use little in addition to the observation that a separable metric space can be written as a countable union of closed subsets of arbitrarily small diameter (for instance, the closed balls of a fixed radius centered at the points of a countable dense set). They were developed initially in order to resolve questions about the measurability of the projection onto a one-dimensional subspace of a Borel subset of R2 . We develop enough of the theory of standard spaces to display some of the desirable features of measure theory on such spaces. We start with the idea of isomorphism between measurable spaces. Definition 11.1. Two measurable spaces (X, S) and (Y, T) are said to be isomorphic if there is a bijection f : X → Y such that, for E ⊂ X, we have E ∈ S if and only if f (E) ∈ T. We say that (X, S) embeds in (Y, S) if there exists F ∈ T such that (X, S) is isomorphic to (F, TF ). Another way to state the first condition on the function f above is to say that it is measurable [S, T] and the inverse map f −1 is measurable [T, S]. The following result is an analog of the Cantor-Bernstein theorem of set theory (see [I, Theorem 4.1]). Theorem 11.2. Assume that the measurable spaces (X, S) and (Y, T) are such that (X, S) embeds in (Y, T) and (Y, T) embeds in (X, S). Then (X, S) and (Y, T) are isomorphic. Proof. Let E ∈ S and F ∈ T be such that (X, S) and (Y, T) are isomorphic to (F, TF ) and (E, SE ), respectively, and let f : X → F and g : Y → E be bijections realizing these isomorphisms. It suffices to find sets A ∈ S and

© Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1 11

265

266

11 Standard measure spaces

B ∈ T such that f (A) = Y \ B and g(B) = X \ A. Indeed, once such sets are found, an isomorphism between (X, S) and (Y, T) is given by the bijection h : X → Y defined by

f (x), x ∈ A, h(x) = g −1 (x), x ∈ X \ A. To find the set A, consider the map u : S → S defined by u(D) = X \ g(Y \ f (D)),

D ∈ S.

The map u is monotone increasing (that is, D1 ⊂ D2 implies u(D1 ) ⊂ u(D2 )), and it carries countable unions to countable unions. It follows that the set A ∈ S defined by ∞  u(u(· · · u( ∅) · · · )) A= 7 89 : n=1 n times

satisfies the equation A = u(A). To conclude the proof, set B = Y \ f (A) and verify that these sets satisfy the desired conditions.   Definition 11.3. A measurable space (X, S) is said to be standard if there exists a complete, separable metric space Y such that (X, S) is isomorphic (as a measurable space) to (Y, BY ). A measure space (X, S, μ) is said to be standard if (X, S) is a standard measurable space. Another way to state this definition is to say that (X, S) is standard if there is a complete metric d on X such (X, d) is separable and S = BX . Observe that the definition of the Borel σ-algebra BX only requires the topology of the metric space X, not its metric. There are usually many different topologies on X which generate the same Borel σ-algebra. See, for instance, Theorem 11.9 below. Example 11.4. The space ((0, 1), B(0,1) ), where (0, 1) is given its usual topology as a subset of R, is standard. Indeed, (0, 1) is homeomorphic to R so ((0, 1), B(0,1) ) is isomorphic to (R, BR ). Similarly, [0, 1) is homeomorphic to [0, +∞), and therefore ([0, 1), B[0,1) ) is standard. Definition 11.5. A topological space (X, τ ) is called a Polish space if it is homeomorphic to a complete, separable metric space. The topology τ is said to be Polish if (X, τ ) is a Polish space. Remark 11.6. Example 11.4 shows that a subspace of a separable, complete metric space (X, d) may be Polish even though it is not complete in the metric d. Such subspaces are precisely the Gδ sets in X. See Problem D. ∞ Example 11.7. .∞ If {(Xn , τn )}n=1 is a sequence of Polish spaces, then the space X = n=1 Xn endowed with the product topology is also a Polish

11 Standard measure spaces

267

space. Indeed, it is easy to see that X is separable. Moreover, if τn is defined by a complete metric dn on Xn , the metric ∞ d((xn )∞ n=1 , (yn )n=1 ) =



2−n min{1, dn (xn , yn )},

∞ (xn )∞ n=1 , (yn )n=1 ∈ X,

n=1

is complete and this metric topology is the product topology. Two important particular cases are the space S = NN of all sequences of positive integers and the Cantor space C = {0, 1}N . Both of these are Polish spaces when endowed with the product topology, where every factor is given the discrete topology. The reason for this terminology is that C is homeomorphic to the standard ternary Cantor set C ⊂ [0, 1] via the continuous bijection f : C → C defined by ∞

2xn , (xn )∞ f ((xn )∞ n=1 ) = n=1 ∈ C. n 3 n=1 (Recall that the inverse of a continuous bijection between compact Hausdorff spaces is necessarily continuous.) The space S is also homeomorphic to a set of real numbers, namely the set [0, 1] \ Q. The identification g : S → [0, 1] \ Q is obtained by use of continued fractions: g((xn )∞ n=1 ) =

1 x1 +

1 x2 +

, 1

..

(xn )∞ n=1 ∈ S.

.

Of course, C is a compact subset of S. The space S is also homeomorphic to a subspace of C via the homeomorphism h((xn )∞ n=1 ) = (0, . . . , 0, 1, 0, . . . , 0, 1, 0, . . . , 0, 1, . . . ). 7 89 : 7 89 : 7 89 : x1 times

x2 times

x3 times

Thus, via Theorem 11.2, C and S are isomorphic as measurable spaces. Another useful Polish space is the Hilbert cube Q = [0, 1]N with the product topology, where [0, 1] is given its usual topology as a subset of R. Theorem 11.8. Let (X, τ ) be a Polish space. Then (1) X is homeomorphic to a subspace of Q. (2) If X is uncountable, then X contains a subspace homeomorphic to C. (3) There exists a continuous surjective map h : S → X. In particular, if X is not countable then it has cardinality c = 2ℵ0 . Proof. Fix a complete metric d defining the topology τ and a dense sequence {xn }∞ n=1 in X. The map f : X → Q defined by f (x) = (max{1, d(x, xn )})∞ n=1

268

11 Standard measure spaces

is easily seen to be a homeomorphism of X onto f (X). To prove (2), assume that X is uncountable. We claim that there exist closed, disjoint, uncountable subsets X0 , X1 ⊂ X each of which has diameter at most 1. Indeed, assume that such sets cannot be found, and let ε be a positive number. Denote by Bn,ε the closed ball of radius ε centered at xn . Then any two uncountable sets of the form Bn,ε must intersect. Therefore the union of all uncountable sets Bn,ε is contained in a closed ball of the form Bmε ,3ε . In other words, X \ Bmε ,3ε is countable for some m. This argument, applied to εk = 4−k , yields ; a ball< so X \ Bk of radius 3εk such that X \ Bk is at most countable, k∈N Bk is at most countable as well. The intersection k∈N Bk consists of at most one point, thus leading to the conclusion that X itself is countable, contrary to the hypothesis. We conclude that there exist closed, disjoint uncountable sets X0 , X1 ⊂ X of diameter at most 1. We can now construct, by induction, for each N ∈ N, closed, uncountable, pairwise disjoint sets Xn1 ,n2 ,...,nN for (n1 , . . . , nN ) ∈ {0, 1}N such that Xn1 ,n2 ,...,nN ⊂ Xn1 ,n2 ,...,nN −1 and the diameter of Xn1 ,n2 ,...,nN is less than 2−N . Indeed, we just apply the preceding observation to the uncountable Polish space Xn1 ,n2 ,...,nN −1 . Once this is done, a function g : C → X such that ∞ 

Xn1 ,n2 ,...,nN = {g(t)},

t = (nj )∞ j=1 ∈ C,

N =1

exists because d is a complete metric. The map g is a homeomorphism from C to g(C). The proof of (3) is similar to that of (2). We construct, for every k ∈ N k and every (n1 , . . . , nk ) ∈ N  , a closed subset An1 ,...,nk = ∅ with diameter at −k most 2 such that X = n1 ∈N An1 and An1 ,...,nk =



An1 ,...,nk ,n ,

k, n1 , . . . , nk ∈ N.

n∈N

The function h : S → X is defined by the requirement that {h(n)} =

∞ 

An1 ,...,nk

k=1

for every sequence n = (n1 , n2 , . . . ) ∈ S.

 

It is convenient in what follows to write Bτ for the Borel σ-algebra of a topological space (X, τ ). Thus (X, Bτ ) is a standard measurable space if (X, τ ) is a Polish space. There are usually different Polish topologies τ, τ  on a set X such that Bτ = Bτ  , and which one is used is immaterial for questions of measure theory. Theorem 11.9. Let (X, S) be a standard measurable space, and fix a set F ∈ S. There exists a Polish topology τ on X such that S = Bτ and F is both open and closed in (X, τ ).

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Proof. Let d be a metric on X that defines a Polish topology σ on X such that S = Bσ . Denote by T the collection of those sets E ∈ S for which there exists a Polish topology τ on X finer than σ such that S = Bτ and E is both open and closed in τ . We prove the theorem by showing that T = S. To do this it suffices to show that T contains all the open sets in σ and that it is closed under the formation of finite intersections and countable unions. Assume first that E is an arbitrary open set in σ. The subspace X \E ⊂ X, endowed with the restriction of d, is again a complete metric space. We now show that E is also a Polish space in the relative topology on E induced by σ. A complete metric d on E is obtained by selecting a dense sequence {xn }∞ n=1 in X \ E and setting d (x, y) = d(x, y) +



2−n |hn (x) − hn (y)|,

x, y ∈ E,

n=1

where

,

1 hn (x) = min 1, d(x, xn )

,

n ∈ N.

