Real Analysis Lecture Notes

8/3/2019 Real Analysis Lecture Notes

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REAL ANALYSIS LECTURE NOTES

Contents1. Introduction and Preliminaries 12. The Structure ofR 73. Lebesgue Measure 113.1. Lebesgue Measure and Measurable Sets 113.2. Measurable Functions 174. Lebesgue Integration 235. Differentiation and Integration 325.1. Motivation 325.2. Derivatives and Derivates 34

5.3. Vitalis Covering Lemma 375.4. Variation and Absolute Continuity 395.5. Convexity 466. Lp Spaces 497. Metric Spaces 597.1. The Basics 597.2. Compactness 607.3. Continuous Functions 627.4. The StoneWeierstrass Theorem 638. General Measure and Integration 67

8.1. Introduction 678.2. Measurable Functions 708.3. Integration 738.4. Lp Spaces 798.5. Signed Measures 828.6. Radon-Nikodym Theorem 909. Outer Measure and Extensions of Measures 929.1. Outer Measure 929.2. The Extension Theorem 959.3. Baire Measures on R 999.4. Product Measures 1019.5. Integral Operators 11010. The AscoliArzela Theorem 113Appendix A. Convergence in Measure 116References 118Index 119


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c 2009, Jonathan R. Kish


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REAL ANALYSIS LECTURE NOTES 1

1. Introduction and Preliminaries

The following notes are an offshoot of a year-long course in graduate analysis that Itook from Judith Packer at the University of Colorado, Boulder in 2005-2006. We usedRoydens text Real Analysis, 3rd edition (see [5]). There are no proofs in what follows,only statements of results as well as definitions, examples and remarks. I wanted alist of results that I could easily and quickly navigate and that was extensively cross-referenced to itself and the original source. I started with my lecture notes from class,added an index and did a bit of reorganizing. It proved enormously helpful while doinghomework and studying for exams. I may one day fill in all the proofs and providedetailed explanations, but until then the book Measure, Integral and Probability byCapinski and Kopp (see [3]) more than suffices. Another book I like is [6]. For anaccount of the history of the subject, see [4]. For undergraduate analysis, I used ARadical Approach to Real Analysis by David Bressoud (see [1]), a book that I have justrecently begun to appreciate and revisit with great interest. I have also just becomeaware that Bressoud has written a book on Lebesgue integration (see [2]). While I

havent looked at it yet, if its anywhere near as good as [1] it will become part ofmy library. Any comments or suggestions are welcome, as are corrections. Errors are ofcourse mine. We begin with a brief motivational review of Riemann integration, followedby some elementary set theory and structure of the real line.

Jonathan KishBoulder, ColoradoNovember 24, [email protected]


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2 REAL ANALYSIS LECTURE NOTES

Definition 1.1. Let a, b R such that a < b. Suppose that f : [a, b] R is boundedin the sense that

< infx[a,b]

f(x) supx[a,b]

f(x) < .

Let P= {a = x0 < x1 < < xn = b} be a partition of [a, b]. For each i, 1 i n, letmi = inf

xi1xxif(x) and Mi = sup

xi1xxif(x).

Define the lower Riemann sum of f with respect to P by

Lp(f) =n

i=1

mi(xi xi1).

Similarly, define the upper Riemann sum of f with respect to P by

Up(f) =n

i=1

Mi(xi xi1).

Note that, by definition,

(1) Lp(f) Up(f).The lower Riemann integral of f over [a, b], written

a

b

f(x)dx,

is defined by

(2)

a

b

f(x)dx = supP

Lp(f)

where sup

Pis the supremum taken over all partitions

Pof [a, b]. Moreover, note that the

set Lp(f) is bounded above and so has a supremum. Similarly, the upper Riemannintegral of f over [a, b], written b

a

f(x)dx,

is defined by

(3)

ba

f(x)dx = infP

Up(f)

.

We always have that

a

b

f(x)dx ba

f(x)dx.

We say that f is Riemann integrable over [a, b] ifa

b

f(x)dx =

ba

f(x)dx =

ba

f(x)dx.


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Remark 1.2. Recall that the Riemann integral has some nice properties:

(i) Every continuous function is Riemann integrable.(ii) The Fundamental Theorem of Calculus holds: If F is differentiable on [a, b] and

if f(x) = F(x) is continuous on [a, b], then

b

a

f(x)dx = F(b)

F(a).

Remark 1.3. There are drawbacks of the Riemann integral, however:

(i) Recall that f has to be bounded. Consider, for example, f : [0, 1] R definedby

f(x) =

0 if x = 0

1x

if 0 < x 1.Note that of course we can get around this with improper integrals.

(ii) Many functions that are easy to define are not Riemann integrable. Consider

f : [0, 1] R defined byf(x) =

0 if x is rational

1 if x is irrational.

For any partition Pof [0, 1], we have Lp(f) = 0 (there is a rational number inany interval), whereas Up(f) = 1. It follows that

a

b

f(x)dx = 0 < 1 =

ba

f(x)dx.

(iii) The Riemann integral does not work well with limiting processes. That is, it is

possible to have a sequence of functions {fn}n=1 defined on [a, b], each Riemannintegrable, and limn fn(x) = f(x) exists, but f(x) might not be Riemannintegrable over [a, b].

Our main aim is Lebesgue integration theory, in which it is easier to impose conditionsunder which

limn

fn(x)dx =

limn

fn(x)

if limn

fn(x) = f(x) exists.

Remark 1.4.We denote the

power setof a set

Xby 2X or

P(X

). The latter shouldcause no confusion with the notation Pthat stands for a partition, as the meaning ofthe symbol Pwill always be clear from the context.

Definition 1.5. Let A be a subset of the underlying set X. The complement of A inX, A, is defined by A = {x X : x / A}.


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Similarly, if A X and B X, we define the relative complement of A in B,written B A, by

B A = {x B : x / A}= B

A.

Definition 1.6. C is called a collection of subsets of X if C P(X).

Lemma 1.7 (Generalized DeMorgans Laws). Let C be a collection of subsets ofX. Then

AC

A =AC

A and AC

A =AC

A.See Exercise 13, p. 16 of [5].

Definition 1.8. Let X be a set and suppose A and B are subsets ofX. The symmetricdifference of A and B, written AB, is defined by

AB = (A B) (B A)= (A B) (B A).

Note that AB = BA by properties of unions.

Definition 1.9. Let X be a nonempty set. Assume that A P(X) is a nonemptycollection of subsets of X. We say that

Ais an algebra of subsets of X if

(i) Whenever A A and B A, then A B A (A is closed under the unionoperation).

(ii) Whenever A A, A A (A is closed under taking complements).Example 1.10. Consider the following examples:

(i) Let X be a set. Then A = P(X) is always an algebra.(ii) Let X = {a, b}. Then A = , {a, b} is an algebra.

Remark 1.11 (Facts about algebras). Let A be an algebra of subsets of X. Then(i) If A, B A, then A B A.

(ii) , X A.(iii) If A1, . . . , An A, then

ni=1

Ai A.

(iv) If A1, . . . , An A, thenn

i=1

Ai A.


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(v) If A, B A, then B A A.(vi) If A, B A, then AB A.

Remark 1.12. One way to construct a nontrivial algebra of sets is to let C be anynonempty collection of subsets of X and let

Abe the smallest algebra of sets of X

containing C. See Proposition 1.13.

Proposition 1.13 (Proposition 1, 1.4, p. 17 of [5]). LetX be a nonempty set and letC be a nonempty collection of subsets of X. Then there is a unique algebra A containingC that is the smallest such algebra in the sense that if B is any other algebra of subsetsof X containing C, thenB also contains the algebra A.

Definition 1.14. The smallest algebra containing

Cas in Proposition 1.13 is called the

algebra generated by C.

Proposition 1.15 (Proposition 2, 1.4, p. 17 of [5]). LetX be a nonempty set and letA be an algebra of subsets of X. Suppose {Ai}i=1 A, that is, Ai A for all i 1.Then there is a collection of subsets {Bi}i=1 A such that

(i) Bm Bn = for m = n.(ii) For every n N,

ni=1

Bi =n

i=1

Ai. (these are in the algebraA)

(iii)

i=1 Bi =

i=1 Ai. (these infinite unions might not be inA, but as subsets of Xthey are equal)Remark 1.16. Proposition 1.15(iii) motivates the following definition.

Definition 1.17. Let X be a nonempty set and let A be an algebra of subsets of X.A is a -algebra of subsets of X if, whenever {Ai}i=1 A,

i=1

Ai A. Thus we areallowing countably infinite unions now, not just finite unions.

Remark 1.18 (Facts about -algebras).

(i) Every -algebra of subsets is an algebra of subsets, but the converse is not truein general.

(ii) Let {Ai}i=1 A for A a -algebra. Theni=1

Ai A.


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Example 1.19. Let X be a nonempty set. Then P(X) is a -algebra of subsets of X.

Proposition 1.20 (Proposition 3, 1.4, p. 19 of [5]). LetX be a nonempty set and let Cbe a nonempty collection of subsets of X. Then there is a unique -algebraA containing

Cthat is the smallest possible in the sense that if

Bis any other -algebra and

C B,

thenA B. Compare to Proposition 1.13, and see Exercise 19, p. 19 of [5].

Definition 1.21. The smallest -algebra containing C as in Proposition 1.20 is calledthe -algebra generated by C.


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2. The Structure ofR

Definition 2.1. Let O R. We say O is open in R if for each x O > 0 suchthat whenever y R and |y x| < , y O.

Remark 2.2. Note that is open as there is nothing to verify. Moreoever, all ofR isopen. Now, ifx (a, b), let = min(x a, x b). Then (x , x + ) (a, b), whichimplies that (a, b) is open.

Proposition 2.3 (Properties of open sets).

(i) LetC be any collection of open subsets ofR. Then OC

O is open (Proposition 7,2.5, p. 41 of [5]).

(ii) If

{Oi

}ni=1 is any finite collection of open subsets ofR, then

n

i=1 Oi is open (Corol-lary 6, 2.5, p. 41 of [5]).Remark 2.4. If we have a countably infinite family of open subsets ofR, {Oi}i=1, theni=1

Oi need not be open (it could be, if the intersection is empty). For example, letOn =

1n ,

1n

for every n N, each of which is open. Then

n=1On =

n=1

1n

,1

n

= {0}.

This is not open since > 0 such that (0 , 0 + ) {0}.

Proposition 2.5 (Proposition 8, 2.5, p. 42 of [5]). Any nonempty, open subset O ofR can be written as a countable union of disjoint open intervals.

