A Primer of Lebesgue Integration, Second Edition

A Primer of Lebesgue Integration Secortd Editiort

H.S. Bear

Department of Mathematics University of Hawaii at Manoa Honolulu, Hawaii

ACADEMIC PRESS A Harcourt Science and Technology Company San Diego New York Boston London Sydney Tokyo Toronto

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This text is dedicated to J. L. Kelley, who taught that in mathematics it is not enough to read the words-you’ve got to hear the music.

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CONTENTS Preface to the First Edition ........................... Preface to the Second Edition ..........................

..

v11

xi

1. The Riemann-Darboux Integral .................... 1 2 . The Riemann Integral as a Limit of Sums ............ 9 3. Lebesgue Measure on (0. 1) ....................... 21

4 . Measurable Sets: The Carathgodory Characterization ................................

27

5 . The Lebesgue Integral for Bounded Functions ....... 43 53 6 . Properties of the Integral ......................... 7. The Integral of Unbounded Functions .............. 61 8 . Differentiation and Integration .................... 73 9. PlaneMeasure .................................. 85 10. The Relationship between /A. and A ................. 93 11. General Measures .............................. 107 12. Integration for General Measures ................. 117 13. More Integration: The Radon-Nikodym Theorem ... 127 14. Product Measures .............................. 135 15 . The Space L2 .................................. 149 Index .............................................

V

161

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PREFACE TO THE FIRST EDITION This text provides an introduction to the Lebesgue integral for advanced undergraduates or beginning graduate students in mathematics. It is also designed to furnish a concise review of the fundamentals for more advanced students who may have forgotten one or two details from their real analysis course and find that more scholarly treatises tell them more than they want to know. The Lebesgue integral has been around for almost a century, and the presentation of the subject has been slicked up considerably over the years. Most authors prefer to blast through the preliminaries and get quickly to the more interesting results. This very efficient approach puts a great burden on the reader; all the words are there, but none of the music. In this text we deliberately unslick the presentation and grub around in the fundamentals long enough for the reader to develop some intuition about the subject. For example, the Carathkodory definition of measurability is slick-even brilliant-but it is not intuitive. In contrast, we stress the importance of additivity for the measure function and so define a set E E ( 0 , l ) to be measurable if it satisfies the absolutely minimal additivity condition: m(E) rn(E’) = 1, where E’ = ( 0 , l ) - E and rn is the outer measure in ( 0 , l ) . We then show in easy steps that measurability of E is equivalent to the Carathkodory criterion, m(E n T ) m(E’ f~ T ) = m(T)for all T . In this way we remove the magic from the Carathkodory condition, but retain its utility. After the measure function is defined in (0, l),it is extended to each interval (n,n 1) in the obvious way and then to the whole line by countable additivity.

+

+

+

vi i

viii

A PRIMER OF LEBESGUE INTEGRATION

We define the integral via the familiar upper and lower Darboux sums of the calculus. The only new wrinkle is that now a measurable set is partitioned into a finite number of measurable sets rather than partitioning an interval into a finite number of subintervals. The use of upper and lower sums to define the integral is not conceptually different from the usual process of approximating a function by simple functions. However, the customary approach to the integral tends to create the impression that the Lebesgue integral differs from the Riemann integral primarily in the fact that the range of the function is partitioned rather than the domain. What is true is that a partition of the range induces an efficient partition of the domain. The real difference between the Riemann and Lebesgue integrals is that the Lebesgue integral uses a more sophisticated concept of length on the line. We take pains to show that both the Riemann-Darboux integral and the Lebesgue integral are limits of Riemann sums, for that is the way scientists and engineers tend to think of the integral. This requires that we introduce the concept of a convergent net. Net convergence also allows us to make sense out of unordered sums and is in any case something every young mathematician should know. After measure and integration have been developed on the line, we define plane outer measure in terms of coverings by rectangles. This early treatment of plane measure serves three purposes. First, it provides a second example of the definition of outer measure, and then measure, starting with a natural geometric concept-here the area of a rectangle. Second, we show that the linear integral really is the area under the curve. Third, plane measure provides the natural concrete example of a product measure and is the prototype for the later development of general product measures. The text is generously interlarded with problems. The problems are not intended as an intelligence test, but are calculated to be part of the exposition and to lure the reader away from a passive role. In many cases, the problems provide an essential step in the development. The step may be routine, but the reader is nevertheless encouraged thereby to pause and become actively

PREFACE TO THE FIRST EDITION

ix

involved in the process. There are also additional exercises at the end of each chapter, and the author earnestly hopes that these will add to the reader’s education and enjoyment. The author is pleased to acknowledge the help of Dick Bourgin, Bob Burckel, and Ken ROSS,all of whom read the manuscript with great care and suggested many improvements in style and content.

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PREFACE TO THE SECOND EDITION

The principal change from the first edition is the new one-shot definition of the Lebesgue integral. The integral is first defined for bounded functions on sets of finite measure, using upper and lower Darboux sums for finite partitions into measurable sets. This approach is designed to emphasize the similarity of the Lebesgue and Riemann integrals. By introducing countable partitions, we then extend the definition to arbitrary functions (bounded or not) and arbitrary sets (finite measure or not). This elegant touch, like many of my best ideas, was explained to me by A. M. Gleason. Many of the errors and crudities of the first edition have been corrected, and the author is indebted to Robert Burckel, R. K. Getoor, K. P. S. Bhaskara Rao, Joel Shapiro, and Nicholas Young for pointing out assorted mistakes. In addition, several anonymous reviewers of the second edition made many helpful suggestions. I feel confident, however, that there remain enough errors to challenge and reward the conscientious reader. Finally, the author wishes to express his gratitude to Susan Hasegawa and Pat Goldstein for their superb work with the typing and proofreading.

H. S. Bear May 2001

xi

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THE RIEMANN-DARBOUX INTEGRAL

We start by recalling the definition of the familiar RiemannDarboux integral of the calculus, which for brevity we will call the Riemann integral. Our later development of the Lebesgue integral will closely parallel this treatment of the Riemann integral. We consider a fixed bounded interval [a, b] and consider only real functions f which are bounded on [a,61. Apartition P of [a, b]isaset P = {xg,X I , x2,. . . , &}ofpoints of [a, b] with

Let f be a given function on [a, b] with

for all x

E

[a, b].For each

i = 1 , 2 , . . . , n let

In the usual calculus text treatment the infs and sups are taken over the closed intervals [ x i - l , xi]. In our treatment of the Lebesgue integral we will partition [a, b]into disjoint sets, so we use the disjoint sets (xi-1, xi) here. We are effectively ignoring a , finite set of function values, f(xo), f(x1), f(xz), . . . , f ( ~ ) and in so doing we anticipate the important result of Lebesgue that

2

A PRIMER OF LEBESGUE INTEGRATION

function values on a set of measure zero (here the set of partition points) are not relevant for either the Riemann or Lebesgue integral. The lower sum L( f, P ) and the upper sum U ( f, P ) for the function f and the partition P are defined as follows: n

n i=1

Clearly m 5 mi 5 Mi 5 M for each i, so

m(b - a ) I U f ,PI I W f ,PI I M ( b - a). For a positive function f on [a,b] the lower sum represents the sum of the areas of disjoint rectangular regions which lie within the region

S = {(x,y ) : a 5 x 5 b,O 5 y 5 f
P

(1)

The integral of f over [a, b] is the common value in (1) and is denoted s,b f . The area of S is defined to be this integral whenever f is integrable and non-negative. The integral is also sometimes written S,b f ( x )dx,particularly when a change of variable is involved. The “x” in this expression is a dummy variable and can be replaced by any variable except f or d. For example,

The last two versions are logically correct, but immoral, since they flout the traditional roles of a , b, c as constants, and the

1

THE RIEMANN-DARBOUX INTEGRAL

3

third integral discourteously uses the same letter for the limit and the dummy variable. Now we make a few computations to derive the basic properties of the integral and show that the integral exists at least when f is continuous. A partition Q is a refinement of the partition P provided each point of P is a point of Q. We will indicate this by writing P + Q without reference to the numbering of the points in P or Q. Clearly Q is a refinement of P provided each of the subintervals of [a,b] determined by Q is contained in one of the subintervals determined by P . Proposition 1. I f P + Q, then L( f, P ) 5 L( f, Q)and U ( f, Q ) i W f , PI. Proof. Suppose Q contains just one more point than P , and to be specific assume this additional point x* lies between the points Q and x1 of P . If

mi = inf{ f ( x ) : xo < x < x*)

my = inf{ f ( x >: x" < x < XI} ml = inf{ f ( x ) : xo < x < XI}, then wil z ml and wii 2 ml so the sum of the first two terms in L( f, Q) exceeds the first term of L( f, P ) :

m;(x* - Q)

+ m'i(x1 - x*) 2 ml(xl - xo).

The remaining terms of L( f, Q) and L( f, P ) are the same, so L( f, Q) 2 L( f, P ) . We can consider any refinement Q of P as obtained by adding one point at a time, with the lower sum increasing each time we add a point. The argument for the upper sums is similar. 1111111 Proposition 2. Every lower sum is less than or equal to every upper sum, as the geometry demands.

Proof. If P and Q are any partitions, and R = P U Q is the common refinement, then

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A PRIMER OF LEBESGUE INTEGRATION

Proposition 3. f is integrable on [a, b] if and only if for each E > 0 there is a partition P of [a,b] such that U ( f, P ) U f ,P ) < E .

Proof. This useful condition is equivalent to sup L( f, P ) = inf U ( f, P ) in view of Proposition 2. 1111111 Proposition 4. I f f is integrable on [a,b] and [a,B ] then f is integrable on [a,83.

c [a,61,

Proof. Let E > 0 and let P be a partition of [a,b] such that U ( f, P ) - L( f, P ) < 1.We can assume that a and /3 are points of P , since adding points increases L( f, P) and decreases U ( f, P ) and makes their difference smaller. If Po consists of the points of P which are in [a,B ] , then Po is a partition of [a,B ] . Note that

u(f, P > - L( f,

n

=C

( M i - mi)(xi - xi-1)

i=l

(2)

and U ( f, Po) - L( f, Po) is the sum of only those terms such that x;-l and xi E Po. Since we omit some non-negative terms from (2)to get f, Po) - L( f, Po),

w

U ( f, Po) - L( f, Po) 5 U ( f, P ) - L( f, P ) < E . 1111111

Problem 1. If f is integrable on [a,b],then -f and I f l are integrable on [a,61, and S,b(-f) = -S,"f, 5 S,b 1 f l . 1111111

IJ:fl

Problem 2. If a < c < b then f is integrable on [a,c] and on [c,b] if and only if f is integrable on [a,b].In this case

Problem 3. If f is integrable on [a,b] and g = f except at a finite number of points, then g is integrable and s,b g = s,b f . lilllll Problem 4. If a = Q < x1 < . . < x,, = 6 and f is defined on [a,61 with f ( x > = yi for x E ( ~ ~ - 1x ,i ) , then f is integrable and .r,b f = Cy=lyi(xj - xi-1). (Note that it is immaterial how f is defined on the xi, by the preceding problem.) 41

1

THE RIEMANN-DARBOUX INTEGRAL

5

Problem 5. We say that g is a step function on [a,61 if there is a partition a = x0 < xl < x2 < - .. < x, = 6 such that g is constant on each (xi-1,xi). (By the preceding problem, step functions are integrable with the obvious value for the integral.) Show that if f is integrable on [a, 61 there are step functions g, and h, with {g,} increasing and {h,} decreasing and g, If 5 h, for all n and all x, and lim J‘ g, = lim J‘ h, = J ‘ f . (Note: We will be able to show 1at“er that g, 2nd h, approach f except possibly on a set of measure zero.) 111111

Proposition 5 . I f f is continuous on [a,61, then f is integrable on [a,61. Proof. If f is continuous on [a,61, then f is uniformly continuous. Hence if E > 0 there is S > 0 so that If(x)- f(x’)I < I whenever 1x - x’I < 6. If P is a partition with xi - xi-1 < 6 for all i , then Mi - mi 5 E for each i, so

u(f, PI - L( f, PI = C ( M i - mi>(xi- xi-1)

Problem 6. (i) If f is continuous on [a,61 except at a (or 6) and f is bounded on [a,b],then f is integrable on [a,b]. (ii) If f is bounded on [a,61 and continuous except at a finite number of points, then f is integrable on [a,61. (iii) Suppose f is bounded on [a,61 and discontinuous on a (possibly infinite) set E . Assume that for each E > 0 there are disjoint intervals ( a l , 61),. . . , ( a N , 6 N ) contained in [a,61 such that E c (al,61)u . . . u ( a N , b N ) and C;”=1(6i - ai) < E . Show that f is integrable. ~~lllll So far we have the integral J‘ f defined only when [a,b] is a bounded interval and f is bounded on [a,61. Now we extend the definition to certain improper cases; i.e., situations where the interval is unbounded, or f is unbounded on a bounded interval. Typical examples of such improper integrals are

/‘Ldx 0&

and

/wL 0 1+x2 dx.

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A PRIMER OF LEBESGUE INTEGRATION

In both these examples the integrand is positive and the definition of the integral should give a reasonable value for the area under the curve. The definitions of the integrals above are

/' 1dx O

f

i

= lim &-+O+

d x = lim 6-m

1'

$dx,

/ b dx. L 0 1+x2

Both these limits are finite, so both functions are said to be (improperly) Riemann integrable on the given interval. The Lebesgue definition of the integral will give the same values. In general, if f is integrable on [a E , b] for all E > 0, but f is not bounded on [a,61 (i.e., not bounded near a),we define

+

provided this limit exists. Similarly, if f is integrable on every interval [ a ,b] for b > a we define

when the limit exists. Similar definitions are made for S,b f if f is unbounded near 6, and for s_", f if f is integrable on [a,b] for all a < 6. These definitions lend themselves to the calculations of elementary calculus, but do not coincide with the Lebesgue definition if f is not always positive or always negative. For example, if f is (-l)'/n on [ n , n + l), n = I,&. . . , then f is improperly Riemann integrable on [l,00). We will see later that f is Lebesgue integrable if and only if I f I is Lebesgue integrable. Hence the above function is not Lebesgue integrable since C 1 = 00.

Problem 7. Show that Jp F d x exists.

1111111

Problem 8. (i) Exhibit a g on [0,00) such that Ig(x)l = 1 and Jpg exists.

1

THE RIEMANN-DARBOUX INTEGRAL

7

(ii) Exhibit a function g on [0,00) such that Ig(x)l -+ 00 as x -+ 00 and J r g exists. (iii) Exhibit a function g on [ O , o o ) so that Ig(x)l -+ 00 and J r g = 0. Hint: Don't think about formulas for continuous functions, think about step functions. 1111111

Problem 9. Let f ( x )= x2on [ O , l ] and let P, = (0, i, i,. . . , l}. Write formulas for L( f, P,) and for U ( f, P,). Show that

Jt x 2 d x = 3.1

41Il(

Problem 20. Write out the proof that U ( f, Q) 5 U ( f, P ) whenever P 4 Q. 1111111

Problem 2 2 . If f is integrable on [a,b] and c is a number, then s,b cf = c s,b f . Show this for the case c < 0. 41 Problem 22. If f and g are integrable on [a,b] and f 5 g on [a,b],then :J f 5 s,b g . 1111111 Problem 23. Let x v y = max{x, y} and x A y = min{x, y}. For functions f and g , ( f v g)(x) = f ( x ) v g(x), ( f A g)(x) = f ( x )A g(x). Show that if f and g are integrable on [a,b],then f v g and f A g are integrable, and s,b f v g 3 s,b f v Jjg, .r,b f A g 5 s,b f A .r,b g. 1'1111

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THE R I E M A N N INTEGRAL AS A LIMIT OF SUMS The engineers and scientists who work with integrals think of the integral as a limit of sums:

where a = Q < x1 < x2 < . . < x, = 6 is a partition of [a,b] and Axi = xi - xi-1. Historically the integral sign is a flattened S for sum, and the “dx” suggests the typical (small) quantity Axi. The limit is the number that is approached as partitions are made finer and finer so that max Ax, tends to zero. Notice that the limit in (1)is a new animal. The sums in question depend on partitions. That is, we want the limit of a function of partitions, rather than a limit of a function of x or n. Assume for the moment that the limit in (1)makes sense for each of two functions f and g defined on [a,b].Then

=

.I”f + lbg.

However we define the limit above, the limit of a sum is certainly the sum of the limits (i.e., the third equality), for that is a basic 9

10

A PRIMER OF LEBESGUE INTEGRATION

property of addition: if a is close to A and b is close to B , then a + b is close to A + B. In this chapter we define a kind of limit, generalizing the idea of the limit of a sequence, which makes precise the ideas above. The objects we use to generalize sequences are called nets, and convergence of nets is sufficiently general to describe any kind of limit which occurs in analysis or topology. A directed set is a non-empty set D equipped with a partial ordering + satisfying the following conditions: (i) a + a for all a E D; (ii) i f a + B a n d B + y , t h e n a + y ; (iii) for any elements a and /!? in D there is y a + y andB + y .

E

D so that

We will write B > a to mean the same as a + B, and say that B is farther out than a when this relation holds. A net is a function defined on a directed set. A sequence is a type of net, with D being the set of natural m. We will adopt numbers directed as usual: n 4 rn means n I the sequence notation for nets and write {xa}for the net consisting of the real-valued function x defined on some directed set D of elements a. We use some index other than n, for instance a, to emphasize that D need not be the set N of natural numbers. The net {xu}converges t o t , denoted xa-+t, or lim x,=C, provided that for every E > 0 there is a0 E D such th; lxa - tI < E whenever a > ao. Of course the limits of nets are unique if they exist, justifying the notation lim x, = L . The uniqueness is a consequence of the fact that thereOLis some y beyond any given a and B. Hence if tl and t 2 were different limits, with C2 - L , = E > 0, andIxa-tl < ~ / 2 w h e n a> aoandIxp-t21 < ~ / 2 w h e n B> Po, then if y > a0 and y > Po, the two contradictory conditions would both hold. The familiar limit of the calculus, x+a lim f ( x ) = L , is another instance of a net limit. Here D consists of the points x near a and x > y means x is closer to a than y: 0 < J x- a1 5 ly - al. ~

THE RIEMANN INTEGRAL AS A LIMIT OF SUMS

2

11

Problem 1. Describe D and 4 so these limits are limits of nets: (i) x-00 lim f ( x ) = C; (ii) lim f ( x ) = C; n+a+

Proposition 1. Let {x,} and {y,} be real-valued nets on the same directed set D, with lim, x, = C, lim, y, = m. Then

+

+

(i) lip(x, y,) = C m ; (ii) lim(x, - y,) = C - m ; a (iii) lim(x,y,) = em; a (iv) lim(x,/y,) = C/m i f m a

+ 0, 3/01 $: 0 for all a.

Proof. The proofs of these statements, which are basically just familiar properties of addition and multiplication, are virtually the same as the corresponding statements for sequences or functions. We prove (iii) by way of illustration. Assume x, -+ C and y, + m, or equivalently, x, - C --+ 0, y, - m -+ 0. Let r, = x, - C and s, = y, - m for all a. Then r, + 0 and s, -+0 and

Fix E > 0 and pick a1, beyond which Ira/ < 1 and in addition Ir,l is so small that Ir,ml < E . We similarly pick a2 so that beyond a2, Is,( < E and (s,Cl < E . There is a0 so that a0 + a1 and a0 + a2. Hence if a + ao, we have

so

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A PRIMER OF LEBESGUE INTEGRATION

Problem 2. Prove parts (i) and (iv) of Proposition 1. 1111111 Problem 3 . (i) If x, > 0 for all a in a directed set D and x, --+ t , then t > 0. (ii)If x, 5 y, 5 z, for all a E D, and x, + t , z, --+t , then Ya

--+

t.

11111(

Problem 4. Let D be the set of all pairs (m,n) of positive integers. Partially order D as follows: (m,n) > (m’,n’) if and only if m+ n 3 m’ + n’. (i) Describe geometrically what (m,n) > (m’,n’) means. = t , then lim G,, = t for all m, (ii) Show that if (m,n) lim G,, ,---too and m-cc lim G,,= t for all n. (iii) Let G,, = m n / ( d + n2).Show lim xm,,= 0 for all n and m-m lim G,, = 0 for all m but lim h,,fails to exist. nlllll! n-cc

(m?)

Problem 5. Let D be the set of all pairs (m,n) of natural numbers, with the partial ordering (m,n) > (WQ, no) iff max{m, n} 3

m a x i m , no}.

(i) Describe geometrically the set of (m,n) such that (m,n) > (mo, no) for a fixed (WQ, no). (ii) Does lim G,, = C imply lim xm,, = C for all m? 4~ (m,fl)

n+m

Problem 6. Let D be as above with the ordering (m,n) > (m’,n’) iff mn 2 m’n’. Give examples of nets {G,,} which

converge and nets which diverge. What is the connection, if any, between convergence in the ordering of D and the limits lim G,,, ,l@cc h,,? 41 m+cc

The nets we want to consider here, and later for the Lebesgue integral, are nets of Riemann sums. Again let f be any real function on [a,61 and let P = {Q, XI, . . . , x,} be any partition of [a,b].A choice function c for the partition P is a finite sequence c1, c 2 , . . . , c, with c, E ( ~ ~ - xi) 1 , for each i. The Riemann sum for f , P , and c is

R f ,p , c ) =

c n

i=l

f(Ci)CG

- xi-1).

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THE RIEMANN INTEGRALAS A LIMIT OF SUMS

13

The Riemann sum R( f, P , c ) , for fixed f , is a real-valued function which depends on the partition P and the choice function c. Thus {R(f, P , c)} becomes a net when we put an appropriate partial ordering 4 on the pairs ( P , c). We do this as follows:

Thus the pairs are ordered by the partitions themselves in the sense that ( P I , cl) is farther out than ( P z , c2) if PI is a refinement of P2. We say that a real-valued net {x,} is increasing provided xp 2 x, whenever B > a. A decreasing net is defined similarly. The net {x,} is bounded provided there are numbers b and B so that b 5 x, 5 B for all a.

Problem 7. Show that an increasing bounded net {xa}con-

verges to c = sup{x, : a

E

01.

1111111

The lower sums L( f, P ) and the upper sums U ( f, P ) for a bounded function are nets, with the partitions ordered by refinement. If f is the function in question and m If ( ~ 5) M, then the lower sums and upper sums are bounded. The lower sums form an increasing bounded net, and hence converge, and similarly for the upper sums. Clearly f is Riemann integrable if and only if P I,( f, P ) = l i p U ( f, P ) . lim

(2)

If ( P , c) is a partition of [a,61 with a choice function c, then

for each i, so

It follows immediately from (2),(3) that if f is integrable, then

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A PRIMER OF LEBESGUE INTEGRATION

We want to show that the existence of the limit 1iF R( f , P , c) provides an alternative characterization of integrability. Notice, however, that to talk of lower and upper sums we need to assume that f is bounded. The Riemann sums, on the other hand, are defined even if f is not bounded, as long as f is defined on all of [a,b].Conceivably the limit in (4)could exist for an unbounded function f , and we would have two distinct definitions of the integral. The next proposition resolves this question.

Proposition 2. I f f is defined on [a,b] and l i p R( f , P , c ) exists, then f is bounded.

Proof. Let lim R( f, P , c ) = I and assume that f is not bounP ded above. Let P be a partition of [a,b]such that for all choices c, In particular IR( f , P , c ) - R( f , P , c’)l < 2 for any choices c and c’ for P . Since f is unbounded, f is unbounded on some subinterval, which we will assume is (xg,X I ) . Fix any choice c for P . Let ci = ci for i = 2, . . . , n, and choose c; so f (c’,) > N. Then

R ( f , p , c’) - R( f , p , c ) = ( f (c;) -

f(Cl>>

Ax1

> (N- f ( C l > > A X l .

We can choose N so large that the two Riemann sums differ by more than 2, which is a contradiction. 1111111

Proposition 3. A function f is integrable on [a,b] if and only if lim R( f , P , c ) exists, and of course the limit is the integral in P this case.

Proof. We have only to show that the existence of the limit implies that the function is integrable. To do this, fix E > 0 and choose a partition P so that IR( f , P , c ) - I ) < E for all choices c, which implies that for any two choices c, c’ for this P , IR( f,P , c)R ( f , p , c’)l < 2 ~We . will choose c and cf so that

2

THE RIEMANN INTEGRALAS A LIMIT OF SUMS

15

This implies that U ( f , P ) - L( f, P ) < 3 E . To see how c and c’ can be chosen to satisfy (5)and (6),let

For each i , choose ci so that

Then

We similarly choose c’ so that for each i ,

and we have (6) by the same argument as above. 1111111 Proposition 4. I f f and g are integrable on [a,b] and k is a

constant, then

Proof. For every partition-choice pair ( P , c),

A PRIMER OF LEBESGUE INTEGRATION

16

Hence (i) and (ii) follow from the general results for nets in Proposition I. Similarly, R( f, P , c ) 3 0 for all ( P , c) if f 3 0, so lim R( f, P , c ) 3 0. 1111111 P

Problem 8. If F is a continuous function on R and {x,} is a net such that x, -+ x, then {F(x,)} is a convergent net and F(x,)+F(x). As a special case, if x,+x then lx,l--+I~l. Apply this to show that if f and hence I f l are integrable, so that s,b f= l i p R( f, P , c ) and S,b I fl= l i p R(I fl, P , c), then IS," f l 5 s,b 1 f 1 . ~l~llll

Proposition 5. I f f is integrable on [a,b]and F is a continuous function on [a,61, and differentiable on ( a , b) with F'(x) = f ( x ) on ( a , b), then

lbf

=

F ( 6 ) - F(a).