We define a metric dE on X by setting ⎧ ⎪ x, y ∈ X \ E, ⎨d(x, y),  dE (x, y) = d (x, y), x, y ∈ E, ⎪ ⎩ max{1, d(x, y)}, x ∈ E, y ∈ X \ E or x ∈ X \ E, y ∈ E. The topology τE on X defined by this metric is Polish, S = BτE , and E is both open and closed in τE . (See Problem 11C.) This argument shows that T contains all open sets in σ. Suppose next that {En }∞ n=1 ⊂ T, and dn is a metric on X defining a Polish topology τn finer than σ such that En and X \ En are open in τn and Bτ = S for all n ∈ N. Define a new metric d on X by setting n

 y) = d(x,



2−n min{1, dn (x, y)},

x, y ∈ X.

n=1

The inequality d ≥ 2−n min{1, dn } shows that the topology τ generated by d is finer than τn for every n. In particular, En and X \ En are open in τ  for all n ∈ N. A sequence {xk }∞ k=1 is Cauchy in the metric d if and only if it is Cauchy in each dn . Since each dn is complete, there exists an ∈ X such that limk→∞ dn (xk , an ) = 0. Moreover, since τn is finer than σ, limk→∞ d(xk , an ) = 0 as well. We conclude that there exists a ∈ X such that  k , a) = 0. Thus d is a complete metric. an = a for all n ∈ N and limk→∞ d(x We show next that the space (X, τ) is second countable. Assume that {Bn,j }∞ j=1 is a base of open sets for τn and note that each Bn,j ∈ S. We show

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that the finite intersections of sets in {Bn,j }∞ . Let n,j=1 form a base for τ G ⊂ X be open in τ. For every x ∈ G, we have  y) < ε} G ⊃ {y ∈ X : d(x, for some ε > 0, and  y) < ε} ⊃ {y ∈ X : d(x,

N 

{y ∈ X : dn (x, y) < ε/2}

n=1

 y) < ε} contains provided that 2−N < ε/2. It follows that {y ∈ X : d(x, ∞ a finite  intersection Wx of sets in {Bn,j }n,j=1 such that x ∈ Wx and thus G = x∈G Wx . Since Bn,j ∈ S for all n, j ∈ N, we conclude that Bτ = S. Moreover, En is both open and closed in τ for n ∈ N. We see immediately that E1 ∩ E2 is both open and closed in τ, and this establishes the fact that T is closed  under the formation of finite intersections. It also follows that the union E = n∈N En is open in τ. The first part of the proof provides a Polish topology τ on X finer than τ such that Bτ = Bτ = S and E is closed as well as open in τ . This shows that T is closed under the formation of countable unions and thus concludes the proof of the theorem.   The last part of the proof of Theorem 11.9 yields the following strengthening of the theorem. Corollary 11.10. Let (X, S) be a standard measurable space, and fix a sequence {Fn }n∈N ⊂ S. There exists a Polish topology τ on X such that S = Bτ and each Fn is both open and closed in (X, τ ), n ∈ N. Corollary 11.11. Let (X, S) be a standard measurable space, and let E ∈ S. Then the measurable space (E, SE ) is standard. Proof. Let τ be a Polish topology on X such that S = Bτ . By Theorem 11.9 we may assume that E is closed in X, and therefore is a Polish space in the relative topology. The conclusion follows now because SE is the Borel σ-algebra of the topology induced by τ on E.   Corollary 11.12. Let (X, S) be a standard measurable space. There exists an injective map f : X → C which is measurable [S, BC ]. Proof. Let τ be a Polish topology on X such that Bτ = S, and let {Gn }∞ n=1 be a base of open sets for τ , so S is the σ-algebra generated by {Gn }∞ n=1 . By Corollary 11.10 there is a Polish topology τ  on X finer than τ such that Bτ  = S and such that the sets Gn are also closed in τ  . If x, y ∈ X and / Gn . Thus the map x = y, there exists n ∈ N such that x ∈ Gn and y ∈ f : X → C defined by f (x) = (χGn (x))∞ n=1 is injective and continuous on (X, τ  ), and hence measurable [S, BC ].

 

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We have seen in Theorem 11.8 that the Cantor space C is homeomorphic to a closed subset of any uncountable Polish space. This implies that (C, BC ) embeds into any standard space (X, S) such that X is uncountable. Proposition 11.13. Let (X, S) and (Y, T) be standard measurable spaces and let f : X → Y be measurable [S, T]. Then there exist Polish topologies σ and τ on X and Y , respectively, such that S = Bσ , T = Bτ , and f : (X, σ) → (Y, τ ) is continuous. Proof. Let τ be a Polish topology on Y such that T = Bτ , and let {Bn }n∈N be a base of open sets for this topology. By Corollary 11.10 there exists a Polish topology σ on X such that S = Bσ and f −1 (Bn ) is open for every n ∈ N. The map f : (X, σ) → (Y, τ ) is obviously continuous.   Theorem 11.14. Let (X, S) and (Y, T) be two standard measurable spaces, and let f : X → Y be a measurable function. Assume that A, B ∈ S and satisfy f (A) ∩ f (B) = ∅. Then there exist disjoint sets C, D ∈ T such that f (A) ⊂ C and f (B) ⊂ D, that is, f (A) and f (B) are separated by measurable sets. Proof. Let d1 and d2 be complete metrics on X and Y that define Polish topologies σ and τ on X and Y , respectively, such that Bσ = S, Bτ = T, and f : (X, σ) → (Y, τ ) is continuous. This is possible by Proposition 11.13. Suppose, to get a contradiction, that f (A) and f (B) cannot be separated by measurable sets. We claim that there exist decreasing sequences A = A(0) ⊃ A(1) ⊃ · · · ⊃ A(n) ⊃ · · · and B = B (0) ⊃ B (1) ⊃ · · · ⊃ B (n) ⊃ · · · of closed sets such that for each n ∈ N, A(n) and B (n) have diameter less than 1/n and f (A(n) ) cannot be separated from f (B (n) ) by measurable sets. (k) These sequences of sets are constructed inductively. Suppose that A  and (k) (n−1) = j∈N Aj B have been constructed for k = 0, . . . , n − 1, and write A  and B (n−1) = j∈N Bj as unions of closed measurable sets Aj and Bj with diameter less than 1/n. If, for all j, k ∈ N, f (Aj ) and f (Bk ) can be separated by measurable sets Cj,k ⊃ f (Aj ) and Dj,k ⊃ f (Bk ), then the sets *∞ + ∞   C= Cj,k , D = Y \ C, j=1

k=1

separate f (A(n−1) ) and f (B (n−1) ), contrary to the inductive hypothesis. Thus there exist j, k ∈ N such that f (Aj ) and f (Bk ) cannot be separated by measurable sets. We complete the inductive process by setting A(n) = Aj and B (n) = Bk .

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11 Standard measure spaces

Since d1 is a complete metric, there exist two unique points x1 ∈

∞ 

A(n) , x2 ∈

n=1

∞ 

B (n) .

n=1

We have f (x1 ) = f (x2 ) because x1 ∈ A, x2 ∈ B, and f (A) ∩ f (B) = ∅. Therefore f (x1 ) and f (x2 ) have disjoint neighborhoods. In other words, there exist p, q ∈ N so Gp ∩ Gq = ∅, f (x1 ) ∈ Gp , and f (x2 ) ∈ Gq . Using the continuity of f at x1 and x2 , we see that f (A(n) ) ⊂ Gp and f (B (n) ) ⊂ Gq for sufficiently large n, showing that f (A(n) ) and f (B (n) ) are separated by measurable sets. This contradiction concludes the proof.   Repeated application of Theorem 11.14 shows that it is possible to separate more than two sets. Corollary 11.15. Let (X, S) and (Y, T) be two standard measurable spaces, and let f : X → Y be a measurable function. Assume that the sets {En }∞ n=1 ⊂ S are such that the sets {f (En )}∞ n=1 are pairwise disjoint. Then there exist pairwise disjoint sets {Fn }∞ n=1 ⊂ T such that f (En ) ⊂ Fn for all n ∈ N. Proof.  By Theorem 11.14, we can select sets Cn ∈ T such that f (En ) ⊂ Cn and f ( m =n Em ) ⊂ Y \ Cn , for n ∈ N, and set Fn = C n ∩



(Y \ Cm ).

m =n

The sequence {Fn }∞ n=1 has the two desired properties.

 

Theorem 11.16. Let (X, S) and (Y, T) be two standard measurable spaces, and let f : X → Y be an injective measurable function. Then for every E ∈ S we have f (E) ∈ T. Proof. For every E ∈ S, the space (E, SE ) is standard, and thus we may, and do, assume with no loss of generality that E = X. By Proposition 11.13, there exist Polish topologies σ and τ on X and Y such that Bσ = S, Bτ = T and f is continuous from (X, σ) to (Y, τ ). Assume thatσ is defined by the ∞ complete metric d on X. Construct a partition X = n=1 An into Borel sets of diameter less than ∞ 1/2. Moreover, for each n1 ∈ N, An1 can be partitioned as An1 = m=1 An1 ,m where each An1 ,m is a Borel set of diameter less than 1/4, and continuing by induction, we construct, for each finite sequence n1 , n2 , . . . , nk of natural numbers, a (possibly empty) Borel set An1 ,n2 ,...,nk of diameter less than 2−k such that the nonempty sets in the collection {An1 ,n2 ,...,nk−1 ,n : n ∈ N} form a partition of An1 ,n2 ,...,nk−1 . The sets {f (An1 ,n2 ,...,nk ) : n1 , n2 , . . . , nk ∈ N} are pairwise disjoint, and Corollary 11.15 yields pairwise disjoint sets {Bn1 ,n2 ,...,nk : n1 , n2 , . . . , nk ∈ N} ⊂ T such that An1 ,n2 ,...,nk ⊂ Bn1 ,n2 ,...,nk , n1 , n2 , . . . , nk ∈ N.

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273

Replacing Bn1 ,...,nk by Bn1 ∩ Bn1 ,n2 ∩ · · · ∩ Bn1 ,...,nk we may also assume that Bn1 ,...,nk ⊂ Bn1 ,...,nk−1 for all n1, n2 , . . . , nk ∈ N and for all k ∈ N \ {1}. The theorem will follow from the equality ⎡ ⎤ ∞   ⎣ (Bn1 ,n2 ,...,nk ∩ f (An1 ,n2 ,...,nk )− )⎦ , f (X) = k=1

n1 ,n2 ,...,nk ∈N

because the right-hand side of the equation is clearly a Borel set. Here we denote by f (An1 ,n2 ,...,nk )− the closure of f (An1 ,n2 ,...,nk ) in the topology τ . To prove this equality, it suffices to show that each point y in the intersection above is in f (X). Indeed, given such a point y, there exists (see Problem 11I) a sequence of integers {nk }∞ k=1 such that y ∈ Bn1 ,n2 ,...,nk ∩ f (An1 ,n2 ,...,nk ),

k ∈ N.

In particular, the sets An1 ,n2 ,...,nk corresponding to these indices are not zero as k → ∞. empty. In addition the diameter of An1 ,n2 ,...,nk tends to ∞ Since d is a complete metric, there is a unique point x ∈ k=1 An1 ,n2 ,...,nk . ∞ Continuity of f at x implies that f (x) ∈ k=1 f (An1 ,n2 ,...,nk ) and also that the diameters of the sets in this intersection tend to zero. We conclude that f (x) = y.   A consequence of the above results is that there are very few equivalence classes of standard measurable spaces with respect to the equivalence relation of isomorphism. Theorem 11.17. Two standard measurable spaces are isomorphic if and only if they have the same cardinality. All uncountable standard measurable spaces are isomorphic to (C, BC ). Proof. It suffices to verify the last statement. Let (X, S) be an uncountable standard measurable space. By Theorem 11.8, (C, BC ) embeds into (X, S), while Theorems 11.12 and 11.16 show that (X, S) embeds into (C, BC ). Theorem 11.2 then applies to yield the desired conclusion.   It was shown in Problem 1Z that analytic sets in a complete metric space X (see Definition 1.24) are precisely the images of continuous functions defined on some other complete metric space Y . (As noted in that problem, the space Y can be taken to be the space S.) The following result allows us to define the concept of an analytic set in an arbitrary standard measurable space. Proposition 11.18. Let (X, S) be a standard measurable space and let A = ∅ be a subset of X. The following assertions are equivalent. (1) There exists a Polish topology τ on X such that S = Bτ and A is analytic in (X, τ ). (2) There exist a standard measurable space (Y, T) and a measurable function f : Y → X such that f (Y ) = A.