Definition 2.6. Let E R. We say that x R is a point of closure for E if forevery > 0, (x , x + ) E= .

Definition 2.7. Let

E = {x R : x is a point of closure of E}.E is called the closure of E. Note that E E, for if x E, then > 0, (x , x +

) E= . However, E E in general.


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Example 2.8. Consider a, b R with a < b. Let E = (a, b). It can be shown that(a, b) = [a, b].

Proposition 2.9 (Properties of closures).

(i) If A, B

R, then A

B = A

B .

(ii) If A B, then A B (Proposition 10, 2.5, p. 43 of [5]).(iii) Note, however, that in general, A B = A B.(iv) If E R, then E = E (Proposition 11, 2.5, p. 43 of [5]).

Definition 2.10. We say a set F R is closed if F = F.

Proposition 2.11 (Properties of closed sets).

(i) Let

Cbe any collection of closed subsets ofR. Then FC F is closed (Proposition

13, 2.5, p. 44 of [5]).(ii) If {Fi}ni=1 is any finite collection of closed subsets of R, then

ni=1

Fi is closed

(Proposition 12, 2.5, p. 44 of [5]).Compare to Proposition 2.3.

Remark 2.12. If we have a countably infinite family of closed subsets ofR, {Fi}i=1,then

i=1Fi need not be closed. For example, let Fn =

2 + 1n , 3 1nfor every n N,

each of which is closed. Thenn=1

Fn =n=1

2 + 1

n, 3 1

n

= (2, 3),

which is open. Compare to Remark 2.4.

Proposition 2.13 (Proposition 14, 2.5, p. 44 of [5]). The complement of an open setis closed and the complement of a closed set is open.

Example 2.14. F1 = {a} is closed, and F1 = (, a) (a, ) is open. Similarly,F2 = [a, b] is closed and F2 = (, a) (b, ) is open. See Proposition 2.13.Definition 2.15. Let B(R) (or just B) denote the smallest -algebra containing all theopen intervals, that is, containing C = {(a, b) : a < b }. B(R) is calledthe -algebra of Borel sets of R. B(R) also contains every open subset ofR since


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every open set can be written as a countable union of (disjoint) open intervals (seeProposition 2.5). Since B(R) is closed under taking complements, B(R) also contains allclosed subsets ofR (see Proposition 2.11(iii)).

Definition 2.16. A set A

R is said to be a G-set if we can write

A =i=1

Oi,

where each Oi is open ( is from the d in Durchschnitt, German for intersection).Note that a G-set need not be open itself (cf. Remark 2.4)

Definition 2.17. A set B R is said to be an F-set if we can write

B =

i=1 Fi,where each Fi is closed. Note that an F-set need not be closed (cf. Remark 2.12).

Remark 2.18. Note that G-sets and F-sets are all Borel sets. Moreover, we could con-sider (G)-sets (the countable union ofG-sets) or (F)-sets (the countable intersectionof F-sets), and so on, all of which are also Borel sets.

Definition 2.19. Let {xk}k=1 be a bounded sequence of real numbers. For each n N,let

n = inf{xn, xn+1, xn+2, . . . } and n = sup{xn, xn+1, xn+2, . . . }.Thus we have two new sequences {n}n=1 and {n}n=1 where the ns are increasing,the ns are decreasing, and n n for all n.

Define the lower limit of{xk}k=1 bylim xk = liminfxk = sup{n}n=1 = lim

nn

(the limit of an increasing sequence that is bounded above is its supremum).

Similarly, define the upper limit of{xk}k=1 bylim xk = limsup xk = inf{n}n=1 = lim

nn

(the limit of a decreasing sequence which is bounded below is its infimum). Note that

lim xk = limn

n limn

n = lim xk.


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Example 2.20. Consider the sequence

{x1, x2, . . . , x2n1, x2n, . . . } =

1, 3, 12

,5

2, . . . , 1

n, 2 +

1

n, . . .

.

Thenlim {xk} = 0 and lim {xk} = 2.


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3. Lebesgue Measure

3.1. Lebesgue Measure and Measurable Sets.

Remark 3.1. Our ideal situation is to generalize the notion of length of intervals to

every subset ofR so that, letting denote this generalized length,(i) For all A R, 0 (A) . (could have infinite length)

(ii) If I is an interval, then (I) = (I).(iii) Countable Additivity: If {Ai}i=1 is a countable family of pairwise disjoint

subsets ofR, we want

i1

Ai

=i1

(Ai).(iv) Translation invariance: If A R and r R, then

(A + r) =

(A), where

A + r =

{x + r : x

A

}.

We will see that it is possible to define this generalized length , which we will denotem, on P(R), and m will satisfy conditions (i), (ii) and (iv), but we wont be able toget countable additivity in general (condition (iii)). We can satisfy all four conditionsfor Borel sets, but not for every subset ofR.

Definition 3.2. If A R is any subset ofR, the Lebesgue outer measure of A,denoted m(A), is given by

m(A) = infn1 (In) :all countable collections

of open intervals

{In

}n1 suchthat A

n1 In .Here we allow

n1 (In) to equal as some of the intervals might have infinite length.

We also allow, for r [0, ), r + = + r = . Note that it could be the casethat for every countable collection of open intervals {In}n1 such that A

n1 In,

n1 (In) could be . In this case, m(A) = (e.g. A = R).

Remark 3.3 (Basic facts about m).

(i) 0 m(A) for all A R. (note the sums are nonnegative)(ii) m(

) = 0.

(iii) Monotone Property: If A B R, then m(A) m(B).(iv) m

{x} = 0 for every x R.Proposition 3.4 (Proposition 1, 3.2, p. 56 of [5]). Let I be an interval inR. Thenm(I) = (I).


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Proposition 3.5 (Countable Subadditivity - Proposition 2, 3.2, p. 57 of [5]). If{An}n1 is a countable collection of subsets ofR, then

m

n1An

n1

m(An).

Corollary 3.6 (Corollary 3, 3.2, p. 58 of [5]). If E is a countable subset ofR, thenm(E) = 0.

Corollary 3.7 (Corollary 4, 3.2, p. 58 of [5]). The set [0, 1] is uncountable.

Corollary 3.8. R is an uncountable set.

Corollary 3.9. m(Q) = 0.

Corollary 3.10. m

[0, 1] Q = 0.Note that [0, 1] Q is dense in [0, 1], yet m[0, 1] Q = 0 < 1 = m[0, 1].Corollary 3.11. R

\Q is uncountable.

Remark 3.12. Note that we made no assumption of pairwise disjointness of the {An}n1in Proposition 3.5. What if they are pairwise disjoint? Do we get equality? ConsiderA1 = [a, b) and A2 = [b, c]. Then A1 A2 = [a, c] where A1 A2 = . Moreover,m(A1 A2) = c a and m(A1) + m(A2) = b a + c b = c a, so we do get equalityin this particular case. This is not true in general, however. Using the Axiom ofChoice, it is possible to construct A1, A2 R with A1 A2 = and m(A1 A2) 0, there exists an open set U A such that

m(U) m(A) + .(ii) Given A R, there exists a G-set G with G A and m(G) = m(A).

See Exercise 6, p. 58 of [5].

Proposition 3.33 (Proposition 15, 3.3, p. 63 of [5]). LetE R. Then the followingfive conditions are equivalent:

(i) E M.(ii) For every > 0, there exists an open set U E such that m(U E) < .

(iii) For every > 0, there exists a closed set F E such that m(E F) < .(iv) There exists a G-set G E such that m(G E) = 0.(v) There exists an F-set F E such that m(E F) = 0.

If, in addition, the outer measure of E is finite, all of the conditions (i) (v) areequivalent to

(vi) For every > 0, there exists a finite collection{Ui}Ni=1 of open intervals such that

(11) m

Ni=1

Ui

E

< .

See Exercise 13, p. 64 of [5].

Remark 3.34. Part (vi) of Proposition 3.33 is the precise statement of LittlewoodsFirst Principle: Every Lebesgue measurable set is almost a finite union of open

intervals. In other words, you can approximate any finite, measurable E R with afinite union of open intervals.

Remark 3.35 (Existence of nonmeasurable sets). Start with E = [0, 1] R,which is measurable. Define an equivalence relation on E by x y x y Q.Let (E) denote the family of equivalence classes of E with respect to . For eachequivalence class (E), choose x E = [0, 1] (using the Axiom of Choice). Let

P =

(E){x}.

We claim that P is nonmeasurable. If it were measurable, then there are two possibilities:(i) m(P) = 0.

(ii) 0 < m(P) 1 since P [0, 1].Assuming (i), one can show that m(R) = 0, which is a contradiction. Similarly, assuming(ii), one can show that m

[1, 2] = , which is again a contradiction. Thus P must

be nonmeasurable. See 3.5 of [5] for the full details.


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3.2. Measurable Functions.

Remark 3.36. At this point, we wish to generalize the notion of continuous functions,which should of course include the continuous functions. See Proposition 3.37, Definition3.38 and Example 3.39.

Proposition 3.37 (Proposition 18, 3.5, p. 66 of [5]). Let f be an extended real-valued function whose domain E is measurable. Then the following four conditions areequivalent:

(i) R {x E : f(x) > } M.(ii) R {x E : f(x) } M.

(iii) R {x E : f(x) < } M.(iv) R {x E : f(x) } M.

All of the conditions (i) (iv) imply(v)

R

{x

E : f(x) =

} M(actually this is true

R

{,}

).

Note that conditions (i) (iv) can be rewritten as(i) R f1(, ] M.

(ii) R f1[, ] M.(iii) R f1[, ) M.(iv) R f1[, ] M.

Definition 3.38. If any of the conditions (i)(iv) in Proposition 3.37 hold, the functionf is said to be Lebesgue measurable.

Example 3.39. Suppose that f : R R is continuous. Fix R. Considerx R : f(x) > = f1(, ).

Note that (, ) is open. Since f is continuous, f1(, ) is open and hencef1

(, ) M (see Proposition 18, 2.6, p. 47 of [5]). Thus f satisfies condi-

tion (i) of Proposition 3.37 and is therefore Lebesgue measurable. It follows that theLebesgue measurable functions include the continuous functions, as desired (see Remark3.36).

Definition 3.40. Let A R. We define the characteristic function (a.k.a indicatorfunction) A : R R by

A(x) =

1 if x A0 if x A .

Note that if A / {,R}, then range A = {0, 1}. If A = , then range A = {0} and ifA = R, range A = {1}.


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Lemma 3.41. LetA R. Then A is measurable iff A M.

Proposition 3.42. Let E M and suppose that D E is a measurable subset ofE. Let f be a measurable extended real-valued function defined on E. Then f|D ismeasurable.