Proof. Let E > 0 and let Po be a partition such that IR( f, P , c) J;,b f l < E whenever P t Po and c is any choice function for P . That is,

whenever P + Po and c is any choice for P . The hypotheses for the ordinary Mean Value Theorem for F hold on each interval [xi-l, xi].Therefore there is ci E (xi-l, xi) for each i so that

Let c be a choice function for P such that (8) holds. Then from (7)and (8) we get

2

THE RIEMANN INTEGRAL AS A LIMIT OF SUMS

17

Since C(F (xj) - F (xi-1)) = F (6) - F ( a ) ,

Since is arbitrary, the left side is zero, and F ( a ). 1111111

s,b f

= F(b) -

Consider the problem of summing an arbitrary collection of numbers. Say a little boy hands you a basket of numbers and asks you to add them-what do you do? You empty the numbers out on the floor, kick them into a row, and start adding from left to right. If there is a finite number of numbers, then there is no problem. If there is an infinite number of numbers in the basket, then you keep adding from left to right until you determine a limit, and that is the sum. The difficulty with this process is that if you sweep up all the numbers, put them back in the basket, and repeat the process, you will likely get a different answer. Indeed, unless the positive numbers and negative numbers separately have finite sums you will almost surely get different answers on your second and subsequent trials. The point is that a conditionally convergent series is a very artificial thing, unless you have some real reason to want the numbers to appear in a given order. The unordered sum defined next gives a more convincing generalization of finite-sum addition. Let A be any “index set” of elements a , and let x, be a real number for each a E A. For any finite subset F c A define SF to be the finite sum CaEF x,. Partially order the finite subsets F of A by inclusion: F1 > F2 if F1 2 F2. Then { S F } is a net on this partially ordered set. If { S F } converges to L we write ZaEAx,= L and say the x, are summable.

Problem 9. (i) Show that if lim SF = L exists, then at most F countably many x, are non-zero; i.e., there is a countable subset C c A so that x, = 0 if a @ C. (Hint: Suppose first that all x, 2 0. If uncountably many x, > 0, then there is n such that x, 2 for uncountably many a.) Observe that this shows that countable additivity is the most one can ever ask for. There is no such thing as (non-trivial) uncountable addition.

18

A PRIMER OF LEBESGUE INTEGRATION

(ii) Show that if C r r E A=~L, , then the set of positive x, is summable and the set of negative x, is summable and ,€A+

,€A-

,€A

where A+ = {a : x, > 0}, A- = {a : x, < O}. (iii) If {x, : a E A} is summable, then {Ix,I : a mable and

E

A} is sum-

(iv) If xEl x, is absolutely convergent, then {xn : n E N} is summable, and conversely. In either case, 00% - CzN. nlllI Let {x,} be a net on the directed set D. We will say that {xu} is a Cauchy net provided that for each E > 0 there is a0 E D so that Jxp- xyI < E whenever j? > a0 and y > a0. Proposition 6 . I f {x,} is a Cauchy net, then {x,} converges.

- xyl < $ when j?, < a3 .i . . . by replacing a2 if necessary by a; with a; > a2, a; > a1, and a3 by a; with a; > a3 and a; > a;, etc. Then { x , ~ is } a Cauchy sequence, so there is a limit e, and given E > 0 there is N with IxaN - el < E . We can assume < E . Then if j? > a ~ Ixp, - xaNI < E and IxaN- el < E , so Ixp - el < 2~ if j? > a N . 1111111

Proof. For each n pick an so that Ixp

y > a,- We can assume that

a1

<

a2

Problem 10. Every convergent net is a Cauchy net.

1111111

Problem 11. If A is an index set and {x, :a E A} is summable, and A = U A, where the A, are disjoint subsets of A, then {x,: a E A,} is summable for each n and

Problem 12. (i) If {G,} is summable over N x N,then

2

THE RIEMANN INTEGRAL AS A UMlT OF SUMS

19

(ii) Suppose the iterated (ordered) sums both exist. Does it follow that is summable? (iii) Suppose the iterated sums both exist and h, 2 0 for all m, n. Does it follow that {xmn}is summable? ~ 1 1 ~ ~ ~ 1

+

+

+

Problem 23. If F(m,n) = $ ( m n - 2)(m n - 1) n, then F is a one-to-one function on N x N into N. Hence the set of all pairs (m,n) is countable, and any countable union of countable sets is countable. Hint: If f ( x ) = $(x-2)(x- l),then F(m,n) = f ( m n) + n. Show that f ( x 1)- f ( x )= x - 1, a n d c o n c l u d e t h a t i f m + n = i + j + l , F(m,n) > F ( i , j).Does F map N x R onto W

+

+

~1111

Problem 14. Let < and @ be two partial orderings, both of which make D a directed set. Suppose a < /? implies a @ /? for all a, 8, E D. Let {xa}be a net on 0, and let lim xa and limx, 4.

0

denote the limits with respect to the two orderings. Show that if lim x, = C, then lirn x, = l. ~f~llll 0

<

The next problem shows that the Riemann integral can be characterized as a limit of Riemann sums, where the partitions are not directed by refinement but by insisting that the length of the maximum subinterval tends to zero. We will use this result later to characterize the Riemann integrable functions as those which are continuous except on a set of measure zero.

Problem 25. If P = {xo, . . . , xa}is a partition of [a,b],define the norm of P , denoted IjPII, by \(PI1 = max(xi - xi-1). Let P @ Q mean that )I QII 5 11 P 11. Show that @ makes the partitions P and the pairs ( P , c) into directed sets. Let lim stand ll1’llk+O for the limit with the direction 0. Use Problem 14 to show that if lim R( f, P , c) = I, then f is Riemann integrable with /I p lI+O integral I . Conversely, lim R( f, P , c) = I implies lim R( f, P , c) = 1. P I1 P II+o Prove this. Hint: Let Po = {xo,. . . , x,} be a partition of [a, b] such that U( f , PO)- L( f, Po) < E , and so J R (f, Po, co) - I I < E . Let Q = {yo, yl, . . . , yk} be another partition with 11 QII = 6 < min(xi - xi-1). Let J be the set of all indices j for Q such that (yj-1, y j ) is contained wholly within some (xi-1, xi) of Po. If Mj

A PRIMER OF LEBESGUE INTEGRATION

20

and mj are the sups and infs of f for Q, then

C ( M j -mj)nyj 5 1 €1

u - L(f,PO>-

If K denotes the Qindices not in J contains some x i ) , then

(SO

all j such that (yj-1, y j )

C ( M j - m j ) A y j < n ( M - m)6

i€K

where M and m are bounds for f on [a, b]. Consequently, if IIQII = 6 5 E / ~ z ( M - m), then

u(f, Q) - U f, Q) IU( f, PO>- L( f, PO)+ E < 2 ~ 7 and so 1 R( f, Q, c) - I I < 2 ~ .1111111

LEBESGUE MEASURE ON (0, 1)

Let f be the characteristic function of the rational numbers in (0,l); i.e., the function which is one on the rationals and zero elsewhere. Then for any partition P = {xg,XI, . . . , x,} of [0,1], mi = 0 and Mi = 1for all i. Here, as usual, mi and Mi are the inf and sup of the function values on (xi-1, xi). Since L( f , P ) = 0 and U ( f , P ) = 1 for all partitions P , f is clearly not Riemann integrable. Now recall the geometric interpretation of the integral as the area under the graph of f . If f is the characteristic function of the rationals in [0,1], then the region under the graph of f is a very simple one; it is just the “rectangle” Q x I where Q is the set of rationals in [0,1] and I = [0,1]. The area of this rectangle obviously ought to be the length of Q times the length of I . Once we have a sensible definition for the length of Q we will have a reasonable value for the integral of f . In this chapter we extend the idea of length from intervals to all subsets of R. This generalized length, called the (Lebesgue outer) measure of a set, will assign measure zero to Q (and all other countable sets) so that we will have no difficulty agreeing that J f = 0 when f is the characteristic function of the rationals. The difference between the definitions of the Riemann and

Lebesgue integrals consists in just this fact: for the Lebesgue integral we allow partitions into sets more general than intervals,

and this requires that we can assign a length to these partitioning sets. With this one variation, the definition of the Lebesgue integral will be the same as the definition of the Riemann integral. 21

22

A PRIMER OF LEBESGUE INTEGRATION

We now proceed with the definition of the measure function m. We restrict our attention initially to subsets of the open unit interval U = (0,l). This will ultimately give us also the measure of subsets of any interval (n,n + I), since we want measure to be invariant under translation. Countable sets will turn out to have measure zero, so we will finally define the measure of any set E to be the sum of the measures of the sets En(n, n l),n = 0, f l ,f 2 , . . . . For any interval I we let l ( I ) denote the length of I . An interval can be open, closed, or half-open, but not just a single point. Hence l ( I ) > 0 for every interval I. Roughly speaking, we define the measure of a set E to be the minimum of the sums of the lengths of families of intervals which cover E . To make this precise, we say a finite or countable family { I,} of intervals is a covering of E if E c U1,. The family is an open covering if all the I, are open intervals, and a closed covering if all the I, are closed intervals. The total length of the family {Ij} is C l ( I i ) . Finally, we define the measure of E to be the infimum of the total lengths of all coverings of E :

+

xl(Ij): E

c UIj

Let us check that it does not matter in this definition whether we use open intervals or closed intervals or a mixture. For each open covering { I i } of E there is a closed covering { T i ) of the same total length, so using closed coverings might conceivably give a smaller value for m(E).However, for each closed covering { I k } of E there is an open covering { J k ) whose total length is less than C l (I k ) E . (Let J k be an open interval containing I k with .t(Jk) < l ( I k ) ~ / 2 ~Hence . ) to each closed covering { I j } of E there are open coverings with total lengths arbitrarily close to the total length of { I j ) , so the infima of total lengths of coverings are the same. If E is a compact set in (0, 1)-i.e., a closed bounded setthen every open covering of E has a finite subset which covers E ; this is the Heine-Bore1 Theorem. It follows that if E is compact, m(E)is the inf of the total lengths of finite open coverings.

+ +

3

LEBESGUE MEASURE ON (0, 1 )

23

By the discussion above, we could also use finite closed coverings to find m(E) if E is compact. These remarks are perhaps unnecessarily complicated for defining measure on the line, since we could simply stick with coverings by open intervals. However, when we consider plane measure in Chapter 9 it will be convenient to know that the covering sets can be open or closed (rectangles), or anything in between. The number m(E) is the Lebesgue outer measure of E , although we will refer to m(E)simply as the measure of E . Here are some immediate properties of m. Proposition 1. (i) 0 5 m(E) 5 1 for all E c (0,l); (ii) m(E) 5 m(F) if E c F (mis monotone); (iii) m(0) = 0; (iv) m ({x}) = 0 for all x.

Problem 1. Write out the very short proofs of parts (i), (ii), (iii), (iv) of Proposition 1. Note that subsets of sets of measure zero have measure zero. 1111111

Since m(E) is to be a generalization of length, we need to know that m ( I ) = C(1) for every interval I c ( 0 , l ) . That is the content of the next proposition and problem.

Proposition 2. I f J = [a,b] c (0, l),or if J = ( a , 6) c (0, l), then m ( J )= C ( J ) .

Proof. First assume J is a closed interval [a,b].Clearly m ( J )5 C ( J ) since { J } is a one-interval covering of J of total length C ( J ) . We show by induction that if 11, . . . , I, is any finite covering of J by intervals, then C ( J ) 5 C(Ik). If J is covered by one interval 11, then clearly C ( J ) 5 t(Il).Suppose as our inductive hypothesis that whenever J is a closed interval covered by n or fewer open intervals 11, . . . , Im ( m 5 n), .t( J ) 5 CF'Ll C( I k ) . Let J = [a,61 c 11 U . - . U In+l, and assume no n of these intervals cover J . If any 1k is disjoint from J we are done. Let us assume, to be definite, that 1,+1 = (c, d ) with a < c < d < b. Let J 1 = [a,c] and J 2 = [ d ,61 be the two subintervals of J not covered by 1,+1. No interval I k , k = 1,. . . , n, can intersect both J 1 and 1 2 , for such an interval would cover In+l, and n of the Ik

24

A PRIMER OF LEBESGUE INTEGRATION

would cover J. Therefore some of the intervals 11, . . . , 1%cover J 1 and the rest cover 1 2 . By the inductive assumption applied separately to J 1 and J 2 we see that

k= 1

Since

we then have

k=I

This shows that m ( J )= t ( Jfor ) every closed interval. The cases where I,, = (c, d ) with c 5 a or d 2 b are treated similarly. If J = ( a , b), then we pick a closed interval [c, d ] c ( a , b) with d - c > b - a - E . Hence, by monotonicity,

m(a,b) 2 m[c,d ] = d - c > b - a

- E.

We already know that m(I) 5 t(1) for any I so m(a, b) = t ( a , 6). The remaining cases are left as an exercise. 1111111

Problem 2. Show that m ( 0 , l ) = 1and that m(a,b]= m[a,6 ) = b - a for all half-open intervals in (0, 1). 1111111 Proposition 3. rn is countably subadditive; i.e., for any finite or countable family { Ei}of subsets of (0, l),

Proof. Let E > 0 and let { I i i } be a covering of intervals so that

Ei

by open

3

Then u E ~c ~

LEBESGUE MEASURE ON (0, 1 )

I i and j

25

hence

=

c m(EJ +

E.

i

Since this holds for all E > 0,

Problem 3. Show that countable sets have measure zero.

1111111

Problem 4. If Q is the set of rationals in ( 0 , l ) then we know from Problem 3 that m(Q) = 0, and hence for any I > 0 there and C f ( I j )< 1. Show are open intervals { I j } so that Q c uI~ that if Q c I1 u - .. u I, with 11, . . . , I, open intervals in (0, l), then C ( I 1 ) + . . + l ( I , ) 2 1. Hence, although finite coverings suffice to approximate m(E) for compact sets E , arbitrary sets

require countable coverings.

1111111

Problem 5. Let E = EIUE2 with m ( E 2 ) = 0. Show that m(E) = m(E1).Make precise and prove the assertion that if two sets differ by a set of measure zero, then the two sets have the same measure. 1111111

Problem 6. Let El c I1 and E2 c 12 where 11 and I2 are disjoint intervals in (0,l). Show that m(E1 U E2) = ~ ( E I ) m(E2). 41 Generalize to a finite number of sets E l , . . . , E,.

+

+

Problem 7. (i) Let E c (0, l),and let E, = {x r : x E E } . If E, c (0, l),then m(E) = m(E,). (ii) Let r E ( 0 , l ) . For x E (0, l),define x @ r=

(:::-I

if

x+r

if

x+r>l.

E

(0,l)

26

A PRIMER OF LEBESGUE INTEGRATION

We need not define x @ r if x + r = 1, since we want x @ r to lie always in (0, 1).Let E, = { x e r : x E E } , so that E, is now the r-translate of E with the points that fall outside ( 0 , l ) put back at the left end of the interval. Show that m(E) = m(E,). 1111111

Note. We will show in the next chapter that the measure function m is countably additive on a usefully large family of sets (the measurable sets) but not on all sets. To construct a nonmeasurable set we will need the kind of translation invariance in part (ii) above.

MEASURABLE SETS: THE CARATHEODORY CHARACTERIZATION The critical property of the measure function m is that it be additive. Ideally we should have an identity like

for all finite or countable disjoint families { Ei}. Unfortunately, m is not countably additive over all sets, and we must sort out the so-called measurable sets on which (1) does hold. If E is any set in ( 0 , l ) and E’ is its complement in (0, l),then a minimal requirement for additivity is certainly

m(E)+ m(E’) = m ( 0 , l ) = 1.

( 2)

It turns out that this condition is sufficient to distinguish the sets E on which m is countably additive, i.e., sets on which 1 holds. Hence we make the following definition: A set E c (0, 1) is measurable if and only if

m(E)+ m(E’) = 1,

(3)

where E’ = ( 0 , l ) - E . It is immediate from the definition that E is measurable if and only if E’ is measurable. Moreover, since m is subadditive, we automatically have

m(E)+ m(E’) 2 m(0,l) = 1.

Hence E is measurable if and only if

m(E)+m(E’)5 1. 27

28

A PRIMER OF LEBESGUE INTEGRATION

The measurable sets include the intervals and are closed under countable unions and intersections. The measure m is additive on any finite or countable family of disjoint measurable sets. The verification of these facts is the program for this chapter. A cautionary word about notation and nomenclature: most texts use m* for our function m and refer to it as Lebesgue outer measure. The unadorned letter m is used by these authors for the restriction of m* to the measurable sets, and this restricted function is called Lebesgue measure. We will stick to m, defined on all subsets of (0, l),and call m(E)the measure of E whether E is measurable or not. In practice (i.e., after this chapter) we only consider m(E) for measurable sets since measurability is essential for m to have the critical property of additivity. Proposition 1. (i) I f m ( E ) = 0, E is measurable.

(ii) Intervals are measurable. Proof. (i) If m(E) = 0, then

+

m(E) m(E’) = m(E’) I 1, and this inequality is equivalent to measurability. (ii) Let J = ( a , b) be a proper subinterval of ( 0 , l ) and let J’ = J 1 U J 2 where J 1 and J 2 are the two complementary intervals to J . (One of J 1 , J 2 will be empty if J = ( 0 ,b) or J = ( a , l).)Since the measure of an interval is its length, m(J1)

+ m<J>+ m(J2) = 1.

Therefore,

+

m<J’> I m<Jd m(J2)= 1 - m<J> m<J> +m(J’>I 1. This argument works for closed and half-open intervals too. 1111111 J2

We saw in Problem 6 of the preceding chapter that if are disjoint intervals in (0, l),then for any set E ,

m(E n (11 u J 2 ) ) =

n J1> + m(E n J 2 ) .

We now extend this to finite or countable families {Ji}.

J1

and

4

MEASURABLE SETS

29

Proposition 2. I f { J i } is a finite or countable family o f disjoint intervals in (0,l)ythen for any set E ,

Proof. If { J 1 , J 2 , . . . , J n } is a finite family of disjoint intervals, and E c J 1 U . U J n , then m(E) = C m(E n Ji) by Problem 6 , Chapter 3 . Now let { J k } be a countable disjoint family of inter-

vals. Then using subadditivity in the first inequality below and monotonicity in the last we get /

c o \

co

= lim n+cc

C m(E n J k )

k= 1

Hence the first inequality is an equality and we are done. 1111111 The next result is a formalization of the statement that measurability is a local property.

Proposition 3. E is measurable if and only if for every interval

J c (01 11,

Proof. Suppose, for example, that J = ( a , b) with 0 < a < b < 1. Let J 1 = (0,a ) , J z = J = ( a , b), J 3 = (b, 1).Since the two-element set { a , b} has measure zero, m(E - { a , b}) = m(E) and similarly for E’, so

A PRIMER OF LEBESGUE INTEGRATION

30

Adding columnwise we get

for each i . If J is of the form (0, b) or ( a , l),the same argument works by considering the single complementary interval. 1111111

Problem 2 . Carry out the proof of Proposition 3 in the case J = (0, b). 1111111 Our definition says E is measurable if E splits ( 0 , l ) additively. Proposition 3 shows that E is measurable only if E splits every subinterval of (0, 1) additively. We next show that E is measurable if and only if E splits every subset T additively.

This result is due to Carathkodory and has become the modern definition of measurability in all general settings. That is, given any countably subadditive non-negative monotone function m defined on all subsets of any set, the function m will be countably additive when restricted to the sets E which satisfy

m(E n T ) + m(E’ n T ) = m(T)

for all subsets T . Proposition 4. E is measurable if and only if for every “test c (0, l),

set” T

m(E n T> + m(E’ n T ) = m(T).

Since m is subadditive, E is measurable if and only if for every set T m(E n T>

+ m(E’ n T ) 5 m(T).

(4)

Proof. The condition (4)is clearly sufficient since we can take T = ( 0 , l ) . To show that (4)holds for every measurable set E ,

4

MEASURABLESETS

31

let T be any set in (0, l), E > 0, and { I j } a covering of T by open intervals such that

C m ( I j )< m(T) +

E.

Then

If E is measurable, then by Proposition 3

m(E n I j ) + m(E’ n I j ) = m ( I j )

(5)

for each j . Hence, using monotonicity, subadditivity, and (5),

m(E n T ) +m(E’n T ) 5 C m ( E n I j ) + Cm(E’n Ij)

+

= C m ( I j )< m(T)

1.

Since E is arbitrary we have the desired inequality (4). 1111111

Problem 2. If E l , E2 are measurable sets then m(E1 - E2) = m(El)-m( ElnE2)and m ( E 1 U E 2 )=m(El)+m(E2)-m( El nE2). Do you need all the hypotheses? 1111111 We show next that m is countably additive on measurable sets, and that the measurable sets are closed under countable unions and intersections. The discerning reader will notice, with Carathkodory, that the next three propositions make no use of the fact that the sets are subsets of (0, l),or of how the function m is defined. We use just these facts:

m(0) = 0 0I m(E) 5 00 m(E) >_ m(F) if E 2 F

m (U~

i )I Cm(Ei)

and for measurable sets E , and all T ,

m(E n T ) + m(E’ n T ) = m(T).

32

A PRIMER OF LEBESGUE INTEGRATION

Proposition 5. I f { E i } is a finite or countable family of disjoint measurable sets, then m (U Ei) =

C m(Ei).

Proof. Let E l , . . . , E, be disjoint measurable sets. The set El cuts the test set T = El u . - u E, additively, so m(E1) + m(E2 U - . . U E,) = m(E1 U . . . U E,).

The measurable set E2 cuts E2 u . . u E , additively, so

m(E2) + m(E3 U - . . U E,) = m(E2 U - . U E,),

and hence

m(E1) + m(E2) + m(E3 U .

U E,) = m(E1 U . . . U E,).

In a finite number of steps we have

Now let { E i } be a countable family of disjoint measurable sets. For each n,

and hence

The opposite inequality is automatic by subadditivity, so equality holds. 1111111

Problem 3. Show that if { E i } is a countable disjoint family of measurable sets and T is any set, then m (T n

uE l )

=

m(T n Ei).

i

i

~

~

~

~

~

Proposition 5 tells us what m (U E i ) is if the Ei are measurable (and pairwise disjoint), but does not say that U Ei i s itself

4

33

MEASURABLESETS

El

E2

T

Fig. 1

measurable. This we show next, proving first that El U E2 is measurable.

Proposition 6 . I f El and Ez are measurable, then El U E2 is

measurable. Proof. Let E l and E l be measurable sets and let T be any test set. Let T = TI U T2 U T3 U T4 as indicated in Fig. 1. What we must show is

m [(El u E2) n TI

+ m [(El u Ed’ n

= m(T) ;

or, in terms of Fig. 1,

Cutting the test set % U Similarly, cutting

5U

Cutting T with El gives

with the measurable set E2 gives with

E2

gives

34

A PRIMER OF LEBESGUE INTEGRATION

Combining (7),(8), (9)we can write

Now cut Ti U

T2 U

with El and then use (7):

From (11)and (10)we have the desired equality

Corollary. Finite unions and finite intersections of measurable sets are measurable. El - E2 is measurable i f El, E2 are.

Proof. Notice that E satisfies the characterizing equation m(E n T ) + m(E’ n T ) = m(T) for all T if and only if E’ does. Therefore

El n E2 = (E; U Ei)’ is measurable whenever El, E2 are. The inductive proof from two sets to a finite number is immediate. Since

El - E2 = El n E; , differences are measurable. I1 1 1

Proposition 7. I f { E;} is a countable family of measurable sets, then UEi is measurable and nEi is measurable. O p e n sets and closed sets are measurable.

Proof. We can assume the Ej are disjoint by replacing E2 by E2 - El, E3 by E3 - ( E l U El), etc. Let Fn = El U U En, SO Fn is measurable and by Proposition 5

4

MEASURABLE SETS

35

For any test set T , by Problem 3,

+ m(T n F;) = C m(T n E ~ +) m(T n F;).

m(T)= m(T n F,) n

i=l

If E = U E i , then F, c E for all n so m(T n E’) and

Fi

3 E’, m ( T n FL) 3

m(T) 2 x m ( T n Ei) + m ( T n E’). n

i=l

This last inequality holds for all n, so, again by Problem 3,

m(T) 2 C m ( n~ 00

i=l

=m(Tn E )

+m

( n~E’)

+m(Tn E’),

which shows that E = UFl Ei is measurable. The remainder is left as a problem. 1111111

A a-algebra of subsets of any set X i s a family of subsets which contains X and 0 and is closed under finite or countable unions, finite or countable intersections, and complementation. Since El - Ez = E,nE;, o-algebras are closed under differences.

Problem 4. Show that the measurable subsets of ( 0 , l ) form a

o-algebra, and that every open or closed set is measurable.

1111111

Problem 5. Let E be a measurable subset of ( 0 , l ) . Show that for each E > 0 there is an open set U and a closed set F such that F c E c U and

+

m(E)- F 5 m(F) 5 m ( U ) < m(E)

E.

Show that the existence of such F and U for every E > 0 is also sufficient for E to be measurable. 1111111 Now we extend the definition of Lebesgue measure to include arbitrary subsets of R.We use p for the extended measure function, so p is defined on all subsets of R.For a set E c (n,n 1) define p ( E ) = m(E - n), where E - n = { x - n : x E E } .

+

36

A PRIMER OF LEBESGUE INTEGRATION

If E c (0, l),then of course p ( E ) = m(E). For any set E c R, define

c 00

P(E)=

n=-cc

p(E

n (n,n + 1)).

If E contains integer points, that will not affect the value of p ( E ) since we still want countable sets to have measure zero. Notice that p ( E ) = GO is now a possibility. We will say that E is measurable if and only if E n (n,n 1) is measurable for each n, i.e., if

+

p(E

n (n,n + 1))+ p(E’ n (a, n + 1)) = 3

for each n. We could alternatively have used the Carathkodory criterion to define measurability, as the next proposition shows.

T

Proposition 8. A subset E c R,

c R is measurable if for every set

Proof. By definition,

c 00

PU-7 =

p(TnE)= p ( T n E’) =

n=-cc cc

n=-cc 00

n=-m

p ( T n (n,n

+ 1))

p(TnEn(n,n+l)) p ( T n E’ n (n,n

+ 1)).

If E is measurable then

p(Tn E n (n,n+ 1))+ p ( T n E’n (n,n+ 1))= p(Tn (n,n+ 1)) for each n, and addition gives (12). On the other hand, if (12) holds for all T , then we see that each E n (n,n+ 1)is measurable by letting T = (n,n + 1). 1111111

As we remarked earlier, the properties of m on the measurable subsets of ( 0 , l ) which are proved in Propositions 5, 6, and 7

4

MEASURABLE SETS

37

depend only on these facts:

m(0) = 0

0 5 m(E)5

m(E)

(13) 00

for all E

m(F) if E 3 F

(14)

(15)

E is measurable if and only if

m ( E n T ) + m(E’ n T ) = m(T) for all T.

(17)

Properties (13)-( 16) obviously hold for p and subsets of R, and Proposition 8 shows that (17)also characterizes measurable sets in R. Therefore we have the properties of Propositions 5 , 6 , and 7 for p and measurable subsets of R. Specifically,

UEi) = C p ( E i )

for disjoint families El U E2 is measurable if E l , E2 are Ei is measurable if all Ei are.

p(

u

The measurable subsets of are obviously closed also under complementation and so form a a-algebra, and p is countably additive on this o-algebra.