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11 Standard measure spaces

(3) There exists a measurable function f : S → X such that f (S) = A. (4) There exist sets {Fn1 ,...,nk : k, n1 , . . . , nk } in S such that  A= (Fn1 ∩ Fn1 ,n2 ∩ · · · ∩ Fn1 ,...,nk ∩ · · · ). (n1 ,n2 ,... )∈NN

Sets A that satisfy these equivalent conditions are said to be analytic in (X, S). Proof. The equivalence of (1) and (4) follows immediately from Definition 1.24 and Corollary 11.10. Indeed the sets {Fn1 ,...,nk : k, n1 , . . . , nk } from condition (4) are closed in some Polish topology τ such that S = Bτ . Similarly, the equivalence between (2) and (3) follows from Problem 1Z and Proposition 11.13 because the map f in (2) is continuous when X and Y are endowed with appropriate topologies. Finally, the equivalence of (2) and (3) follows from Theorem 11.8(3).   For our next result about standard spaces it is useful to consider the lexicographical order on the space S = NN . Given m = (m1 , m2 , . . . ) and n = (n1 , n2 , . . . ) in S, we write m < n if m = n and for the first index j such that mj = nj we have mj < nj . We also write m ≤ n if either m = n or m < n. The set S is totally ordered by this relation, but it is not well ordered as illustrated by the decreasing sequence {nk }k∈N , where nk = (0, . . . , 0, 1, 1, . . . ), 7 89 :

k ∈ N.

k times

Proposition 11.19. Every closed subset A = ∅ of S contains a least element. Proof. Define m1 ∈ N to be the smallest integer with the property that there is some sequence in A with first entry equal to m1 . Then define inductively mk ∈ N for k ∈ N \ {1} to be the smallest integer with the property that there exists a sequence in A starting with m1 , . . . , mk−1 , mk . We show that m = (m1 , m2 , . . . ) is an element of A. Indeed, the definition of this sequence ensures the existence of elements nk ∈ A starting with (m1 , . . . , mk ). Clearly m = limk→∞ nk . The inequality m ≤ n, n ∈ A, follows immediately from the definition of m.   Theorem 11.16 implies that a measurable bijection between standard measure spaces is automatically an isomorphism. The following result is a substitute of this fact for noninjective maps. Theorem 11.20. Let (X, S) and (Y, T) be standard measurable spaces, and let f : X → Y be a measurable map. There exists a function g : f (X) → Y such that f (g(y)) = y for all y ∈ f (X) and, for every A ∈ T, g −1 (A) belongs to the σ-algebra generated by the analytic sets in Y .

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275

Proof. We may, and do, assume that X = ∅. Let σ and τ be Polish topologies on X and Y , respectively, such that S = Bσ , T = Bτ , and f : (X, σ) → (Y, τ ) is continuous. We show first that it suffices to prove the theorem in the special case in which X equals the sequence space S. To do this, choose a surjective continuous function h : S → X. The existence of such a function is guaranteed by Theorem 11.8(3). Then f ◦ h : S → Y is surjective. If the theorem is true for X = S, then there exists a map u : f (X) → S, measurable when f (X) is endowed with the σ-algebra generated by the analytic sets, such that f (h(u(y))) = y for every y ∈ f (X). Therefore the map g = h ◦ u satisfies the conditions in the statement of the theorem. For the remainder of the proof, we assume that X = S. For each y ∈ f (X), the set f −1 ({y}) is closed in S and therefore it has a smallest element g(y). To conclude the proof, it suffices to show that g −1 (F ) is analytic in Y for F in a collection of sets that generates the σ-algebra BS . Such a collection is provided by the open sets Fm = {n ∈ S : n < m},

m ∈ S.

We show that g −1 (Fm ) = f (Fm ), and these sets are analytic by Theorem 11.18. The inclusion g −1 (Fm ) ⊂ f (Fm ) follows from the equality f (g(y)) = y, y ∈ g −1 (Fm ). On the other hand, if y = f (n) for some n ∈ Fm , then g(y) ≤ n < m so g(y) ∈ Fm The fact that the sets Fm generate BS is left as an exercise (see Problem 11O).   Standard measure spaces are also easy to classify. The appropriate concept of isomorphism is as follows. Definition 11.21. Two measure spaces (X, S, μ) and (Y, T, ν) are said to be isomorphic if there exist sets X0 ∈ S, Y0 ∈ T, and a bijection f : X0 → Y0 such that: (1) μ(X \ X0 ) = ν(Y \ Y0 ) = 0, (2) f and f −1 are measurable relative to the σ-algebras SX0 and TY0 , and (3) μ(f −1 (E)) = ν(E) for every E ∈ TY0 . A measurable map f : X → Y (injective or not) satisfying condition (3) is said to be measure preserving. Example 11.22. Endow the Cantor space (C, BC ), where C = {0, 1}N , with the product measure μ = ρ × ρ × · · · , where ρ({0}) = ρ({1}) = 1/2. The map f : C → [0, 1] defined by f ((tn )∞ n=1 ) =



tn , n 2 n=1

(tn )∞ n=1 ∈ C,

is a bijection with range [0, 1] when restricted to the complement of a (countable) set of measure zero [μ]. (The countable set in question consists of those

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11 Standard measure spaces

sequences (tn )∞ n=1 such that tn = 1 for all sufficiently large n.) This map realizes an isomorphism between (C, BC , μ) and ([0, 1], B[0,1] , λ1 |B[0,1] ). Example 11.23. The map h : [0, 1) → [0, 1) defined by

2t, 0 ≤ t < 12 , h(t) = 2t − 1, 12 ≤ t < 1, is measure preserving on ([0, 1), B[0,1) , λ1 |B[0,1) ), but it is not an isomorphism because it is two-to-one. Proposition 11.24. Let (X, S, μ) be a standard measure space and let A ∈ S be an atom for μ. Then there exists x ∈ E such that μ({x}) = μ(A). Proof. Fix a Polish topology τ on X such that S = Bτ , and assume that τ is defined by a complete metric d. We show that for each n ∈ N we can find a set An ∈ S such that An ⊂ An−1 , μ(An ) = μ(A), and the diameter of An is less than 1/n. This follows by induction because, given n ∈ N and supposing that An−1 with the desired properties has been constructed, An−1 can be written as a union of countably many pairwise disjoint Borel sets of diameter less than 1/n. Only one of these sets can have positive ∞ measure to be that set. Then B = (equal to μ(A)) and we define A n=1 (A \ An ) n∞ satisfies μ(B) = 0 and A \ B = n=1 An has diameter zero, hence it consists of a single point x with μ({x}) = μ(A).   Theorem 11.25. Let (X, S, μ) be a σ-finite standard measure space of positive measure such that μ({x}) = 0 for every x ∈ X (that is, such that μ is atom-free). Then (X, S, μ) is isomorphic to ((0, μ(X)), B(0,μ(X)) , λ1 |B(0,μ(X)) ). Proof. Since a σ-finite measure space is isomorphic to a countable direct sum of finite measure spaces, it suffices to prove the theorem for finite measure spaces. Assume for simplicity that μ(X) = 1. The space X cannot be countable, and by Theorem 11.17 we can assume that X = R and S = BR . The function f : R → [0, 1] defined by f (t) = μ((−∞, t)) is easily seen to be monotone increasing and continuous, limt→−∞ f (t) = 0, and limt→+∞ f (t) = 1. Thus the range of f contains (0, 1). Denote by G ⊂ R the open set consisting of those points t ∈ R for which μ((t − ε, t + ε)) = 0 for some ε > 0. We can then write G as the disjoint union of its components  G = (αn , βn ), n

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each of which satisfies μ((αn , βn ]) = 0. Setting  X0 = R \ (αn , βn ], n

we see that μ(X \ X0 ) = 0 and f is a bijection from X \ X0 to (0, 1). Clearly, f |(X \ X0 ) and its inverse are Borel functions, and the equality μ(f −1 (E)) = λ1 (E) is easily verified for intervals E ⊂ [0, 1]. The theorem follows from the   regularity of λ1 (see Proposition 7.10). Isomorphic models for arbitrary standard σ-finite measures spaces are easily deduced from the preceding result. Indeed, if (X, S, μ) is such a space, the set A = {x : μ({x}) > 0} is countable, and the space (X \A, SX\A , μ|(X \A)) is a standard atom free space, so Theorem 11.25 can be applied. See Problem 11N for the precise statement. We turn now to the study of conditional expectations in the context of standard spaces. Assume that (X, S, μ) is a probability space and T ⊂ S is a σ-algebra. Fix, for every E ∈ S, a conditional expectation fE for the function χE relative to T, that is, fE is measurable T and  fE dμ = μ(F ∩ E), F ∈ T. F

Given pairwise disjoint sets E, F ∈ S, the function fE + fF is a conditional expectation for χE∪F relative to T, and therefore the equality fE∪F (x) = fE (x) + fF (x) holds for [μ]-almost every x ∈ X. This equality may not hold for every x ∈ X, so in general we cannot conclude that the map E → fE (x) is finitely additive for any given x ∈ X. This issue can be overcome for standard probability spaces. Theorem 11.26. Let (X, S, μ) be a standard probability space, and let T ⊂ S be a σ-algebra. There exists a map ν : X × S → [0, 1] with the following properties: (1) For each x ∈ X, the map νx (E) = ν(x, E), E ∈ S, is a probability measure on S. (2) For each E ∈ S, the map x → ν(x, E) is a conditional expectation of χE relative to T. If ν  : X × S → [0, 1] is another map with these properties, then νx = νx almost everywhere [μ]. Proof. The result is easily verified when X is countable, so we consider only the uncountable case, and we assume without loss of generality that X = R and S = BR . For every r ∈ Q, fix a Borel function fr : R → [0, 1] which is

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11 Standard measure spaces

a conditional expectation of χ(−∞,r] relative to T, that is, fr is measurable [T] and  fr dμ = μ(E ∩ (−∞, r]), E ∈ T. E

For two rational numbers r ≤ s, the inequality fr (x) ≤ fs (x) must hold for x outside some [μ]-null set Er,s ∈ T. The functions f−∞ (x) = inf fr (x), f+∞ (x) = sup fr (x) r∈Q

r∈Q

are conditional expectations for the constant functions 0 and 1, and therefore set F ∈ T. Redefine f−∞ (x) = 1 − f+∞ (x) = 0 for x outside > =some [μ]-null by setting, for example, the functions fr on the [μ]-null set F ∪ E r≤s r,s

fr (x) =

0, 1,

r < 0, r ≥ 0,

and set gt (x) = inf{fr (x) : r ∈ Q ∩ (t, +∞)},

t, x ∈ R.

Then gt is a conditional expectation of χ(−∞,t] relative to T for every t ∈ R. In addition, for fixed x, the map t → gt (x) is monotone increasing, right continuous, and has limits 0 and 1 at −∞ and +∞, respectively. Therefore there exists a Borel probability measure νx on R satisfying νx ((−∞, t]) = gt (x),

t, x ∈ R.