Proposition 3.43 (Proposition 19, 3.5, p. 67 of [5]). Let c R and let f and g bemeasurable, real-valued functions defined on E M. Then

(i) f + g is measurable.(ii) c f is measurable.

(iii) g f is measurable.(iv) gf is measurable (product, not composition).

Definition 3.44. We say that a property P holds almost everywhere (abbreviateda.e.) on a measurable subset E ofR if

m{x E : P is not true for x} = 0.

Proposition 3.45 (Proposition 21, 3.5, p. 69 of [5]). If f and g are defined onE Msuch that f = g a.e. on E and f is measurable, then g is measurable (Note f and g canbe extended real-valued functions).

Theorem 3.46 (Theorem 20, 3.5, p. 68 of [5]). Let{fi}i=1 be a sequence of measurableextended real-valued functions defined on E M. Then

(i) sup{fi}ni=1 is measurable n N.(ii) sup{fi}i=1 is measurable.

(iii) inf{fi}ni=1 is measurable n N.(iv) inf{fi}i=1 is measurable.(v) Hence lim fn and lim fn are measurable extended real-valued functions.

Corollary 3.47. If{

fn}n=1 is a sequence of measurable functions defined on E

Mand if f(x) = limn

fn(x) exists in the extended real numbers for every x E, then f ismeasurable.

Definition 3.48. Let E M. We say that : E R is simple if is Lebesguemeasurable and the range of is finite. The simple functions are the building-blockfunctions.


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Proposition 3.49. Let be a simple function defined on E M. Then we can write

(12) (x) =n

i=1

ciAi(x)

where the {Ai}ni=1 are pairwise disjoint measurable sets such thatn

i=1Ai = E, we have

c1 < c2 < < cn and range = {ci}ni=1.

Remark 3.50 (Converse to Proposition 3.49). If

g(x) =n

i=1

ciAi(x)

where the Ai M are pairwise disjoint, then g is a simple function defined on R.

Definition 3.51. Equation (12) is called the canonical representation of the simple

function . Note that corresponds one set to each element in the range of .

Remark 3.52. If1 and 2 are simple functions and c, d R, then c1 +d2, c1 d2and 12 (the product) are all simple.

Definition 3.53. Let {fn}n=1 be a sequence of real-valued functions defined on E R,and let f be a real-valued function defined on E R. We say that fn convergespointwise to f on E if for every x E, lim

nfn(x) = f(x). That is, > 0, N N

(depending on and on x) such that

|fn(x) f(x)| < whenever n N.

Definition 3.54. Let {fn}n=1 be a sequence of real-valued functions defined on E R,and let f be a real-valued function defined on E R. We say that fn convergesuniformly to f on E if > 0, N N (depending only on and not on x) such that

|fn(x) f(x)| < for every n

N and for every x

E (the same N

N works for all x

E). This means

limn

fn(x) = f(x) and fn(x) f(x) at a rate that can be made independent of x E.

Theorem 3.55. Letf : E [0, ] be a measurable, nonnegative extended real-valuedfunction defined on E M. Then there is an increasing sequence {n}n=1 of simple functions with lim

nn(x) = f(x) for all x E. If f is bounded, this convergence is

uniform. See Exercise 4, p. 89 of [5].


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Definition 3.56. For a measurable extended real-valued function f defined on E M,let

P = {x E : f(x) 0}and

N = {x E : f(x) < 0}.Note that P, N

Mby Proposition 3.37, P

N =

and P

N = E. Let

(13) f+ = fP =

f(x) if x P0 if x N = E P

(where we define 0 = 0) and

(14) f = fN =

f(x) if x N0 if x P = E N

(where we define 0 = 0). f+ and f are called the positive and negative parts off, respectively.

Remark 3.57. Note that,

(15) f(x) = f+(x) f(x)and

(16) |f(x)| = f+(x) + f(x).Furthermore, f+ and f are measurable whenever f is (note that P and N aremeasurable, now use Proposition 3.43 with (13) and (14)). Therefore |f| is measurablesince it is the sum of two measurable functions (see (16) and Proposition 3.43). This

motivates the following corollary.

Corollary 3.58. Let f : E [, ] be a measurable function defined on E M. Then there exists a sequence {n}n=1 of simple functions defined on E such thatlimn

n(x) = f(x) for every x E. If f is bounded, this convergence is uniform.

Remark 3.59. An intuitive way of stating Corollary 3.58 is that Every measurablefunction is nearly a simple function. Technically this means that if f is a measurablefunction defined on E

Mwith m(E) 0 there exists a simple

function such that

m{x E : |f(x) (x)| } < .

That is, if we let

B = {x E : |f(x) (x)| },then for all x E B, |f(x) (x)| < . In other words, approximates f very wellon E B. Related to this is Littlewoods Second Principle: Every measurable


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function is nearly a continuous function. The technical form of this principle is givenin Proposition 3.60.

Proposition 3.60 (Proposition 22, 3.5, p. 69 of [5]). Letf be a measurable functiondefined on a closed interval [a, b]. Assume f takes on the values

on a set of measure

0. Then given any > 0 there exists a step function g and a continuous function h suchthat

m{x [a, b] : |f(x) g(x)| } <

and

m{x [a, b] : |f(x) h(x)| } < .

That is,

|f(x) g(x)| < except on a set of measure less than and

|f(x) h(x)| < except on a set of measure less than .Moreover, if f is bounded, m f(x) M for every x [a, b], then we can choose g andh such that m g(x) M and m h(x) M for all x [a, b]. See Exercise 23, p. 71of [5].

Remark 3.61. The intuitive form of Littlewoods Third Principle states that ev-ery sequence of measurable functions that is pointwise convergent is nearly uniformlyconvergent. See Proposition 3.65 for the technical form of this principle.

Proposition 3.62 (Proposition 23, 3.6, p. 72 of [5]). LetE be a measurable subset ofR such that m(E) < . Suppose that {fn}n=1 is a sequence of measurable real-valued

functions defined onE that converges pointwise to some real-valued function f definedon E. Then given > 0 and > 0 there is a measurable set A E with m(A) < and N N such that |fn(x) f(x)| < for every x E A and for every n N.Remark 3.63. Note that we are not assuming fn converges uniformly on E

A. The

set A depends on both and so that as you change and the set A changes.

Corollary 3.64 (Proposition 24, 3.6, p. 73 of [5]). In the statement of Proposition3.62, it is enough to assume that fn(x) f(x) a.e. on E. The conclusion of theProposition will still hold.


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Proposition 3.65 (Egoroffs Theorem - See Exercise 30, p. 73 of [5]). Let{fn}n=1be a sequence of measurable, real-valued functions defined on E M with m(E) < .Suppose that {fn}n=1 converges pointwise a.e. to the measurable, real-valued function fdefined on E. Then given any > 0 there exists a measurable set A E with m(A) < and {fn}n=1 converges uniformly to f on E

A.


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4. Lebesgue Integration

Remark 4.1. Recall that if is a simple function, we can write

(17) (x) =n

i=1ciAi(x)

where

Ai = {x R : (x) = ci}and c1 < c2 < < cn (see Proposition 3.49). Note that if ci0 = 0 for some i0,1 i0 n, then

ci0Ai0 (x) = 0 Ai0 = 0,so that this term does not contribute to the sum (17). Thus we will henceforth assumethat all the cis are not equal to zero.

Definition 4.2. We say a simple function is zero outside a set of finite measureor vanishes outside a set of finite measure if

m{x R : (x) = 0} < .

This will happen if and only ifn

i=1

m(Ai) < .

Definition 4.3. Let be a simple function that is zero outside a set of finite measure.Then the Lebesgue integral of over R, written

or

R

is defined by

(18)

=

ni=1

cim(Ai)

where is given by (17). Similarly, for E M, the Lebesgue integral of overE R is defined by

(19) E

=n

i=1

cim(Ai E)

Lemma 4.4. If E M, then E

=

E.


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Lemma 4.5 (Lemma 1, 4.2, p. 78 of [5]). Let be a simple function that vanishesoutside a set of finite measure. Suppose that we can write

(20) (x) =m

j=1

bjEj(x)

where the{

Ej}mj=1 are measurable, pairwise disjoint subsets ofR. Note that (20) might

not be the canonical representation of (the bjs might not be increasing, or they mightrepeat if we split an Ai into two parts). But then

(21)

=

mj=1

bjm(Ej).

So as long as the Ejs are pairwise disjoint, we may use the same formula for the integralas in (18).

Proposition 4.6 (Proposition 2, 4.2, p. 78 of [5]). Let and be simple functionsthat both vanish outside a set of finite measure. Then(i) a, b R,

(a + b) = a

+ b

.

(ii) If (x) (x) onR, then

.

Definition 4.7. Let f be a bounded, measurable real-valued function defined on E Mwhere m(E) < . We define the Lebesgue integral of f over E, denoted

(22)

E

f

by

(23)

E

f = inf

E

: is simple and (x) f(x) x E

.

Remark 4.8. Note that we could have taken

(24) sup

E


in Definition 4.7. It turns out these two definitions of the integral are the same, but thiswill take some work (see Proposition 4.12).


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Lemma 4.9. If f is a measurable, bounded function defined on E M such thatm(E) < , and if A f(x) B for all x E, then(25) A m(E)

E

f B m(E).

Definition 4.10. Recall Definition 4.7. Iff is a bounded measurable function we define

(26) A = inf

E


,

called the upper envelope of f over E. Similarly, we define

(27) B = sup

E


,

called the lower envelope of f over E.

Remark 4.11. We can always show that B A regardless of whether or not f isLebesgue measurable.

Proposition 4.12 (Proposition 3, 4.2, p. 79 of [5]). Let f be a bounded real-valued function defined onE M with m(E) < . Then

(i) If f is Lebesgue measurable, then

(28)

E

f = A = B

(ii) If A = B, then f is Lebesgue measurable,

where A and B are as defined in Definition 4.10. In other words,A = B if and only if f is Lebesgue measurable.

Corollary 4.13 (Proposition 4, 4.2, p. 81 of [5]). Letf be a bounded function definedon [a, b], < a < b < . If f is Riemann integrable over [a, b], then f is Lebesguemeasurable on [a, b] and

(29) Riem

ba

f(x)dx

= Leb

[a,b]

f

.

Proposition 4.14 (Properties of the Lebesgue integral coinciding with theRiemann integral - Proposition 5, 4.2, p. 82 of [5]). Let f, g be bounded Lebesguemeasurable functions defined on E M such that m(E) < . Then

(i) For every a R, E

af = a

E

f.