Problem 6. Show that p is translation invariant; i.e., if E = {y + x : y E E } , then p ( E ) = p(E + x) for all E and x.

+x 1111111

We finish by constructing a non-measurable subset E of ( 0 , l ) . To show that E is not measurable we show that ( 0 , l ) is the union of a countable number of disjoint translates (modulo 1)of E . If E were measurable and m ( E ) = 0, then we could conclude m(E) = 0. If m(E) > 0, then we could that m(0,l) = C,“=, conclude that m ( 0 , l )= C z l m(E) = 00. The following “construction” of a non-measurable set E de-

pends heavily on the Axiom of Choice or some equivalent logical assumption, such as Zorn’s Lemma. Although some mathematicians are not entirely comfortable with the Axiom of Choice, we will cheerfully accept it here as part of our common logic.

A PRIMER OF LEBESGUE INTEGRATION

38

Pick any x E (0, l), and then any y so y - x is not rational. Then pick z so z - x and z - y are not rational. Continue this process-uncountably many times-until there remains no number in ( 0 , l ) which is not obtained by adding a rational to one of the already chosen numbers. The set E thus obtained is a maximal subset of ( 0 , l ) with the property that all differences x - y for x, y E E are irrational. Hence for all t $- E , t = x + r for some x E E and some rational Y . Let r l , r2, . . . be an enumeration of the rationals in [0, 1) and let E, consist of all numbers x r , (modulo 1)for x E E . That is, if x E E, x r, E E, if x + r , < 1 and x + r , - 1 E E , if x + r , > 1. Since the sets E , are essentially just translates of E, all E , have the same measure. (See Problem 7, Chapter 3.) The E , are disjoint, for if x, y E E and

+

+

x

+r, or

x+r,-l then x - y is rational, so x = y. Clearly ( 0 , l ) = U E , since every t not in E has the form x + r , (modulo 1)for some x E E , some rational r,. Thus ( 0 , l ) is a countable union of disjoint sets with the same measure. If E is measurable then all E , are measurable and m ( 0 , l )= C m(E,), which is zero or infinity.

Problem 7. Use the Carathhodory characterization ( 6 ) to show that if for every F > 0 there are measurable sets A and B such that A c E c B and p ( B ) - p ( A ) < I , then E is measurable. 111111I

Problem 8. (i)If { E i }is a sequence of measurable sets such that p ( E 1 ) < 00 and El II E2 3 E3 II . . . , t h e n p (nE i ) = limp(Ei). Hint: Let E = n Ei, so El - E = ( E l - E2)U(E2 - E3)d . . and ~ ( E -I E ) = Cr1 p(Ei - &+I)(ii) Show that p ( E 1 ) < 00 (or @ ( E x )< 00 for some n) is a necessary assumption. ~ ~ ~ ~ ~ l l Problem 9. Let { E i } be a sequence of measurable sets such c E2 c E3 c . . - . Show that p (U Ei) = lim p ( E i ) . 1111111

that El

39

MEASURABLESETS

4

Problem 2 0. For any two sets E and F , define E A F by E A F =(E-F)U(F -E).

E A F is called the symmetric difference of E and F . Agree to identify sets E and F if E A F has measure zero. (Cf. Problem 5, Chapter 3.) Define a function d on pairs of subsets of ( 0 , l ) as follows:

d ( E , F ) = p(E A F ) . Show that d is a metric on (equivalence classes of) measurable sets. Notice that the triangle inequality-the only non-obvious metric property-implies that the relation E = F , defined by p ( E A F ) = 0, is an equivalence relation, thus providing the justification for identifying sets E and F if E = F . Show that p(E) = p ( F ) if E = F , so that p does not object to the identification of equivalent sets. Show that p is continuous with respect to the metric d; i.e., if d(E,, E ) + 0, then p(E,) + p(E). Is p uniformly continuous? Is the restriction to subsets of ( 0 , 1) necessary? ~ ~ 1 ~ ~ 1 1

Problem 2 2. The operation A has some interesting properties

which might appeal to those with an algebraic bent. For example, is A an associative operation? How does the operation A interact with n, U, ’ ? Show that if intersection is interpreted as multiplication, and symmetric difference as addition, then the subsets of X (or the measurable subsets of X) form a commutative ring with identity. ~1~~111

Problem 12. The Cantor Set. Each number in [0, 11 can be written as a ternary series: x = a113

+

+ ~ 3 1 3+~. . - ,

where all ai are 0, 1, or 2. Some numbers have two such representations, e.g., 2 = 213 3 = 113

-

+ 019 + 0127 + + 219 -+ 2/27 -k

* *

* * *

40

A PRIMER OF LEBESGUE INTEGRATION

5).

Let Ul be open middle third of [O,l]; i.e., U1 = (f, Let U2 be the two intervals which are the middle thirds of the two intervals In general, let U,+1 be in [0,1] - U1; i.e., U2 = ($, $1 u (:, the union of all open middle thirds of the closed intervals in [0,1] - UYIl Ui. The Cantor set is [0,1] - U U,. Show: (i) The Cantor set is a closed set of measure zero. (ii) The Cantor set consists of exactly those points in [0,1] which can be written with a ternary expansion with all ai = 0 or 1 2 1 2 7 8 1 2 7 8 j,3,3,?, ?, z?:z?,z-i, -i7, . . .). Equivalently, 2. (For example, ?, show that U U, consists of those points whose ternary expansion must have some a, = 1. Show that $ is in the Cantor set. (iii) Show that all points of [ O , 1 ] can be expressed as a binary expansion,

g).

where each bi is 0 or 1. (iv) Since both the Cantor set and [0,1] can be put in a 1-1 correspondence with all sequences onto a two-element set, the Cantor set is an uncountable set, and thus is an example of an uncountable set of measure zero. (v) Show that each point of the Cantor set is the limit of a sequence of distinct points of the Cantor set. (vi) Show how to define a closed nowhere dense subset of [0,1] with arbitrary measure between 0 and 1 by modifying the above procedure. For example, to get a set of measure 1/2 we remove open intervals with total length 1/2 as follows: Let U1 be the open interval of length 1/4 centered in [0,1]. Then [0,1] - U1 consists of two closed intervals whose lengths are less than 1/2. From these two closed intervals remove equal centered open intervals with lengths totaling Etc.

a.

1

1

~

~

~

~

Problem 13. Let El and E2 be disjoint measurable sets. Draw the appropriate figure similar to Fig. 1, showing E l , E2 and an arbitrary test set T . Label the subsets of T as follows: El n T = X,E2 n T = Z, ( E l U El)’ n T = Write out the proof that m(Ti U Z) + m(&) = m(T) which shows that El U E2 is measurable in this special case of Proposition 6. llllll!

z.

~

4

MEASURABLESETS

41

Problem 24. Here is the historical definition of measurable set. An open subset of ( 0 , l ) is a countable union of disjoint open intervals. If U = U(ai,bi), then define m(U) = C(bi - ai). If F is a closed subset of (0, l),and U = ( 0 , l ) - F , then define m(F) = 1- m(U). Define outer measure m* and inner measure m, as follows: m*(E)= inf{m(U) : E c U , U open}, m,(E) = sup{m(F) : F c E , F closed}.

A set E is measurable if and only if m*(E) = m,(E). Show that m,(E) = 1 - m*(E’)so m*(E) = m,(E) is the same as m*( E ) + m*(E’) = 1. 1111111

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THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS In this chapter we define the Lebesgue integral of a bounded function on a set of finite measure. The definition is very similar to the definition of the Riemann integral. We partition the finite measure set into a finite number of disjoint measurable sets. There is an upper sum and a lower sum for each such partition, and when the upper and lower sums come together the function is integrable. The only difference between the Riemann and Lebesgue integrals for bounded functions is that now we allow the domain to be a finite measure set rather than an interval, and the partitions consist of measurable sets rather than subintervals. Let S be a finite measure set, by which we will always mean a measurable set of finite measure. A partition of S is a finite family { E l , . . . , E,} of disjoint non-empty measurable sets whose union is S. If S is an interval and P is a partition in the earlier sense, P = {%, XI, . . . , xn}, we will now understand that P denotes the partition of S = [xg,x,] into the disjoint sets ( X O , XI), (XI, XZ),. . . , ( ~ ~ - xn), 1 , and the finite zero-measure set {%, X l , . . X n l . If P and Q are partitions of S, then Q is a refinement of P , denoted Q >P or P 4 Q, provided each F E Q is a subset of some E E P . If P = { E i } and Q is a refinement of P , we will write Q = { F i j } to indicate that Ei = U j F i j for each Ei E P . Notice that if P = { E i }is a partition of S, then p(S) = C p ( E i ) . If f is a bounded function on a set S of finite measure, and P = { Ei} is a partition of S, we define upper and lower sums for * 3

43

44

A PRIMER OF LEBESGUE INTEGRATION

f and P exactly as in Chapter 1: mi

= inf { f(x) : x E E i }

Mi = SUP { f ( x ) : x

E Ei}

n i=l

n i=l

If S is an interval and P is a partition of S into intervals, then U ( f, P ) and L( f, P ) have exactly the same meaning as in Chapter 1.

Proposition 1. I f S is a set of finite measure and m I f(x)5 M for all x E S, and P , Q are partitions of S with Q > P , then

mi = inf { f ( x ) : x mij

E

Ei}

= inf { f ( x ) : x E F i j } .

Then clearly mi 5 mij for all i , j , so

L( f, PI = C mi p ( E i ) =

C mi C pU(Fij) i

< -

i

C mij pU(Fij) i ,i

=U

f ,Q).

The proof that U ( f, Q) I U ( f, P ) is similar, and the remaining inequalities are obvious. 1111111

Problem 1. Show that every lower sum L( f, P ) is less than or equal to every upper sum U ( f, Q). 1111111

5

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS

45

If S is a set of finite measure and f is bounded on S, then f is (Lebesgue) integrable on S if and only if supL( f , P ) =inf P

Q

U ( f, Q). We write Js f for the common value if f is integrable. We observe that f is integrable on S if and only if there is for every E > 0 a partition P such that U ( f, P ) - L( f, P ) < E . Proposition 2. I f f is Riemann integrable on [a,61, then f is Lebesgue integrable on [a,61 and the integrals are the same.

Proof. If f is Riemann integrable then there is a partition P of [a,61 into intervals such that U ( f, P ) - L( f, P ) < E , so f is also Lebesgue integrable. The value of either integral lies between any lower sum and any upper sum, so the Riemann and Lebesgue integrals clearly coincide. 1111111 For the Riemann integral we partitioned the interval [a,61 into a finite number of subintervals ( a , XI), (XI, XZ), . . . ,(%-I, 61, and a zero-measure finite set { a ,XI, x2,. . . , ~ - 1 , b ) For . the Lebesgue integral we partition an arbitrary measurable set S into a finite number of measurable subsets:

S = El U

E2 U . . . U

E,,

Ein E ,

= 0 if i $: j .

Since ,x is countably additive, we might ask whether we should not consider instead partitions of S into countably many disjoint sets:

The problem below asks you to show that countable partitions with their corresponding lower and upper sums would give an equivalent definition of integrable function. Since the sup of lower sums over all countable partitions is larger than the sup over finite partitions, and upper sums similarly could be smaller if countable partitions were allowed, it is a priori easier for f to be integrable if countably infinite partitions are allowed.

Problem 2. Show that the definition of integrability for a

bounded function on a set of finite measure does not change if countable partitions are allowed. 1111111

46

A PRIMER OF LEBESGUE INTEGRATION

The characteristic function of a set E, denoted x E , is the function which is one on the set and zero elsewhere. A simple function is a function (o defined on a set S of finite measure such that (o has a finite range { y l , y2, . . . , yn} and for some partition P = { E l , .. . , En}of S,

i=l

Problem 3. If (o is the simple function above, then qa is inteyi p ( E i ) . 1111111 grable and Js (o = C;=’=, The usual definition of “ f is integrable over S” is

where (o and @ range over simple functions defined on S. This is just different terminology for expressing the same idea as our upper sum-lower sum definition. If f is continuous on an interval [a,61, then f is Riemann integrable on [a,61. The proof consists in showing that since f is uniformly continuous, each M i - mi will be less than any given E > 0 provided P is any sufficiently fine partition of [a,61 into intervals. This implies

U ( f, P ) - L( f, P ) = C(Mj- mj) Axj < ~ ( -6a ) ,

so f is Riemann integrable. A bounded function f will be Lebesgue integrable on a set S of finite measure, by the same argument, if there is a partition P = { Ej} of S so that Mi -mi < E for each i. There will obviously be such a partition provided each set of the form { x E S : a 5 f ( x ) < a E } is measurable; specifically, we can let

+

Ej = {x E S : m+iE 5 f(x) < m+ (i + l ) E } , i = O , l , 2 , . . . , where m is a lower bound for f on S. These sets are disjoint, and a finite number of them will form a partition of S, since f is bounded, provided only that each of these sets is measurable. Accordingly, we agree that f is measurable on S provided

5

47

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS

{x : a 5 f ( x ) < b} is measurable for all a , b. Notice that if f is measurable on S, then S is necessarily a measurable set. The definition of measurable function applies to all functions f , bounded or not, and all measurable sets S, whether or not they have finite measure. In this chapter we are only concerned with the integrals of bounded functions on sets of finite measure, but in Chapter 7 we will consider unbounded measurable functions on sets with infinite measure. Problem 4. If f is measurable on S and g = f except on a zero-measure subset of S, then g is measurable on S. 1111111 Problem 5. Every simple function is measurable. ~ 1 ~ ~ ~ 1 ~ Problem 6. If f is continuous on [a,b],then f is measurable on [a,b].Hint: Show that {x E [a,b] : f ( x ) 2 a } is a closed set (and therefore measurable) for each a. Then use

{x E [a,b] : a 5 f ( x ) < p } = {x E [a,b] : f ( x )2 a } - {x E [a,b] : f ( x )

p}.

1111111

Problem 7. If f is measurable, then I f I is measurable. 1111111 Proposition 3. I f f is a bounded measurable function on a set S of finite measure, then f is Lebesgue integrable on S. Proof. Let - M 5 f (x)< M for all x E S. Let N be a large

integer, and let

Ej = {x : -M

+ (i - 1)/N 5

f (x)< -M

+i / N }

for i = 1 , 2 , . . . , 2 M N . Then P = { E i } is a partition of S and

U ( f, P ) - L( f, P ) 5

cN

1

-p@)

= p ( S ) / N . 1111111

The converse of Proposition 3 is also true for bounded functions (Proposition 6 below), so that measurability is equivalent to integrability for bounded functions on sets of finite measure. The great virtue of this characterization is not the fact that more functions are Lebesgue integrable, but the fact that pointwise limits of measurable functions are measurable, as we show below. Hence if f = lim fa with each fa integrable, then

4a

A PRIMER OF LEBESGUE INTEGRATION

J f makes sense provided only that f is bounded. It is this kind of result which makes it very much easier to deal with limits of integrals and integrals of limits in Lebesgue integration. Problem 8. Show that there is a sequence { f a } of Riemann integrable functions on [0,1], all with the same integral, such that fn(x) --+f ( x ) for all x E [0,1] and f is not Riemann integrable.

1111111

Now we proceed to show that every integrable function is measurable. We show that an integrable function is the pointwise limit of simple functions, which are necessarily measurable, and that every pointwise limit of measurable functions is measurable. We first introduce some useful alternative criteria for measurability.

Proposition 4. Each of the following conditions is necessary and suficient for f to be measurable:

(i) {x:f ( x ) 2 a } is measurable for all a; (ii) {x:f (x)< a } is measurable for all a; (iii) {x:f ( x ) > a } is measurable for all a; (iv) {x:f ( x )I a } is measurable for all a; (v) {x:a < f (x)< b} is measurable for all a, b. Proof. The sets in (i) and (ii) are complements, and similarly for the sets in (iii)and (iv).If f is measurable, then {x: f ( x ) 2 a } is the countable union of the measurable sets {x : a 5 f ( x ) < a n } , n = 1 , 2 , . . . . Conversely, if {x : a 5 f (x)}is measurable for all a , then

+

{x:a 5 f ( x )< b} = {x:a 5 f ( x ) }- {x:b 5 f ( x ) } is measurable for all a , b, and hence f is measurable. The other equivalencies are proved similarly, using the fact that measurable sets are closed under countable unions and intersections, and under complementation. 1111111 Problem 9. Complete the proof of Proposition 4. ~ 1 ~ ~ ~ 1 1 In this chapter we consider the integral only for bounded functions. In later chapters, however, we will consider unbounded

5

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS

49

functions, and indeed functions that take the values +00 or -00, since these values can arise as limits of sequences of integrable functions. Accordingly, we agree that such an extended realvalued function f is measurable provided {x : a 5 f ( x ) < b) is measurable for all a , b, and the sets {x:f ( x ) = +00} and {x:f ( x ) = -00} are both measurable. Proposition 5. I f { f n } is a sequence of measurable functions on a measurable set S, then sup fny inf fny lim sup fn, and lim inf f n are measurable functions. I f lim fn(x)exists for all x, then the limit is a measurable function. Proof. To show sup f n is measurable we verify condition (iii)

of Proposition 4. Since

{x : SUP fn(x>> a>= U{x:fn(x>> a>, n

{x : sup fn(x>> a } is a countable union of measurable sets if

each

fn

is measurable. Similarly,

{ x : inf fn(x>< a ) = U{x: fn(x)< a } , n

so inf f n is measurable. If sup f n takes the value

{x : sup fn(x>=

+00,

then

=

nU{x: fn(x>> NI, N n

and this set is measurable. A similar equality holds for the set where inf fn(x)= -00. Since lim sup fn(x)= inf sup f k ( X ) kzn

lim inf fn(x)= sup inf n

k?n

fk(X),

both lim sup fn and lim inf f n are measurable. If lim fn(x)exists for all x,then lim fn = lim sup fn = lim inf fn. 1111111 We will use the phrase almost everywhere, abbreviated a.e., to mean “except on a set of measure zero.” Hence “ f = g a.e.”

50

A PRIMER OF LEBESGUE INTEGRATION

+

means that {x: f (x) g ( x ) }has measure zero, and " f 3 n a.e." means {x: f (x)< n} has measure zero.

Proposition 6. I f f is a bounded function which is integrable on a set S of finite measure, then f is the a.e. pointwise limit o f

simple functions, artd hence f is measurable. Proof. For each n we let P, be a partition of S such that U ( f , P,) - L( f, P,) < l / n . We can assume that PI + P2 > P3 > . . . by replacing each P, by the common refinement of its predecessors. Let P, = { E,i} and m,i = inf{ f ( x ) : x E E,i} M,i = SUP{ f (x): x E E,i}. Let 40, be the simple function which has the value m,i on E,;, and let @ be M,i on E,i. Then @, - 40, is a simple function and

I(@,

- 40%)

= C ( M n i - ~,i>@(E,i) =W

f ,P,)

-

U f ,P n ) .

The functions p n increase and converge a.e. to some measurf and the @, decrease a.e. to some measurable able function g I h 3 f . We will show that g = h = f a.e., so f is measurable. Suppose on the contrary that h - g > 0 on a set of positive measure. Then (Problem 10) there is p > 0 so that h - g > p on a set A of positive measure. For all n and all points of A,

@12-40,2h-g>

p.

If Eni intersects A, then Mni - m,i > p ; these sets cover A and hence have aggregate measure at least @(A).Thus for each n,

u(f, P x ) - L( f , p a ) = C ( M n i - m,i>@(Li>2 PPU(A). i

This contradicts our assumption that U ( f, P,) 0. 1111111

-

L( f , P,) +

Problem 10. Show that if k is a measurable function and {x : k(x) > 0} has positive measure, then {x : k(x) 3 p } has

positive measure for some

p

> 0.

1111111

5

THE LEBESGUE INTEGRAL FOR BOUNDED FUNCTIONS

51

Problem 11. If f is a bounded function which is measurable on a set S of finite measure, and T is a measurable subset of S, then f is integrable over T. ~ ~ ~ ~ ~ I I Problem 12. Show that if f is integrable over S, then Js f = lim L( f, P ) = l i p U ( f, P ) where the partitions are ordered by P

refinement.

,111111

Problem 23. If f is bounded on [a,61 and continuous except at a finite number of points, then f is measurable and hence

integrable.

1111111

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PROPERTIES OF THE INTEGRAL

We will prove the linearity properties of the integral by showing that the integral is a limit of Riemann sums. First we need to know that k f and f g are measurable if f and g are. Proposition 1. I f f and g are measurable on S, then k f is measurable for every constant k, and f g is measurable.

+

+

Proof. It is clear that k f is measurable if f is, so we consider f + g. The inequality f (x)+ g ( x ) > a is equivalent to f (x)> a - g(x),which holds if and only if there is a rational number r such that

f ( x )> r

and

Y

> a - g(x>.

Hence

{x : f (x)+ g ( x ) > a } = U{x : f (x)> r >n {x : g ( x ) > a r

-r},

where the union is over all rationals r . The right side is a countable union of measurable sets. 1111111 If f and g are bounded measurable functions on a set S of finite measure, then k f and f + g are integrable over S. Now we write J f as a limit of Riemann sums. If S is a set of finite measure and P = { Ei} is a partition of S , then a choice function for P is a finite sequence {ci}with ci E Eifor each i. The Riemann sum for f , P , c is the usual sum

N f , P , c) =

cf(cz>P(Ei>. 1

53

A PRIMER OF LEBESGUE INTEGRATION

54

The partitions P of S form a directed set under the partial ordering of refinement, and the pairs ( P , c) are ordered according to the ordering on the sets P ; i.e., ( P , c) + (P’, c’) means P + P’ ( P is a refinement of P’). The Riemann sums R( f, P , c ) are a net on the partially ordered pairs ( P , c). In this context the limit condition for nets reads as follows: R( f, P , c) -+ I, or lim R( f, P , c) = I , if and only if for each F > 0 there is a partiP tion Po such that JR(f, P , c) - I1 < E whenever P + PO, and c is a choice for P . Although R( f, P , c), for fixed f , is a function of the pair ( P , c), the pairs are ordered only in terms of P , so we write lim R( f, P , c) instead of the correct but more P cumbersome lim R( f, P , c). (P&)

Proposition 2. If f is a bounded function which is integrable on the finite measure set S , then R( f, P , c ) --+Js f .

Proof. If P is any partition of S, and c is any choice function for P , then

If f is integrable then for any E > 0 there is a partition Po so that U ( f, P) - L( f, P ) < F for all P + Po. Hence

for all P

+ Po and all c; i.e., R( f, P , c ) -+ Js f .

1111111

Notice that if { U ( f, P ) } , { L (f, P ) } , {R(f, P , c ) } were nets on the same directed set, the consequence R( f, P , c ) -+ Js f would follow immediately from the inequality (1)and the fact that lim L( f, P ) = lim U ( f, P ) = Js f . As it is, the directed set P P for the Riemann sums is the much larger directed set consisting of all pairs ( P , c) instead of just all partitions P . The following proposition shows that the net of Riemann sums { R( f, P , c)} cannot distinguish between f and a function which equals f almost everywhere.

6

PROPERTIES OF THE INTEGRAL

55

Proposition 3. I f f and g are arbitrary functions on a finite measure set S, and f = g a.e., then lim R( f , P , c) = lim R(g, P , c); P

P

i.e., one limit exists if and only if the other does, and then the limits are equal. The functions f and g are not assumed to be bounded or measurable. Proof. Let f = g except on A c S, where p ( A ) = 0. Assume lim R( f , P , c) = L. Let E > 0 and choose a partition Po P so that IR( f , P , c) - LI < E if P > Po. Let PI be the refinement of Po obtained by replacing each set Ei of Po by the two sets Ei n A and Ei n A’. Many of these sets may be empty, but that doesn’t matter. Let P > PI. If P = { F i } , then each Fi is a subset of some Ej n A, so Fi c A and p(Fi) = 0, or Fi is a subset of some E i n A’, so f = g on Fi. Therefore, if P > PI + PO,

Since IR( f , P , c) -LI < E if P >Po, it follows that IR(g, P , c)-L( < E if P > PI; i.e., if lim R( f , P , c ) exists, then so does lim R(g, P , c) and the limits are equal. The situation is symmetric in f and g , so we are done. 1111111 Notice from Proposition 3 that Riemann sums can converge for an unbounded function, which is unlike the situation for the Riemann integral. For example, let g be a bounded measurable function on S, so R(g, P , c) -+ Jsg. Let f = g except on some countable set {xn},and let f ( G ) = n, so f is unbounded, but f = g a.e. By Proposition 3, lim R( f , P , c) = lim R(g, P , c) = Js g. The next two problems point out that the only way Riemann sums can converge is for the function to be essentially equal to a bounded integrable function.

Problem 1. If g is a bounded function on a set S of finite measure, and R(g, P , c) -+ I , then g is integrable (hence measurable) and Js g = I . Hint: Cf. Proposition 3, Chapter 2. r l ~ ~ ~ l ~

A PRIMER OF LEBESGUE INTEGRATION

56

Problem 2. If f is any function on a set S of finite measure, and R( f , P , c ) -+I , then there is a bounded function g such that f = g a.e. and R(g, P , c ) -+ 1. Hence R( f , P , c) -+I if and only if f = g a.e. for some bounded integrable function g . 1111111 The following proposition is now a simple consequence of the fact that the integral is a bona fide limit.

Proposition 4. 1f f and g are bounded measurable functions on a set S of finite measure, and k is a constant, then (4 ss k f = kSs f ; (4 S s < f + g > = ss f +ssg;

(4ISS f I I ss I f I.

Proof. (i)

S, k f = l i p R ( k f , P , c ) = lim k P

C f (Cj)p(Ei)

= lim kR( f , P , c ) P

= klim R( f , P , c ) P

=k

.1! f .