The function ν(x, E) = νx (E) obviously satisfies condition (1) in the statement of the theorem. We show now that it also satisfies condition (2). Indeed, the collection C of those sets E ∈ S which satisfy (2) is a σ-algebra containing all intervals of the form (−∞, t], and therefore C = S. Assume finally that ν  is another map satisfying properties (1) and (2) (with ν  in place of ν). Given a rational number r, we must have a [μ]-null set Er . It follows that νx = νx νx ((−∞, r]) = fr (x) for x outside  for x outside the [μ]-null set r∈Q Er .   The function ν constructed above is called a system of conditional measures or a disintegration of the measure μ relative to T. Example 11.27. Consider standard measurable spaces (X, S) and (Y, T), and let μ be a probability measure on the product space (X × Y, S × T). The collection C = {E × Y : E ∈ S} is a σ-algebra contained in S × T, and therefore Theorem 11.26 produces numbers νx,y (G) ∈ [0, 1], x ∈ X, y ∈ Y , G ∈ S × T. Since the map (x, y) → νx,y (G) is measurable [C], it follows that νx,y (G) is independent of y ∈ Y . We

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can thus define νx (G) = νx,y (G). The fact that x → νx (G) is a conditional expectation relative to C means that  μ(G ∩ (E × Y )) = νx (G) dμ(x, y), E × Y ∈ C, G ∈ S × T. E×Y

The integral above can be written in terms of the marginal ρ on (X, S) defined by ρ(E) = μ(E × Y ), E ∈ S. Thus,

 μ(G ∩ (E × Y )) =

νx (G) dρ(x),

E ∈ S, G ∈ S × T.

E

Consider now the marginal measures σx : T → [0, 1] defined by σx (F ) = νx (X × F ), Given E ∈ S and F ∈ T, we have

x ∈ X, F ∈ T.





μ(E × F ) = μ((X × F ) ∩ (E × Y )) =

νx (X × F ) dρ(x) = E

σx (F ) dρ(x). E

This extends easily to  μ(G) =

σx (Gx ) dρ(x),

G ∈ S × T,

X

so μ can be viewed as a generalized product measure. The following result is used in the construction of probability measures, other than product measures, on infinite product spaces. The result, due to Kolmogorov, does not hold for arbitrary measure spaces. .∞ Theorem 11.28. Let (X, S) = n=1 (Xn , Sn ) be a product of standard measurable spaces, and let TN ⊂ S, N ∈ N, be the σ-algebra consisting of sets of the form ∞ N # # E× Xn , E ∈ Sn . 

n=1

n=N +1

Assume that μ : T = N ∈N TN → [0, 1] is a set function such that μ|TN is countably additive for all N ∈ N. Then μ is countably additive on T. Proof. We may assume that each Xn is a complete metric space and Sn = E ∈ S and a sequence BXn . Suppose to the contrary that there exist a set  ∞ {En }∞ n=1 of pairwise disjoint sets in S such that E = n=1 En but α = μ(E) −



n=1

μ(En ) > 0.

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∞ n−1 The sets An = m=n Em = E \ m=1 Em belong to T, An ⊃ An+1 for n ∈ N, α. We obtain a contradiction by showing that these conditions and μ(An ) ≥ ∞ imply that n=1 An = ∅. Indeed, Theorem 7.3 applied to the measure μ|TN implies the existence of compact sets Kn ⊂ An such that μ(An \ Kn ) < 2−n α for all n ∈ N. Note that An \ (K1 ∩ · · · ∩ Kn ) ⊂ (A1 \ K1 ) ∪ · · · ∪ (An \ Kn ), and thus μ(An \ (K1 ∩ · · · ∩ Kn )) ≤

n

α < α. 2n

k=1

It follows that ∞the intersections K1 ∩ · · · ∩ Kn are not empty and therefore ∞ A ⊃ n n=1 n=1 Kn = ∅ since the Kj have the finite intersection property.   Theorem 11.20 has a more convenient form in the context of standard measure spaces. Theorem 11.29. Let (X, S) be a standard measurable space, let (Y, T, ν) be a σ-finite standard measure space, and let f : X → Y be a surjective measurable map. There exist a set E ∈ T and a measurable function h : Y \ E → X such that ν(E) = 0 and f (h(y)) = y for every y ∈ Y \ E.  Proof. There is a partition Y = n∈N Yn such that for each n ∈ N, Yn ∈ T, ν(Yn ) < +∞, and Yn is either a finite atom or ν|Yn is atom free. It suffices to prove the theorem for each of the maps fn = f |f −1 (Yn ). Equivalently, we can restrict ourselves to the case in which ν(Y ) < +∞ and either Y is a finite atom or νY is atom free. If Y is a finite atom, simply pick y ∈ Y such that ν({y}) = ν(Y ), set E = Y \ {y}, and define h(y) to be any point in f −1 (y). Suppose therefore that 0 < ν(Y ) < +∞ and ν is atom free. By Theorem 11.25, we may assume that Y = (0, ν(Y )), T = BY , and ν = λ1 |BY . Let g : Y → X be a function that satisfies the conditions in Theorem 11.20, let {An }n∈N ⊂ S be a sequence of sets which generates the σ-algebra S and set Bn = f (An ), n ∈ N. It follows from Proposition 5.33 that each set Bn is measurable [λ1 ], so there exist Borel sets Cn and Dn such that Cn ⊂ Bn ⊂ Dn ⊂ Y and ν(Dn \Cn ) = 0. To conclude the proof, we set E = n∈N (Dn \ Cn ) and h = g|(Y \ E). The measurability of h follows because h−1 (An ) = Bn ∩ (Y \ E) = Cn ∩ (Y \ E) is a Borel set for every n ∈ N.

 

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281

Problems 11A. Show that the metric d defined on a product space in Example 11.7 is complete. Also verify that the functions f, g, h of that example are homeomorphisms onto their ranges. 11B. Show that the map g defined in the proof of Theorem 11.8 is indeed a homeomorphism. 11C. Verify that the metric d defined on the open set E in the first part of the proof of Theorem 11.9 is complete, and show that it induces the same topology as the original metric d. 11D. Let (X1 , d1 ) and (X2 , d2 ) be metric spaces with X2 complete, let A ⊂ X1 be an arbitrary set, and let f : X1 → X2 be a continuous map. Denote by A1 the set of those points x ∈ A− with the following property: for every ε > 0 there exists δ > 0 such that d2 (f (y), f (z)) < ε for any points y, z in X1 satisfying d1 (y, x) < δ and d1 (z, x) < δ. (i) Show that A1 is a Gδ set in X1 . (ii) Show that f extends to a continuous function f1 : A1 → X2 . (iii) Suppose that A is a Polish space with the induced topology. Show that A is a Gδ set in X1 . (Hint: consider the identity map on A.) (iv) Conversely, show that every Gδ set in a Polish space is a Polish space in the induced topology. 11E. Recall that S = NN is endowed with a Polish topology which is the product topology obtained when each factor N is given the discrete topology. (i) Show that every compact subset K ⊂ S has empty interior in S. (ii) Let X = ∅ be a Polish space such that every compact subset K ⊂ X has empty interior in X. Assume, in addition that X is totally disconnected, that is, the topology of X has a base consisting of sets which are closed as well as open. Show that X is homeomorphic to S. (iii) Let X = ∅ be a compact, totally disconnected Polish space without isolated points. Show that X is homeomorphic to C. 11F. (Cantor-Bendixson) Given a Polish space X, denote by A ⊂ X the set consisting of those points x ∈ X which have a countable open neighborhood. Show that A is open and at most countable. Show that the set X \ A has  no isolated points (and hence is a perfect set), and it is equal to the intersection card(α)≤ℵ0 X α , where X α is defined inductively for countable ordinals α as follows: X 0 = X, and if X α has been defined, then X α+1 is the set of those x ∈ X α which  are not isolated in X α . Moreover, if α is a limit ordinal, then X α is defined as β 0 for every n ∈ X2 .

11O. Given k, a1 , . . . , ak ∈ N, show that the set Va1 ,...,ak = {n = (n1 , . . . ) ∈ S : n1 = a1 , . . . , nk = ak } belongs to the algebra generated by the sets Fm used in the proof of Theorem 11.20, with m of the form (m1 , . . . , mp , 1, 1, . . . ) for some p, m1 , . . . , mp ∈ N. (The sets Va1 ,...,ak form a base for the topology of S.) 11P. Let (X, S) be a standard measurable space, and let A ⊂ X be an analytic set such that X \ A is also analytic. Show that A ∈ S. 11Q. Let X be a separable metric space and let {Un }n∈N be a base of open sets in X. (i) Show,that the set  F = (x, n) ∈ X × S : x ∈ ∞ k=1 (X\Uk ), n = (n1 , n2 . . . ) is closed in X ×S and that every closed subset of X is equal to some S-section F n of F . (The S-sections are defined before the statement of Theorem 8.3.) (ii) Let F ⊂ S × S × S be a closed subset such that the collection {F n : n ∈ S} contains all the closed subsets of S × S. Define a set A ⊂ S × S by A = {(p, n) : there exists p ∈ S such that (p, q, n) ∈ F }. Show that A is analytic in S × S and that every analytic subset of S is equal to some S-section An of A. 11R. Suppose that (X, S) is an uncountable standard measure space. Show that there exists an analytic set A ∈ S × S such that every analytic subset of X is equal to some section Ay of A. (i) Deduce that the set B = {x ∈ X : (x, x) ∈ A} is analytic in X but its complement X \ B is not analytic. (Hint: By Problem 11P, B ∈ / S.) (ii) Show that the cardinality of the collection of analytic sets in X equals c = 2ℵ0 . 11S. Let (X, S) and (Y, T) be standard measurable spaces, let A be an arbitrary subset of X, and let f : A → Y be measurable [SA , T]. Show that there exists a measurable function g : X → Y such that g|A = f . (Hint: Consider first the case of a function with countable range. Assume that Y is a separable complete metric space and approximate f uniformly by countably valued functions fn . The set where the corresponding functions gn have a limit as n → ∞ is measurable.)

11 Standard measure spaces

283

11T. Let (X, S) and (Y, T) be standard measurable spaces, let A be an arbitrary subset of X, let B be an arbitrary subset of Y , and let f : A → B be an isomorphism of (A, SA ) to (B, TB ). Show that f can be extended to an isomorphism from (A1 , SA1 ) to (B1 , TB1 ), where A ⊂ A1 ∈ S and B ⊂ B1 ∈ T. (Hint: Apply Problem 11S to f and f −1 to obtain extensions g1 , g2 and consider the set {x ∈ X : g1 (g2 (x)) = x}.) 11U. Suppose that a function f : [0, 1] × [0, 1] → R is such that the map x → f (x, y) is Borel measurable for every y ∈ [0, 1] and the map y → f (x, y) is continuous for every x ∈ [0, 1]. Show that f is Borel measurable. (Hint: f (x, y) = limn→∞ f (x, 2−n [2n y]), where [2n y]denotes the integer part of 2n y. Here we use [x] to denote the integer part of x, that is, [x] is an integer satisfying [x] ≤ x < [x]+1.) 11V. Denote by X the space [0, 1][0,1] of all functions g : [0, 1] → [0, 1] endowed with the topology of pointwise convergence, so X is a compact Hausdorff space. Define fn : [0, 1] → X, n ∈ N, by (fn (x))(y) = max{0, 1 − n|x − y|},

x, y ∈ [0, 1].