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(i) E

(f + g) =

E

f +

E

g.

Hence for every a, b R,

E(af + bg) = a

Ef + b

Eg.

(ii) If f g a.e. on E, then E

f E

g.

(iii) If f = g a.e. on E, then E

f =

E

g.

(iv) If M1 f(x) M2 a.e. on E, thenM1 m(E)

E

f M2 m(E).(v) IfA andB are disjoint measurable subsets ofE with m(A) 0. Then

(i) E

af = a

E

f

where a = in case f is not integrable.(ii)

E

(f + g) =

E

f +

E

g.


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(iii) If f(x) g(x) a.e. on E, thenE

f E

g.

(iv) If E1 and E2 are disjoint measurable subsets ofR, then

E1E2 f = E1 f + E2 f.We use this Proposition to prove Fatous Lemma (Lemma 4.20).

Theorem 4.20 (Fatous Lemma - Theorem 9, 4.3, p. 86 of [5]). Let{fn}n=1 be asequence of nonnegative, Lebesgue measurable functions defined on E M. Supposethat there is a function f defined on E such that fn(x) f(x) a.e. on E (note that fis thus measurable by Corollary 3.47). Then

(34) Ef lim Efn.Note that were not saying the RHS of (34) isnt , or that both sides of (34) arent.

Theorem 4.21 (Monotone Convergence Theorem, or MCT - Theorem 10, 4.3,p. 87 of [5]). Let{fn}n=1 be a sequence of nonnegative, Lebesgue measurable functionsdefined on E M. Suppose also that the fns are increasing, that is,

0 f1(x) f2(x) fn(x) fn+1(x) for every n N and for every x E (hence the monotone). Thenf(x) = lim

nfn(x)

exists as a measurable, extended real-valued function on E and

(35)

E

f = limn

E

fn.

Since f(x) = limn

fn(x), (35) is the same as

(36)

E

limn

fn = limn

E

fn.

Corollary 4.22 (Corollary 11, 4.3, p. 87 of [5]). Let{un}n=1 be a sequence of nonneg-ative, Lebesgue measurable functions defined on E M. Then(37)

E

n=1

un =n=1

E

un.

Note that both sides of the equals sign in (37) might be . Note also that if 0 a band un = anx

n for an 0, then anxn 0 on [a, b].


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Corollary 4.23 (Proposition 12, 4.3, p. 87 of [5]). Let f be a nonnegative integrable function defined onE M. Let{Ei}i=1 be pairwise disjoint measurable subsets of E.Then

(38)

i=1

Ei

f =i=1

Ei

f.

Remark 4.24. Note that, up to this point, we have only been able to integrate functionsthat are nonnegative. The natural question arises: how do we integrate functions thatare not necessarily nonnegative?

Definition 4.25. Suppose that f is a measurable function defined on E M. We saythat f is Lebesgue integrable over E if f+ and f are both integrable over E. Inthis case, we define

(39) E

f = E

f+ E

f.

See Definition 3.56.

Proposition 4.26. Let f be a measurable function defined on E M. Then f isintegrable over E if and only if |f| is integrable over E. See Exercise 10(a), p. 93 of [5].

Proposition 4.27 (Proposition 13, 4.3, p. 88 of [5]). Let f and g be nonnegative,measurable functions defined on E

M. Suppose that f is integrable and g(x)

f(x)

for every x E. Then g and f g are integrable and(40)

E

(f g) =E

fE

g.

Proposition 4.28 (Proposition 14, 4.3, p. 88 of [5]). Letf be a nonnegative, integrableextended real-valued function defined on E M. Then for any > 0, > 0 such thatwhenever A E is measurable with m(A) < , we have

Af < .

This is sometimes referred to as the continuity of the Lebesgue integral. This

shows that

A

f can be as small as you want as long as you take the set A small enough.


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Proposition 4.29 (Proposition 15, 4.4, p. 90 of [5]). Letf andg be integrable functionsover E M. Then

(i) For every c R, cf is integrable over E andE

cf = c

E

f.

(ii) f + g is integrable over E andE

(f + g) =

E

f +

E

g.

(iii) If f(x) g(x) a.e. on E, thenE

f E

g.

In particular,

(41)

E

f

E

|f|.

(iv) If E1 and E2 are disjoint measurable subsets of E, thenE1E2

f =

E1

f +

E2

f.

Compare Proposition 4.19.

Corollary 4.30 (Corollary to Proposition 4.28 and Proposition 4.29 (iii)).If f is an integrable function over E M, then > 0, > 0 such that wheneverA E and m(A) < ,

(42) A

f < .Theorem 4.31 (Lebesgue Dominating Convergence Theorem, or DCT -Theorem 16, 4.4, p. 91 of [5]). Let g be a nonnegative, measurable function definedon E M that is integrable over E. Let {fn}n=1 be a sequence of measurable func-tions all defined on E such that |fn(x)| g(x) x E and n N. Suppose thatlimn

fn(x) = f(x) a.e. for some function f defined on E. Then f is integrable over E

and

(43) E

f = limn

E

fn.

Since f(x) = limn

fn(x), (43) is equivalent to

(44)

E

limn

fn = limn

E

fn.


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Remark 4.32. As an application of DCT, recall that MCT (see Theorem 4.21) neednot hold for a decreasing sequence of functions unless we impose some further conditions(see Exercise 7(b), p. 89 of [5]). This motivates the following corollary.

Corollary 4.33. Let

{fn

}n=1 be a decreasing sequence of nonnegative, Lebesgue mea-

surable functions defined onE M. Suppose further thatf1 is integrable overE. Then,letting f = lim

nfn, we have

(45)

E

f = limn

E

fn.

Note that, since the fns are decreasing and nonnegative (i.e., bounded below by 0),limn

fn exists and is equal to its infimum. See also Assignment 4, Exercise 2(b).

Theorem 4.34 (Generalized Lebesgue Dominating Convergence Theorem,

or GDCT - Theorem 17, 4.4, p. 92 of [5]). Let{gn}n=1 be a sequence of nonnegative,integrable functions defined on E M. Suppose that(i) lim

ngn(x) = g(x) a.e. on E, where g is integrable over E.

Now, let{fn}n=1 be a sequence of measurable functions all defined on E such that(ii) lim

nfn(x) = f(x) a.e. on E.

Suppose also that

(iii) |fn(x)| gn(x) x E and n N.Then, if

(46) limnEgn = Eg,then

(47) limn

E

fn =

E

f.

Since f(x) = limn

fn(x), (47) is equivalent to

(48) limn

E

fn =

E

limn

fn.

Remark 4.35.Note that in Theorem 4.34 each

fnis dominated by

gn, which might

change. If all the gns are equal, we get the regular DCT (see Theorem 4.31).


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5. Differentiation and Integration

5.1. Motivation.

Remark 5.1. Recall the Fundamental Theorem of Calculus: Let f be continuouson [a, b] and suppose that F : [a, b]

R is differentiable and satisfies F(x) = f(x) for

every x [a, b]. Then

(49)

ba

f(x)dx = F(b) F(a) orba

F(x)dx = F(b) F(a).

If F is measurable on [a, b], can we obtain equalities resembling (49)? There are threepossible roadblocks to answering this question in the affirmative:

(i) F(x) might not be defined everywhere, that is,

{x [a, b] : F is not differentiable }

might have positive measure.(ii) Suppose that F(x) is defined a.e. on [a, b]. F(x) might not be integrable over

[a, b], which implies that [a,b]

F

might not be defined.(iii) Even if F exists a.e. on [a, b] and is integrable over [a, b], how would we know

that

(50) [a,b] F = F(b) F(a)holds? In other words, why would (50) be true for more general functions than

just continuous functions?

Indeed, all three roadblocks above can occur, which we illustrate in the following threeexamples.

Example 5.2. (See Remark 5.1(i)) In the late 1800s Karl Weierstrass provided anexample of a function F which is continuous at every x

[a, b] but differentiable at no

point in [a, b].

Example 5.3. (See Remark 5.1(ii)) Note that

F(x) =

x2 sin

1x2

if 0 < x 1

0 if x = 0


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is continuous at every x [0, 1]. Furthermore, one can check that F is differentiable atall points in [0, 1] where

F(x) =

2x sin

1x2

2x

cos

1x2

if 0 < x 1

0 = limx0+

F(x)F(0)x0 if x = 0.

This function F is not integrable over [0, 1], however, so the integral[0,1]

F(x)

is not defined and thus we cant define an analog of the Fundamental Theorem of Cal-culus.

Example 5.4. (See Remark 5.1(iii)) Let

F(x) = 0 if 0 x 0

f(x + h) f(x)h

= inft>0

sup

0


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and

(59) Df(x) Df(x) Left-hand derivates.What if theyre all equal? It turns out that

(60) limh0+f(x + h)

f(x)

h exists and is finite D+f(x) = D+

f(x)

and

(61) limh0

f(x + h) f(x)h

exists and is finite Df(x) = Df(x).Thus

f(x) = limh0

f(x + h) f(x)h

exists and is finite if and

only if Df(x) = Df(x) = D+f(x) = D+f(x)

by (60) and (61). This can be used to determine when a function is not differentiable.

Remark 5.8. If a measurable function f is differentiable a.e. on [a, b], put h = 1n in thedefinition of derivative so that

(62) limn

f(x + 1n

) f(x)1n

= limn

n

f

x +

1

n

f(x)

exists a.e. on [a, b]. Note that fn(x) = n

f

x + 1n

f(x)

is measurable for every

nN, so if the limit in the right-hand side of (62) exists and is finite a.e., the limit

function limn

fn(x) will be Lebesgue measurable. Thus the derivative is measurable in

this case.

Definition 5.9. A function f : [a, b] R is called monotone increasing if wheneverx1, x2 [a, b] and x1 < x2, then f(x1) f(x2).

Definition 5.10. A function f : [a, b] R is called monotone decreasing if when-ever x1, x2 [a, b] and x1 < x2, then f(x1) f(x2).

Remark 5.11. Note that monotone functions are always bounded since f(a) f(x) f(b) for all x [a, b] if f is monotone increasing, whereas f(a) f(x) f(b) for allx [a, b] if f is monotone decreasing.


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Remark 5.12. If f(x) is monotone, then limtx+

f(t) and limtx

f(t) exist. For example, if

f is monotone increasing, then

(63) limtx+

f(t) = inft>x

{f(t)} and limtx

f(t) = supt


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5.3. Vitalis Covering Lemma.

Definition 5.15. Let E R. Let Ibe a collection of nondegenerate intervals ofR.We say that Iis a Vitali cover for E, or that Icovers E in the sense of Vitali, if x E and > 0, I Isuch that x I and m(I) = l(I) < .