IIIIIII

Problem 3. (ij Verify that the net {R(f + g , P , c ) } is the sum of the nets { R (f , P , c ) } and { R ( g ,P , c ) } and use this to prove part (ii) of Proposition 4. (ii) Verify that R(I f l , P , c ) 3 IR( f , P , c)l for all ( P , c), and use this to prove part (iii) of Proposition 4. Why does the net { R ( If 1, P , c ) ) converge? 1111111

Proposition 5. I f f is a bounded measurable function on a finite measure set S, and T is a measurable subset of S, then f is integrable over T, and

6

PROPERTIES OF THE INTEGRAL

57

Proof. The function f XT is clearly bounded and measurable on S, so f x~ is integrable over S. Let Q be all partitions Q of S such that every E E Q is either a subset of T or disjoint from T. Every partition P of S can be refined to get such a partition Q E Q, so all integrals over S can be expressed as limits of sums R( f , Q, c) as Q ranges over Q. Hence if Q E Q and Q = {Ein T , Ein T'},then

=.lr,. f . The last equality holds because every Riemann sum for JT f is one of the sums in (2). 1111111

Corollary. I f f is a bounded measurable function on A where A and B are disjoint measurable sets, then

U By

Proof. f = f . X A + f . X B . 1111111 Problem 4. If f is bounded and measurable on the finite measure set E , and E = l& E,, where the Ei are disjoint and measurable, then JE f = JE, f . 1111111

xi"=,

Most of the functions one wants to integrate are continuousperhaps even analytic. For such functions there is no difference between the Riemann and Lebesgue integrals over a bounded closed interval. The Lebesgue integral, however, is much more accommodating in the matter of limit theorems. For the Riemann integral one must generally know that fn -+ f uniformly to conclude that J f n -+ J f . For the Lebesgue integral, pointwise convergence is enough provided the functions stay uniformly bounded. The reason for this is that pointwise convergence is nearly uniform on finite measure sets. We make this idea precise in the next two propositions.

58

A PRIMER OF LEBESGUE INTEGRATION

Proposition 6. I f { f n } is a sequence of measurable functions on a finite measure set S , and fn + f pointwise on S , then given E > 0 and 6 > 0, there is a measurable set E of measure less than 6 and a number N so that If&) - f (x)l < E for all k 2 Nand all x E S - E .

Proof. Let Fa

= {X

E

s : I fk(x) - f(x)l 2 E

for some k 2 n}.

The sets F, are measurable, and decreasing, and n F n = 8 because fn(x>-+ f ( x )for all x. Since p(F1) < 00, lim p(F,) = 0. Let ~ ( F N <) 6. For x E S - F N , I fk(x) - f(x)l < E for all k 2 N . 1111111

Problem 5. If f n -+

f a.e. the same result holds.

1111111

The next proposition gives the form in which it is easiest to remember and apply the above result.

Proposition 7. (EgoroffS Theorem) I f { f n } is a sequence of measurable functions on a finite measure set S , and f n --+ f pointwise on S , then for every 6 > 0 there is a measurable set E c S o f measure less than 6 so that f n --+ f uniformly on S - E .

Proof. For each n we find a set En of measure less than 6/2n and a number Nn so that I fk(x)- f ( x )I < for k 2 Nn and x E S - E n . Let E = U En, so that p ( E ) < 6. If x E S - E then given E > 0 there is N (any N , with < E ) so that I fk(x)- f (x)l < E if k 3 N. 1111111 For uniformly bounded sequences, the limit of the integrals is the integral of the limit. This follows directly from Egoroff’s Theorem, as we show next.

Proposition 8. (Bounded Convergence Theorem) I f { f n } is a sequence of measurable functions on a finite measure set S , and the functions f n are uniformly bounded on S , and f n ( X ) -+ f ( x ) pointwise on S , then

6

PROPERTIES OF THE INTEGRAL

59

Proof. Let E > 0. Let E be a measurable set of measure less than E such that f, + f uniformly off E . Let If,(x)I 5 M for all n and all x E S. Then

=

L-E

Ifn -

fl

+ /E If, - fl

Since f, -+ f uniformly on S - E,there is N so that I f, - f I < E on S - E if n 2 N. Hence if n 2 N,

Since E is arbitrary and p(S) < 00,the result follows. 1111111 The hypotheses of the Bounded Convergence Theorem require that f, + f pointwise with all f, remaining in some fixed finite area. Specifically, we require that all f, lie in the rectangle S x [-M, MI, where p(S) < 00. If the functions are allowed to wander outside a fixed finite area the result can fail, as the following problem shows.

Problem 6. Let S = ( 0 , l ) . Give an example of bounded measurable functions f, on S so that Js fn = 1 for all n and f n ( x )+ O for all x E S. 1111111

Problem 7. (i) If f is a measurable non-negative function on [0,1] and Jf = 0, then f = 0 a.e. (ii) If f and g are bounded measurable functions on a set S of finite measure, and f 5 g a.e., then Js f 5 Jsg. 1111111 Problem 8. If 0 If, 5 h Ig, 5 M on [a,b] for all n, where { f,}, {g,} are respectively increasing and decreasing sequences of measurable functions with lim Jf, = lim Jg,, then h is measurable and J h = lim Jf,. 41

60

A PRIMER OF LEBESGUE INTEGRATION

Problem 9. Show that almost everywhere convergence is sufficient in the Bounded Convergence Theorem. 1111111 Problem 20. Let f,(x)

= nx/(I

+ n2x2)for 0 5 x 5 1.

(i) Show that f,(x) --+ 0 for all x E [ 0 , 1 ] ,but the convergence is not uniform. (ii) Does the Bounded Convergence Theorem apply? 1111111

a[(

Problem 11. If f and g are measurable functions, then fg is measurable. Hint: fg = f + g)2- ( f - g)2],so it suffices to show that h2 is measurable if h is measurable. 41 Problem 12. If f is bounded on [a,b],then f is Riemann integrable on [a,b] if and only if f is continuous almost everywhere, i.e., the set where f is discontinuous has measure zero. Verify the following arguments. Let { P,} be a sequence of partitions of [a,b] such that IIP,II < and PI + P2 + P3 + . . -.Let I,i be the ith subinterval of P,. Let M,i, m, be the sup and inf of f on I,i. Let g, be the step function which is m,i on I,i, and let h, be M,j on I,i. Then g, and h, are measurable, and g, 5 f 5 h,

for all n. The sequence {g,} increases to a measurable function g If , and {h,} decreases to a measurable function h >_ f . Show that if f is continuous at q,then g ( q ) = h ( q ) . Show conversely that if h ( q ) -g(q) > E , then M,i -m,i > E whenever xo E I,i, and consequently there are points u,, v, E I,i such that f(u,) - f(v,) > 1. Since IIP,II -+ 0, u, --+q and v, -+ xoco, and f is discontinuous at q. Now we know that f is continuous at q if and only if g ( q ) = h ( q ) .Use the fact that Jg, = L( f, P,), J h, = U ( f, P,) to show that f is Riemann integrable if and only if the Lebesgue integral J ( h - g ) = 0. By Problem 6 we know J ( h- g ) = 0 if and only if h = g a.e., so f is Riemann integrable if and only if f is continuous a.e. 1111111

Problem 23. Let C be the Cantor set (Problem 11, Chapter 4) and let D be a Cantor-like set of positive measure (i.e., a nowhere dense closed subset of [ O , l ] of positive measure). Are the characteristic functions xc and xo Riemann integrable? tllllll

THE INTEGRAL OF UNBOUNDED FUNCTIONS

In this chapter we extend the definition of Js f to cases where f is unbounded or S has infinite measure. These are situations where the Riemann integral would be called improper. For the Lebesgue integral the extensions to unbounded functions and infinite measure sets is more natural, and we will not need to stigmatize that situation with the “improper” terminology. We will now also consider extended real-valued functions, which may occasionally take the values +oo or -m. Such functions f will arise naturally as limits of sequences, and indeed as limits of sequences { ffi}such that J f n converges to J f . For integrable functions the sets where f = f o o will of course have measure zero. The general definition of Js f will still coincide with the geometric idea of the net area under the graph, i.e., the area above the x-axis minus the area below the x-axis. Both these areas will be required to be finite, in contradistinction to the improper Riemann integral. For example, the function which is on the interval (n,n + 1)is improperly Riemann integrable over [ 1,oo) since

(-l)ai

- I + - -1- + -1- . . .1 2 3 4 converges, and consequently

61

62

A PRIMER OF LEBESGUE INTEGRATION

converges. This function is not Lebesgue integrable since the positive area is

-1+ -1+ - 1 +... 2 4 6 which is infinite. A Lebesgue integrable function is always absolutely integrable in the sense that if f is measurable, then f is integrable if and only if I f l is integrable. We use the same upper sum-lower sum approach to define Js f for f unbounded or S of infinite measure, only now we allow countable partitions. We saw earlier (Problem 2, Chapter 5 ) that countable partitions make no changes for bounded functions and sets of finite measure, so none of our earlier results change. As before, integrability will be equivalent to measurability, provided both the positive and negative areas are finite. The integral is a limit of Riemann sums as before, and the linearity properties follow speedily from that fact. Upper and lower sums are defined as usual, but these sums are now generally infinite series, and we require that all such series converge absolutely. It would not do if a lower sum changed its value just because the partition sets were listed in a different order. For the partition P = { E i } we again let m, and Mi be the inf and sup of f on Ei, and now define

Mj = SUP{[f ( x ) l : x

E

Ei}.

If I f l has a finite upper sum, then we again let

Since

the series in (2)will converge absolutely if I f l has the finite upper sum (1).If (1)holds for some partition P of S, then we say f

7

THE INTEGRAL OF UNBOUNDED FUNCTIONS

63

is admissible over S, and P is an admissible partition for S. The existence of an admissible partition implies that the graphs of f and 1 f l are contained in a countable union of rectangles Ei x [ - M i , M i ] of finite total area. We can write the sums C m i p ( E i ) and C M j p ( E i ) whether or not f is admissible, but we write L( f, P ) , U ( f, P ) only if P is admissible for f .

Problem 1. If P is an admissible partition for f over S, and Q t P , then Q is admissible and

U f ,0I U f ,Q) I W f , Q) I U < f ,P I . All lower sums are less than or equal to all upper sums.

1111111

A function f is integrable over the measurable set S provided sup L( f, P ) = inf U ( f, P ) . The sup and inf are over all admissible partitions of S. An integrable function is necessarily admissible, and has finite integral. Since functions are no longer required to be bounded, some mi or Mi may be infinite. If Ei is a set in an admissible partition ) As with M i = 00, then necessarily p ( E i ) = 0 so M ~ F ( E=~ 0. usual, we define

5

Problem 2. Show that f ( x ) = is admissible on [0, 11.Hint: Let P = { E i }with Ei = ($, &)for i = 1 , 2 , 3 , .. . , and Eo = (0, I, $, $, . . .>.U ( f, P I is a geometric series. Proposition 1 and Problem 3 below show that integrability 1111111

and measurability are again equivalent, given that the function is admissible.

Proposition 1. An admissible measurable function on S is integrable over S.

Proof. Let P = { Ei be an admissible partition of S so that z M i ~ ( E i<) 00. We can assume that all zero-measure sets of P have been lumped together as Eo, so Mj = 00 only if i = 0. There is N so that CZ"=,+, M i p ( E i ) < E , and hence CEN+I(Mj - mi) x k ( E i ) < 2 ~The . set T = El U . . . U E N has finite measure and f is bounded on T by max { X I ,M z , . . . , MN}.

A PRIMER OF LEBESGUE INTEGRATION

64

Hence (Prop. 3, Ch. 5) f is integrable over T and there is a partition PT of T with U ( f, P T ) - L( f, P T ) < & . The sets of PT together with Eo, E N + * , EN+2,. , . form a partition Q of S with U ( f, Q) - L( f, Q) < 3 ~ .1111111 Problem 3. If f is integrable, then f is measurable. Hint: See the proof of Proposition 6 , Chapter 5. 1111111 Now we show that the general integral is a limit of Riemann sums, and prove the linearity properties. A Riemann sum for f defined on the measurable set S is a sum C f ( c i ) p ( E i ) ,where P = { E i } is a partition of S, and c, E Ei for each i. We will use the notation R( f, P , c) for such a sum only if P is an admissible partition of S. Riemann sums are now generally infinite series, and the admissibility condition ensures that the series converge absolutely. The pairs ( P , c) are directed as usual, with ( P , c) > (Q, c’) if P t Q, and with this agreement { R( f, P , c)} is a net on the pairs ( P , c) with P admissible. Recall that lim R( f, P , c) = I means that given E > 0 there is Po so that I R( f, P , c ) - I1 < E if P > Po.

Proposition 2. f is integrable over S with Js f = I if and only i f R ( f, P , c ) -+I. Proof. Assume f is integrable, so given E > 0 there is a partition Po with U ( f, Po) - L( f, Po) < E . If P > Po, then for any choice c for P ,

-

From (3) it follows that IR( f, P , c) - Js f l < E if P > PO, so R( f, P , c ) ss f . Now assume that { R( f, P , c)} converges, with R( f, P , c) -+ I. The admissibility of f is implicit in the notation R( f, P , c). Let E > 0 and let I R ( f ,P , c ) - I1 < E if ( P , c ) > (Po,co). In particular, since (Po, c) > (Po, co) for any choice c, any two Riemann sums for Po are within 2~ of each other. If PO = { E i } , we find for each i with p ( E i ) # 0 a ci E Ei so that

Mi - ~ / 2 2 p ( E j ) f ( c i ) p ( E i )> M i p ( E i ) - ~ / 2 ’ * f(ci) >

7

THE INTEGRAL OF UNBOUNDED FUNCTIONS

65

For this pair (Po, c), One can similarly choose ci for Po so Now we have

Hence f is integrable, and from the first part of the proof we know the integral is the limit of the Riemann sums. 1111111 Notice that nothing in the argument above precludes the possibility that some M ior mior f (ci)might be infinite, provided the corresponding Ei has measure zero.

f

Proposition 3. If f and g are integrable over S, then kf and g are integrable over S, and

+

Jskf=kJ

s,(f +

g) =

S

f

J’ f + J’ g. S

S

Proof. The partitions which are admissible for f may not be admissible for g , so we need to restrict our attention to those partitions which are admissible for both f and g. If Po is such a partition (Problem 4) then all the nets {R(f, P , c ) } , { R ( g ,P , c)}, {R(kf,P , c)}, {R(f + g , P , c)} are defined for P > PO,and

It follows from general net theorems that P

A PRIMER OF LEBESGUE INTEGRATION

66

and

Since

we have immediately that

The convergence of { R( f , P , c)} implies the convergence of { k R ( f , P , c)}, and hence of { R ( k f ,P , c)}, which says that k f is integrable. Similarly, &+g)=limR(i+g,

~

c

)

+ K g , P , c>l = lim R( f, P , c) + lim R(g, P , c) = lim[R(f, P , c)

=

s, f + J!g.

The limit of a sum is still the sum of the limits. 1111111

Problem 4. Show that if f and g are admissible over S, then + g.

so is f

1111111

Problem 5. If f and g are integrable over S, and f 5 g on S, then Js f 5 Js g. Be explicit about what facts you are using. 1111111 Problem 6. Define f + ( x )= max{f(x),O},f-(x>= min{f(x), 0}, so f + ( x )3 0, f - ( x ) 5 0, and f ( x ) = f + ( x )+ f - ( x ) . Show that f is integrable if and only if f + and f - are integrable, and in this case J f = J f + + J f - . 41

7

THE INTEGRAL OF UNBOUNDEDFUNCTIONS

67

Proposition 4. I f f is integrable over S and T is a measurable subset of S, then f is integrable over T, f X T is integrable over s, and ST f = SS f XT. Proof. See the proof of Proposition 5 , Chapter 6. 1111111 If f is integrable over T and T c S, we will now feel free to consider f as defined on all of S, with f = 0 off T , and know that ss f = JT f . Problem 7. If f is integrable over A and integrable over B , where A and B are disjoint, then f is integrable over A u B and

L"Bf / f + / f . =

B

A

11'1111

The usual approach to defining J f for unbounded functions or sets of infinite measure is to consider first f 2 0 and define J f to be the sup of integrals JT g where g is a bounded function under f and T has finite measure. Simultaneous approximation by areas inside and outside is not required. Conceivably this could yield more integrable functions, but the next problem shows it does not.

Problem 8. Show that if f is a positive measurable function on S, and f has no finite upper sums, then f has infinite lower sums, and there are bounded integrable functions O s g i f with g = 0 off a set of finite measure and S g arbitrarily large.

Hint: Note that we necessarily consider inadmissible partitions here. Let P = { E ; }where E; = { x : 2i 5 f ( x ) < 2"') for i = 0, fl, f2,. . . . Show that & , N m i p ( E i ) ? $ Cl;l,NM;p(Ei). You had better also consider what happens if p ( E ; )= 00, since that isn't allowed in partitions. 1111111

Proposition 5. I f f is integrable over S, then there is a finite measure subset T of S such that f is bounded on T and SS-T

If I <

Proof. Let P = { E i } be an admissible partition of S, so M ; p ( E ; )< E for some N,and f is bounded on T = El U . . U E N (by max {MI, . . . , M N } ) ,and 1111111 SS-T I f I < &.

C M i p ( E i ) < 00. Then

+

68

A PRIMER OF LEBESGUE INTEGRATION

The approximation result of Proposition 5 allows us to prove the general Lebesgue convergence theorem below as an easy consequence of the bounded convergence theorem (Proposition 8 of Chapter 6 ) .

Proposition 6. (The Lebesgue Dominated Convergence Theorem) I f ( f n } is a sequence of measurable functions and f n --+ f a.e. on a set S and there is a n integrable function g on S such that 1 fnl 5 g for all n, then f is integrable and Js f n -+Js f .

Proof. Notice that all f n are integrable since I fnl 5 g and g is integrable. Since f n -+ f , f is measurable, and f is integrable since 1 f I 5 g. Let E > 0 and let T be a finite measure subset of S such that g is bounded on T and JS-Tg < E . Clearly ( f a } is uniformly bounded (by a bound for g ) on T , so JT f n -+ JT f by the bounded convergence theorem. Since I fnl i g and I f I i g ,

Therefore, if we choose then for n 2 N ,

N so that I JT

f n - JT

fI

< E for 12 2 N ,

The dominated convergence theorem says in essence that pointwise convergence of { f n } implies convergence of the integrals, provided only that the graphs of the fn all lie in some fixed finite area. If the fn are positive and their graphs are not all in some fixed finite area, then the integrals J f n may get too big, but they are always at least as big as J f in the limit. That is what Fatou's lemma says.

Proposition 7. (Fatou's Lemma) I f { f n } is a sequence of nonnegative integrable functions on S , and f n --+fa.e., and f is integrable over S , then

7

THE INTEGRAL OF UNBOUNDEDFUNCTIONS

69

I f f is not integrable (i.e., not admissible), then liminf

Sf n

= lim

Jf n

= 00.

Proof. Let h be a bounded measurable function, with 0 5 h 5 f and h = 0 off a finite measure set T c S , and JT h > Js f - E . Let gn = fn A h, so the g, are uniformly bounded on T , and gn --+ h on T. By the bounded convergence theorem,

Since E is arbitrary, lim inf Js fn 3 Js f . The second part of the proof is the following problem. 1111111

Problem 9. Let { f n } be a sequence of non-negative measurable functions such that f n -+ f a.e. Assume that f , which

is measurable, is not admissible. Show that lim J fn = 00. Hint: See Problem 8. 1111111

Problem 10. State and prove a Fatou-type theorem for negative f n . 1111111

Here is a slight generalization of the Dominated Convergence Theorem.

Problem 11. Let { fn}, {gn}be sequences of non-negative measurable functions such that fn 5 gfi for all n, fn -+ f a.e., and g, -+ g a.e. If g is integrable, and Jgn -+ Jg, then J f n -+ J f . Hint: Use Fatou’s lemma. alllll Problem 12. (The Monotone Convergence Theorem) If { f n } is a sequence of non-negative measurable functions, with fn 5 f for all n, and f n -+ f a.e., then j’ f n --+ J f if f is integrable. Note: The usual situation is the case where { f n } increases to f ; hence the name.

1111111

We have seen that some functions which are not Lebesgue integrable are nevertheless improperly Riemann integrable because of a cancellation of positive and negative areas. If a non-negative function is Riemann integrable, however, then it certainly is Lebesgue integrable. Improper Riemann integrals

70

A PRIMER OF LEBESGUE INTEGRATION

are of two basic types: either f is unbounded at one end of a finite interval, or f is bounded on an infinite interval. We will consider the second case next and leave the first case as a problem.

Proposition 8. I f the improper Riemann integral JFf converges, and f 2 0 on [O, CQ), then f is Lebesgue integrable on [0,CQ), and the integrals are the same.

Proof. We are assuming that f is Riemann integrable, hence lim f converges. bounded, on each interval [0, n],and that x-co It follows that f is integrable, in both senses, on each interval [n,n 11, and

Jt

+

n=O

where the last integrals are in the Lebesgue sense. For each of the Lebesgue integrals J,n,n+ll f there is a partition P, = { E,i} of [n,n + l ] such that U ( P n ,f ) - L(Pn,f ) < ~ / 2 For ~ . convenience assume that the singletons {n}and { n 1)are elements of each P,. Let Q be the partition of [ O , o o ) consisting of all the sets in all the P,. We have

+

Now add and get

7

THE INTEGRAL OF UNBOUNDED FUNCTIONS

71

Using (4)we conclude

and the integrals are the same. 1111111

Problem 13. Let f be non-negative and Riemann integrable on [ E , 11 for each E > 0 and assume lim :J f = I. Show that f is E.-jq+ Lebesgue integrable over [0,1]with integral I . sllllll

+

Problem 24. Let f,(x) = n3I2x/(l n2x2)for x E [ 0 , 1 ] . (i) Show that f,(x) -+ 0 for all x E [ 0 , 1 ] . (ii) Show that { f,} is not uniformly bounded. (iii) Show that g(x) = l / f i is Lebesgue integrable on [ 0 , 1 ] , and fn 5 g for all n. (iv) Conclude that J f, -+ 0. 41

Problem 15. (i) Let {E,} be a sequence of measurable sets in [0,1]with p(E,) < F1 , and let g, = xE,. Show that g, + 0 a.e. E , and F = n, F N . Show p ( F ) = 0 and Hint: Let F N = g, -+ 0 off F . (ii) Show that assuming p(E,) -+ 0 is not sufficient to conclude that g, -+ 0 a.e. Hint: Let El = [0, $1, 1 3 1 1 1 1 E2 = [z,11, E3 = [o, 71, E4 = [, 21, E5 = [, 71, E6 = [:, 11, E7 = [0,$1, . . . . Problem 16. If f is integrable on S, then for every E > 0 there is 6 > 0 so that JE 1 f I < E whenever E c S and p(E) < 6. Hint: Assume without loss that f > 0 and suppose the statement false. Then there is E > 0 and a sequence {E,} of measurable subsets of S with p(E,) < 1/2" and JE, f 3 E . Use Problem 15 and the Lebesgue Dominated Convergence Theorem with g, = xE, . f to reach a contradiction.

UEN ~11111)

1111111

Problem 17. (Differentiation under the Integral Sign) Let f ( x ,t ) be defined on [a,b]x (c, d ) , with to E (c, d). Assume that for each fixed t E (c, d), f ( x , t ) is a measurable function of x on [a,b],and for each fixed x E [a,b], f (x,t ) is a differentiable function of t on (c, d); i.e., ft(x,t ) exists for (x,t ) E [a,b] x (c, d ) . Assume that f (x,t o ) is integrable on [a,b],and that there

72

A PRIMER OF LEBESGUE INTEGRATION

is an integrable function g on [a,b] such that I f t ( x , t)l 5 g(x) for all (x,t ) E [a,b] x (c, d). Show that: (i) ft(x,t ) is a measurable function of x for each t E (c, d), and hence ft(x, t ) is integrable on [a,b]. (ii) For each t E (c, d), f ( . , t ) is integrable over [a,b]. (iii) If F ( t ) = s,b f ( x , t ) d p ( x ) for t E (c, d), then F’(to) = S,b ft(x,to)dp(x) for each to E (c, d). Hints: (i) Let tn + t E (c, d), so for x E [a, b]

f<.,

tn) -

f @ ,t )

tn - t

-

h(x,t ) .

(ii) For each t E (c, d) there is s E (c, d ) such that f ( x ,t ) = f@,to) h(x,s>(t - to), and hence 1 f<x,t>l I1 f < x ,to)( I s)llt - tol. (iii) Let t, + to, so

+

+

The difference quotients are dominated by g(x) and converge pointwise to ft(x, to). 4l

DIFFERENTIATION AND INTEGRATION

In this chapter we make some connections between differentiation and integration. In particular, we see how the Fundamental Theorem of Calculus fares with respect to Lebesgue integration. The results of this chapter are necessarily germane only to real-valued functions defined on the real line. This is in contrast to our earlier work, where the results are stated for functions and sets on the line, but the techniques and proofs have quite general application. The theorems of this chapter are standard useful results, but the proofs tend to be difficult and of limited applicability. The reader may be well advised to study the statements of the theorems and problems, and leave the proofs for a long rainy day. The two forms of the Fundamental Theorem are

In calculus one shows that (1) holds for a Riemann integrable function f which is continuous at x . This is also true for a Lebesgue integrable function, with essentially the same proof (Problem 1).It is considerably more difficult (Proposition 5 ) to show that if f is bounded and measurable, and so Lebesgue integrable, then (1)holds a.e.

Problem 1. Show that if f is Lebesgue integrable on an interval containing a and %, and f is continuous at xo, and a < %, 73

A PRIMER OF LEBESGUE INTEGRATION

74

and

then

F’(x0)=

f ( ~ ) .

1111111

We will now show that (2) does not hold in general by constructing a continuous increasing function f o n [O,1] with f ( 0 ) = 0, f ( 1 ) = 1, and f’(x) = 0 a.e. Thus

jol f’ = 0 + f ( 1 )

-

f ( 0 ) = 1.

Define f on [0,1] as follows: let f = on [+, $1; then let f = 1 2 on [g, 91 and f = on In the next step, take the middle third out of each of the four remaining intervals, and definef 3 5 to be g1 , i, s, and on these middle thirds. The picture thus far is Fig. 1. Continuing in this way we get f defined on a union of disjoint closed intervals with total length

[g, g].

1 2 4 + - + -+ 3 9 27

-

1 1

27

9

2 9

1 -

= 1.

2 3

3

Fig. 1

7 9

8 9

1

8

DIFFERENTIATION AND INTEGRATION

75

Clearly f’(x) = 0 on the union of the interiors of these intervals, so f’(x)= 0 a.e. The inductive scheme outlined above defines f on a dense subset of [0,1]. To show that this definition can be extended to the rest of [0,1] with the result being continuous, it suffices (Problem 2) to show that f is uniformly continuous on the union of the closed intervals. Notice that

and so on, so f is uniformly continuous on its domain. The function f is known as the Lebesgue singular function.