Show that each fn is a continuous function but the pointwise limit f of {fn }n∈N is not Borel measurable. 11W. Let (X, S) be a measurable space. (i) Suppose that there exists a sequence {An }n∈N ⊂ S that generates S and separates the points of X, that is, for every x, y ∈ X such that x = y there is n ∈ N such that x ∈

An and y ∈ / An . Show that (X, S) embeds into ([0, 1], B[0,1] ). (Hint: f = n∈N 3−n χAn .) (ii) Suppose that (X, S) is standard and {An }n∈N ⊂ S is a sequence that separates the points of X. Show that {An }n∈N generates S. 11X. Show that the σ-algebra generated by the sets {x}, x ∈ [0, 1], is not countably generated. (Thus a sub-σ-algebra of a countably generated σ-algebra (namely, B[0,1] ) need not be countably generated.) 11Y. (Dieudonn´ e) Consider the standard measure space ([0, 1], T, λ1 |T) where T = B|[0, 1], and let A ⊂ [0, 1] be a set that is not Lebesgue measurable (Example 5.27). (i) Show that there exist sets A+ , A− ∈ T such that A− ⊂ A ⊂ A+ , λ1 (A+ \ A− ) > 0, and for every B+ , B− ∈ T satisfying B− ⊂ A ⊂ B+ we have λ1 (A+ \ B+ ) = λ1 (B− \ A− ) = 0. The set A+ can be chosen to be a Gδ set and A− can be chosen to be an Fσ set. (ii) Fix sets A+ and A− as in (i) and denote by S the σ-algebra generated by T ∪ {A}. Show that S consists of all sets of the form E ∪ (F ∩ A) ∪ G where E, F, G ∈ T satisfy E ⊂ A− , F ⊂ A+ \ A− , and G ⊂ [0, 1] \ A− . (iii) Define μ : S → [0, 1] by setting μ(E ∪ (F ∩ A) ∪ G) = λ1 (E) + λ1 (F ) + λ1 (G), where E, F, G are as in (ii). Show that μ is a probability measure. (The measure μ is a restriction of the outer measure λ∗1 used to define Lebesgue measure.) (iv) Show that there does not exist a disintegration of μ relative to T. (Hint: Suppose such a disintegration ν : [0, 1] × S → [0, 1] exists. Show that there exists a set E ∈ T such that λ1 (E) = 0 and ν(x, F ) = δx (F ) for every x ∈ [0, 1] \ E and F ∈ S. Derive a contradiction when F = A. As usual, δx denotes the unit point mass at x.)

284

11 Standard measure spaces

11Z. Show that there exist subsets {An }n∈N of [0, 1] with the following properties. (i) An is not Lebesgue measurable, n ∈ N.  (ii)  If E ⊂ [0, 1]\ n A k=1 k is a Borel set, then λ1 (E) = 0. (iii) n∈N An = ∅. (Hint: The equivalence relation ∼ on [0, 1] defined by the requirement that s ∼ t precisely when t − s ∈ Q has countably infinite equivalence classes. Write each equivalence class E as E = {t1E , t2E , . . . } in such a way that tnE − tnF is constant modulo 1 for any two equivalence classes E, F . Define Bn = {tnE : E an equivalence class} and set An = [0, 1]\Bn .) 11AA. Let {An }n∈N be a sequence of sets satisfying properties (a-c) of Problem 11Z. For each n ∈ N, set Xn = [0, 1] and let Sn be the σ-algebra generated by B[0,1] and An . Using the notation of Theorem 11.28, define a set function μ : T → [0, 1] as follows. For every set E ∈ T, define DE = {t ∈ [0, 1] : (t, t, . . . ) ∈ E} and μ(E) = λ∗1 (DE ) = inf{λ1 (F ) : F ∈ B[0,1] , F ⊃ DE }. Show that μ|Tn is countably additive for each n ∈ N but μ is not countably additive on T. (Hint: The  sets En = A1 × · · · × An × X × X × · · · satisfy En ⊃ En+1 , μ(En ) = 1, and μ( ∞ n=1 En ) = 0.) 11BB. Show that there exists a Borel set E ⊂ [0, 1] × [0, 1] such that the set {y ∈ [0, 1] : there exists x ∈ [0, 1] such that (x, y) ∈ E} is not Borel. (Hint: Replace [0, 1] by S and consider the graph of a continuous function f : S → S whose range is not a Borel set.)

Subject index

Abel summability, 248 absolute continuity, 78 -uniform, 100 absolutely continuous measure, 75 additivity, 52, 185 algebra, 73 -associative, 208 -of continuous functions, 102 -of sets, 1 σ-algebra, 3, 4, 19, 20, 30, 39, 45, 51, 53, 55, 58, 69, 108, 114, 122, 129, 130, 143, 162, 196, 199, 200, 236, 270, 274, 279, 284 -Borel, 131, 167, 186 -countably generated, 283 -of sets, 1 -product, 183 algebra of functions, 36, 39 almost everywhere, 80, 88 almost everywhere [μ], 80 almost surely, 80 analytic, 17 analytic set, 8, 118, 273, 274, 282 -in a standard measurable space, 274 arc, 251 arithmetic mean, 162 atom, 53, 276 -finite, 53 -infinite, 53, 197, 209, 214 average, 252 Baire -first category, 18 -second category, 18 -third category, 18 Baire category theorem, 13

Baire function, 32, 168 Baire measure, 168 Baire set, 167 ball, 17, 145, 155, 156, 161, 192, 200, 227 -closed, 173, 268 -unit, 205 Banach space, 204, 206, 214, 220, 234, 237 -complex, 204, 221 base of open sets, 269, 282 Bernstein, 29 Bernstein approximation, 181 Bessel inequality, 222, 243 bijection, 105, 200 -continuous, 267 -measurable, 274, 282 binary expansion, 18 binomial coefficient, 29 Boolean homomorphism -complete, 3 Borel measure, 146, 151, 161, 168, 172, 178, 180, 181, 251 -complex, 259, 260 -moment of, 176 -positive, 180 Borel separated, 17 Borel set, 4, 6, 148, 157, 159, 276, 280 Borel σ-algebra, 183, 266, 268, 270 Borel-Lebesgue measure, 187 calculus, 50 Cantor function, 152 Cantor set, 7, 73, 79, 116, 132, 162, 267 Cantor space, 267, 275 Carath´ eodory outer measure, 123, 126, 130

© Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1

285

286 Carath´ eodory procedure, 125, 130, 131 cardinal number, 10 cardinality, 180, 198, 199, 267, 282 Catalan number, 176 Cauchy criterion, 47, 68 Cauchy property, 70, 223 Cauchy-Schwarz inequality, 176 cell, 112, 183 -closed, 39, 122 -nonoverlapping, 112 cellular partition, 112 Ces` aro summability, 261 change of variables formula, 189 character -of a group, 263 characteristic function, 95 closed strip, 22 closure, 168, 179, 273 collection -countable, 68 -empty, 52 -pairwise disjoint, 53 combinatorial coefficient, 38 common refinement, 91 compact set, 119 compact space, 15 complement, 26, 36, 72 complete metric, 271, 272 completeness, 98, 220 completion, 99 -of a measure, 99, 187 complex conjugation, 33, 34 complex measure, 78, 78 -absolutely continuous, 77 complex measure space, 75 complex plane, 26 complexification, 50, 64 composition, 23, 34 concentration, 128 conditional expectation, 143, 143, 162, 277, 282 conjugate exponent, 213, 230, 261 conjugation, 208 connected component, 159 continued fraction, 267 continuity, 42, 43 -absolute, 76 -uniform, 76 continuum, 12 convergence -almost everywhere, 80, 88, 90, 100, 101, 209, 238, 258, 262 -in measure, 101, 234, 237 -in the mean, 44, 90, 100, 102

Subject index -pointwise, 17, 28, 32–44, 51, 56, 59, 61, 62, 64, 68, 70, 100, 242, 283 -sequential, 37 -unconditional, 222 -uniform, 29, 38, 40, 59, 70, 89, 90, 101, 167, 179, 224, 251, 253, 258 convolution, 254, 263 -of functions, 243 coset, 81 countable additivity, 76 counting measure, 53, 65, 100, 122, 129–131, 161, 186, 197, 235, 243, 262 covering, 111, 117 -countable, 126 -open, 15, 121 cycle, 241 cylinder, 196, 199, 200 decreasing rearrangement -of a function, 128 definition -transfinite, 6 dense, 224 density, 176 density point, 158, 161 derivative, 18, 153, 154, 159, 189, 262 determinant, 189 diagonal, 197, 199 diameter, 15, 17, 118, 131, 173, 190, 268, 273, 276 difference -symmetric, 10 difference quotient, 152 differentiability, 42 dilation -of a function, 256 dimension -Hausdorff, 132 Dirac measure, 127 direct sum, 276, 282 -of measure spaces, 200 Dirichlet kernel, 244, 255 discontinuity, 153, 159 discrete measure, 65 discrete topology, 267, 281 disintegration, 278, 283 disjointification, 122 distance, 26 distribution -joint, 127 -of a numerical function, 127 distribution function, 193 distributive, 200 doubleton, 3

Subject index dual, 219, 234, 237 -topological, 234 dyadic fraction, 35 elementary figure, 14, 132 epicycle, 241 equality -almost everywhere, 80, 140, 144 equidistributed sequence, 103 equivalence class, 116 equivalence relation, 116, 144, 284 essential supremum, 97, 213 essentially bounded function, 97 Euclidean space, 10, 73, 187, 221 -of dimension d, 131 Euler characteristic, 132 eventually, 26 existence -almost everywhere, 153 extreme point, 237 -of a convex set, 237 F -space, 204, 234 face of a convex set, 237 family -indexed, 6, 46, 47, 49, 52, 67, 72 -empty, 53 -of nonnegative extended real numbers, 65 -summable, 47, 68 -monotone increasing, 41 -real, 47 -sum of, 47 -summable, 47, 48, 50 -uniformly absolutely continuous, 77 -uniformly integrable, 77 Fej´ er kernel, 261 finitely additive set function, 69, 73 fixed point, 37, 39 Fourier coefficient, 225, 242, 250, 262 -of a measure, 259 Fourier expansion, 224, 242 Fourier inversion formula, 253, 255, 257, 258, 263 Fourier series, 242, 249, 261 Fourier transform, 253, 257 -of a function in L2 (λ1 ), 258 -of a measure on R, 260 -on Z/nZ, 262 fractional part, 103 function -Baire, 33, 35 -Euler beta and gamma, 192 -Lipschitz, 18, 160