Example 5.16. Let E = R and let

I= {(a, b) : a < b, a, b R}.This is a Vitali cover for R since if x R and > 0 is given, let

I =

x 3

, x +

3

.

Then x I and l(I) = 23 < .

Example 5.17. Let E = [0, 1] and letI= {(s, t) : 0 s < t 1, s ,t Q}.

You should convince yourself that this Iis a Vitali cover for [0, 1].

Lemma 5.18 (Vitalis Covering Lemma - Lemma 1, 5.1, p. 98 of [5]). SupposeE R with m(E) < (E need not be measurable here). LetIbe a Vitali cover forE.Then > 0 there exist pairwise disjoint intervals {In}Nn=1 with{In}Nn=1 I, IiIj =

for 1 i j N, i = j and such that

(66) mE Ni=1

Ii < .Theorem 5.19 (Theorem 3, 5.1, p. 100 of [5]). Letf be monotone increasing on [a, b].Then f is differentiable a.e. on [a, b]. Moreover, f is measurable (see Exercise ??) andintegrable over [a, b], with

(67) 0

[a,b]

f f(b) f(a).

Note that the left-hand inequality in (67) follows from the fact that all the derivatives

are nonnegative.

Corollary 5.20. Let f : [a, b] R be monotone decreasing. Then f exists a.e. on[a, b], f is integrable over [a, b] and

(68) < f(b) f(a)

[a,b]

f.


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Note that f is monotone increasing. The result follows by applying Theorem 5.19 tothe function f.

Remark 5.21. Theorem 5.19 and Corollary 5.20 show that all monotone functions fare differentiable a.e. on [a, b] and that f is integrable. Furthermore, these two results

show that all finite linear combinations of monotone functions are differentiable a.e. on[a, b] and that their derivatives are integrable.


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5.4. Variation and Absolute Continuity.

Definition 5.22. Let f : [a, b] R. Let(69) P= {a = x0 < x1 < x2 < < xn = b}be a partition of [a, b]. Define the total variation of f with respect to P, T(f,

P),

by

(70) T(f, P) =n

i=1

f(xi) f(xi1) < (note that (70) is finite since its a finite sum). We say that f is of bounded variationover [a, b], written f BV[a, b], if(71) sup

P partitionsof [a,b]

T(f, P) < .

If f

BV[a, b], let

(72) Tba(f) = supPT(f, P),

Tba(f) is called the total variation of f over [a, b].

Example 5.23. Let

(73) f(x) =

sin 1x if 0 < x 10 if x = 0

What happens when you take the supremum of all of these sums (70) over all possiblepartitions P of [0, 1]? Surely f(x) is not of bounded variation as defined above (seeDefinition 5.22).

Remark 5.24 (Properties and examples of functions of bounded variationover [a, b]).

(i) If f is monotone on [a, b], then f BV[a, b].

(ii) There exist examples of functions that are continuous on [a, b] yet are not of

bounded variation over [a, b]. Thus continuity alone is not enough to ensure afunctions is of bounded variation over [a, b]. See Exercise ??.

(iii) If f and g are of bounded variation over [a, b] and c R, thenf + g BV[a, b]

and

cf BV[a, b].


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This implies that the class of functions of bounded variation over [a, b] is a vectorspace over R.

(iv) If f exists at all x [a, b], and if f is bounded on [a, b], then f BV[a, b]. SeeExercise ?? and use the Mean Value Theorem.

(v) If f BV[a, b], then f is bounded on [a, b], i.e., M > 0 such that |f(x)| M.

Remark 5.25. We fix a bit of notation. For r R, let

(74) r+ =

r if r 00 if r < 0

and

(75) r =

0 if r > 0

r if r

0

.

Note that

(76) r = r+ r,(77) |r| = r+ + r,(78) 0 r+ |r|,and

(79) 0 r |r|.Note the similarity to Definition 3.56.

Definition 5.26. Let f : [a, b] R. If P= {a = x0 < x1 < < xn = b} is somepartition of [a, b], let the positive variation of f with respect to P be

(80) p(f, P) =n

i=1

f(xi) f(xi1)

+.

Similarly, define the negative variation of f with respect to P by

(81) n(f, P) =n

i=1 f(xi) f(xi1)

.

Remark 5.27. Note that by (78) and (79) we get that

(82) 0 p(f, P) T(f, P)and

(83) 0 n(f, P) T(f, P).


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This is true for all partitions Pof [a, b]. Note that, using (76), one can easily derive that(84) p(f, P) n(f, P) = f(b) f(a),that is,

(85) p(f, P) = f(b) f(a) + n(f, P)

for any partition Pof [a, b]. Similarly, using (77), we see that(86) p(f, P) + n(f, P) = T(f, P)for every partition Pof [a, b].

Definition 5.28. Let f BV[a, b]. We define the positive variation of f over [a, b]as

(87) Pba(f) = supP

p(f, P).

Similarly, we define the negative variation of f over [a, b] as(88) Nba(f) = supP

n(f, P).

Remark 5.29. Letting f BV[a, b], since(89) p(f, P) T(f, P) Tba(f) < ,(where the last inequality on the right comes from the assumption that f BV[a, b]),we see that

(90) 0 Pba(f) < since Tba(f) is an upper bound on p(f, P) for any partition Pof [a, b] by (89), hencethere exists a least upper bound on

p(f, P) for all partitions P of [a, b], which we

have defined as Pba(f) (see Definition 5.28). Similar to (89) and (90), we see that

(91) 0 Nba(f) Tba(f) < where again the last inequality on the right follows from the hypothesis that f BV[a, b].

Lemma 5.30 (Lemma 4, 5.2, p. 103 of [5]). Letf : [a, b] R be of bounded variationover [a, b]. Then(92) Pba(f) Nba(f) = f(b) f(a)and

(93) Pba(f) + Nba(f) = T

ba(f).


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Definition 5.31. Let f BV[a, b]. For a x b, considerTxa (f) : [a, b] [0, )

defined by

(94) Txa (f) = 0 if x = a

Tx

a

(f) if a < x

b

Similarly, define

Pxa (f) : [a, b] [0, )and

Nxa (f) : [a, b] [0, )in the corresponding way. Note that if f BV[a, b], then f is of bounded variation

over [a, x] for all x (a, b]. Thus, by Lemma 5.30 we have, for every x [a, b],(95) f(x) f(a) = Pxa (f) Nxa (f)and

(96) Txa (f) = Pxa (f) + Nxa (F).

If x = a, everything is zero.

Proposition 5.32. If f BV[a, b], then Txa (f), Pxa (f) and Nxa (f) are monotone in-creasing functions on [a, b].

Theorem 5.33 (Theorem 5, 5.2, p. 103 of [5]). Let f : [a, b] R. Then f is

of bounded variation on [a, b] if and only if f can be written as the difference of twomonotone increasing (real-valued) functions on [a, b].

Corollary 5.34 (Corollary 6, 5.2, p. 104 of [5]). Suppose that f : [a, b] R and thatf is of bounded variation over [a, b]. Then f exists a.e. on [a, b] and f is integrableover [a, b]. Note that f integrable over [a, b] implies that f is measurable over [a, b].

Remark 5.35. Note that we still dont have an analogue of the Fundamental Theoremof Calculus. Recall Exercise 5,

4.3, p. 89 of [5]: If f : [a, b]

[0,

], and f is

integrable over [a, b], then

(97) F(x) =

[a,x]

f(t)dt or F(x) =

[a,x]

f

is continuous on [a, b]. It turns out this is true in more generality. We have the followingLemma 5.36.


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Lemma 5.36 (Lemma 7, 5.3, p. 105 of [5]). Suppose that f : [a, b] [, ] isintegrable over [a, b]. Then

(98) F(x) =

xa

f(t)dt

is continuous on [a, b] and F BV[a, b].

Remark 5.37. Note that by Theorem 5.33 and its Corollary 5.34, F is differentiablea.e. on [a, b] and F is integrable over [a, b] where

F(x) =

xa

f(t)dt.

The natural question arises: What is F(x)? It turns out that F(x) = f(x), but thisis not as easy to show as in the Riemann case. We need to develop some further toolsfirst.

Lemma 5.38 (Lemma 8, 5.3, 105 of [5]). Suppose that f : [a, b] [, ] isintegrable over [a, b], and suppose that

(99) F(x) =

xa

f(t)dt = 0

for every x [a, b]. Then f(x) = 0 a.e. on [a, b].

Lemma 5.39 (Lemma 9, 5.3, p. 106 of [5]). Letf : [a, b] R be Lebesgue measurableand bounded on [a, b] (note that f is then integrable over [a, b]). Suppose that

(100) F(x) = F(a) + xa

f(t)dt.

Then F(x) = f(x) a.e. on [a, b].

Theorem 5.40 (Theorem 10, 5.3, p. 107 of [5]). Letf be a Lebesgue integrable functionover [a, b] and suppose that

(101) F(x) = F(a) +

xa

f(t)dt

for F(a)

R. Then F(x) = f(x) a.e. on [a, b].

Definition 5.41. Let g : [a, b] R. We say that g is absolutely continuous on[a, b] if > 0, > 0 such that whenever (ci, di)ni=1 is a pairwise disjoint collectionof intervals in [a, b] with

(102)n

i=1

(di ci) < ,


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then

(103)n

i=1

g(di) g(ci) < .Remark 5.42. It turns out that g is absolutely continuous on [a, b] if and only if g(x)exists a.e. on [a, b], g is integrable over [a, b], and

(104) g(x) g(a) =xa

g(t)dt

for every x [a, b]. We will see shortly how to prove this. This result extends theFundamental Theorem of Calculus to the class of absolutely continuous functions. Notethe integrand in (104) need not be continuous as in the Riemann case, it only needs tobe Lebesgue integrable.

Remark 5.43. Note that any absolutely continuous function is uniformly continuous on[a, b] (take n = 1 in Definition 5.41 and you get the definition of uniformly continuous).The converse is not true in general, however. We will shortly characterize all absolutelycontinuous functions. See Theorem 5.50.

Remark 5.44. If

F(x) =

xa

f(t)dt + F(a)

where f is Lebesgue integrable over [a, b], then F is absolutely continuous over [a, b].To prove this, use the continuity of the Lebesgue integral (see Proposition 4.28). The

converse of this will turn out to be true, but this is harder to prove. We will show thatany absolutely continuous function is differentiable first, and the easiest way to do thisis to show it is of bounded variation over [a, b].

Lemma 5.45 (Lemma 11, 5.4, p. 108 of [5]). Let f : [a, b] R be absolutelycontinuous on [a, b]. Then f is of bounded variation over [a, b].