Problem 2. Let f be defined and uniformly continuous on a dense subset E of [0,1]. Specifically, given E > 0 there is 6 > 0 so that if XI, x;?E E and 1x1 -x21 < 6 , then I f ( x l )- f(x2)l < E . Let x E [0,1] and let {x,}be a sequence in E such that x, +-x. (xmay or may not be in E.) Show that { f ( x , ) } converges, and the limit is independent of the sequence {xa}converging to x. If x E E , show that lim f,(x) = f ( x ) . If x 6 E , define f ( x ) = lim fn(x),where x, E E and x, +-x. Show that f is continuous on [O, 11. ~ l l l l A Vitali covering of a set E is a family V of proper intervals (open or closed or half-open, but not points) such that every point of E lies in intervals of V of arbitrarily small length. That is, given x E E and q > 0 there is I E V such that x E I and l ( I ) < q. The proof that an increasing function has a derivative a.e. depends on the following proposition. Proposition 1. (Vitali’s Theorem) I f p ( E ) < 00 and V is a Vitali covering of E , then for each E > 0 there is a finite disjoint set { I l , . . . , I N } of intervals of V such that ,x(E-(Il U - - . UIN)) < E , and hence p ( E n U I j ) > p ( E ) - E .

Proof. Let U be an open set containing E with p ( U ) < 00. Discard from V all intervals not contained in U , and all intervals which do not intersect E . What remains is of course still a Vitali covering of E . We can assume without loss that

A PRIMER OF LEBESGUE INTEGRATION

76

all intervals of V are closed, because closing them would affect neither the hypothesis nor the conclusion. Suppose we let I1 be any interval from V of maximum length, and 12 any interval of maximum length which does not intersect 11. Of course there may not be such maximum length intervals, but for the moment we assume there are to see how the argument goes. Later we will correct the argument by replacing these by intervals which are nearly of maximal length. Define inductively a disjoint sequence I1,12, . . . of intervals of V such that each I,+l is of largest possible length and disjoint from 11, . . . , I=. If at any stage E is covered by 11 u . . . u I, we are done. Otherwise, since I1 u . . u I, is closed, each x E E - ( I 1 u . . uI,) is in some intervals of V which are so small they miss I1 u . . . u I=. Thus we have a sequence {I,} of disjoint intervals, all contained in U , with Cl(I,) < 00 and L(I,) + 0. Let E > 0 and pick N so large that C;+, l ( I , ) < E . Consider any point x of E - ( I 1 U. . .U I N ) and any interval I of V which contains x and is disjoint from I 1 u - . u I N . The interval I must intersect some I, with m > N,for otherwise since I, is of maximal length among unchosen intervals, l(I,) >_ L ( I ) for all m, and this is impossible since t(I,) -+ 0. Let I, be the first interval that I intersects, so m > N.Since l(I,) 2 L ( I ) , x lies in the interval 1, which has the same center as I, but is three times as long (Fig. 2). Hence every point x of E - ( I 1 U . . U E N ) is in U,"=,+, J m , which has measure less than 3r . Jm

I

I

I

I

I I I

a

X

I I

I

Im

Fig. 2

8

DIFFERENTIATION AND INTEGRATION

77

Now instead of assuming that each I,+, has maximum length, let L, = sup{l(I) : I E Vand I n (11 u . . . u I,) = 0). Then pick I,+1 so that l(I,+l) 2 ;La. The arguments above all proceed as before except we must make J , five times as long as I,, so that E - (11 U . . U I N ) has measure less than SE. 1111111

Problem 3. Let q be any function and assume that { x : lim inf q(t) < lim sup q ( t ) } t+x+

t+x+

has positive (outer) measure. Show there are rationals r and s such that { x : liminfq(t) < r < s < limsupq(t)) t+x+

t+x+

has positive (outer measure). Hint: The measurability of the sets is not relevant here. 1111111

Proposition 2. I f f is increasing on [a,61 then f’(x) exists a.e.

Proof. The four Dini derivates of f at x are D+ f < x ) = lim sup h+O+

D+ f ( x ) = liminf

f<.+

h) h

f @+ h) - f<.>

h D- f ( x ) = lim sup fcx + h) h h+Oh+O+

D- f ( x ) = lim inf h+O-

f<.>

f<.>

f @+ h) - f<.> h

The derivative f’(x) exists if and only if the four derivates are equal and finite at x. The proof consists in showing that the set where any two derivates differ has measure zero. We illustrate by showing that the set { x : D + f ( x ) > D + f ( x ) }has measure zero. If this set has positive measure, then there are (Problem 3) rational numbers r and s such that

A = { x : D+ f ( x ) < r < s < D+ f ( x ) >

78

A PRIMER OF LEBESGUE INTEGRATION

has positive measure. Assume p ( A ) = p > 0, and let U be an open superset of A so that p ( U ) < p E . For each x E A there are, by definition of D+ f ( x ) ,arbitrarily small numbers h such that [x,x h] c U and

+

+

+

f ( x h) - f ( x >< rh.

+

These intervals [x,x h] form a Vitali covering of A, so there is a finite disjoint collection [xi,xi hi],i = 1,2, . . . , N,with

i=l

+

+

Let B = A n UE1(xi,xi hi), so that p ( B ) > p - E and each point of B is the left endpoint of an interval [ y , y k ] ,contained in some (xi,xi hi), such that

+

+

f(Y

+ k) - f ( Y > > sk.

This last inequality follows from the fact that D+ f ( x ) > s for each x E B. Invoke Vitali's Theorem again to get a finite disjoint family [ y j , yj k j ] , j = 1,. . . , M, such that

+

Bn

u[ y j , yj + k j ] ) > p ( ~ ) M

M

It follows that C kj > p 1=1

and

-

j=l

-2

>

~Now . we have

p

-2E.

8

79

DIFFERENTIATION AND INTEGRATION

Since each interval [ y i , y; + K j ] is contained in some [xi , and f is increasing,

+hi],

Thus, for every E > 0

so

and we have a contradiction. 1111111 Proposition 3. I f f is increasing on [a,b], then f‘ is measurable and s,b f’ If ( b ) - f ( a ) .

Proof. Let

n -> 6. Then f n is meawhere we interpret f ( x + :) = f ( 6 ) if x+ 1 surable for each n, and fn(x) + f ’ ( x ) a.e., so f’ is measurable. Each fn is non-negative, so Fatou’s Lemma gives

= liminf =

(nr’ f

f ( b ) - limsupn

- n / aa + f f

la+:

)

A PRIMER OF LEBESGUE INTEGRATION

80

The last inequality follows from the fact that f is increasing, so for all n,

Proposition 4. I f f is integrable on [a, b] and

x E [ a ,b],then f = 0 a.e.

st f

= 0 for all

Proof. Suppose :J f = O for all x , so Jf f = O for all c, d E ( a , b). Suppose, to be specific, that f is positive on a set E of positive measure. Then there is a closed subset F c E so that p ( F ) > 0 and JF f > 0. Let U be the open set ( a , b ) - F , and write U in terms of its components: U = U(ai,6j). Since U U F = ( a , b),

Therefore

+

and so J$ f 0 for some interval (ai,bi),which contradicts the assumption. 1111111 The following result is true for functions which are integrable on [ a ,b],and so not necessarily bounded. The proof of the more general theorem requires additional machinery, and so we stick with the version below, which contains the essential ideas. Proposition 5. I f f is bounded and measurable on [ a ,b] and Ja

then F is continuous and F ( a ) = 0 and F'(x) = f ( x ) a.e.

Proof. Let f + = f v 0 and f - = (- f ) v 0, so f + and f - are non-negative functions and f = f f - f - . Hence F(x) =

LXf' / -

a

X

f-

=Fl(Xj

- FZ(X),

8

DIFFERENTIATION AND INTEGRATION

81

where F1, F2 are increasing functions. It follows from Proposition 2 that F’(x) exists for almost all x. Moreover, if M is a bound for f ,

so F is continuous. Let

=n

lX+: f,

so that 1 fn(x)(I M for all x, and fn(x) + F’(x) a.e. By the Bounded Convergence Theorem and the continuity of F ,

I”

F’

= lim

1”

fn

= F(x)=

/

a

X

f

Now we have JoX(F’- f ) = 0

for all x, and consequently F’ = f a.e. by Proposition 4. 1111111 For convenience we will call the points x where F’(x) = f ( x ) the L-points of f . The L-points play critical roles in many theorems on integral representations. We give one example from potential theory. The problem is the so-called Dirichlet problemnamely, to find a harmonic function u ( r , @ )on the open unit disc in the plane with a specified boundary value f ( 4 p ) on the

a2

A PRIMER OF LEBESGUE INTEGRATION

unit circle. Here we will let f be a bounded measurable func, we regard as the unit circle. The Poisson tion on [ 0 , 2 n ] which kernel is the function

defined for (7, 8) in the open unit disc (polar coordinates), and p on the unit circle. For each fixed po, P ( r , 8; po) is a harmonic function on the open disc (0 5 r < 1 , 0 5 8 5 2 n ) . For each fixed (7, e), P ( r , 8; q ) is a continuous function of p E [ 0 , 2 n ] . Hence f ( p ) P ( r ,8 ; p) is an integrable function of p for each (r, 0) in the disc. We let

&

The is so the total measure of the circle is 1. We can regard the integral as a limit of Riemann sums, so N

1

The sums have the form N i=l

&

where pi(r, 8) = P ( r , 8 ; p i ) and ai = f ( p i ) p ( E i ) . The sums (3) are linear combinations of harmonic functions, and hence are harmonic. The limit of these sums, u(r, e), is also harmonic, although not obviously so. The function u(r, 8 ) solves the Dirichlet problem in the following sense: if po is an L-point for f (and that includes all points where f is continuous) then

In particular, (4)holds for almost every po. Problem 4. Let F ( x ) = f . Show that F’(0) can exist even though f is not continuous at 0. Hint: Let f be defined on [ 0 , 1 ]

8

DIFFERENTIATION AND INTEGRATION

[i,i] [i, i)

83

as follows: f = 1on and f = -1 on (:, 11, so 0 5 J1"z f 5 up into four equal intervals, for 5 x 5 1. Divide J1"7 f 5 and let f be alternately +1 and -1 on these, so 0 I for 5 x 5 etc. Show that f is integrable-even properly Riemann integrable since the set of discontinuities has measure zero-and that ( F ( x ) - F(O))/x+. 0 as x +. O+. 1111111

i,

&

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PLANE MEASURE

In this chapter we develop Lebesgue measure for sets in the plane, R2. Our purpose is threefold. First, by developing another specific example of a measure we show how the same techniques used on the line can be used to define countably additive measures in quite general situations. Our second purpose is to have so we can show that two-dimensional measure-area-defined Jfdp really is the area under the graph of f . Thirdly, the development of plane measure provides a template and example for defining general product measures. We take the rectangle as our basic plane figure, with the rectangle playing the role played by the interval on the line. Here we will use “rectangle” to mean rectangle with sides parallel to the axes and having positive length and width. Thus a rectangle is a set I x J with I and J intervals, which may be open or closed or half-open. Single points and line segments are not rectangles. If R = I x J , then the area of R is a ( R ) = C ( I ) C ( J ) . For any set E c EX2,

where {Ri}is a finite or countable family of rectangles. h ( E ) is the Lebesgue outer measure of E which we will call simply the measure of E . The same arguments used when we defined m(E) on the line will show that it is immaterial whether we use coverings of E by open rectangles or closed rectangles or partially closed rectangles (Problem 2). We can also invoke the plane version of 85

86

A PRIMER OF LEBESGUE INTEGRATION

the Heine-Bore1 Theorem and notice that if E is a compact set, then h ( E ) is the inf of sums C a(Ri)for finite coverings of E by rectangles of any sort.

Problem 1. (i) Show that line segments have plane measure zero; for example, the segment from (0,O) to (1,2) has measure zero. (ii) Show that lines have plane measure zero.

tllllll

It will occasionally be convenient to use coverings by squares k kfl of the form d x e, where d = [z;;, e = [ eF , Te + 1I . We will call an interval of the form [ Fk , k + l with or without endpoints, a dyadic interval of length $. A square of the form d x e will be called a dyadic square of side Any rectangle R can obviously be covered by a finite number of dyadic squares, S1, S2, . . . , S,, with

&.

Any countable covering { R,} of any set can be covered by countably many dyadic squares { &}, where S,1, Sn2, . . . cover R,, and

so

Thus we could replace coverings of E by rectangles in our definition of h with countable coverings of E by dyadic squares. Notice that a finite covering of E by non-overlapping dyadic squares can be replaced by a finite covering of non-overlapping dyadic squares all of the same size, with the same total area. (Rectangles are called non-overlapping provided their interiors are disjoint.)

Problem 2. Show that the infimum of sums C a ( R i )for families { Ri } of closed rectangles covering E is the same as the

9

a7

PLANE MEASURE

infimum of sums C a( Qi) for coverings { Qi} of E by open rectangles, and the same as the infimum of sums C a(Sj) for coverings { S i } of E by dyadic squares. 4I

Problem 3 . h is translation invariant.

llllilI

The following elementary properties are immediate from the definition.

Proposition 1. (i) h ( 0 ) = 0 ; (ii)h(C) = 0 for every countable set C ; (iii)h ( E ) > 0 for all E ; (iv) if E c F , then h ( E ) 5 h ( F ) ; (v)h is countably subadditive; i.e., if { E i } is any countable or finite family, then h (U Ei) 5 C h ( E i ) .

Problem 4. Prove parts (ii) and (v) of Proposition 1. 1111111 Since h ( E ) is to represent the area of E we need to know that

h gives the right answer for rectangles.

Proposition 2. I f Q is a rectangle, h( Q) = a(Q).

Proof. First assume that Q is a closed rectangle. Clearly h(Q) 5 a ( Q ) since { Q }is a 1-rectangle covering of itself. Fix E > 0 and let { Sk} be a finite covering of Q by dyadic squares of the same size so that XU(&) < h(Q)+ E . w e may assume all Sk intersect Q. Since Q is a rectangle, say Q = I x J , the squares Sk will consist of all squares di x ej(i = 1,.. . , n; = 1,.. . , m) where the di form a non-overlapping covering of I and the ej form a non-overlapping covering of J . Hence U S , = U d i x ej k

and

. .

t,1

= (dl

u - .- u d,J x (el u . u ern), a

.

A PRIMER OF LEBESGUE INTEGRATION

88

Since this holds for all E > 0, and we already have h(Q) I p ( I ) p ( J ) ,we conclude that h(Q) = p ( I ) p ( J ) .The remaining cases, where Q is not necessarily closed, are left to the reader as an exercise. 1111111

Problem 5. Complete the proof of Proposition 2 by showing that h ( Q ) = a ( Q ) for an open rectangle Q, and consequently, by monotonicity, for rectangles containing some but not all of their boundary points. 41 We saw in Chapter 4 that the Carathkodory criterion for measurability, while not in itself very intuitive, does get us speedily to the additivity properties that outer measure has when restricted to the measurable sets. We therefore adopt this condition forthwith as our definition of measurability. A set E c R2 is measurable if and only if h(E nT )

+ h(E’ n T ) = h ( T )

for every set T . Here E’ = R2 - E . Since h is subadditive, we always have h(E nT )

+ h(E’ n T ) 2 h ( T ) ,

so E is measurable if and only if, for all T , h(E nT )

+ h(E’ n T ) 5 h ( T ) .

(2)

Clearly we need only consider sets T of finite measure.

Problem 6. (i) Sets of measure zero are measurable. (ii) If E2 = El U Eo with E l measurable and h(E0) = 0, then is measurable. Problem 7. Translates of measurable sets are measurable.

E2

1111111

lllN1

Proposition 3. E is measurable if and only if h ( E f~R)

+ h ( E ’ n R) = h ( R )

for every rectangle R. Proof. (Cf. the proof of Proposition 3, Chapter 4.) The condition is obviously necessary, so assume that E splits all rectangles

9

PLANE MEASURE

89

additively, and let T be any set of finite measure. Let { Rj} be a covering of T by rectangles with

+

C a ( R j )< h ( T ) Then E n T c U ( E n R j ) and E ’ n T monotonicity and subadditivity,

E.

c U ( E ’ n Ri). Hence, by

h ( E n T ) + h(E’ n T ) 5 C h ( E n Rj) + C h ( E ’ n Rj)

+ h(E’ n Rj)] = C h ( R j ) < h ( T )+ = C [ h ( En Rj)

E.

Therefore h ( E n T ) + h(E’ n T ) 5 h ( T ) and E is measurable. 1111111 Proposition 4. Rectangles are measurable.

Proof. Let Q be the rectangle that we want to show is measurable. We show that Q splits any rectangle R additively. R is the union of Q n R and at most eight other non-overlapping rectangles &, . . . , & whose areas total a(R). (See Fig. 1.) Since Q’ n R c S1 U . . . U Sg, h(Q’ n R) 5 a(&)

Therefore, h ( Q n R)

R

+ - - + a(&>. *

+ h(Q’ n R) 5 a ( Q n R) + a(&) + . + a(&) * .

= a ( R ) = h(R). R

I

I

s3

;

s4

I I I I

S1

Fig. 1

:

s5

90

A PRIMER OF LEBESGUE INTEGRATION

Subadditivity gives the other inequality, so Q splits any R additively, and hence Q is measurable. 1111111

Proposition 5. I f { E i } is a finite or countable family of disjoint measurable sets, then h(U E , ) = C h ( E i ) .

Proof. The proof of Proposition 5, Chapter 4,applies verba-

tim to this case. 1111111

Problem 8. If T is any set and { Ei} is a finite or countable family of disjoint measurable sets, then

Proposition 6. I f E l , . . . ,En are measurable, then El U. . is measurable, El n ... n En is measurable, and El measurable.

U En E2

is

Problem 9. Prove Proposition 6. Hint: Cf. Proposition 6 , Chapter 4. ~41 Proposition 7. I f { E,} is a countable family ofmeasurable sets, then U Ei is measurable.

Proof. We can assume the sets are disjoint because the differences E2 - E l , E3 - ( E l u Ez), etc., are measurable by the preceding proposition. If F, = El U . . U E n and E = UEl Ei, then for any T ,

h ( T )= h(T n F,) =

i=l n

>

+ h(T n FL)

C A(Tn E ~+) h(T n FL) n

i=l

h(T n Ei) + h(T fE’). l

Since this holds for all n,

h ( T ) > C h(T n Ei) + A(T n E’) co

i=l

2 h(T n E ) + h(T n El). 1111111

9

PLANE MEASURE

91

Problem 10. If A and B are measurable, A c B , and h ( B ) < 00, then B - A is measurable and h ( B - A) = h ( B ) - h(A). nllNl of

Proposition 8. Every open set in R2 is a countable union open squares, so every open set and every closed set is

measurable.

Proof. Every point (x,y ) of an open set U lies in arbitrarily small dyadic squares. Since some open disc around (x,y ) is contained in U , a sufficiently small dyadic square containing (x,y ) will lie in U . The union of all these squares is obviously U , and there are only a countable number of dyadic squares altogether. Squares are measurable, so open sets are measurable. If F is closed, then F’ = U is open, so measurable, and hence F = U’ is measurable. 1111111 Problem 11. For any measurable set E , h ( E ) = inf{h(U) : E c U , Uopen} = sup{h(F) : F c E , F compact}.

Hint: Show this first for a set E c [0,1] x [0,1] and then write E as a countable union of sets E n S, where the S, are nonoverlapping unit squares. ~lllill Any measure u on a topological space X is called regular provided that all open and closed sets are measurable, and for every measurable set E ,

u ( E ) = inf{u(U) : E c U , Uopen} u ( E ) = sup{u(F) : F c E , F compact}. Problem 11is the statement that h is regular on R x R. We have already seen that p is regular on R.

Problem 12. If for each E > 0 there are measurable sets A and B such that A c E c B and h ( B ) < h(A) E , then E is measurable. Show that the condition h ( B ) 5 h(A) E , which allows h(A) = h ( B ) = 00,does not suffice. 1111111

+

+

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10 THE RELATIONSHIP BETWEEN p AND X

Since R2 = R x R there is naturally a relationship between plane measure and linear measure. In this chapter we will investigate the connection, and thereby preview the development of general product measures. Proposition 1. I f A and B are subsets of R of finite outer measure (but not necessarily measurable sets), then h ( A x B ) 5 P ( A )L o ) .

Proof. Let E > 0 and let { I j } , { J k } be coverings of A and B by intervals, with

This holds for all E > 0, and p (A), p ( B )are finite, so h (A x B ) I p (A) p ( B ) . 1111111

Problem 1. Let A, B be subsets of R with p ( A ) = 0. Show that h ( A x B ) = 0. Hint: If p ( B ) < 00 this is clear. Otherwise, write B = U B, where B, = B n [n,n + l),n = 0, f l ,. . . . 1111111

93

94

A PRIMER OF LEBESGUE INTEGRATION

Proposition 2. I f A and B are compact subsets of R, then h ( A x B ) = P(A)Pu(B). Proof. Since A and B are closed and bounded sets they are automatically measurable with respect to p, and have finite measure. By Proposition l we need only show that h ( A x B ) 2 P(A)Pu(B). Since A x B is compact we can approximate h( A x B ) by the total areas of finite coverings by dyadic squares all of the same size. Let { S i j } = {di x e,} be such a covering, where all ei and dj are dyadic intervals of length 1 / 2 N ,and

Ca(Sij)< h ( A x B ) + I . . .

r 3 1

We may assume that { S i j }is exactly the set of all dyadic squares of side 1/2Nwhich intersect A x B , and hence that

where { d l , . . . , d,} is a non-overlapping covering of A and {el, . . . , em}is a non-overlapping covering of B. Thus

The use of dyadic squares of the same size is necessary to ensure that we have a finite covering of A x B of the form {Ii x J j : i = 1, . . . , n ; j = 1, . . . , m}. 1111111

Proposition 3. I f A and B are measurable subsets of E,then

Proof. First assume that A and B have finite measure, so there are compact sets F and G, with F c A, and G c B , and

10

THE RELATIONSHIP BETWEEN p AND h

95

Hence

Therefore the desired equality holds if A and B have finite measure. If p (A) = 00 or p ( B ) = 00, we consider two cases: (i)p (A) = 00 and 0 < p ( B ) < 00, with p ( A ) p ( B )= 00, and (ii)p ( A ) = 00 and p ( B ) = 0, with p ( A ) p ( B )= 0. In case (i), for each n there is a compact set F c A with p ( F ) > n. Therefore, for every n,

so h ( A x B ) = 00 = p ( A ) p ( B ) .If p ( A ) = p ( B ) = 00 then clearly h ( A x B ) = 00. In case (ii), we let A, = A n [n,n 1) for -00 < n < 00,so p(A,) < 00 and

+

SO h ( A

x

B ) = 0 = p ( A ) p ( B ) . 1111111

Proposition 4. I f A and B are measurable subsets of R, then A x B is a measurable subset of R x R.

Proof. We check that h ( ( A x B ) n R) + h ( ( A x B)’n R) 5 h ( R ) for every rectangle R = I x J which is sufficient by Proposition 3, Chapter 9. Since A and B are measurable,

A PRIMER OF LEBESGUE INTEGRATION

96

Let

Rl = ( A n I ) x ( B n J ) = ( A x B ) n R R2=(AnI)x(B’nJ)

R3 = (A’ n I ) x ( B n J ) R4=(A’nI)x(B’nJ). R1, R2, R3, R4 are disjoint, their union is R, and R2 U R3 U R4 = ( A x B)’ n R. By Proposition 1,

with similar inequalities for R2, R3, R4 in place of RI. Hence from (l),

+

+

+

h ( R ) 3 h [ ( A x B ) n R] h(R2) h(R3) h(R4) h [ ( A x B ) n R] + h [ ( Ax B ) ’ n R]. 1111111 In the general study of measure theory one starts with a 0algebra of subsets of some set X , and a countably additive measure function u on these specified “measurable sets.” To define the product measure u x u on X x X one starts with the basic sets of the form A x B , where A and B are measurable subsets of X . The product measure of such sets is of course u x u ( A x B ) =

44 L o ) .

Our plane measure h is of course the same as the product measure p x p, although we did not define it that way. There is a fundamental intuitive idea of area, based on the area of a genuine rectangle, which is more or less independent of linear measure. We therefore used a(I x J ) = t ( I ) t ( j ) as our basic notion in defining h on plane sets. This procedure was also designed to provide a second example of how an outer measure

10

THE RELATIONSHIP BETWEEN p AND h

97

is defined in terms of some more basic notion. In the following proposition we show that h could equally well have been defined startingwithh(AxB) = p ( A )p ( B ) formeasurablesets Aand B rather than intervals. Proposition 5 . For any set E

c R2,

h ( E ) = inf {Ch(A,x Bi) : E = inf {Cp(A,)p(Bi): E

c U A, x Bi}

c U A, x

Bi}

where { A,} and { Bi} are countable families of measurable subsets of R. Proof. Let h,(E) denote the right side above; i.e., h,(E) would be the outer measure of E starting with measurable sets A x B rather than rectangles I x J . Clearly h,(E) I h ( E ) since

the inf is over a larger collection of coverings. On the other hand, E c U A, x Bi and subadditivity and Proposition 3 imply that Hence h ( E ) = A,( E ). 1111111 Now we have the following: (i) If A and B are measurable subsets of R, then A x B is a measurable subset of R2 and h(A x B ) = p(A) p ( B ) . (ii) If E c R2, then

where the {A,} and { B i } are countable families of measurable subsets of R. We finish this chapter by showing how plane measure is related to linear measure through integration. In calculus one defines the area of the plane region

S = {(x,y ) : a I x 5 b, 0 I y If<x>>

to be the integral S,b f d p . We now have a second definition of this area as h ( S ) . It is necessary, and not hard, to show that these definitions agree.

98

A PRIMER OF LEBESGUE INTEGRATION

Problem 2. If f is measurable and non-negative on [a,61, and S is the region under the graph of f as given above, then S is measurable and h ( S ) = s,b f d p . 1111111

We want to establish the connection between area and integration for more general plane regions, and thus pave the way for multiple integrals and the Fubini Theorem. Let E be a measurable subset of the plane, and to start we will assume that E c [0,1] x [0,1]. Let Ex be the cross section of E over the point x E [0,1];i.e., Let f ( x ) = p(E,), so f ( x )is the “length” of the x cross section of E . We proceed to show that f is measurable and, as one would expect, h(E)=

/’ f d p . 0

Problem 3 is an essential lemma for Proposition 6.