287 -Riemann integrable, 51, 94 -absolutely continuous, 76, 104, 115, 132, 150, 152 -α-H¨ older, 18 -bounded, 30, 70, 72, 95 -bounded away from zero, 70 -characteristic, 21, 27, 30, 31, 51, 56, 62, 95, 102, 128, 184, 186, 251 -complex, 27 -complex-valued, 22, 23, 26–28, 35, 40, 70, 71, 77, 95, 161 -Baire, 33 -integrable, 65, 82, 104 -measurable, 29, 37, 43, 45, 67, 70, 80 -constant, 29, 34, 35, 278 -almost everywhere, 199 -continuous, 18, 31–33, 39, 50, 102, 162, 224 -continuous almost everywhere, 95 -continuously differentiable, 95 -convex, 152, 162 -continuous, 154 -strictly, 152 -coordinate, 23 -countably determined, 179 -countably-valued, 282 -defined almost everywhere, 93 -defined everywhere, 88 -differentiable, 18, 257, 261 -almost everywhere, 150 -continuously, 251 -discontinuous, 111, 241 -elementary symmetric of degree k, 132 -essentially bounded, 97, 212 -everywhere defined, 98 -extended real-valued -measurable, 82 -nonnegative, 57 -infinitely differentiable, 262 -integrable, 44, 45, 50, 51, 58, 60, 66–68, 72, 73, 78, 82, 96, 97, 102, 154, 157, 160, 168, 186, 197, 219 -with respect to L, 44, 50 -with respect to μ, 57 -integrable [μ], 57, 70, 71, 80, 84, 88, 90, 131 -integrable [μ] over E, 57 -locally bounded, 40 -measurable, 23, 26, 27, 31, 67, 78, 83, 144, 160, 193, 199, 271, 272, 282 -Borel, 35, 36, 92, 93, 186, 230, 283 -Lebesgue, 191 -complex-valued, 58, 59, 80 -injective, 272

288 -nonnegative, 98 -real-valued, 72 -measurable [μ], 88 -measurable [μ], 81, 85, 88 -monotone, 36 -monotone decreasing, 261 -monotone increasing, 151, 153, 163 -continuous, 159 -multiplicative, 189 -nonnegative, 50, 62, 63, 67, 72, 131, 168, 178 -almost everywhere [μ], 82, 84, 85 -measurable, 72 -of Baire class α, 32, 39 -of Baire class less than α, 40 -of Baire class one, 31, 39 -of Baire class two, 32, 39 -of bounded variation, 137, 158 -of finite variation, 162, 259 -oscillation of, 92 -periodic, 241 -positive, 157 -real-valued, 21, 27, 29, 31, 35, 50, 51, 56, 61–63, 67, 78, 94 -bounded, 90 -continuous, 35 -extended, 36, 83–86 -integrable, 64, 79, 97 -measurable, 24, 37, 43, 50 -monotone, 23 -monotone increasing, 110 -right continuous, 113 -semicontinuous, 31 -right-continuous, 128 -scalar-valued, 27, 33, 35, 39, 40, 98, 101 -continuous, 33 -measurable, 36, 39, 97 -semicontinuous, 22, 93 -simple, 27, 27, 28, 31, 56, 60, 62, 73, 85, 93, 99, 128, 139, 186, 217, 219, 233 -integrable, 72, 232 -measurable, 56, 60, 63, 70, 160, 185 -strictly convex, 162 -total variation of, 137, 162 -truncation of, 228 -uniformly continuous, 30 -variation of, 159 function algebra, 27, 31, 35 -complex, 27 -real, 27 function lattice, 27 functional, 48, 58, 65, 71 -linear, 47, 50, 219 -continuous, 234

Subject index -monotone increasing, 50, 61 -positive, 67, 73 -real, 50 -real-valued, 61 -self-conjugate, 61–63, 71 -zero, 44 gauge, 109, 109, 110, 114, 125, 126, 129, 130 -countably additive, 197 -countably subadditive, 111–113, 119, 131 -extendible, 117, 121, 170 -finitely additive, 117 Gaussian measure, 176 generalized Cantor set, 73 geometric mean, 162 group -additive, 262 -locally compact -Abelian, 263 Hahn decomposition, 140, 141 Hamburger moment problem, 176 Hamel basis, 222 Hankel matrix, 177 harmonic mean, 162 Hausdorff dimension, 132 Hausdorff measure, 132 Hausdorff measure of dimension p, 130 Hausdorff moment problem, 181 Hausdorff outer measure, 125 Hausdorff outer measure of dimension p, 130 Hausdorff-Young inequality, 243, 262 Hilbert cube, 267 Hilbert space, 222, 238 -complex, 221, 235 -linearly isometric, 223 H¨ older’s inequality, 213 homeomorphism, 267, 268, 281 homomorphism -Boolean, 15 ideal, 16, 36, 70, 71, 130 -generated, 16 σ-ideal, 16, 42, 44, 57, 80, 98, 171 image measure, 127 improper integral, 255 inclusion-exclusion formula, 132 indefinite integral, 77, 78, 87, 89, 90, 98, 100, 101 -concentrated on sets of finite measure, 98

Subject index index family -countable, 101 -countably determined, 101 index set -countably determined, 101 induction, 23 inductive construction, 11 inequality -Markov, 97 -Tchebysheff, 97 infinite series, 45 inner measure, 130 inner product, 221 inner regular set function, 169 integrable, 44 integrable function, 44 integral, 43, 46, 50, 60, 64, 71, 83, 191 -Cauchy, 50 -Darboux, 90, 95 -lower, 91, 93 -upper, 91 -Lebesgue, 43, 44, 46, 48, 50, 51, 56, 58–60, 62, 64, 65, 68, 69, 72, 91, 105 -complex, 43, 45, 50, 60, 64, 66, 67 -real, 43, 50, 51, 67 -restriction of, 58 -Lebesgue-Borel, 96 -Riemann, 51, 94, 95, 102 -improper, 96 -indefinite, 77 -iterated, 197 -with respect to μ, 57 Lebesgue, 1 integrand, 95 integration, 85 -by parts, 165 -theory, 81, 85 -with respect to a measure, 57 interior, 113 interior point, 18, 161 interpolation, 231 intersection, 54, 107, 268 interval, 23, 90, 95, 282 -closed, 54, 129 -compact, 152, 190 -finite, 150 -half-open, 54, 72, 111, 122 -open, 169 -finite, 145 inverse image, 21, 22, 28, 40 isomorphism -of measurable spaces, 265, 283 -of measure spaces, 275 iterated integral, 161, 197

289 Jacobi identity, 262 Jensen inequality -generalized, 162 Jordan decomposition, 137, 139 jump discontinuity, 151 Korovkin, 29 L0 (μ), 208 L∞ (μ), 212 Lp,q (μ), 226 Lp (μ), 212 ∞ (Γ ), 216 p (Γ ), 216 λ-system, 7 L0 (X, S, μ), 208 L0 (μ), 208 L∞ (μ), 212 Lp (μ), 212 lattice, 13, 14, 16, 40, 44, 159 -complemented, 13, 40 -generated, 13 -of elementary figures, 132 -of sets, 13, 119 σ-lattice, 14 -generated, 14 least element, 274 Lebesgue constants, 247 Lebesgue decomposition, 144, 146, 151 Lebesgue integral, 1, 43, 168 Lebesgue measure, 115, 128, 131 Lebesgue outer measure, 112, 126, 188 Lebesgue point, 158 Lebesgue-Borel measure, 66, 73, 79, 85, 92, 96, 122, 186, 193 -d-dimensional, 115 Lebesgue-Stieltjes measure, 115 level set, 24 lexicographical order, 274 limit, 88, 90, 93, 100, 156 -pointwise, 61, 86, 101, 195, 283 -sequential, 92 -uniform, 261 limit inferior -of a sequence of sets, 69 limit superior -of a sequence of sets, 69 linear combination, 27, 31, 78, 224 linear functional, 45, 50, 60–62, 67, 73, 81, 169, 177, 205 -complex, 43 -continuous, 218 -positive, 73, 121, 170, 178, 180, 181 -self-conjugate, 56

290 linear manifold, 18, 60, 69, 228, 231, 232, 235 -dense, 225, 235, 243, 254 -of measurable functions, 66 -trivial, 58 linear map -continuous, 205 linear transformation, 43, 188, 189 -invertible, 189, 200 Lorentz space, 226 lower bound, 91 lower envelope -of a function, 93 manifold -linear, 27 map -continuous, 17 -diffeomorphism, 189 -identity, 281 -injective, 270 -measurable, 274 -surjective, 280 -subadditive, 228 -surjective, 267 mapping, 23, 36, 39, 131 -Baire, 41 -constant, 21, 41 -continuous, 22, 23, 34, 39 -coordinate, 37, 41 -elementary, 38 -measurable, 42 -identity, 39 -measurable, 21, 23, 36–39, 70 -Borel, 37 -net of, 37 -of Baire class α, 41 -onto, 39 -simple, 38 -measurable, 39 -with property (MSR), 41 Markov’s inequality, 97 mass, 53 maximal, 235 maximal element, 178 maximal function, 226, 231 mean, 38 measurability, 1, 19, 24, 73, 175 measurable -Borel, 21–23, 35, 42 -direct sum, 36 -mapping, 20 -subspace, 20 -with respect to, 19

Subject index measurable cylinder, 194 measurable rectangle, 160, 183, 184–186, 197, 198 measurable space, 75, 78, 105, 107, 133, 135, 140, 144, 159, 162, 163, 198, 265, 268, 270, 283 -embedded, 265 -isomorphic, 265 -product, 183 -standard, 266 measure, 51, 52, 53, 55–58, 64, 69, 71, 76, 80, 84, 97, 99, 105, 194, 198 -Baire, 168 -Borel, 168, 170 -finite, 173 -inner regular, 169 -outer regular, 170 -signed, 159 -Borel-Lebesgue, 66 -d-dimensional, 66 -Dirac, 58, 127, 259 -Gaussian -standard, 176 -Hausdorff -of dimension p, 130 -Lebesgue, d-dimensional, 115 -Lebesgue-Stieltjes, 115 -absolutely continuous, 143, 251 -absolutely continuous [μ], 75 -associated, 59, 64, 65 -associated with L, 56 -complete, 107, 108 -completely additive, 53, 66 -completion, 99 -complex, 75, 133, 142, 144, 160, 162 -concentration of, 79 -counting, 65, 122, 131 -defined by an outer measure, 107 -discrete, 53, 53, 65, 71, 100 -equivalent, 144, 236 -finite, 60, 140, 173, 184, 185, 211, 219 -finite-valued, 53 -finitely additive, 54–56, 69, 73 -inner regular, 171, 172 -marginal, 279 -normalized arclength, 242 -outer, 105 -outer regular, 170 -positive, 180 -problem, 198 -product, 183 -push-forward, 127 -real, 135 -representing a functional, 170