Corollary 5.46. If f is absolutely continuous on [a, b], then f exists a.e. on [a, b] andf is Lebesgue integrable over [a, b].

Proof. See Corollary 5.34.

Remark 5.47. Linear combinations of absolutely continuous functions are absolutelycontinuous, but compositions are not necessarily. See Exercise 17(a), p. 111 of [5] (andbe careful - the Exercise as stated in the book is incorrect. Can you come up with acounterexample?).


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Lemma 5.48 (Lemma 13, 5.4, p. 109 of [5]). If f is absolutely continuous on [a, b]and f(x) = 0 a.e., then f(x) = f(a) is constant for every x [a, b].

Remark 5.49. Note that Lemma 5.48 is not true in general for functions of boundedvariation. Consider the Cantor ternary function (see Exercise 48, p. 50 and Exercise 15,

p. 111 of [5]). Call this function f. Then f : [0, 1] [0, 1], f is monotone increasing, fexists a.e. and f(x) = 0 a.e., but f is not constant. The problem is that f is continuousbut not absolutely continuous.

Theorem 5.50 (Characterization of absolutely continuous functions -Theorem 14, 5.4, p. 110 of [5]). A function F : [a, b] R can be expressed as anindefinite integral of a Lebesgue integrable function over[a, b] if and only ifF is absolutelycontinuous on [a, b].

Corollary 5.51 (Corollary 15, 5.4, p. 110 of [5]). Every absolutely continuous functionF on [a, b] can be written as the indefinite integral of its derivative plus a constant, F(a).


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5.5. Convexity.

Definition 5.52. Suppose that : (a, b) R is a function defined on an open interval(a, b) in R. We say that is convex if for each x, y (a, b), x < y and for every [0, 1],(105) x + (1 )y (x) + (1 )(y).This relates to concave up in the Calc 1 sense of the term (whereas concave meansconcave down in Calc 1 language). Consider the chord joining

x, (x)

and

y, (y)

.

The point on this chord whose first coordinate is x + (1 )y has second coordinateequal to (x) + (1 )(y). Hence is convex iff the graph of lies on or below anychord joining two points on the graph.

Lemma 5.53 ((Lemma 16, 5.5, p. 113 of [5]). Let be convex on (a, b). Supposec1, d1, c2, d2 are points in (a, b) such that c1 c2 < d2 and c1 < d1 d2, then

(106) (d1) (c1)d1 c1

(d2) (c2)d2 c2 .

The inequality in (106) is equivalent to

(107) (d2 c2)(d1) + (d1 c1)(c2) (d1 c1)(d2) + (d2 c2)(c1).

Definition 5.54. Suppose f : (a, b) R and let x0 (a, b). IfDf(x0) = Df(x0)exist, are finite and equal to each other, we say that f is left differentiable at x0 andwrite the common value Df(x0) = Df(x0) as fL(x0). Thus

(108) limxx

0

f(x) f(x0)x x0 = f

L(x0).

Similarly, if we have D+f(x0) = D+f(x0) and both are finite, we say that f is right

differentiable at x0 and write the common value D+f(x0) = D+f(x0) as f

R(x0). Thus

(109) limxx+

0

f(x) f(x0)x x0 = f

R(x0).

Remark 5.55. Note that, if perchance m

R is such that L(x0)

m

R(x0), then

we always have

(110)(x) (x0)

x x0 m (y) (x0)

y x0whenever a < x < x0 < y < b. We will need this to prove Jensens inequality (seeProposition 5.61).


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Proposition 5.56 (Proposition 17, 5.5, p. 113 of [5]). Let : (a, b) R be convex.Then on every closed subinterval of(a, b) is absolutely continuous. Moreover, the right-and left-hand derivatives L and

R exist at each point in (a, b) and

L(x) =

R(x)

except on a countable subset of [a, b] (Note that in general if a function is absolutelycontinuous then the derivative exists a.e. on its domain. See Corollary 5.46). Further-more, L and

R are monotone increasing functions and

L(x)

R(x) at each point

x (a, b).

Proposition 5.57 (Proposition 18, 5.5, p. 114 of [5]). Suppose is continuous on(a, b) and suppose that one derivate, say D+ is finite and monotone increasing on(a, b). Then is convex on [a, b].

Definition 5.58. If is convex on (a, b), m R and x0 (a, b), we say the line(111) y = m(x

x0) + (x0)

through the point (x0, (x0)) is a supporting line for if the graph of is always onor above the line in (125). This is equivalent to saying that (x) m(x x0) + (x0)for every x (a, b). In other words, (125) is a supporting line if

(x) (x0)x x0 m

for x > x0 and(x) (x0)

x x0 mfor x < x0. Thus if is differentiable and m

L(x0),

R(x0)

, then (125) is a

supporting line for . Thus convex functions always have supporting lines (althoughthis is not true for more general functions). See Proposition 5.56 and Remark 5.55.

Corollary 5.59 (Corollary 19, 5.5, p. 115 of [5]). Let : (a, b) R be twicedifferentiable on (a, b). Then is convex on (a, b) if and only if (x) 0 for everyx (a, b).

Example 5.60. To illustrate Corollary 5.59, consider

(i) Let (x) = e

x

= exp(x) for every x R. Since (x) = ex

> 0 for every x R, is convex.(ii) Let (x) = x2. Since (x) = 2x and (x) = 2 > 0 for every x R, is convex.

Proposition 5.61 (Jensens Inequality - Proposition 20, 5.5, p. 115 of [5]). Letbe convex on (, ) = R and suppose that f : [0, 1] R is Lebesgue integrable over[0, 1]. Suppose that either


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(i) f is integrable over [0, 1], or(ii) (x) 0 for all x R.

Then

(112)

[0,1]

f(t)

dt

[0,1]

f(t)dt

.

Corollary 5.62 (Corollary 21, 5.5, p. 116 of [5]). Let f be Lebesgue integrable over[0, 1]. Then

(113)

[0,1]

exp

f(t)

dt exp

[0,1]

f(t)dt

.

Remark 5.63. The concave version of Jensens inequality and its corollary (See Propo-sition 5.61 and Corollary 5.62) are as follows: If is concave and h is Lebesgue integrableover [0, 1], then

(114)

[0,1]

h(t)

dt

[0,1]

h(t)dt

.

Otherwise, let h(x) be positive and (x) = ln x. Then

(115)

[0,1]

ln

h(t)

dt ln

[0,1]

h(t)dt

.

Remark 5.64. Here is a useful fact about convex functions. Let : (a, b) R beconvex. Suppose that t1, t2, . . . , tn [0, 1] satisfy

ni=1 ti = 1. Ifx1, x2, . . . , xn (a, b),

then

ni=1

tixi

ni=1

ti(xi).


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6. Lp Spaces

Definition 6.1. Let X be either [0, 1] or R. Let f : X [, ] be Lebesguemeasurable. Recall f is said to be integrable over X if f+ and f are integrable overX (see Definition 4.25). This is equivalent to |f| being integrable over X (see Exercise10, p. 93 of [5]) and we say in this case that f is in L1(X,m), and write

(116) f1 =X

|f(x)|dm.

Remark 6.2. Recall if h : X [, ] is Lebesgue measurable and finite a.e.,and g : R R is continuous, then g h is Lebesgue measurable. Now, suppose thatf : X [, ] is Lebesgue measurable on X and finite a.e. Then |f| is Lebesguemeasurable on X and finite a.e. as well. Fix p [1, ). Let gp : [0, ) [0, )be defined by g

p(t) = tp. Note that g

pis continuous, thus g

p|f| = |f|p is Lebesguemeasurable and finite a.e.Definition 6.3. Let f be Lebesgue measurable on X, f : X [, ]. Fix p [1, ). We say f is in Lp(X,m) if

(117)

X

|f(x)|pdm(x) < ,

and we write

(118) fp = X |f(x)|pdm(x)1/p

(note that we will need the 1/p in (118) to prove the triangle inequality later).

Example 6.4. Let X = [0, 1] and let

f(x) =

if x = 0

1x

if 0 < x 1.It can be shown that f is in L1

[0, 1], m

. f is not in L2

[0, 1], m

, however, since

|f(x)|2 = if x = 01x if 0 < x 1

and [0,1]

1

xdm(x) = .


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Remark 6.5. Example 6.4 shows that, even on [0, 1], just because |f(x)| < a.e., itdoes not necessarily follow that f is in Lp[0, 1] for all p.

Example 6.6. Let X = R and let

g(x) = 0 if < x 11x

if 1 x < .Recall that g is not integrable over R since

R

|g(x)|dm(x) =

[1,)

1

xdx = lim

tln x

t1

= ,

so that g / L1(R, m). However,

|g(x)|2 =

0 if < x 11x2

if 1 x <

and R

|g(x)|2dx < so that g L2(R, m).

Remark 6.7. Powers of 1x are useful for constructing examples, as in Examples 6.4 and6.6.

Definition 6.8. We turn to the p =

case. Let f : X

[

,

] and suppose that

f is Lebesgue measurable. We say that f is essentially bounded on X if M 0such that

|f(x)| M a.e. on (X, m),that is,

m{x : |f(x)| > M} = 0.

M is called an essential bound for f in this case.

Example 6.9. Let X = R and let

f(x) =0 if x = 01 if x R\Q1

x if x Q\{0}.This function is not bounded since lim

nf( 1n) = limn

11n

= . Note, however, that f isessentially bounded by 1 since |f| = 1 a.e.


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Definition 6.10. We say that f is in L(X,m) iff is essentially bounded on (X, m).In this case we define

(119) f = M = inf{M 0 : M is an essential bound for f}.Note that the set in (119) is nonempty and bounded below by 0, so it has an infimum.

Proposition 6.11. If f L(X, m), thenf = M is an essential bound for f. Inother words, M is an element of the set in (119).

Remark 6.12. For 1 p , we want to show that Lp(X, m) is a normed vectorspace. Thus the key elements of the norm we want to show are

(i) For p : Lp(X, m) [0, ), fp = 0 f = 0 (this shows uniqueness ofthe zero vector).

(ii) For every R and for all f Lp

(X, m), f Lp

(X, m) and fp = ||fp.(iii) If f1, f2 Lp(X, m), then f1 + f2 Lp(X, m) and (Minkowskis inequality)f1 + f2p f1p + f2p.

The problem that arises is that iff = 0 a.e., then fp = 0 but f is not identically zero.To remedy this situation, if fp < and gp < , we say f and g are equivalent,written f g, if(120) m

{x : f(x) = g(x)} = 0,that is, if f(x) = g(x) a.e. on X. You should check that is an equivalence relation.We are now ready to give the proper definition of Lp(X, m).