Problem 3. Let F be a compact subset of [0,1] x [0,1], and let F, be the 3c cross section of F for each x E [0,1].For x E [0,1], let L,(x) be the total length C t ( e i ) of all dyadic intervals of length 2-” which intersect F,. Let a,(F) be the total area C a(d; x e i ) of all dyadic squares of side 2-” which intersect F . (i) F is h-measurable, and each F, is p-measurable. (ii) a,(F) decreases to h ( F ) as n -+ 00. (iii) L,(x) decreases to p(F,) as n + 00. (iv) What goes wrong if F is not compact? 1111111 Proposition 6 . I f F is a compact subset of [0,1]x [0,1]and F, is the x cross section, and f ( x ) = p( F,), then f is measurable and =

/

0

1

f (X)dP(X).

Proof. Let {di x e j } be all the dyadic squares of side 2-“ which intersect F . For x in the interior of di, let cp,(x) be the total length C t ( e i ) of all e , such that d; x ei intersects F . That is, cp,(x) is

THE RELATIONSHIP BETWEEN

10

AND h

99

the total length of the column of dyadic squares in the covering which lie over di. If x is an endpoint of di there may be two columns over x, so define pn(x)only for interior points of the di, and let pn(x) = 0 elsewhere. Thus pfl is a simple function and

Moreover, pn(x) -+ p(F,) for all x which are not dyadic points; thus pfi(x) -+p ( F J a.e. It follows that f ( x ) = p ( F x ) is measurable, and li?

1'

pndp =

If n is so large that

1'

dp(x)=

p ~ ( ~ x )

1fdp. 1

0

C a ( d i x e,) < h ( F ) + E

for the dyadic covering sets di x e , of side 2-", then h ( F ) 5 C a C d i x ej>

Hence

Recall that h is a regular measure, so for any measurable subset E of finite measure there is an open set U and a compact set F so that F c E c U and h(E)- I < h(F)5 h(E) 5 h(U) < h ( E )

+E.

If E c ( 0 , l ) x ( 0 , l ) then of course U and F can also be taken as subsets of ( 0 , l ) x (0,l). Proposition 7. If E is a measurable set, then p( E x ) is measurJR p(E,) d p ( x ) = h ( E ) .

able and

A PRIMER OF LEBESGUE INTEGRATION

100

Proof. First assume that E c (0, 1)x (0, 1).Let Fn c E c Un with each F, compact and each U, open, and h(U, - F n ) < ;1. Assume that F1 c F 2 c . . - and U1 3 U2 3 Let K , = [0, 11 x [0, 11 - U,, so Kn is compact, and p ( F f l x )p(KK,,) , are measurable functions, with .

A(&)

.

a

.

= 1 - h(K,) r1

It follows that lim fn(x)= limg,(x) = p(E,)

a.e.,

and so p(E,) is a measurable function and

= h ( E ). 1111111

Problem 4. Extend the above proof of Proposition 7 so that

it applies to general measurable sets.

111111

Notice that the preceding results are proved first for subsets of the unit square. The final result is obtained by adding up

1o

THE RELATIONSHIPBETWEEN p AND

a

101

what happens on a finite or countable number of squares. This is possible because both the line and the plane can be written as countable unions of finite measure sets. A general measure u with this property is called o-finite. Thus u is a o-finite measure on X if X = u Ei with u ( E i ) < 00 for each i. The hypothesis of o-finiteness is essential in the general discussion of product measures. Integrals with respect to A are defined just as they are for p, and the same integration theorems hold. Integration for general measures will be treated in Chapter 12, but for now the reader should take on faith that A-integrals have the standard properties. Proposition 7 provides the basis for the proof of the Fubini Theorem, which states that A-integrals (“double integrals”) can be evaluated as iterated p-integrals. In practice, there is no other way to evaluate most double integrals. One of the most useful aspects of this connection between A-integrals and p-integrals is the fact that generally speaking the order of integration in iterated integrals can be reversed. The “Fubini Theorem” is the name commonly associated with the results of the next proposition, but the reader is cautioned that the name Tonelli is also given to this form. Proposition 8. (Fubini’s Theorem) Let f ( x , y ) be a nonnegative measurable function, and define f x by f x ( y ) = f (x,y). Then f x is a measurable function on IW for almost all x, and the function F (x)defined by

is measurable and non-negative, and

I f f is integrable, f x is integrable for almost all x and F is integrable; conversely, if the iterated integral is finite, then f is A-integrable. Of course the same statements apply to the iteration in the other order, so the order of integration for iterated integrals (of a positive function) is immaterial.

102

A PRIMER OF LEBESGUE INTEGRATION

Proof. First we give an outline of the proof. For characteristic functions the result follows directly from Proposition 7. By linearity of the integrals-all three of themthe result then holds for simple functions. A non-negative measurable function f is the limit of an increasing sequence q n of simple functions. This fact depends on the o-finiteness of R2. (If R2 were not o-finite we could not guarantee the existence of such functions qn which are zero off a set of finite measure; cf. Problem 8 below.) The theorem extends from simple functions to f by the Monotone Convergence Theorem. Now we fill in the details. Let f ( x , y ) = a x ~ ( xy,) where E is a measurable set and h ( E ) < 00. Then

By Proposition 7,

Integrating both sides of (2),and using ( 3 ) ,

i.e., for f ( x ,y ) = ~

X E ( X y, ) , we

have the result:

If q(x,y ) is a simple function, then massive applications of linearity give the result. We illustrate the argument for the sum of two very simple functions. Let El and E l be disjoint measurable sets of finite measure, and let f = f~ f2 where f~= alXE,,

+

1o f2

THE RELATIONSHIPBETWEEN

p AND a

103

= a 2 x E 2 . Then

Now let f be measurable and non-negative on R2. There is an increasing sequence { p n ( x ,y ) } of simple functions so that pn --+ f a.e. By the Monotone Convergence Theorem

That is,

The right side of (4)could be +oo. For each fixed x,po,(x,y ) is an increasing sequence of simple functions, and

104

A PRIMER OF LEBESGUE INTEGRATION

For each n,

i=l

By Proposition 7, it follows that each cDn is a measurable function of x , since p ( E x ) is measurable if E is. Since cD,(x) increases to F(x), F is measurable. ( F might take the value +cm.)Hence the integral

is a measurable function of x. Finally, by the Monotone Convergence Theorem,

By (4) we have the desired result for non-negative measurable functions f . The A-integral is finite if and only if the iterated integral is finite. If f is a not necessarily non-negative function, but is integrable, then the result holds for f + and f - and hence for f . 1111111 The following problem illustrates the way the Fubini Theorem is usually used to justify changing the order of integration in an iterated integral.

Problem 5 . Let f ( x , y) be measurable. If

// I

f < x ,Y)IdP(4dP(Y) < 00

10

THE RELATIONSHIPBETWEEN

p AND a

105

then f ( x , y) is A-integrable and the order of integration can be reversed. 1111111

Problem 6. Let X be an uncountable set, and S the a-algebra of all subsets of X . Let p be counting measure on X , so p ( E ) is the number of elements in E if E is finite, and p ( E ) = 00 otherwise. Show that the function f which is identically one on X is measurable, but there is no sequence {qn} of simple functions which increases to f a.e. 1111111

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11 GENERAL MEASURES

The basic additivity property of Lebesgue linear or planar measure holds when the outer measure is restricted to the a-algebra of measurable sets. We axiomatize the general theory of measure by starting with a a-algebra S of subsets of some set X , and a countably additive non-negative extended real valued function u defined on the sets in S. This triple ( X , S, u ) is called a measure space. In practice, one says “ u is a measure on X.” The a-algebra S generally disappears from the discussion, and we write “ E is measurable” to mean u ( E ) makes sense-i.e., that E E S. Measures are automatically monotone in the sense that if A and B are measurable and A c B , then u(A) 5 u(B). This is clear because B - A is measurable and

u ( B ) = u ( A )+ u ( B - A) 3 u ( A ) .

Countable subadditivity is also an automatic property of measures, as we show next. Proposition 1. If { E i } is a finite or countable family of mea-

surable sets, then

Proof. Let F, = En - ( E l U . . . U E,-l), so the F, are measurable and disjoint, and U F, = U En. Clearly u(F,) 5 v(E,) for all n, so

107

A PRIMER OF LEBESGUE INTEGRATION

108

Problem 1. Let X be any set and for E c X, let u ( E ) be the number of elements in E if E is a finite set, and u(E) = 00 otherwise. Show that u is a measure on the a-algebra of all subsets of X. This measure is called counting measure. 41 Problem 2. Let X be an uncountable set. Let S be all subsets of

X which are either countable or have a countable complement. Show that S is a a-algebra. Let u ( E ) = 0 if E is countable and u ( E ) = 1 if E is uncountable. Show that u is a measure on X. Hint: “Countable” means “countably infinite or finite.” nl~l~!

Problem 3. Let f be a non-negative function on a set X. Let u ( E ) = CxEE f ( x ) , where the sum is the unordered sum in the sense of Problem 9, Chapter 2. Show that u is a measure on all subsets of X. If Xis uncountable and f is strictly positive, show that u ( X ) = 00. [If f is the characteristic function of a single point x, then u is called point-mass at x, and Jgdu = g ( x ) for all g .I 41

Proposition 2. Let { E,} be a sequence of measurable sets.

(i) I f El 3 E2 2 . . , and u ( E 1 ) < 00, then 1

u

(ii) I f El c

E2

(nE,)

= limv(E,).

c . , then

Proof. (i) Let El 3 E2 2 . . and let E = n E,. Then El = E U ( E l - E2) U (E2 - E3) U . . . , and the summands are disjoint. Hence

+

v(Ei) = ( v ( E i ) - u(E2)) ( ~ ( 4 5 2 ) u(E3)) (U(En-1) - v(E,)) * . v(E) = u(E1) - lim u(E,) u(E).

+

+

+ +

+

* * *

*

Since u(E1) < 00, subtraction is legitimate and gives the result. (ii) Let El c E2 c . . and let E = U E,. Then

E = El

U (E2 - E l ) U (E3 -

El) U

* * .

,

11

GENERAL MEASURES

109

and hence

The two measures p on IW and h on IW x IW already studied have the property that every set of (outer) measure zero is a measurable set, and consequently every subset of a set of measure zero is measurable. Measures with this property are called complete. In a general measure space ( X , S , u ) it is possible to have subsets AandBof XwithAcBES,andu(B)=O,butA$S,sov(A) makes no sense. However, for any measure space (X, S , u ) we can extend S to a larger a-algebra SO by throwing in all subsets of zero-measure sets. The measure u can then be extended to a complete measure UO on So by defining uO(A) = u ( B ) if B E S and A differs from B by a subset of a zero measure set; i.e., if AAB c C with u(C) = 0.

Problem 4. Define a non-complete, non-trivial (i.e., not identically zero) measure on a a-algebra of subsets of the threeelement set X = { a , b, c}. 41 Proposition 3. Any measure can be extended to a complete

measure.

Proof. Let (X, S , u ) be a measure space. For brevity let us say that a subset A of X is a null set provided A is a subset of some B E S with measure zero. Let So consist of all sets of the form (E U A) - B where E E S and A and B are null sets. Clearly S c SO since 0 is a null set. We can assume that the sets in So have the form (E U A) - B where A and B are null sets such that E n A = 0 and B c E. In this case, (E U A) - B = ( E - B ) U A [(E U A) - B]’ = (E’ U B ) -A, so SO is closed under complementation. Let F, = (E,u A,) - B, SO,with E, E S , and A,, B, null sets. Then

E

110

A PRIMER OF LEBESGUE INTEGRATION

where C c U B,, and C is consequently a null set. Therefore

and U An, C are null sets. Hence So is a a-algebra. We define

UO[(EU A)

-

B ] = u(E).

To show that uo is well defined, let

(El U Al)

-

B1 = (E2 U

A2) - B2

with El, E2 E S and Al, A2, B1, B2 null sets. Let A1 be a zero measure set in S which contains Al. Then

ElUA, c E ~ U A ~ U B I , El UA1 c E2 U A2 U B1 UKl. Since A2 u B1 UK1 is a null set, there is C E S with u ( C ) = 0 and C 2 A2 U B1 U A-1. Hence El U A1 c E2 U C, and both of these sets are measurable. Hence

v(E1) = v(E1 U A1) 5 u ( E U~ C) = u(E2). The argument is symmetric, so u(El) = u(E2) and uo is unambiguously defined. It is left as an exercise to show that uo is countably additive on So. 1111111

Problem 5. (i) Show that every set in So can be written F U C with F E S and C a null set. (ii) Show that So.

ug

is complete and countably additive on

1111111

Problem 6. (i) What are So and uo for the completion of your example in Problem 4? (ii) Are the measures of Problems 1 and 2 complete? 41 Since any measure can be extended to a complete measure, u is henceforth assumed to be a complete measure. We will say that u is finite if u ( X ) < 00, and u is a-finite if Xis a countable union of finite-measure sets. If u1 and u2 are two measures on the same a-algebra of subsets of X , then it is clear that ul + u2 is also a measure on X .

11

GENERAL MEASURES

111

More generally, any finite or countable sum C a i u i with all ai > 0 is again a measure on X if the ui are measures on the same a-algebra. If we consider a difference u1 - u2 of two measures we again get a countably additive set function provided both measures are finite. However, if there is a set E such that u l ( E ) = u2(E) = 00, then (u1 - uZ)(E) makes no sense. More complicated atrocities can happen. Since ul - u2 could take both positive and negative values, it would be possible for CE1(u1- u2)(Ei) to converge conditionally for some disjoint sequence { Ei}. Conditional convergence means that the sum depends on the order of the summands. Since U Ei obviously does not depend on the arrangement of the Ei, the countable additivity condition

would be an impossibility in this case.

Problem 7. Show that if u1 and u2 are finite measures on the same a-algebra of subsets of X , then u1- u2 is countably additive on this o-algebra. 4l Now let us consider a countably additive set function u, defined on some o-algebra, with u taking both positive and negative values. We assume that u ( 0 ) = 0, and allow the possibility that some sets have measure +00, but allow no sets to have measure -m. The other choice of allowing -00 but not +00 would work equally well. We call such a function u a signed measure. The countable additivity assumption requires that if { E i } is a disjoint family, then u ( U E i ) = C u ( E i ) and the sum of the negative terms u ( E i ) is finite. Otherwise the union of these Ei would be a set of measure -00. Thus for any disjoint family { E i } the sum C u ( E ; ) converges absolutely, or diverges to +oo as an unordered sum. From Problem 7 we know that some signed measures arise as the difference of two positive measures, and we show next that in fact every signed measure is of this form. A measurable set A is called a positive set provided u ( E ) 2 0 for every measurable subset E of A, including A itself.

A PRIMER OF LEBESGUE INTEGRATION

112

A measurable set B is called a negative set provided u ( E ) 5 0 for all measurable E c B. If A is both a positive set and a negative set, so that u ( E ) = 0 for all E c A, then A is called a null

set. Notice that “null set” in this context is different from “null set” in the earlier completion argument.

Problem 8. If u is a signed measure on X and A is a positive set for u , and u l ( E ) = u ( E n A), then u1 is a positive measure on X . If B is a negative set for u and u2(E) = - u ( E n B ) , then u2 is a positive measure on X . 1111111

We proceed to show that every signed measure u can be written as the difference of positive measures as indicated in the preceding problem, where B is a “largest” negative set for u and A = X - B. The problem is to show that there is a “largest” negative set B , so that X - B is necessarily a positive set. We find B first rather than A because of the possible complications stemming from the fact that u(A) might be +oo. Proposition 4. Let u be a signed measure on X . I f u ( E ) < 0, then E contains a non-null negative set.

Proof. If E contains no subsets of positive measure, then E is

a non-null negative set, and we are done. Otherwise let

pi

= s u p ( ~ ( F :) F

c E},

0 < pi 5 00. Let be the largest of the numbers 1, such that < p , , and pick F1 c E with

SO

d

i, i, -

If nl > 1, then l / n l < p1 5 l / ( n l - 1).If E - F1 has no subsets of positive measure, then E - F1 is a negative set, and

u ( E - F 1 ) = u ( E ) - v ( F 1 ) < 0, so E

- F1

is non-null. If E

- F1

is not a negative set, we let

113

GENERAL MEASURES

11

&

so 0 < p 2 5 00. Again let be the largest of the numbers 1, 1 3 ' . . such that < p 2 , and pick F 2 c E - F1, with

&

i,

Continuing this way we either find a non-null negative set of the form E - ( F 1 U . - U F N ) , or we find a disjoint sequence { F k } such that for all k 1

and if nk > 1, then

The

Fk

are disjoint, so

Since u ( E - U F k ) f: -co and u ( E ) < 0, the positive series C u ( F k ) must converge to a finite number. Therefore C $ converges, so n k --+ 00, and hence p k +0. w e claim that E - u F k is a negative set. Otherwise, there is G c E - IJ Fk with u(G) > 0. For some N,p N < u(G),and this contradicts the definition of p N . The relation shows that u ( E - U F k ) < 0, and hence E

-

U F k is non-null. 1111111

Problem 9. Does the proof of Proposition 4 with the appropriate changes work to show that if u ( E ) > 0, then E contains a non-null positive set? 41 Proposition 5. I f u is a signed measure on X , then there is a positive set A and a negative set B so that A and B are disjoint and A U B = X . Proof. If there are no measurable sets of negative measure, then we take B = 0, A = X , and we are done. Otherwise, there

114

A PRIMER OF LEBESGUE INTEGRATION

are sets of negative measure, and hence negative sets of negative measure. Let

q = inf{u(E) : E a negative set}. Let {B,} be a sequence of negative sets such that u(B,) +q < 0. The union of negative sets is a negative set, so we can assume that B1 c B2 c B3 c ... Let B = U B,, so that B is a negative set and

u ( B ) = limu(B,) = q.

By our assumption that u does not take the value -00, we have u ( B ) > -00, so there is no set of negative measure disjoint from B , and X - B is a positive set. 1111111

A decomposition of X into disjoint sets A and B , with A a positive set and B a negative set, is called a Hahn decomposition of X . Such a decomposition is not unique since any null set can be thrown into either A or B. However, apart from null sets, “the” Hahn decomposition is unique.

Problem 10. Let u be a signed measure on X and let A1 and A2 be positive sets for X and B1, B2 negative sets for X , with X = A I U B 1 = A 2 U B Z a n d A l n B 1= A 2 n B 2 = s . S h o w t h a t A1 A A2 is a null set, which proves that the Hahn decomposition is unique except for null sets. 1111111

If A and B form a Hahn decomposition of X for the signed measure u, then we know from Problem 8 that u can be written as the difference of two measures, u = u+ - v - , where u+(E) = u(E n A), V ( E ) = v ( E n B ) . The two measures u+ and u- are supported on the disjoint sets A and B in the sense that u + ( B ) = 0 and u-(A) = 0. We will say that two (positive) measures ul and u2 on X are singular (or mutually singular, or singular with respect to each other), provided there are disjoint sets A and B with A U B = X and u1 ( B ) = v2( A) = 0. The canonical representation u = u+ - u- of a signed measure u as the difference of two singular measures is

11

GENERAL MEASURES

115

called the Jordan decomposition of u . The Hahn decomposition of X into a positive set A and a negative set B is only unique up to null sets. However, the null sets do not affect the Jordan decomposition u = u+ - u- of u into singular measures, because null sets do not affect the definition of u+ and u - . Therefore the Jordan decomposition u = u+ - u- is unique. If u is a signed measure, then we define the absolute value of u or total variation of u by

+

I u ~ ( E= ) U+(E> u - ( E ) . Observe that IuI is again a measure on X , and Iu(E)I 5 Iul(E) for all E .

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INTEGRATION FOR GENERAL MEASURES

In this chapter we consider a general measure space ( X , S , u ) , and define the integral of an extended real-valued function with respect to u. Measures are non-negative and all measures are complete. Measures are at least a-finite, which means that X can be written as the countable union of finite-measure sets, but this does not rule out the possibility that u ( X ) < 00. Earlier we defined the integral in two steps-first bounded functions on finite measure sets, then general functions and sets. Here we do it in one swift swoop-the functions are not necessarily bounded and the sets may have infinite measure. A function f is measurable provided {x : a 5 f ( x ) < b} is measurable (i.e., a set in S , and therefore a set in which u is defined) for all a and b. Since S is a a-algebra, all the usual equivalent conditions for measurability still hold. (See Proposition 4, Chapter 5.) For functions whose graphs lie in a finite area, U Ei x [ - A&, M i ] with C M i u ( E i ) < 00, we again show that measurability is necessary and sufficient for integrability. The reader is encouraged to notice that the development here is basically the same as for integrals on the line (Chapter 7),and to consider specifically what each statement means for integrals with respect to plane measure h. Most of our earlier theorems, and their proofs, will carry over verbatim to the present setting. Let S be a measurable set. A partition P = { E i } of S is a finite or countable collection of disjoint measurable sets, all of finite measure, whose union is S . Measurable sets, including X, have partitions because Xis a-finite. Another partition Q = { P i } 117

118

A PRIMER OF LEBESGUE INTEGRATION

is a refinement of P , denoted P 4 Q o r Q + P , provided each set F , of Q is a subset of some set Ei in P . The partitions of S form a directed set under the partial ordering +. Problem 1. Show that all measurable sets have partitions. 1111111 For the partition P = { E j }of S, we again let mi and Mi be the inf and sup of the function values on E i , and let

Mj = SUP{] f ( x ) I : x

E

Ej}.

If C M j u ( E i ) < 00, then P is an admissible partition for f on S, and f is an admissible function on S. If P is an admissible partition, and only then, we write

U f ,P ) = Cmiv(Ei) U ( f, PI = C M i V ( E i ) .

(1)

For admissible partitions, the lower and upper sums (1)are either finite sums or absolutely convergent series. Since f can be unbounded, and even occasionally take the values f00,some mi and Mi may now be infinite. If mi or Mi is infinite, then necessarily u ( E i ) = 0, so M j u ( E i ) = 0. When convenient we can lump all zero measure sets in a given partition together in a single set Eo, and effectively ignore it. The lower sums L( f, P ) and the upper sums U ( f, P ) are nets on the admissible partitions P . Lower sums increase and upper sums decrease when a partition is refined, and all lower sums are less than or equal to all upper sums. The lower sums { L( f, P ) } form an increasing net which is bounded above by any upper sum, and the upper sums { U ( f, P ) } form a decreasing net which is bounded below. Both nets { L( f, P ) } and { U ( f, P ) } converge, and lim L( f, P ) = sup L( f, P ) 5 iyf U ( f, P ) = lip U ( f, P ) . P

P

If the limits are equal, f is integrable over S, and we write Js f for the common limit.

Problem 2. If f and g are integrable over S and f 5 g a.e. on S, then ss f 5 ssg. 1111111

12

INTEGRATION FOR GENERAL MEASURES

119

Problem 3 . If g is constant, k, on the finite-measure set T , and g = 0 on S - T , then Js g = ku(T). Hint: The sets T and S - T may not form a partition of S since sets of a partition have finite me a sure.

1111111

Proposition 1. If f is admissible and measurable on the measurable set S, then f is integrable over S.

Proof. Let P={Ei} be an admissible partition of S, so C Miu(Ei) < 00.Let E > 0 andchoose Nso that CEN+lMiu(Ei) < E . The set T = Eiu . . u E N has finite measure, and I f 1 is bounded on T by K = rnax { M I , . . . , &IN}. Let { 11, . . . , I,} be a covering of the interval [-K, K] by disjoint intervals each of length less than &/u(T).Let F , = { x E T : f ( x ) E I j } , SO that Q = { F j } is a partition of T , and

The sets of Q, together with EN+^, EN+^, - . . form a partition Po of S and rn

u- U f , Po) < u i=N+1 < E + 2 E = 3 E . 1111111 Next we show that measurability is also necessary for integrability. It is convenient to modify the definition of simple function to include now all functions which can be written p = C ai X E , , where { E i }is a finite or countable family of disjoint finite measure sets, and C JaiIu(Ei)< 00. If P = { E i } is an admissible partition for f on S and mi, Mi are as usual, and

then pp and

@p

are simple functions with pop I f I @p.

Problem 4. Show that if p p and @ p are as in (2),then pop and @P are integrable and their respective integrals are L( f, P ) and u( f, P ). 4l

120

A PRIMER OF LEBESGUE INTEGRATION

Problem 4 shows that approximating by upper and lower sums is equivalent to approximating by the integrals of upper and lower simple functions. Thus a function is integrable over S if and only if for each I > 0 there are simple functions q and $ on S withq 5 f 5 $ and J $ - J q < E .

Problem 5. Show that a simple function is measurable. 41 Problem 6. Show that the a.e. pointwise limit of a sequence of measurable functions is measurable. Hint: The proof hasn’t changed-see if you remember it. Start with sup f,. 1111111 Problem 7. Let {g,} be a decreasing sequence of measurable functions which are non-negative and integrable on S. If Js g,, -+ 0, then g, --+ 0 a.e. Show that “a.e.” is necessary. 1111111 Proposition 2. I f f is integrable over S , then f is measurable.

Proof. Let PI + P2 + P3 + . - . be a sequence of admissible partitions of S , with U( f, P,) - L( f, P,) < f. Let p, @, be simple functions such that J q, = L( f, P,) and J @, = U ( f, P,).

Then (9,)is an increasing sequence of measurable functions since PI + P2 + . . . , and q, 5 f for all n. Similarly, {$,} is a decreasing sequence with $, >_ f . The functions {$, - 9,) are non-negative and decreasing, and measurable by Problem 5, and -

=

w f, P,)

-

L( f, Pa)

-

0.

By Problem 7, $,-pa -+ 0 a.e., so q, -+ f a.e. (and $,, + f a.e.), and f is measurable. 1111111 The integral is of course a linear functional, and we want to prove that next. Linearity is not particularly obvious from our definition, so we again show that the integral is the limit of Riemann sums. Linearity is then obvious since R( f g , P , c ) = R( f, P, c) R(g, P , c), and the limit of a sum is the sum of the limits. If f is an admissible function on S, and P = { E i } is an admissible partition for S, we define R( f, P , c) = f(ci)v(Ei), where ci E Ei for each i. We will write R( f, P , c) only for admissible partitions, so the notation implies the admissibility of P .

+

+

z

12

INTEGRATION FOR GENERAL MEASURES

121

The pairs ( P , c ) form a directed set as usual with ( P , c ) 4 (P’, c’) meaning the same as P 4 P’. For all P and c

L ( f ,PI 5 R ( f , p , c ) 5 W f , P ) . It follows almost immediately that the net { R( f, P , c ) }converges to the integral if f is integrable. Problem 8. Show that if f is integrable over S then {R(f, P , c ) } converges to Js f . Hint: Take note of the fact that { R( f, P , c ) } is a net on a different directed set than the nets { L (f, P ) } and { U ( f, P ) } . Problem 9. If f and g are integrable over S, then f g and kf are integrable over S, and 1111111

+

.I,kf=kJ

S

f.