Subject index -restriction of, 59 -semi-finite, 53, 197, 198 -semicircle, 176 -semifinite, 209, 213 -σ-finite, 131, 144, 163, 184, 211, 220, 234 -σ-finite, 187 -signed, 75, 79, 133, 139 -finite-valued, 78 -singular, 144, 151 -translation invariant, 113, 131 -variation of, 136 -negative, 136 -positive, 136 measure algebra, 98 measure preserving map, 275 measure problem, 198 measure space, 51, 52, 58–60, 70–72, 75–80, 84–88, 90, 97–101, 104, 108, 110, 127, 131, 143, 162, 197, 210, 213, 214, 217, 220, 221, 225, 232, 235, 236, 238 -atom free, 199 -complete, 98, 130, 209 -complex, 75 -finite, 53, 70, 72, 90, 236 -isomorphic, 282 -product, 185 -σ-finite, 130, 142, 162, 185, 187, 193, 199, 200, 210, 219, 235, 237, 276 -signed, 75 mesh, 15, 93 metric, 207, 266, 269, 281 -complete, 18, 267, 268, 273, 281 -product, 37, 41, 42 -relative, 20 -space, 39 metric space, 4, 6, 118, 123, 125, 126, 130, 131, 281 -complete, 98, 173, 269, 273, 282 -separable, 130–132, 173, 282 -complete, 118, 173, 266 metrizable, 237 minimal element, 103 Minkowski inequality, 216, 235 moment, 198 moment of a measure, 176 moment problem, 176 -Hamburger, 176 -Hausdorff, 181 -Stieltjes, 180 -trigonometric, 180 moment sequence, 176

291 multiplicity -of a root, 177 negative part, 24, 44, 56 negative variation of a measure, 136 negligible, 85 neighborhood, 40, 205, 252, 281 neighborhood base -countable, 234 net, 37, 47, 91, 145, 156, 206 -Cauchy, 47 -bounded, 101 -convergent, 47, 204 -of functions -monotone increasing, 101 -real-valued, 101 nonnegative definite, 177 norm, 204 normalized arclength measure, 242 normed space, 204, 205, 206, 220 null set, 79, 85, 98, 101, 144 number -Catalan, 176 -cardinal, 129 -ℵ0 , 87 -ideal, 19 -integer -positive, 88 -irrational, 85 -normal, 18 -ordinal, 6, 12 -countable, 12, 32, 34, 35, 40, 41, 171, 281 -even, 12, 14 -infinite, 12 -odd, 12, 14 -positive, 58, 60, 72, 97 -rational, 18, 35, 54, 110 -positive, 148 -real, 22, 25, 36, 54 -extended, 49, 82, 87, 146, 156 -fractional part of, 103 -nonnegative, 67 -positive, 130 open cube, 161 order -inductive, 178 order topology, 171, 179, 196 ordinal -limit, 179 ordinal number, 6, 12 -countable, 14

292 orthogonal -vector, 221 orthogonal complement, 235 orthonormal basis, 222, 223, 235, 236, 262 orthonormal sequence, 237 orthonormal set, 222 outer measure, 105, 107, 108, 110, 111, 114, 117, 119, 120, 128–130, 170 -Carath´ eodory, 123 -Hausdorff -of dimension p, 130 -Lebesgue, 112, 123 -concentration of, 106 -generated by a gauge, 109, 130, 131, 197 -restriction of, 106 outer regular set function, 169 π-system, 7 p-norm, 212 parallelogram identity, 238 Parseval’s identity, 222, 257 partial sum, 48 partition, 3, 4, 10, 36, 54, 68, 72, 77, 79, 91–93, 107, 131, 135, 220 -Borel, 15 -cellular, 112 -countable, 15, 36 -finite, 134 -indexed, 5, 20 -measurable, 140 partition of unity, 121 permutation matrix, 201 point mass, 53, 58, 176, 260 Poisson kernel, 249 -for R, 253 polar coordinates, 192 polarization identity, 221 Polish space, 266, 268, 281 -totally disconnected, 281 Polish topology, 268, 270–273, 275, 276, 282 -finer, 269 polynomial, 24, 29, 102, 177, 180, 181 -Bernstein, 29, 181 -Laurent, 180 -complex, 24, 261 -nonconstant, 199 -real, 24, 31, 102 -nonnegative, 177 -trigonometric, 103, 260 positive extension, 178 positive linear functional, 170 positive part, 24, 44, 56

Subject index positive variation of a measure, 136 power, 23 power class, 3, 16, 98, 128 power series, 102 preimage, 162 probability -theory, 38 probability measure, 261, 277, 283 -Borel, 278 probability space, 53, 128, 162, 194, 199, 200, 216 product -countable infinite, 41 -of measure spaces -infinite, 194 -of probability spaces, 199 product measure, 183, 185, 187, 197, 198 product σ-algebra, 196 product space, 281 product topology, 168, 171, 179, 207, 266, 267, 281 projection, 37 quasi-seminorm, 203 quasinorm, 203, 211, 216, 234, 236 quasinormed space, 204 quasiring, 7 -σ-quasiring, 7, 122, 185 quotient space, 81 Radon-Nikod´ ym density, 162 Radon-Nikod´ ym derivative, 140, 145, 236 rearrangement, 52 rectangle, 23 -measurable, 183 refinement, 92 regularity, 157, 277 relatively compact set, 121 restriction, 39 Riemann -integral, 51 Riemann-Lebesgue lemma, 243, 254 ring, 1, 10–13, 55, 70, 75, 108, 117, 172, 194, 195 -Boolean, 11 -coarsest, 4 -complemented, 1 -direct sum, 4, 15 -countable, 5 -finite, 5 -full, 5, 20 -finer, 3 -strictly, 5 -generated, 10

Subject index -of sets, 117 -σ-ring, 1, 10, 12, 36, 72, 120, 194 -complemented, 36 -generated, 6 -trivial, 3, 5 σ-algebra, 1 ring measure, 117 rotation invariance, 242 scalar field, 47 scalar product, 221, 221, 238 Schwarz inequality, 221 X- or Y -section of a function, 184 X- or Y -section of a set, 184 semicircle measure, 176 semicontinuity, 142 semimodularity, 120 seminorm, 204, 206, 234 sequence, 25, 28, 29, 37, 61 -Cauchy, 51, 63, 269 -Cauchy in the mean, 44, 45, 48, 51, 62–64, 68, 70, 88 -bounded, 206 -locally uniformly, 40 -bounded above, 51 -complex, 46 -convergent, 26, 27, 51 -dense, 269 -equiconvergent, 64 -equidistributed, 103 -increasing, 57 -monotone increasing, 28, 36, 51, 55, 72 -numerical, 26, 64, 69 -bounded, 51 -of bounded variation in the mean, 88 -of complex-valued functions, 59 -of functions, 31, 88, 99, 101, 179, 210, 234, 238 -Cauchy in measure, 210 -Cauchy in the mean, 100 -complex-valued, 90 -convergent almost everywhere, 88 -convergent in the mean, 88 -monotone, 25 -monotone increasing, 51 -monotone increasing almost everywhere, 85, 88, 93 -pointwise bounded, 25 -real-valued, 100 -of indefinite integrals -equicontinuous from above, 90 -of intervals, 113 -of measurable sets -decreasing, 69

293 -increasing, 69 -of partitions -nested, 93 -of real numbers, 181 -of real-valued functions, 60 -of sets -convergent, 69, 98 -decreasing, 134, 136 -increasing, 10, 184 -metrically nested, 130 -monotone increasing, 69, 78 -of simple functions, 63 -pairwise disjoint, 56 -summable, 180 -transfinite, 6, 34 -monotone increasing, 6 -two-way infinite -of real numbers, 72 -unbounded, 85, 206 -uniformly absolutely continuous, 90 -uniformly concentrated on sets of finite measure, 90 sequence space -c0 , 220 series, 52, 76 -absolutely convergent, 50, 52, 72, 237, 248 -convergent, 45, 102 -unconditionally, 52 -divergent, 83 -infinite, 45 -multiple, 224 -numerical -convergent, 87 -partial sum of, 96 -telescoping, 88 set, 25 -Borel, 4, 6, 9, 14, 18–20, 28, 35, 73, 76, 85, 118, 131, 188, 190, 196, 273 -Cantor, 7, 41, 95, 116, 132, 152 -generalized, 73 -Fσ , 6, 9, 13, 73, 283 -Gδ , 6, 73, 281 -compact, 168, 172 -almost equal [μ], 98, 99 -analytic, 8, 118 -at most countable, 53, 195, 198 -bounded, 109, 130 -closed, 4, 6, 9, 13, 17, 22, 26, 173, 174 -unbounded, 171 -compact, 119, 148, 170, 227, 280 -convex, 200 -congruent, 16 -convex, 125, 234, 237

294 -countable, 3, 20, 72 -dense, 35 -difference, 6, 11 -directed, 36, 46, 145 -disjoint, 20 -empty, 13, 123 -exceptional, 93 -finite, 3, 45, 47, 49, 235 -index, 46 -level, 21 -limit, 16 -meager, 18 -measurable, 19, 21, 23, 24, 27, 36, 37, 39, 52, 55, 57, 59–62, 70, 71, 78, 83, 84, 88, 97, 107, 117, 119, 129, 130, 135, 140, 158, 184, 188, 200 -Borel, 163 -Lebesgue, 189, 190 -measurable with respect to an outer measure, 107 -nonmeasurable, 21 -of finite measure, 89 -of full measure, 98 -open, 4, 6, 14, 18, 20, 22, 23, 26, 29, 31, 35, 40, 73, 119, 121, 124, 148, 157, 162, 170, 173, 174 -perfect, 281 -relatively compact, 121 -σ-finite, 71, 72, 197 -with respect to μ, 71 -splitting, 15 -totally bounded, 119, 173 -totally ordered, 274 -unbounded, 179 -uncountable, 53, 168, 179, 199, 234, 268, 282 -well ordered, 274 set function, 51, 56, 57, 69, 77, 109 -complex-valued, 51 -additive, 69 -concentrated on sets of finite measure, 77 -countably additive, 52, 55, 75, 107, 139, 194, 219, 279, 284 -countably subadditive, 52, 55, 56, 105 -extended real-valued, 51 -additive, 52 -family -uniformly concentrated on sets of finite measure, 77 -finitely additive, 52, 57, 70, 71, 75, 78, 108, 120, 173, 180, 194, 235 -regular, 173 -finitely subadditive, 105

Subject index -increasing, 174 -inner regular, 169 -modular, 120 -monotone, 55, 55, 56, 105, 130 -outer regular, 169, 174 -real-valued, 51 -semicontinuous, 55, 56, 57, 69 -subadditive, 174 -subtractive, 55 -variation of, 138 signed measure, 135 signed measure space, 75 singleton, 20, 36, 53, 66, 71, 72, 79 -not measurable, 199 singular measure, 144 space, 55, 130 -Banach, 204 -Euclidean, 10, 73 -Lebesgue, 97 -compact, 15 -complete, 205 -continuous functions, 169 -discrete, 168 -measurable, 19, 20–24, 26–28, 30, 35–39, 41, 44–46, 50–53, 55, 56, 58, 67, 69 -infinite, 36 -measure -complex, 75 -semi-finite, 54 -signed, 75 -metric, 4, 6, 13–15, 17–23, 32–34, 36, 38–41, 110, 125, 130, 204 -complete, 279 -countable, 7 -discrete, 7 -product, 37 -separable, 15, 37, 39, 41, 70, 237 -σ-compact, 38, 39 -metrizable, 204, 234 -normed, 204 -of all sequences of complex numbers, 204 -of bounded continuous functions, 167 -of compactly supported continuous functions, 167 -of continuous functions, 18, 50 -of functions that vanish at infinity, 167 -quasinormed, 204 -quotient, 81 -self-conjugate, 44, 50 -σ-finite, 71 -topological, 7, 167, 196 -Hausdorff, 21, 168, 173