Definition 6.13. The elements ofLp(X,m) are equivalence classes of functions

(121)

[f] : f : X [, ] is Lebesgue measurable and fp <

,

where the equivalence relation is as defined in Remark 6.12. It can be shown easily thatLp(X, m) thus defined is still a vector space. We get a unique zero vector in this case,fixing the problem from Remark 6.12. We will immediately become less formal and usemostly function notation (i.e. f and g) to denote elements ofLp(X, m) instead of [f], [g].

This should cause no confusion.

Proposition 6.14 (See 1, 6.2, p. 120 of [5]). Fix p [1, ]. Let f, g Lp(X, m).Then f + g Lp(X, m). Note that this actually works for all p (0, ], but we dontget a norm for p (0, 1).


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Proposition 6.15. We now show that for 1 p , Lp(X, m) is a normed vectorspace with norm

(122) fp =

X

|f(x)|pdx1/p

if 1 p <

ess sup |f(x)| if p = .That is, p : Lp(X, m) [0, ) defined in (122) satisfies

(i) fp = 0 f = 0.(ii) For every R and for every f Lp(X, m), fp = ||fp.

(iii) Minkowskis inequality: If f1, f2 Lp(X, m), thenf1 + f2p f1p + f2p.

(See 1, 6.2, p. 120 of [5] and Theorem 6.16).

Theorem 6.16 (Minkowskis Inequality - See 1, 6.2, p. 120 of [5]). This is theproof of Proposition 6.15(iii). If f, g Lp(X, m) with 1 p , then(123) f + gp fp + gp.Furthermore, if 1 < p < , then equality will hold in (123) if and only if , Rsuch that f = g.

Remark 6.17. Thus for 1 p , Lp(X, m) is a normed vector space. In fact, we willshow that Lp(X, m) is a Banach space for 1 p , that is, Lp(X, m) is complete in

the metric induced by its norm (see Remark 6.22 and Theorem 6.34). First we establishthe next important inequality for Lp-spaces, the Holder inequality.

Definition 6.18. Ifp [1, ], we defined the conjugate exponent q to p as follows:

q =

if p = 1

pp1 if 1 < p <

1 if p =

.

Note that 1 < pp1 < in the middle case above and limp1+

pp1 = . Furthermore,

limp

pp1 = 1 so our definition is a continuous choice of conjugate exponents. Note that

1

p+

1

q=

1

p+

1p

p1=

1

p+

p 1p

= 1.

The special case where p = q occurs when p = q = 2.


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Proposition 6.19 (Youngs Inequality - See Exercise 8, p. 123 of [5]). Leta 0,b 0 and p (1, ). Let q be the conjugate exponent to p. Then(124) ab a

p

p+

bq

q

with equality holding in (124) if and only if ap = bq.

Theorem 6.20 (Holders Inequality - See 4, 6.2, p. 121 of [5]). Let (X, ,m)be

[0, 1], M, m or (R, M, m). Letp [1, ] and let q be the conjugate exponent top. Letf Lp(X, m) and g Lq(X, m). Thenf g is Lebesgue integrable over X (i.e.,f g L1(X, m)) and(125) f g1 fpgq,with equality holding in (125) if and only if , 0 with |f(x)|p = |g(x)|q a.e.Corollary 6.21 (Cauchy-Schwarz Inequality). If f, g L2(X, m), then f g L1(X, m) (i.e., f g is Lebesgue integrable over X) and

f g1 f2g2.This is why L2 is an important space - we are able to define an inner product.

Remark 6.22. Note that the norm p in (122) makes Lp(X, m) into a metric spacewith metric

dp

[f], [g]

= f gp.

Minkowskis inequality can be used to prove that, given [f], [g], [h]

Lp(X, m),

dp

[f], [g] dp[f], [h]+dp[h], [g],

the triangle inequality in the metric space.

Definition 6.23. We say that a sequence {fn}n=1 in a normed linear space is a Cauchysequence if for any > 0 there is an N N such that for all n N and all m N,fn fm < . It is easy to show that every convergent sequence is a Cauchy sequence.

Definition 6.24. Recall a metric space (Y, d) is said to be complete if every Cauchysequence {yn}n=1 Y converges to some y Y.

Example 6.25. Note that (R, d = | |) (i.e. R with the normal absolute value) iscomplete, whereas (Q, | |) is not complete.


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Definition 6.26. Let (Y, ) be a normed vector space over R. We say that (Y, ) isa Banach space over R if (Y, d) is a complete metric space (where d is the metricinduced by the norm. See Remark 6.22 for an example). We will show that Lp(X, m) isa Banach space for 1 p .

Remark 6.27. For the record, everything we do from now on will be over R. Ourresults generalize to over C, but this is not the focus of our course.

Definition 6.28. Let (Y, ) be a normed vector space (over R). Let {yk}k=1 Y bea sequence. We say that

k=1 yk is summable in (Y, ) if s Y such that taking

sn =n

k=1 yk, then the sequence of partial sums {sn}n=1 converges to s in (Y, ), thatis,

limn

s sn = 0.

Example 6.29. Consider the following:

(i) Let (Y, ) = (R, | |). Let

yk =(1)k

k

k=1

R. Thenk=1

(1)kk is summable

(the alternating harmonic series).

(ii)

yk =(1)kk2

k=1

R is also summable.

Definition 6.30. Let {yk}k=1 be a sequence in (Y, ), a normed vector space. Wesay that the sum k=1 yk is absolutely summable if(126)

k=1

yk < ,

that is, if the sum in (126) converges.

Remark 6.31. Absolutely summable does not imply summable in general, althoughit does in R. Take (Q, | |), for example, and let y1 = 3, y2 = 110 , y3 = 4100 , y4 = 11000 ,y5 =

510000 , etc. That is, let

yn = nth digit of

10n.

Thenk=1

yk =k=1

|yk| <

converges to in R but

k=1 yk is not summable in (Q, | |).


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For the converse, i.e. that summability does not imply absolute summability, consider

the sequence from Example 6.29(i),

yk =(1)k

k

k=1

(R, | |). Recall that k=1 yk issummable, but

k=1

|yk| =k=1

1

k=

so the sum is not absolutely summable. Note that the series from Example 6.29(ii) isabsolutely summable.

Theorem 6.32 (Proposition 5, 6.3, p. 124 of [5]). A normed vector space (Y, ) overR is a Banach space (in the metric induced by its norm) if and only if every absolutelysummable series in (Y, ) is summable in (Y, ).

Remark 6.33. As an application of Theorem 6.32, we will show that Lp(X, m) isa Banach space for 1 p . Suppose that {yk}k=1 Lp(X, m) is such that

k=1 fkp < . It must be shown that there exists an f Lp(X, m) such that

limn

fn

k=1

fk

p

= 0.

This is not easy, but we prove it in the following Theorem 6.34.

Theorem 6.34 (Riesz-Fischer Theorem - Theorem 6, 6.3, p. 125 of [5]). Letp [1, ]. Then L

p

(X, m) is a Banach space.

Remark 6.35. Recall that we showed in Section 3 that for f a Lebesgue integrablefunction over X, for every > 0 there exists a step function and a continuous function such that

X

f(x) (x)dm < and X

f(x) (x)dm < with |(x)| < |f(x)| and |(x)| < |f(x)| (see Proposition 3.60). We want to do the same

thing in L

p

(X, m), but we will need p < . We first do the bounded case.

Lemma 6.36 (Lemma 7, 6.4, p. 127 of [5]). Let 1 p < and let f Lp(X, m).Then there exists anM > 0 and a bounded Lebesgue measurable function fM Lp(X, m)such that |fM(x)| |f(x)| for every x X and f fMp < for any > 0.


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Proposition 6.37 (Proposition 8, 6.4, p. 128 of [5]). Letf Lp[0, 1], p [1, ). Given > 0 there exists a step function Lp[0, 1] and a continuous function Lp[0, 1]such that

f p < and f p < .Note that we already proved this for the p = 1 case (see Proposition 3.60). We provethis for X = [0, 1] and leave the case X = R as an exercise (Hint: Use DCT).

Definition 6.38. Let (Y, ) be a normed linear space over R. A map F : Y R iscalled a bounded linear functional if:

(i) F(f1 + f2) = F(f1) + F(f2) for all , R and for all f1, f2 Y (linearity)(ii) There exists an M > 0 such that |F(f)| Mf for every f Y (boundedness).

Proposition 6.39. The bounded linear functionals on (Y, ) are precisely the contin-uous linear functionals on Y.

Proof. (=) If F is a bounded linear functional on Y, then M > 0 such that for anyf1, f2 Y, |F(f1 f2)| Mf1 f2 |F(f1) F(f2)| Mf1 f2, which canbe made arbitrarily small if f1 and f2 are close together.(=) The proof of this direction requires functional analysis and so we will assume itfor now.

Definition 6.40. Given a bounded linear functional F on (Y, ), we define the normof F, written F, by

(127) F = sup |F(f)|f : f Y {0} .Note that, by definition,

|F(f)|f F f Y {0},

which implies that

(128) |F(f)| F f f Ysince (128) is also true for f = 0.

Example 6.41 (See Proposition 11, 6.5, p. 131 of [5]). Fix g L2

[0, 1]. Note thatby Holders inequality, gf L1[0, 1] (i.e., gf is Lebesgue integrable over [0, 1]) for allf L2[0, 1] (recall 2 is the conjugate exponent of itself). Define Tg : L2[0, 1] R byTg(f) =

[0,1]

gf dx and note that Tg is linear (which follows from the linearity of the

integral). Tg is also bounded:

|Tg(f)|

[0,1]

gf dx

[0,1]

|g||f| dx

[0,1]

|g|2dx 12

[0,1]

|f|2dx 12


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by Holders inequality again, which implies that |Tg(f)| g2f2. Thus for [f] = 0 inL2[0, 1],

|Tg(f)|f2 g2 = Tg g2

since g2 is an upper bound on the set in (127), but Tg is the least upper bound onthat set by Definition 6.40, thus

Tg

g

2.

Remark 6.42. Example 6.41 is true more generally for conjugate exponents p and q.The same argument shows that for any fixed g Lq(X, m), Tg : Lp(X, m) R definedby

Tg(f) =

X

gf dx

is a bounded linear functional on Lp(X,m, p) with(129) Tg gq.It turns out that

Tg

=

g

q, which we now show. First we need a definition.