Hint: An admissible partition for f may not be an admissible partition for g. What is the domain of the net { R( f g , P , c)}? Proposition 3. If f is defined on S, then l i p R( f, P , c ) = I if and only if f is integrable (hence measurable) over S and Js f = I. Proof. The “if” part is Problem 8 above, so we assume that R ( f , P , c ) -+ I . The notation R( f, P , c ) presumes that there are admissible partitions for f . Let E > 0 and let Po = { E i } be a partition such that 1 R( f, P , c ) - I I < E whenever P > Po, so in particular any two Riemann sums R( f, Po, c ) and R( f, Po, c’) for Po are within 2~ of each other. We can choose ci E Ei with f ( c i ) > M i- E / ~ ~ u ( & )so , R( f, P , c ) is within E of U ( f, Po). Similarly, if f (ci) is within ~ / 2 ~ u ( Eofi )mi,then R( f, Po, c’) is within E of L( f, Po). It follows that U ( f, Po) - L( f, Po) < 48, and f is integrable. By Problem 8, lim R( f, P , c ) = J f . The details are the same as in Proposition 2 of Chapter 7. 1111111

+

1111111

The limit theorems for the Lebesgue integral depend basically on these facts:

122

A PRIMER OF LEBESGUE INTEGRATION

I. If fn +f uniformly on a finite-measure set T, then JT f n * JT f .This is the convergence theorem for Riemann

integrals, and of course it still holds here. 11. If f n +f pointwise on a finite measure set S, then f n +f uniformly off sets of arbitrarily small measure. 111. If g 2 0 is integrable over S, then there are finite-measure subsets T of S such that g is bounded on T and J T g is arbitrarily close to Js g. It follows that if I fnl Ig and f n + f , then

and all these T integrals are close to integrals over S.

Proposition 4. (Egoroff's Theorem) I f { fn} is a sequence of measurable functions and f n -+ f a.e. on a finite-measure set S, then given 6 > 0 there is an exceptional set E c S with u ( E ) < 6 such that f n + f uniformly off E .

Proof. We can assume (Problem 10) that fn(x)-+ f (x)for all x E S. For a given E let AN = { x E S : I fn(x)- f(x)l 2

E

for some n 2 N}.

Clearly A1 2 A2 II A3 II . . . Since fn(x) + f ( x ) for all x, for each x there is some N with x # AN;i.e., n AN = 8. Since u(A1) < 00, and the AN are nested, lim U ( AN)= 0, and for any given 6 there is N with u(AN) < 6. Thus we have for each pair E > 0, 6 > 0, an integer N a n d a set AN of measure less than 6 such that I fn(x)- f (x)l < E for all x $ AN and n 2 N. Now let {&k} be a sequence of E'S with &k -+ 0. Let 6 > 0. For each pair &k, 6/2kwe find as above an integer and a set Tk of measure less than 6/2k such that I f n ( X ) - f ( x ) l < E k if n >_ A& and x # q.We can assume the A& increase. For any given E > 0 there is & (corresponding to some &k < E ) such that 1 f n ( X ) - f(x)l < &k < E if n 2 and x # q.Let T = U so ~(7') < 6 and if x # T , then x # Tk. Hence if x # T and k 2 hlk, I fn(x)- f ( x ) l < E . That is, f n + f uniformly off T and u ( T ) < 6. 1111111 ~

z,

12

INTEGRATION FOR GENERAL MEASURES

123

Problem 10. Show that Egoroff’s Theorem holds if f n + f a.e. on S (instead of for all x E S as in the proof). 1111111 Problem 11. Prove the Bounded Convergence Theorem. If { f a } is a uniformly bounded sequence of measurable functions on the finite measure set S , and f n -+ f a.e. on S , then Js f n + Js f . 1111111 The next proposition takes care of the cases where the function is unbounded or the set has infinite measure. Proposition 5. I f g is integrable on S and E > 0, then there is a finite measure set E such that JSPE lgl < E and g is bounded on E .

Proof. Let P = { Ei} be an admissible partition of S for g , so E M i u ( E i ) < 00. If M i u ( E i ) < E , and E = UEl Ei7 then JSPE lgl < E and lgl 5 max{Ml,. . . , M N } on E , and u( E ) < 00. 1111111

ErN+l

Problem 12. Prove the Lebesgue Dominated Convergence Theorem: If { f n } is a sequence of measurable functions on S, and I fnl 5 g on S , and g is integrable on S , and f n + f a.e. on S , then Js f n + Js f . Hint: Do not forget to show that f is integrable.

111111/

The Dominated Convergence Theorem says that integrals converge nicely provided all the functions stay in a fixed finite area (between -g and g ) . If they don’t-i.e., if the f n are allowed to wander out into pastures of infinite area-then Fatou’s Lemma says what can go wrong. If 0 5 f n + f , then the integrals J f n can of course approach J f ,but if they don’t it’s because they have included too much area outside f . Proposition 6. (Fatou’s Lemma) I f { f n } is a sequence o f nonnegative integrable f u n c t i o n s , and fn -+ f a.e. on S, then lim inf J f n 2 J f if f is integrable, and lim J f n = 00 i f f is not integrable.

Proof. We suppose f is integrable over S with Js f > 0, and leave the other case as a problem. Let P = { E i } be a partition

124

A PRIMER OF LEBESGUE INTEGRATION

such that L( f, P ) = C miu(Ei) is within E of J f . Pick N large enough so miu(Ei) is within 2~ of J f . Let h = mixE, be the corresponding simple function. The function h is bounded f. and non-zero on the finite measure set T = UE, E i , and h I Since f n -+ f , f n A h + h, and { f a A h} is a uniformly bounded sequence on a finite-measure set. Therefore

xi”,l

and lim inf J f n 2 J f . IIIIIII The “monotone” convergence theorem is immediate from f n 5 f and f n + f , then Jfn -+ Sf. Fatou’s Lemma: if 0 I

Problem 23. Give two examples where f is integrable and lim inf Js f n > Js f : one example where S has finite measure so the f n can’t be uniformly bounded, and one example where the f n are uniformly bounded, so u( S) must be infinite. 1111111 Problem 14. (Second Part of Fatou’s Lemma) Let { f n } be a sequence of non-negative integrable functions on S such that f n -+ f on S but f is not integrable. Show that lim Js f n = 00. Hint: Use the partition En = { x : 2” If ( x ) < 2”+’} for n = O , f l , f 2 , ... to find a simple function hN If such that hN is bounded and non-zero on a finite measure set and J hN > N. lilllll Problem 25. Define f + ( x ) = max{f(x),O} and f - ( x ) = min{ f ( x ) ,0) so f + ( x )3 0 and f - ( x ) 5 0. (i) Show that f + and f - are measurable if and only if f is measurable. (ii) If f is measurable, then f is integrable if and only if f + and f - are integrable, and then Jf = J f + + J f - . 1111111 It is conventional to write Js f = 00 if f is non-negative and measurable on S, but not integrable (i.e., not admissible). With this convention Fatou’s Lemma can be stated thus: i f { f n } is a

12

125

INTEGRATION FOR GENERAL MEASURES

sequence of non-negative measurable functions and fn -+ a.e., then

f

This is now correct even if f is not integrable.

Problem 16. Show that if 0 5 f n -+ f and the surable, then

fn

are mea-

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13 MORE INTEGRATION: THE RADON-NIKODYM THEOREM We continue to consider a complete, non-negative measure u on a a-algebra S of subsets of X . X will always be at least ofinite, and sometimes finite. Since X is o-finite it can be written as a countable disjoint union of finite-measure sets, so Xand its measurable subsets have partitions. We have seen that if f is integrable over Xand T c X, then f is integrable over T and JT f = Jx f . xT. Hence if f is integrable over T we can extend f to X, with f = 0 off T , and write JT

f

= sx

f.

We will show next that if f is integrable over X, and B ( E ) = JE fdu, then B is a new (signed) measure on X. If f 2 0 then /3 is a positive o-finite measure on X. We will then characterize l can be represented as integrals. those measures #which If Aand B are disjoint measurable sets and f is integrable over A U B , then X A U B = X A X B and hence

+

L U B

f

=/f(xn+xd =

.I,f +/

B

f.

Thus if B ( E ) = JE fdu, B is finitely additive:

for disjoint sets E l , . . . , E,.

Problem 1. (i) If f is non-negative and integrable over X , and B ( E ) = JE f d u for every measurable set E , then B ( E ) = 0 whenever u ( E ) = 0. 127

A PRIMER OF LEBESGUE INTEGRATION

128

(ii) Is /3 necessarily complete? Hint: Subsets of u-measure zero sets are measurable, but are subsets of B-measure-zero sets measurable ? 41

Proposition 1. I f f is a non-negative integrable function on of X,then /? is a finite measure on X.

X,and B ( E ) = JE f for each measurable subset E

Proof. From Problem 1 we know that B(s) = 0. The monotone property of /I is clear since f is non-negative. Since j3 is defined on the same sets as u by definition, B is defined on a

o-algebra. We need only show that B is countably additive. Let { E i } be a countable disjoint family of measurable sets, with E = U Ei. By finite additivity we have

c N

i=l

B(Ei) = =

+.

5

/fxEldV

If i=l

(XE1

+

*

. . -k

X E N ) ~ ~ .

(1)

We let h N = f . ( x E 1 . .+xEN)so { h N } is an increasing sequence of non-negative measurable functions, and { h N } converges to f X E , where E = U Ei. From (1)we have

and therefore by the Monotone Convergence Theorem N

03

i=l

=liml N

= =p

X

hNdu

fXEdv

(uE i ) .

1111111

Problem 2. Suppose f is non-negative and measurable on X, but not integrable. If B ( E ) = J' fdu for measurable sets

13

MORE INTEGRATION

129

E on which f is integrable, and B ( E ) = 00 otherwise, is /3 a measure ? ~llldl For any two finite measures B and u on X we will say that B is absolutely continuous with respect to u, written B << u, provided B ( E ) = 0 whenever u ( E ) = 0. Problem 1 shows that if B is an integral with respect to u, then B is absolutely continuous with respect to u. We now show that absolute continuity characterizes measures B which are u-integrals. That is, if B attaches measure only to sets E where u ( E ) is positive, then there is a weighting function f for u which gives B. Problem 3. Show that if /3 and u are finite measures on X , then /3 << u if and only if given E > 0 there is 6 > 0 so that u( E ) < 6 implies B ( E ) < E . Hint: The “if” part is just logic, since if u ( E ) = 0, then u(E) is less than any 6. To show “only if,” assume the contrary condition that B << u and there is a sequence {E,} with u(E,) < 1/2” and /!?(En)2 E . If E = &(En U E,+l U . . -) then u ( E ) F: 1/2”-l for all n. If F, = E, U E,+1 U then F1 3 F 2 3 . . . and n F, = E , SO p ( F , ) -+p ( E ) . -

.

a

,

~111111

Proposition 2. I f ,d and u are finite measures with ,6

<< u and

B $ 0, then there is a set A with u(A) > 0 and a positive number E

such that E U ( E )I B ( E ) for every measurable E c A.

Proof. For every n let A,, B, be a Hahn decomposition of X

iu.

iu

for the signed measure Thus /3 2 on all subsets of A, and /3 I i u on subsets of B,. The A, increase and the B, decrease; let B = n B, and A = U A,. Then for all n,

1 1 B(B) I B(B,) I -@,) 5 -v(X>. n n Since u ( X ) < 00, B ( B ) = 0. Since B 0, B(A) > 0, and hence u(A,) > 0 for some n. Thus we have a set A, of positive umeasure and a positive number E ( = such that i u ( E ) 5 B ( E ) for all E c A,. 1111111

+

A)

Corollary. I f

B and u are finite measures with B << u, there

are non-negative functions g such that for all E

130

A PRIMER OF LEBESGUE INTEGRATION

Proof. The function g = ; x A , with A, as in Proposition 2 is such a function, since for any set E ,

Proposition 3. (The Radon-Nikodym Theorem) I f B and v are finite measures on X and ,6 << v, then there is a non-negative v-integrable function f on X such that

for every measurable set E . Proof. Assume B and u are defined on the same a-algebra in X , with B << v. We know there are non-negative functions f > 0 such that for all E , JE f d v 5 B ( E ) . If fi and f 2 are two such functions, then f 1 v f 2 is again such a function (Problem 4).

Let .Fbe the set of all such functions, and let

p

= sup

{L

fdv

:

f

E

F).

Clearly p > 0, and since B ( X ) < 00,p < 00. Let f , E .F with Jf,dv + p. Let g, = f 1 v f 2 v . . . v f,, so the g, increase and Jxg,dv + p . Let f = sup f , = limg,, so

L

and for any set E ,

fdv

fdv = P,

= limJ1gndv

(3)

5 B(E).

(4)

Now we have to show that equality holds in (4).Let a be the measure defined by

a ( E ) = ,B(E) -

E

fdv.

13

MORE INTEGRATION

131

Clearly a is finite, and positive, and a << u. Either a = 0, which is what we want, or by Proposition 2 there is a set A and I =- 0 such that A is a positive set for a - E U . If a $ 0, and E c A,

Hence for any set E ,

which is a contradiction, and hence a

Problem 4. Show that

fi

v

f2 E

= 0.

.F if

fi,

1111111 fz E .F. ~1~~111

Problem 5. Let B be Lebesgue measure on [0,1], so B is defined on the a-algebra of Lebesgue measurable subsets of [0, ll. Let u be counting measure on the same o-algebra; i.e., if E is measurable, u ( E ) is the number of elements in E , so u ( E ) is usually +GO. Obviously B << u . Show there is no measurable f on [0,1] such that B ( E ) = JE fdu. Where does the proof of the Radon-Nikodym Theorem fail? 1111111

When there is more than one measure in sight, the concepts of “zero measure” and “almost everywhere” become ambiguous. We therefore write a.e.[u] for “almost everywhere with respect to u.” So “ f = g a.e.[u]” means that f = g except on sets of u-measure zero.

Problem 6. Show that the f of Proposition 3 is unique; i.e., if f and g are two such functions, then f = g a.e.[u]. 1111111

132

A PRIMER OF LEBESGUE INTEGRATION

The weighting function f for u which gives B is called the Radon-Nikodym derivative of B with respect to u, and we write dp = f d u , or = f.

2

Problem 7. Extend the Radon-Nikodym Theorem to the case where B and u are a-finite, and B << u. Hint: Recall the convention that we write JE f d u = 00 if f is non-negative and measurable but not integrable over E . 4l The Radon-Nikodym theorem characterizes measures B which have positive mass only where u has positive mass. The antipodal relationship would be a measure all of whose mass is carried on some set of u-measure zero. For example, Lebesgue measure p on [0,1] is a uniform mass distribution where mass is the same as length. A point mass at ~0 E ( 0 , l ) would be the measure B on measurable sets such that B ( E ) = 1 if xo E E and B ( E ) = 0 otherwise. This B is clearly a finite measure, and all of B’s mass is carried on the set {xg} which has p-measure zero. Recall that two measures /3 and u are singular with respect to each other if there exist disjoint measurable sets A and B such that AUB = Xand u(A) = B ( B ) = 0. We say that is singular with respect to u or u is singular with respect to B depending on the emphasis. The notation is B 1u. Proposition 4. I f /3 and u are two o-finite measures on X (ie., on the same o-algebra of subsets of X), then B can be written as a sum B = Po with Bo << u and 1u.

+

Proof. Let h = B + u , so that h is o-finite and both B and u are absolutely continuous with respect to A. Let f be a non-negative measurable function on X such that u(E)= for all measurable sets E . Let

A = {x : f ( x ) > 0 } , B = {x : f ( x ) = O } , so A and B are disjoint measurable sets and X = AUB. We define

13

Clearly B = Bo

MORE INTEGRATION

+ PI. To see that U(B) =

S, fdh

133

Iu observe that =

JB Odh = 0.

To see that BO << u, let u ( E ) = 0. Then f = 0 a.e. [h]on E ; i.e.,

Since

Bo(E) = B ( E n A) 5 h ( E n A) = 0 Bo(E) = 0 if u ( E ) = 0, and

<< u.

IIIIIII

Problem 8. If B and u are finite measures on X, and and /3 Iu, then B ( E ) = 0 for all E . 1111111

B << u

Problem 9. Suppose /?and u are finite measures on X, and ,i? << u. Let h = B + u, so B << h, and hence B ( E ) = JE gdh for some g 3 0 and all measurable sets E . Show that 0 5 g < 1 almost everywhere with respect to u ; i.e., g(x) < 1 except possibly on a set E of u-measure zero. ~1~~~~~

Problem 10. Let S and 7 be a-algebras on the same set X, with 7 c S . Let u be a finite measure on S and let f be a non-negative u-integrable function. Let uo be the restriction of u to 7. The function f is necessarily S-measurable but not necessarily 7-

measurable, and therefore not necessarily uo-integrable. Show that there is a 7-measurable, uo-integrable function g on X such that JAgduO= JA fdu for all A E 7. 1111111

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PRODUCT MEASURES

Let ( X , S, p ) and (Y,J , u ) be two measure spaces. We want to define the product measure a on Z = X x Y so that sets A x B have measure A(A x B ) = p ( A ) u ( B ) .The process is roughly parallel to that used to develop plane measure h in Chapters 9 and 10. Perhaps the most useful result of this chapter is the theorem which allows one to change the order of integration in an iterated integral:

The most effective path to this result involves showing that both iterated integrals equal the “double” integral with respect to the product measure A:

In many cases the product integral above is little more than a curiosity except for its relationship with the iterated integrals. Plane measure h on R2 provides our model in a limited sense. Recall that the outer measure of a set E c R2is the inf of all sums C a(Rj), where { Ri} is a sequence of rectangles which cover E , and a(Rj) is the area. One of the first problems with plane measure was to show that the outer measure of a rectangle is its area; i.e., that outer measure really extends the basic idea of area for rectangles. The same problem had to be faced for linear measure: namely, to show that the outer measure of an interval is 135

136

A PRIMER OF LEBESGUE INTEGRATION

its length. For both linear and plane measure we could use compactness to reduce some countable coverings to finite coverings, and thereby simplify the arguments. In the general setting of this chapter we have no such simplification, which accounts for the distressing amount of set theory which is necessary. Our plan in outline is as follows. We will call a set A x B , with A a measurable subset of X and B a measurable subset of Y , a rectangle. We will define the “area” of a rectangle by

a(A x B) =P ( A ) O ) , and the area of a finite disjoint union of rectangles as the sum of their areas; thus, if E = Al x B1 U. . . U A, x B, and the A, x Bi are disjoint, then we define n

n

The problem immediately arises to show that (1)gives a consistent definition, since the representation of E as a finite disjoint union is not unique; indeed, a rectangle can be written as a finite disjoint union of rectangles, and in many different ways. Once we have shown that a is consistently defined by (1)on the set R of all finite disjoint unions of rectangles, we show that R is an algebra of sets; i.e., a family of sets which is closed under finite unions and finite intersections, and differences and complementation. Since a is consistently defined on R, a is finitely additive on R. R is not necessarily a a-algebra, so countable unions of sets in R need not belong to R. However, a countable disjoint union of sets of R might be in R, and if so, say E = U Ej E R, we need to know that a is countably additive for this family:

Equation (2) is essential if U Ei E R because our final measure h is to be an extension of a,so of course (2)must hold for a if it is to hold for A. Once we have a defined on the algebra R, with a countably additive on R in the above sense, we define h on all subsets of

14

137

PRODUCT MEASURES

2 in terms of countable coverings by sets in R: h(E) = inf{xa!(Ei) : E

c U E i , Ei

E

R}.

Measurability with respect to h is defined by the Carathkodory criterion, and as before the measurable sets form a o-algebra. The sets inRmust be shown to be measurable, and then we know that the a-algebra of measurable sets contains the o-algebra generated by the rectangles A x B. Now the details. Throughout this chapter ( X , S , p ) and (Y,3,u ) will be measure spaces, and Z = X x Y . All subsets A x B of Z with A and B measurable subsets of X and Y respectively are called rectangles, or sometimes measurable rectangles. We let R denote all sets in Z which are finite unions of rectangles. We define a! on rectangles by a ( A x B ) = p(A)u(B). Proposition 1. Every set in R can be written as a finite disjoint union of rectangles. R is an algebra; i.e., R is closed under finite unions and finite intersections, and differences and complementation. Proof. The two-dimensional picture (of genuine rectangles) shows that the union, intersection, or difference of any two rectangles can be written as a finite disjoint union of rectangles. The complement of a rectangle is a finite disjoint union of rectangles, since

( A X B)’ = (A’ x B’) U (A’ x B ) U ( A x B’). To show by induction that any finite union of rectangles is a finite disjoint union, suppose this is true for any n rectangles. Let R1, . . . , R,+1 be n 1 rectangles. Let

+

R1 U . U R, = S1 U . . U S, * *

*

where the Si are disjoint rectangles. Then each set Si a finite disjoint union of rectangles, so

R1 u .

* *

u R,+1

= R,+1

u (S1 - &+I) u

* * *

u (S,

-

-

R,+1 is

R,+1)

and the right side is a finite disjoint union of rectangles. Of course any finite intersection of rectangles is again a rectangle.

138

If E l , E2

A PRIMER OF LEBESGUE INTEGRATION

E

R,

with the Ri and the S, disjoint rectangles, then the Ri n Sj are disjoint rectangles and

so R is closed under finite intersections. 1111111

Problem 1. Show that the intersection of any family of alge-

bras is again an algebra. Hence there is a smallest algebra of sets containing the rectangles; this is called the algebra generated by the rectangles. 111111 Now we want to show that a is countably additive on R. The critical step is the next proposition.

Proposition 2. I f A x B is a finite or countable union of disjoint rectangles, A x B = U Ai x Bi, then

Proof. The characteristic function of E = U Ai x Bi = A x B 1s

Integrate both sides of ( 3 )with respect to u, using the Monotone Convergence Theorem in the case of a countable sum:

Now integrate both sides of (4)with respect to p, again using the Monotone Convergence Theorem for a countable sum:

14

PRODUCT MEASURES

139

This is the same as

a ( A x B ) = C a ( A i x Bi). 1111111

R and E is written as a finite or countable disjoint union of rectangles in two different ways, Proposition 3. I f E

E

E = U R j = USj,

then C a ( R j > = Ca(Sj>. Proof. Each Ri and S j can be written as a disjoint union of

rectangles as follows:

Rj = U Ri n S j i

si = U R~n S i . 1

By Proposition 2,

C ~ (n SRj ) ~ i a ( s j )= C ~ ( nRsi). ,

a ( & )=

i

Therefore,

Now we can define a unambiguously on the algebra R by

for any finite disjoint family { R i }of rectangles whose union is E . Moreover, using Proposition 3, it is easy to verify that a is countably additive on R.

Problem 2. Show that a is countably additive on R; i.e., if Ej = U j Rii E R, where {Rij}is a finite collection of disjoint rectangles for each fixed i , and E = U Ei E R, where { E i } is a countable disjoint family, then a ( E ) = C a(Ei). 1111111

140

A PRIMER OF LEBESGUE INTEGRATION

The outer measure h is defined on all subsets of Z by h ( E ) = inf

{Ccr(Ej): E c UEj, Ei E R } ,

where the inf is over all countable coverings {Ei}. This differs formally from the definition of plane measure in Chapter 9 in that here we use coverings by sets which are finite unions of rectangles (i.e., sets in R) rather than coverings by rectangles. The difference is not material because a is additive on the sets in R. The formal change to R rather than sticking to rectangles A x B is desirable because R is an algebra of sets, and we will use this fact.

Problem 3. Verify that h is an outer measure; i.e., that h(0) = 0, and h is a monotone, non-negative, and countably subaddi-

tive function defined on all subsets of 2. ~ l ~ N ~

As before, a set E

c 2 is measurable provided

h ( E n T ) + h(E’ n T ) = h ( T )

for every set T . The proofs of Chapter 4 apply without change to show that the h-measurable sets form a o-algebra, and that h is countably additive on this o-algebra. Proposition 4. The sets in R are measurable; in particular,

rectangles are measurable. Proof. (cf. Proposition 4, Chapter 9) Let E show that

h ( n~T )

E

R. We must

+ h(E’ n T ) F h ( T )

for every set T . Assume that h ( T ) < 00 and let { E i }be a countable covering of T by sets from R, with

Then

E ’ n T c u ( E i n E’),

14

PRODUCT MEASURES

141

and since R is an algebra, the sets E n Ei, E’ n Ei are all in R. Therefore, since a is additive on R,

h ( E n T ) + h(E’ n T ) 5 C a ( E n E j ) + C a ( E ’ n Ej) = C [ a ( E n Ei)

+ a(E’ n Ej)]

=C a ( E i )

< h(T)

+

1.

Since E is arbitrary, E is measurable by the Carathkodory criterion. 1111111 We will let R, denote all sets which can be written as countable unions of sets in R, and R,a will denote all countable intersections of sets in R,. We will find for every measurable set E a set A in R, which includes E and has approximately the same measure. We then find a set B in Raswhich includes E and has the same measure as E . We can show that sets in R,, and then those in R,s, have the basic property relating h to p and u : = /u(A,)dP(x)

*

Here A is of course a subset of X x Y and for each x the x-section of A:

E

X , A, is

A, = { y E Y : (x, y ) E A}. We then get the result for arbitrary measurable sets E by limit theorems on the integrals, and from this we can show the iterated integrals are equal. First we need the following lemma.

Proposition 5. Every set A E R, can be written as a union of increasing sets in R, and every set B E Ras can be written as the intersection of a decreasing sequence of sets in R,.

Proof. Let A E R,, with A = U Ei and each Ei E R. If F, El U . . . U E,, then F, E R and F1 c Fz c . , and A = U F,. Now let A1 and A2 be sets of R,, with

=

142

A PRIMER OF LEBESGUE INTEGRATION

and all the sets

E l i , E2j E

R. Then

and all the sets Eli n E z j are in R, so A1 n A2 E R,. Hence R, is closed under finite intersections. If B E R,a, with B = n A, and each A,, E R,, then let

A;= A l n . . . n A , , so each A:

E

R,, AT 1 A; 2 . . . , and B =

A:. 1111111

Proposition 6. I f E is any measurable set with h ( E ) < 00,and E > 0, there is a set A E R, with E c A and h(A) < h ( E ) E. There is also a set B E Ros with E c B and h ( B ) = h ( E ) .