Subject index -Polish, 266 -locally compact, 167, 171 -normal, 173 -second countable, 269 -totally disconnected, 281 standard deviation, 38 standard measurable, 273, 282 standard measurable space, 266, 268, 270–273, 278, 279, 282 standard measure space, 274, 276, 283 -σ-finite, 280, 282 -uncountable, 282 standard probability space, 277 standard representation, 70 Stieltjes moment problem, 180 strong type, 227, 232 subspace -of a measure space, 59, 84 sum -Darboux, 91 -lower, 91, 92 -upper, 91 -Riemann, 94 -empty, 46 -indexed, 46, 65, 101, 129 -divergent, 83 -partial, 48, 254 -convergent, 48 summability, 47 summable family, 47 summand, 4, 112 support, 56, 60, 70–72, 96 -closed, 167 -finite, 180 -relatively compact, 121 -σ-finite, 72, 82, 96 surjective, 219 Suslin operation, 8, 17, 118 symmetric difference, 10, 98 symmetry, 84, 140, 191 system of conditional measures, 278 tail σ-algebra, 200 Tchebysheff inequality, 97, 226 the interval, 242 theorem -Baire category, 13 -Baire functions form an algebra, 34 -Baire functions of class α form an algebra, 35 -Banach-Alaoglu: the dual unit ball is weak* compact, 207 -Banach-Steinhaus on uniform boundedness, 205

295 -Beppo-Levi, series with positive terms, 86 -Borel functions are Baire on a metric space, 35 -Borel separation of disjoint sets, 271 -Borel sets are analytic, 9 -Borel sets are open in a different Polish topology, 268 -Cantor-Bernstein, 116 -Cantor-Bernstein for measurable maps, 265 -Egorov:pointwise convergence implies uniform convergence on a large set, 59 -Fatou’s lemma, 88, 99 -Fubini and Tonelli: double vs. iterated integrals, 185 -Hahn decomposition of signed measures, 135 -Hardy-Littlewood maximal, 227 -Heine-Borel, 113 -Jensen’s inequality, 154 -Jordan decomposition of a signed measure, 136 -Kre˘ın-Milman, 237 -Lebesgue differentiation in dimension d, 157 -Lebesgue differentiation in one dimension, 150 -Lusin:a measurable function has a continuous restriction to a large set, 169 -Marcinkiewicz interpolation, 228 -Polish spaces as subspaces and quotient spaces, 267 -Rademacher: Lipschitz functions on Rn are differentiable almost everywhere, 163 -Radon-Nikod´ ym, 140 -Riesz representation, 169 -Riesz representation for normal spaces, 174 -Riesz-Fisher: Lp is complete, 214 -Riesz-Thorin interpolation, 232 -Stone-Weierstrass, 102 -Tonelli for complete measures, 187 -Urysohn’s lemma, 168 -Weierstrass approximation, 29 -Weierstrass approximation for trigonometric polynomials, 103 -Zorn’s lemma, 178 -analytic sets are measurable, 118 -any Lebesgue integral yields a measure, 56

296 -bounded convergence, 90 -change of variable, 189 -classification of standard measure spaces, 276 -convergence of the integrals of uniformly integrable functions, 88 -disintegration:existence of conditional measures, 277 -dominated convergence, 90 -every measure determines a Lebesgue integral, 60 -existence and uniqueness of product measure, 184 -existence of generated ring, 3 -existence of measurable selections, 280 -existence of probability measures on infinite products, 279 -existence of right inverses, measurable relative to the analytic sets, 274 -function algebras determine σ-algebras, 29 -inductive construction of generated σ-ring, 6 -injective image of a measurable map between standard spaces is measurable, 272 -measurable functions form an algebra, 27 -measures that agree on a generating collection are equal, 122 -monotone class, 8 -monotone convergence, 85 π-λ, 8 -product of infinitely many probability spaces, 194 -solution of the Hamburger moment problem, 177 -uncountable standard spaces are isomorphic, 273 topological dual, 205 topological space, 167 topological vector space, 203, 204, 205, 238 -Hausdorff, 204 -infinite dimensional, 207 topology, 6, 13, 14, 17, 203, 204, 234, 266, 281 -Polish, 266 -induced, 281 -of convergence in measure, 209 -relative, 269, 270 -weak, 206 -weak*, 206 total variation, 161, 162, 260

Subject index -finite, 235 -of a complex measure, 135 total variation of a function , 137 trace, 15, 20, 122 transfinite construction, 12, 14 transfinite definition, 6, 32 transfinite induction, 6, 35, 40 transfinite procedure, 41 transformation, 200 -linear, 43 translate -of a function, 225, 236, 256 triangle inequality, 216 trigonometric moment problem, 180 trigonometric polynomial, 103 truncate, 87 truncation, 80, 238 uniform integrability, 77, 88 uniform limit, 179 union, 56, 129, 184, 219 -countable, 1 -finite, 54 unit circle, 242 unit cube, 224 unit square, 186 upper bound, 25, 160, 178 -least, 211 upper envelope -of a function, 93 value, 28 variation of a measure, 136 vector space, 44, 45, 47, 64, 70, 71, 81, 98, 132 -complex, 50, 67, 208, 221 -of bounded functions, 73 -of continuous functions, 121 -of functions, 43 -of integrable functions, 56 -of measurable complex-valued functions, 68 -of measurable functions, 51, 203 -of real polynomials, 177 -real, 50, 51, 67, 237 -topological, 203 Vitali covering, 146, 155, 157 Vitali covering lemma, 155 volume, 190 -of a ball, 192 weak Lp (μ), 226 weak topology, 206, 237 weak type, 227, 231

Subject index weak* topology, 206, 235, 237 Weierstrass, 29 weight, 53, 71 weights, 65

297 Young inequality, 245, 257 zero functional, 58 zero measure, 58

Notation index

(A, SA ): measurable subspace, 20 (a, b), (a, b], [a, b): intervals in R, 3 A, B, E, F : sets, 1 A \ B: set difference, 1 A × B: Cartesian product, 23 ℵ0 : the cardinality of N, 3 ℵα : the αth cardinal number, 3 A, B, C, S, T: collections of sets, 1 B(p, q): the Euler beta function, 192 Bn : Bernstein polynomial, 29 BX : Borel σ-algebra, 4 B(V, W): space of continuous linear maps, 205 BaX : Baire σ-algebra, 167 B(x, r): ball of radius r centered at x, 155 C: the set of complex numbers, viii C(X): the algebra of complex continuous functions on X, 18 Cα : Baire class α, 32 Cc (X): the algebra of complex, compactly supported continuous functions, 167 c = 2ℵ0 : the cardinality of the continuum, 3 ck (μ): moment of μ, 176 card(X): the cardinality of X, 3 χE : characteristic function of E, 27 C0 (X): the algebra of complex continuous functions that vanish at infinity, 167 Cb (X): the algebra of bounded complex continuous functions on X, 167 C{θn } : generalized Cantor set, 73 DN : Dirichlet kernel, 244 δx : unit point mass at x, 58 det: .determinant, . . 176 ⊕[ , (f ) , (ω) ]: direct sum [full, finite, countable], 5

dμ/dν: Radon-Nikod´ ym derivative, 140 DP (f ), dp (f ): upper and lower Darboux sums, 91 E(f < a), E(a < f ≤ b): Lebesgue sets, 22 EF : symmetric difference, 18 E, E − : the closure of E, 18 ∅: empty set, 2 ess sup: essential supremum, 97 f : the complex conjugate of f , 33 |f |p : norm of f in Lp (μ), 208 f , μ : Fourier transforms, 241 f  : the derivative of a function f , 18 f ∗ g: convolution, 241 f ∨ g: the maximum of two functions, 24 f ∧ g: the minimum of two functions, 24 f + , f − : positive and negative parts, 24 fx , f y , Ex , E y : sections of f, E, 184 Γ (p): the Euler gamma function, 192 γ:  gauge, 109 ∩[ ]: intersection [of a family], 2 f dμ: integral of f , 57 X Kr : Fej´ er kernel, 261 L: a Lebesgue integral, 43 Λα : the collection of α-H¨ older functions, 18 p : space of p-summable sequences, 97 lim inf, lim sup: lower and upper limits, 25 L: the domain of a Lebesgue integral, 43 λd : Lebesgue measure on Rd , 66 Lp (μ), Lp (X, S, μ): Lebesgue space, 97 Lp (μ), Lp (X, S, μ): space of classes of p-integrable functions, 208 M f : maximal function, 226 M⊥ : orthogonal complement, 235

© Springer International Publishing Switzerland 2016 H. Bercovici et al., Measure and Integration, DOI 10.1007/978-3-319-29046-1

299

300 μ, ν: measures, 53 μ ⊥ ν: mutually singular measures, 144 μ × ν: product measure, 184 μ± : positive and negative variations, 136 |μ|: total variation measure, 136 m: normalized arclength measure, 241 MR [MC ]: algebra of real [complex] measurable functions, 27 μ∗ : outer measure, 105 μE : concentration of μ, 79 μd : discrete part of μ, 151 mesh(P): mesh of a partition, 92 Nf = E(f = 0): support of f , 56 N = {1, 2 . . . }: the set of natural numbers, viii N0 = {1, 2 . . . }: the set of nonnegative integers, viii νf : indefinite integral, 77 Ω: the smallest uncountable ordinal, 6 2X : power class (all subsets of X), 3 P r : Poisson kernel, 249 : Cartesian product of sets or infinite product of numbers, 37 ψ ◦ ϕ: composition of maps, 23 ϕ−1 : inverse function or inverse image under a function, 3 ϕ  μ: absolute continuity, 75 Q: the set of rational numbers, viii Q(C): quasiring generated by C, 7 Qσ (C): σ-quasiring generated by C, 7

Notation index Q: Hilbert cube, 267 R: the field of real numbers, viii R : the extended real line, 19 Rd : d-dimensional Euclidean space, 10 R(C): ring generated by C, 5 S(C): σ-algebra generated by C, 5 S × T: product σ-algebra, 183 SA : trace σ-algebra, 20 S = NN : the space of sequences of natural numbers, 17 sup, inf: least upper bound, greatest lower bound, 24 T: the unit circle, 180 T |: operator norm, 205  ∪[ ]: union [of a family], 1 V ∗ : topological dual, 205 W (α): initial segment of ordinals, 34 (X, S), (Y, T): measurable spaces, 19 [x]: integer part, 283 x, y: scalar product, 221 ξd∗ : Hausdorff outer measure, 126 {x}: singleton, 3 x ⊥ y: orthogonal vectors, 222 (X, S, μ): measure space, 59 |x|2 : Euclidean norm of x ∈ Rd , 189 Y X : collection of all functions f : X → Y , 41 z, z: real and imaginary parts of z, 22 Z: the set of integers, viii