+

Proof. Let { E i } be a countable covering of E by sets in R with

If A = U Ei, then A E R,, E

c A, and

+ :.

Now let A, E R, with E c A, and h(A,) < h ( E ) We may assume that A1 3 A2 1 . . . . Let B = n A,, so B E &a and E c B . Since h(A1) < 00 and the A, are nested, h ( B ) = limh(A,) = h ( E ) . 1111111

Proposition 7. All sections of sets in Ros are measurable; i.e., if B E Ras then B, = { y : (x,y ) E B } is a measurable subset of Y for each x and BY = {x : (x,y ) E B } is a measurable subset of X for each y. Since R c R, c the sections of R and R, sets are measurable.

Proof. If E is a rectangle, or a finite union of rectangles (i.e., a set in R ) then sections are obviously measurable. If A E R,, with A = U Ej, Ej E R, then for each x

14

143

PRODUCT MEASURES

so A, is measurable. If B E Raswith B = n A, and each A, R,, then each (A,), is measurable so

E

is measurable. 1111111 Proposition 8. I f B

E

ksand h ( B ) < 00,

and

then f and g are measurable functions, and

Proof. If B is a rectangle, the result is clear. If B E R,, then B is a countable disjoint union of rectangles, B = U R,. Hence

so f is the sup of a sequence of measurable functions. Moreover, by the Monotone Convergence Theorem,

= lim

c n

i=l

Now let B E k s , with h ( B ) < 00. Then, as in the proof of Proposition 6, we can write B = n A, with A, E R,, h(A1)< 00, and A1 2 A2 3 . . We know

144

A PRIMER OF LEBESGUE INTEGRATION

so u ( A l ) , is measurable and finite-valued for almost all x. For these x,u(A,), is a decreasing sequence of integrable functions and, since u(Al), is integrable, / u ( Az),dP

But

-

/u

(B,) dF (XI.

and so

h(B) = /u(B,)dp(x).

1111111

To show that

holds for all measurable sets E of finite measure, we surround E with a RaJ set B with the same measure. The result follows once we show that u(B,) = u(E,) a.e. [ p ] . Proposition 9. I f E is measurable and h ( E ) < 00, then E, and EY are measurable for almost all x and almost all y, and

f (XI =

g(y) = P(EY)

are measurable functions defined a.e., and h ( E )= /f(X)dP(X) = /g(y)dv(y).

Proof. Let h ( E ) < 00 and use Proposition 6 to select B E RaJ withE c Bandh(B) = h(E).LetC= B-E,soCismeasurable and h(C) = 0. For all x, C, = B, - E,. Let H E Ra8 where H 3 C and h ( H ) = h ( C ) = 0. Then by Proposition 8

Jv(H,)dp(x)

= h ( H ) = 0.

14

PRODUCT MEASURES

145

It follows that u(H,) = 0 a.e., and since H, 3 C,, u(C,) = 0 a.e. Therefore Ex = B, a.e., so u(E,) is measurable and

The companion result for y-sections is proved similarly. 1111111

Proposition 10. (Fubini) Let ( X , S, p ) and (Y,J , u ) be two measure spaces, and let A be the product measure on X x Y. For any integrable function f (x,y ) on X x Y let fx(Y> =

f<., y )

fY<.>

>

=

f @ , y).

Then f , is a n integrable function on Y for almost all x and is a n integrable function on X for almost all y, and F(x)=

/ f kY W ( Y ) , Y

G(y)=

/

X

f<%

f Y

Y)dP((X)

are integrable functions on X and Y, respectively, and /xF(x)dP(x) =

/ G(y)dv(y)

=

Y

a.e.,

s,,,

f b ,Y ) d W , y ) ;

Proof. By symmetry we need only show that f , is integrable for almost all x,and F ( x ) = Jf,(y)du(y) is integrable, and

It is sufficient (Problem 4 below) to consider only non-negative functions f (x,y ) . Since f (x,y ) is integrable and non-negative, there is an increasing sequence {qB}of simple functions on X x Y so that pn(x,y ) + f ( x , y ) a.e. [A]. It follows that (VB>X(Y)

-

fx(Y>

a.e.[AI.

146

A PRIMER OF LEBESGUE INTEGRATION

If ga is simple, ga is a finite or countable sum of functions @(x, y ) of the form

where E is measurable and h ( E ) < 00. Notice that

@x(y)= aXE,(y>. Since Ex is a measurable subset of Y for almost all x, x E , is a measurable function of y for almost all x, and each summand $ x ( y ) of gax(y) is measurable. Therefore ga, is measurable for almost all x if ga is a simple function. If pOn(x, y ) increases to f ( x ,y ) , then ( 4 0 ~ ) ~increases to fx, so fx, is measurable for almost all x. By the Monotone Convergence Theorem,

-

/(gan)x(Y)dv(Y)

/ f . ( Y ) d U ( Y ) = F (x)

for almost all x. Since each gan is simple, each integral on the left is a measurable function of x by Proposition 9, so F is a measurable function. By the Monotone Convergence Theorem (twice), and Proposition 9 again,

-

- Jx,v

f ( x , y ) d h ( x , y ) . IIIIIII

Problem 4. Show that it suffices to prove the Fubini Theorem for non-negative functions. *111111 Problem 5. Let X and Y be uncountable sets, for instance X = Y = [0,1]. Let p { x } = 1 for all x E X a n d u { y } = 1 for all y E Y , so both p and u are counting measure. If h = p x u, then h{(x,y ) } = 1 for all (x,y). Show that no sequence of simple

functions increases to the function f which is identically one on X x Y , so the proof of Fubini’s Theorem fails. (Of course this f is not integrable.) 1111111

14

PRODUCT MEASURES

147

Proposition 11. (Tonelli’s Theorem) I f ( X , S, p ) and (Y,3,u ) are a-finite measure spaces and f is a non-negative measurable function on X x Y, then the necessary measurability conditions hold as in the Fubini Theorem, and

Proof. If X = U X,, Y

=

U Y, with each X, and Y, of finite

measure and X I c X2 c . . . , Yl c Y2 c . . . , then X x Y = U X,x Y, and X, x Y, has finite h-measure. There is a sequence {q,j} of simple functions supported on X,x Y, which increases to f on X, x Y,:

Therefore there is a sequence lCr,(x, y ) of simple functions which increases to f ( x , y ) on X x Y . The point is that the @, are zero off finite-measure sets, which was the critical point where Proposition 9 is used in the proof of Fubini’s Theorem. The rest of the proof of Proposition 10 proceeds as before. 1111111 To use Tonelli’s Theorem to change the order of integration in an iterated integral

you first check that p and u are a-finite measures. If you are an analyst, X and Y are probably both the real line or the plane, so this is no problem. (If you are a probabilist, then you need more help than you can get here; I suggest prayer.) Given that X and Y (or p and u ) are a-finite, you check that f is measurable. Generally this is because f is continuous. If all you have is facts such as f (x,y ) is continuous in y for each x and measurable in x,then you are in the realm of Polish topology, which is beyond the scope of a primer. If you know that f is measurable, and either iterated integral of 1 f ( x , y ) 1 is finite, then you can change the order of integration.

148

A PRIMER OF LEBESGUE INTEGRATION

Problem 6. If f ( x ) is integrable on X and g ( y ) is integrable on Y and h(x, y) = f ( x ) g ( y ) ,then h is integrable on X x Y and

Do you need to assume that h is non-negative or that X and Y are a-finite? 41

15 THE SPACE L2

The study of Fourier series in the early 1800s gave rise to many fundamental advances in analysis. The question was this-what functions can be represented by Fourier series? It soon became clear that this question could not be answered without a better understanding of the basic ideas of analysis, including what is meant by “function” and “represent.” After the Lebesgue integral was introduced in 1904, the space L2 of square-integrable functions and the space l 2of square-summable sequences emerged as heroes. We will start with 12,which we introduce as the infinite dimensional analogue of Euclidean space. We are familiar from elementary physics and calculus with the treatment of R2 and R3 as vector spaces. If x = ( X I , x2) and y = ( y l , y2) are elements of R2, and a is a real number, then we define vector sum and scalar multiplication as follows:

x + y = (Xl

+,

+ y1, x2 + y2)

a x = (ax1,ax2).

(1) (2)

The sum, of (1)makes R2 an abelian group, and the scalar multiplication (2)then makes R2 a vector space, or linear space. Similarly, R3 is a linear space under coordinate-wise addition and scalar multiplication. If x = ( X I , x2, x3) and y = ( y l , yz, y3), then x + y = (XI

+

y1, x2

+ Y2,x3 + Y3)

ax = (ax1,ax2, ax3). 149

150

A PRIMER OF LEBESGUE INTEGRATION

The absolute value of a vector (point) of R or R2 or R3 is its distance from the origin. For x E R, 1x1 is the distance from the origin, and Jx- yI is the distance between x and y . For x = (XI, x2) in R2, 1x1 = is the distance from x to the origin, and

d m

is the distance from x to y. Each R" becomes a linear space with coordinate-wise addition and scalar multiplication. For x = (XI, . . . , a)E R"with n 2 3 the distance from the origin is called the norm instead of the absolute value, and the notation acquires two additional vertical bars:

For any R"the norm satisfies the following relation with scalar multiplication:

The dot product or scalar product of vectors in R2 or R3 is familiar, and we make the analogous definition for vectors in R" : if x = (XI, . . . , xn)and y = (yl, . . . , y,), then

Problem 1. Verify that the following identities hold in R".

(Parts (iii), (iv),(v)say that the inner product is a linear function of each variable.) 1111111

15

THE SPACE L

151

In R,R2, and R3 the angle 8 between two vectors x and y is determined by the formula

We can define the angle 8 between vectors in any R” by ( 3 ) provided I(x, y)( 5 llxll IIyJI for all x and y. The definition of 8 is not important, but the inequality is.

Problem 2. (Cauchy’s Inequality.) If x = (XI, . . . , x,) and y = (Yl, , y,>, then * * *

i.e.,

Hint: First show that it is sufficient to prove this when llxll = IIyII = 1. Notice that lab] 5 (a2 + b2)/2 for all a, 6, since (a - b)2 2 0, and therefore (xiyil 5 (xi” y:)/2 for each i. Summation, using llxll = IIyII = 1, gives the result. 4I

+

The other elementary but important inequality for norms in R”is Minkowski’s inequality, which gives us the triangle inequality for the metric IIx - yII. Proposition 1. I f x = (XI, . . . , x,) and y = (y1, . . . , y,), then

Ilx+ YII 5 i.e.,

IIXII

+ IIyII;

152

A PRIMER OF LEBESGUE INTEGRATION

so

Taking the square root of both sides gives us the inequality we wanted to prove. 1111111

Problem 3. Let d(x,y) = 11x - yII be the distance between x and y in R". Verify the three conditions which characterize a metric: (i) d(x, y) 2 0 and d(x, y) = 0 if and only if x = y; (ii) d(x, y) = d(y, x) for all x , y; (iii) d(x, y) d(y, x) 2 d(x, z ) for all x, y, x.

+

1111111

The family of all sequences x = (XI, x2,. . . , x", . . .) again forms a linear space under coordinate-wise addition and scalar multiplication. To get a norm analogous to that for R" we restrict our attention to sequences x such that Ex," < 00. This space is called f2, and in f 2 we define the norm

and the inner product

Problem 4. (i) The sum in (4)converges absolutely and I (x, y) I I Ilxll IIyII for all x, y E 12. (ii) Given that x,y E f2, show that x + y E l 2 and that Ilx+ YII I llxll + IIYII. 11'' 11 Now fix a measure space ( X , S, u ) , and let L2 be the space of all square-integrable functions on X ; i.e., f E L2 if and only if J f 2 d u < 00. We will primarily be interested in just two measure spaces: (i) X = [-n,n], which is the natural home for the trigonometric functions cos kx, sin kx; and (ii) X = N,with

15

THE SPACE L

*

153

u equal to counting measure. Notice that L2[N] is exactly the

space C 2 of square-summable sequences. For our first few results we let ( X , S,u ) be any measure space and define the norm and inner product in L2 as follows:

II f I1 = (/f2dU)l’*

Proposition 2. I f f , g

I1 f I1 + llgll;

E

L2, then (i) f +g

E

L2, and II f+gll I

(ii) fg is integrable, so ( f , g ) makes sense, and J I fgl I II fll llgll. Hence K f , g)I 5 I1 fll llgll for all f , g. Proof. The function x2 is convex, which means that (Ax1

+ (1 - h)x2)25 Ax; + (1 - h)x;

(5)

for all numbers x1 and x2 all h E ( 0 , l ) . Assume f, g E L2 and let fo = f/ll fll,go = g/llgll, so II foII = IlgoII = 1. Estimate as follows:

I f<x> + g(x>I25 (I

= (I1

.r

+ lg(x)ll2 II

fll + llg1D2(,Ifll f I1 +

2

fo(x) II f II + llgll

go(x)) .

+

The second parenthesis in the last line above is a convex combination of fo(x)and go(x), of the form ( 5 ) ,so

I f ( x >+ g(x>I2

= (I1

fll + 11g11>(11fll fOW2+ llgllgo(x>2>.

+

The right side above is integrable, so f g E L2. Integrating both sides above and using I( foil2 = llgo112 = 1, we get

II f

+ g1I2 I(I1 fll + llg11>2.

154

A PRIMER OF LEBESGUE INTEGRATION

Part (ii) is proved just as before. Using

I f (x)g(x)l 5 ( f (XI2 + € m 2 > / 2 , we see that fg is integrable, and

/I

1 z"l f 112 + llg1I2>.

fgl I

The nice connection between the functions in L2[-n, n ] and the Fourier series with coefficients in t 2= L2[N] depends on the fact that every L2 space is complete in its norm. The following apparent digression is necessary for the completeness proof. In any normed space the convergence of infinite series is defined just as for series of reals. If {x,} is a sequence of vectors (points 0 in a normed space) then C % = x means IlC,"=,x,- x(1 as N + cm.The series C % is absolutely convergent provided the real series C Ilx,C,l converges.

Proposition 3. A normed space is complete if and only if every absolutely convergent series converges to a point of the space.

Proof. First assume that {x,} is a sequence in a complete normed space and C llxll converges. If s, = x1 . . . x,, then

+ +

Since C 11 x, 11 converges, the right side above is less than any given for all sufficiently large m and n. Thus {s,} is a Cauchy sequence in a complete space, so s, converges; that is, C x,,converges. Now assume that every absolutely convergent series converges. Let {%} be a Cauchy sequence; we must show that {%} converges. For each k pick Mk so that llxi -xi (1 < 2-k if i, 1 2 nk. We can assume that the nk are strictly increasing. Consider the E

15

series

+

THE SPACE I

155

+ ( X n 3 - xn2)+ .

(7) whose kth partial sum is xZk.This series is absolutely summable since )IX,~+,. - ~ ~ < 1 1/2k. 1 Therefore the series (7) converges; i.e., there IS some x such that xnk + x as k + 00. Since {xn}is Cauchy and has a subsequence which converges to x,the sequence {xn}itself converges to x,and the space is complete. 1111111 Xrtl

(&z2

- xn,)

*

.

7

Proposition 4. (The Riesz-Fischer Theorem) The space L2 is complete. Since l 2 is L2 for X = N and v equal to counting

measure, t2is complete. Proof. We show that every absolutely convergent series in L2 is convergent. Let { f,} be a sequence in L2 such that C 11 fall = M < 00. Let gn(x) = Cyx, I f ; ( x ) ( ,so g, is the sum of a finite number of L2 functions, and hence g, E L2. Moreover, n

i=l

For each fixed x, {g,(x)}is an increasing sequence of real numbers, so {g,} converges pointwise to an extended real-valued function g , which is measurable if it is finite almost everywhere. For all n,

so J g2 I M2 and g E L2. Since g2 is integrable, g is finite almost everywhere. For those x for which g(x) = C I f; (x)l is finite, the series C fi(x) converges; let n

i=l

The function s is measurable. Moreover, for all n,

cI h(x>I n

Isn(x>l 5

i=l

=g n w ,

A PRIMER OF LEBESGUE INTEGRATION

156

so Is(x)l I g(x). Therefore s E L2, and the absolutely convergent series fi converges pointwise a.e. to s E L2. We must show that the series converges to s in the L2 norm; i.e., I(s, - s 11 + 0. Notice that

c

Is,(x> - s(x>I25 (2g(xN2= 4g(xI2. The functions ( s , ( x ) - s ( x ) ) converge ~ pointwise a.e. to zero, and )~. they are dominated by the integrable function 4 g ( ~ Therefore

/(s,

-

s)2 + 0;

i.e., /Is, - s 112 + 0 and the absolutely summable series converges to s in the L2 norm. 1111111

C fi

Now we return to Fourier series and give one pleasant answer to the question of what functions can be represented by Fourier series. A trigonometric series is a series of the form

1 -ao 2

+

03

ak k= 1

cos kx

+ bk sin kx.

If s,(x) is the nth partial sum of (8), then of course S, E L2 [-n, n], and we will need the formula for the L2 norm calculated in the following problem.

Problem 5. If s,(x) = iao + Ci=,ak cos kx + bk sin kx, then I I s , ~ ~=~ n [ $ a ; + a; + i.e., the L2 norm of s, is essen-

xizl

q];

tially the same as the C 2 norm of the sequence of coefficients: 1 Zao, a l , bl, a2, b2, . . . Hint: The requisite orthogonality relations are:

1:

cos kxcos l xdp(x) =

sl

s:

sin kx sin C xdp(x) =

{

n if k = l O if k + l

n if k = l {O if k + C

sin kx cos l xdp(x) = 0.

1111111

(9)

15

157

THE SPACE I

If the trigonometric series (8) converges pointwise to an integrable function f , and if the partial sums are dominated by an integrable function so that the Lebesgue convergence theorem applies, then the coefficients {a,}, {b,} are determined by f as indicated in the next problem.

Problem 6. Let s,(x) be the nth partial sum of the trigonometric series (8). If there is an integrable function g on [-n,n] such that Is,(x)l 5 g(x) for all x and n, and if {s,} converges pointwise a.e. on [-n,n]to the integrable function f , then

Unfortunately, the kind of bounded or dominated convergence required in Problem 6 is not easy to come by, so pointwise convergence of Fourier series is less than ideal. However, the formulas (10)make sense for any integrable function f on [-n,n], and hence for any L2 function. If f is an integrable function which happens to have a trigonometric series that converges to it nicely, then we know what the series must be. Accordingly, the numbers {ak}, {bk}are called the Fourier coefficients of f . The trigonometric series with these coefficients is the Fourier series for f , and no assumption is made about the convergence of the series. The mapping from integrable functions (or L2 functions) to sequences of Fourier coefficients is linear; this is a simple consequence of the fact that the integral formulas ( 9 ) are linear functions of f . The mapping from integrable functions, and therefore from L2-functions, to sequences of Fourier coefficients is also one-to-one; i.e., if f # g, then f and g have different sequences of Fourier coefficients. The proof of this last fact is unfortunately out of our path, so this will have to be an article of faith. Given this fact, we will show that the mapping from L2-functionsto their sequences of Fourier coefficients is a oneto-one mapping on L2 onto .t2.Moreover, the mapping nearly preserves the norm in the sense that 11 f 1 I 2 = n [+a; C(at 631

+

+

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A PRIMER OF LEBESGUE INTEGRATION

if {ah},{bk}are the Fourier coefficients of f . The calculations of the following problem give us most of the information we need.

Problem 7. Let f E L2[-n, n]and let { a k } , {bk}be the Fourier coefficients of f . Let

be any nth-degree trigonometric polynomial. Show that

k= 1 n

Hint: This is nothing more than a moderately unpleasant exercise in algebra. The simplest way may be to verify the case n = 1, which involves completing a square, and then use induction. 1111111 Now we milk the identity (11).One obvious but significant consequence of (11) is that the nth partial sum of the Fourier series for f is the trigonometric polyomial of degree n or less which best approximates f in the L2 norm. Any other nthdegree trigonometric polynomial Pn(x)gives a strictly larger value for I( f - Pall. We also see from (11)that if Pn is the nth partial sum of the Fourier series for f (i.e., ci!k = ak, B k = 6 k for k = 0 , 1 , . . . , n) then, since 11 f - Pall2 2 0,

This is called Bessel’s inequality. It follows immediately that if f E L2, the sequence iao, a1, bl, a2, b2,.. . of its Fourier coefficients is in 12.Moreover, (12)shows that the mapping from functions in L2 to their Fourier coefficients in C 2 is continuous. Our final result shows that Bessel’s inequality (12) is an equality, and that the mapping from L2 to C 2 is onto. That is, every l2sequence makes a Fourier series which represents an

THE SPACE I

15

159

L2-function, and the norms are preserved from L2 to C 2 with a constant factor, so 00

k= 1

Proposition 5. (Sometimes called the Riesz-Fischer Theorem because the completeness of L2 is the essential ingredient in this proof.) I f {an}y{b,} are sequences in 12,then

1 2

-ao

+

c 00

ak

k= 1

cos kx

+ bk sin kx

is the Fourier series of a function f E L2. I f s,(x) is the nth partial sum of the series (14), then s, +f in L2; i.e., (IS, - f 11 +0. The equality (14) holds between the l2norm of the coefficients and the L2 norm of f . Proof. We show first that if s,(x) is the nth partial sum of (14),then {s,} is a Cauchy sequence in L2, and hence has a limit f . Then we show that (14) is the Fourier series of this limit f . A calculation like that of Problem 5 shows that

all the cross product integrals being zero. Since {an},{b,} E C 2 , the identity above shows that {s,} is a Cauchy sequence in L2. Since L2 is complete, there is f E L2 so (Is, - f l l ---+ 0. Since 11 f - s,I + 0 it follows that the inner product ( f - s, g ) -+ 0 for every g E L2. In particular, if g(x) = cos kx, we have

O = lim ( f n+m

- s,

cos kx)

= lim [( f, cos kx) - (s, n-00

=(

f , cos kx) - n a k .

cos kx)]

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A PRIMER OF LEBESGUE INTEGRATION

Therefore ak is the kth Fourier cosine coefficient, and a similar calculation shows that bk is the kth sine coefficient. That is, the general trigonometric series (14)formed from l 2sequences {am}, {bn},is the Fourier series of the L2 function which is the L2 limit of its partial sums. We know from Bessel's inequality that

Since ]Isn- f l l --+ 0, it follows that IIsmll + 11 f l l ; i.e.,

To sum up: Each l2sequence defines a trigonometric series whose partial sums converge in L2 to an L2 function f . The Fourier coefficients of this f are the given sequence in 12.The L2 functions all determine sequences of Fourier coefficients in 12, and the above result says this mapping from L2 to l 2is onto. The mapping is obviously linear, since integrals are linear. The mapping is also one-to-one. This is our one article of faith. There is, therefore, a one-to-one linear mapping cp on L2 onto l 2such that ~ 1 1 q f ~) 1 I(2 = 11 f 1 I 2 for all f E L2.

INDEX

A

B C

absolutely continuous 129 absolutely convergent 18, 154 absolute value of a vector 150 admissible function 63, 118 admissible partition 63, 118 algebra of sets 136 algebra generated by a family 138 almost everywhere (a.e.) 49 Axiom of Choice 37 basket of numbers 17 Bessel’s inequality 158 Bounded Convergence Theorem 58,123 Cantor set 39, 60 Carathkodory criterion 30 Cauchy net 18 Cauchy’s inequality 151 characteristic function 46 choice function 12,53 compact set 22 complete measure 108, 110 complete normed space 154 convergent net 10 convex function 153 countable additivity 31 countable subadditivity 24, 87 -for signed measures 111 countable partition 62 counting measure 105, 108 covering of a set 22 cross section 98 161

162

A PRIMER OF LEBESGUE INTEGRATION

D

dyadic interval 86 dyadic square 86 differentiation under the integral sign 71 Dini derivates 77 directed set 10 distance in R" 151 Dirichlet problem 81 dominated convergence theorem 68,123 double integral 135

E

Egoroff's Theorem 58, 122

F

Fatou's Lemma 68,123,124 Fourier coefficients 157 Fourier series 157 Fubini's Theorem 101,145 Fundamental Theorem of Calculus 73

H

Hahn decomposition 114 harmonic function 81 Heine-Bore1 Theorem 22

I

improper integral 6 inner measure (m,) 41 inner product in L2 152 integrable function, Riemann 2 integrable function, Lebesgue 45, 11 8 integral, Riemann 2 -, improper Riemann 6 -, Lebesgue 45 iterated integral 135

J

Jordan decomposition 115

L

L ( f >P ) 2 L2 152 C2 149,152 Lebesgue Dominated Convergence Theorem 68, 123 Lebesgue integrable function 45 Lebesgue (outer)measure 21,22, 85

INDEX

-,

in the plane 85 Lebesgue singular function 75 linear space 152 little boy 17 lower sum, upper sum 2

M

N

0

p

Q

R

m*,m,28,41 measure 21,22 -, complete 109 measurable function 46,117 measurable set 27, 88, 140 measure space 107 metric 152 Minkowski’s inequality 151 Monotone Convergence Theorem 69, 124 monotonicity of measures 23, 167 negative set 111 net 10 nonmeasurable set 37 norm 19,153,156 null set 109, 112 outer measure (m*)28 partition 1, 43 point mass 108 Poisson kernel 82 positive set 111 the rationals) 21,25

Q (

R ( f ,P , c > 13 Ro, Ros 141

Radon-Nikodym Theorem 130 Radon-Nikodym derivative 132 rational numbers 21,25 rectangles 85, 136, 137 refinement 3 , 4 3 regular measure 91 Riemann integrable 2, 13, 60 Riemann integral 1 Riemann sum 12,53 Riesz-Fischer Theorem 155, 159

163

164

S

T

u V

Z

A PRIMER OF LEBESGUE INTEGRATION

scalar multiplication 149 scalar product 1.50 section of a set 141 u-algebra 35, 107 a-finite 101, 110 signed measure 111 simple function 46, 119 singular measures 114, 132 step function 5 subadditive 24 summable 17 symmetric difference 39 Tonelli’s Theorem 101, 147 total length of a covering 22 total variation of a measure 115 trigonometric series 156 UCf, p > 2 unordered sum 1 7 upper and lower sums 2 -, for Lebesgue integral 43

vector space 149 Vitali covering 75 Vitali’s Theorem 7.5 Zorn’s Lemma 37