Definition 6.43. Define

sgn g(x) =

1 if g(x) > 0

0 if g(x) = 0

1 if g(x) < 0Note that sgn g(x) is bounded and measurable (its a simple function). Note also thatg(x) [sgn g(x)] = |g(x)| and [sgn g(x)] |g(x)| = g(x).

Proposition 6.44 (Proposition 11, 6.5, p. 131 of [5]). If p and q are conjugate expo-nents in [1, ], thenTg = gq.

Remark 6.45. Our main aim of this section is to show that if p [1, ), any boundedlinear functional on Lp[0, 1] must be of the form Tg, where g Lq[0, 1] and q is theconjugate exponent to p (Tg is as defined in Remark 6.42). This g is in fact unique,which establishes a one-to-one correspondence between the bounded linear functionalson Lp[0, 1] and the elements of Lq[0, 1], a result known as the Riesz RepresentationTheorem (see Theorem 6.47). This is also true for Lp(R, m), but this is beyond thescope of this course.

Lemma 6.46 (Lemma 12, 6.5, p. 131 of [5]). Let p [1, ). Letg be a real-valued,Lebesgue measurable function on [0, 1] with the following property: M > 0 such that

for every bounded, Lebesgue measurable function f on [0, 1], fg is integrable with[0,1]

fg

Mfp.


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Then g Lq[0, 1] and gq M, where q is the conjugate exponent to p.

Theorem 6.47 (Riesz Representation Theorem - Theorem 13, 6.5, p. 132 of[5]). LetF be a bounded linear functional on Lp[0, 1], 1 p < . Letq be the conjugateexponent to p. Then there exists a unique g

Lq[0, 1] (up to equivalence class) such that

F(f) = [0,1]

gf dx

for every f Lp[0, 1]. Moreover, F = gq.

Corollary 6.48. LetF be a bounded linear functional on L2[0, 1]. Then there exists aunique g L2[0, 1] such that

F(f) =

[0,1]

f(x)g(x) dx.

Proof. Take p = q = 2 in Theorem 6.47.


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7. Metric Spaces

7.1. The Basics.

Definition 7.1. Let X be a nonempty set. We say that a map d : X

X

[0,

) isa metric on X if

(i) d(x, y) = 0 x = y, x, y X.(ii) d(x, y) = d(y, x) for every x, y X (symmetry)

(iii) If x,y,z X, then d(x, z) d(x, y) + d(y, z) (triangle inequality).In this case, (X, d) is said to be a metric space.

Example 7.2 (Examples of metric spaces).

(i) X = R, d(x, y) = |x y|.(ii) X = L

p

[0, 1], d(f, g) = f gp, p [1, ].(iii) X = Rn, d(x, y) =

ni=1

(xi yi)21/2

, where x = (x1, . . . , xn) and y = (y1, . . . , yn)

are in Rn.

Definition 7.3. Let {xn}n=1 X.(i) We say {xn}n=1 converges to x0 X if for every > 0 there exists an N N

such that d(x0, xn) < whenever n N.(ii) The sequence {xn}n=1 X is Cauchy if and only if for every > 0 there exists

an N N such that if m > n N, d(xm, xn) < .(iii) (X, d) is called a complete metric space if every Cauchy sequence converges tosome x0 X.

Definition 7.4. Let U X. We say U is open in (X, d) if, given any x U, thereexists a > 0 such that whenever y X and d(x, y) < , then y U.

Remark 7.5 (See Proposition 1, 7.2, p. 142 of [5]). Note that and U = X are bothopen subsets of X.

Example 7.6. Let x X, fix > 0 and define B(x, ) = {y X : d(x, y) < }. ThenB(x, ) is open in X (prove this - it doesnt simply follow from the definition), calledthe open ball at x with radius .


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7.2. Compactness.

Definition 7.7. Let (X, d) be a metric space and let K X. We say that {U}I isan open cover of K if

I

U K and each U is open in X.

Example 7.8. Let X = R, d(x, y) = |x y|. Let K = (0, 1) R and Un =

1n+1 ,

nn+1

.

Each Un is open in R andn=1

Un = (0, 1) = K so {Un}n=1 is an open cover of K.

Definition 7.9. Let (X, d) be a metric space and K X. We say that K is compactin (X, d) if every open cover {U}I of K admits a finite subcover U1, . . . , U N withN

j=1Uj K. X is said to be a compact metric space ifX is compact in (X, d) itself.

Example 7.10. (See Example 7.8) Consider (0, 1) (R, d||). Let Un =

1n+1

, nn+1

.

Thenn=1

= (0, 1) but {Un}n=1 contains no finite subcover since U1 U2 U3 (0, 1) and thus

Ni=1

Ui = UN =

1

N + 1,

N

N + 1

(0, 1)

for every N N.

Remark 7.11. It turns out that [0, 1] R is compact however (see Example 7.10).More generally, [a, b] R is compact. In fact, any closed and bounded subset of Ris compact, which follows from the HeineBorel Theorem: a subset K ofRn withEuclidean metric (see Example 7.2(iii)) is compact if and only ifK is closed and boundedin Rn (see Theorem 15, 2.5, p. 44 of [5] for the proof in R). This is not true for moregeneral metric spaces.

Definition 7.12. Let (X, d) be a metric space. We say that (X, d) is sequentially

compact if for any sequence {xn}n=1 X there is a convergent subsequence {xnk}k=1 {xn}n=1 (i.e., there exists an x0 X with lim

kxnk = x0 in the d-metric). This will turn

out to be equivalent to a metric space being compact (see Theorem 7.18).

Remark 7.13. [a, b] R are all sequentially compact, which follows from the Bolzano-Weierstrass Theorem (see Lemmas 16 and 17, 7.7, pp. 153-154 of [5]).


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Definition 7.14. We say that K (X, d) is bounded if there exists an M [0, )such that for every x, y K, d(x, y) M. Note that [a, b] R is bounded. We saythat (X, d) is bounded if X is bounded as a subset of itself.

Example 7.15. Let > 0. Fix x0

X and let K = B(x0, ) (see Example 7.6). Then

if x, y K,d(x, y) d(x, x0) + d(x0, y) by the triangle inequality

= d(x, x0) + d(y, x0) by symmetry

< +

= 2.

Thus B(x0, ) is bounded.

Definition 7.16. We say that K

(X, d) is totally bounded if for every > 0 there

exist x1, . . . , xn X such that ni=1

B(xi, ) K, that is, we can cover K with a finitenumber of -balls.

Remark 7.17. Note that totally bounded implies bounded but not conversely.

Theorem 7.18 (Theorem 21, 7.7, p. 155 and Proposition 25, 7.7, p. 156 of [5]).Let(X, d) be a metric space. Then the following are equivalent:

(i) (X, d) is compact (open covers have finite subcovers).(ii) (X, d) is sequentially compact (every sequence has a convergent subsequence).(iii) (X, d) is complete and totally bounded.


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7.3. Continuous Functions.

Definition 7.19. Let (X, d) be a metric space and let f : X R be a function. Wesay f is continuous at x0 X if for every > 0 there exists a > 0 such that wheneverx X and d(x, x0) < , then |f(x) f(x0)| < . We say that f is continuous on X iff is continuous at each point x

X.

Remark 7.20. Let (X, d) be a compact metric space. Let f : X R be continuouson X. Then the range of f, f(X) = {f(x) : x X} is a compact subset ofR. Inparticular, the range off is bounded. Hence there exists an M > 0 such that |f(x)| Mfor every x X.

Definition 7.21. If (X, d) is a compact metric space, we let C(X) denote the familyof all real-valued continuous functions on X.

Remark 7.22. Note that C(X) is a vector space over R:

(i) C(X) has a zero function f0 where f0(x) = 0 for every x X.(ii) If R and f C(X), then f C(X) where (f)(x) = f(x).

(iii) If f, g C(X), then (f + g)(x) = f(x) + g(x) is continuous on X.Note also that C(X) has a norm, defined by

f = supxX

|f(x)| = maxxX

|f(x)|.

Note that this definition makes sense since the supremum exists (the fs are bounded

see Remark 7.20) and continuous functions achieve their maximum. We have the usualproperties of the norm:

(i) f = 0 f = f0 = 0.(ii) If R and f C(X), then f = ||f.

(iii) If f, g C(X), then f + g f + g (triangle inequality).Thus C(X) is a normed vector space.

Proposition 7.23. Let (X, d) be a compact metric space. Let {fn}n=1 C(X) andf

C(X). Then lim

n fn

f

= 0 iff fn

f uniformly on x.

Proof. See Exercise 10, p. 136 of [5], Assignment 1. We can use this result to prove thatC(X) is complete just like L (see Exercise 11, p. 126 of [5], Assignment 1).

Proposition 7.24. Let (X, d) be a compact metric space. Then C(X) is complete inthe metric determined by its norm (see Definition 7.21 and Remark 7.22).


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7.4. The StoneWeierstrass Theorem.

Remark 7.25. A natural question to ask is what subspaces are dense in C(X)? Thisis the aim of the StoneWeierstrass Theorem (see Theorem 7.35).

Definition 7.26. Let S C(X). We say S separates the points of X if for everyx1, x2 X with x1 = x2, there exists an f S with f(x1) = f(x2). In other words, fdistinguishes between the points of X.

Example 7.27. Let S = C(X). Then C(X) separates the points of X. See Exercise 2,Assignment 2.

Definition 7.28. Let

A C(X) and suppose

Ais a subspace (over R) of C(X). We

say that A is an algebra of functions in C(X) if whenever f, g A, then f g Awhere (f g)(x) = f(x)g(x).

Remark 7.29. Note that C(X) is not only a normed vector space, but it is also analgebra: if f and g are continuous, then so is f g.

Example 7.30. Let X = [0, 1]. Let

A=

{p : [0, 1]

R : p is a polynomial of arbitrary degree with R coefficients

}.

Then A is an algebra of C[0, 1] since A is a subspace and is closed under multiplication.

Definition 7.31. Let L be a nonempty subset of C(X). We say that L is a lattice inC(X) if whenever f, g L, then fg and fg are in L where fg(x) = max{f(x), g(x)}and f g(x) = min{f(x), g(x)} for every x X.

Remark 7.32. Note that if L C(X) is a lattice and if {f1, . . . , f n} L, it followseasily by induction that f

1 f

2 fn

(x) = minf1(x), . . . , f n(x) and f1 f2 fn(x) = maxf1(x), . . . , f n(x) define functions f1 f2 fn and f1 f2 fnin L.

Remark 7.33. C(X) is a lattice in itself since if f, g C(X), then

(130) f g(x) = maxf(x), g(x) = f(x) + g(x) + |f(x) g(x)|2

C(X)


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since the absolute value of a continuous function is continuous (prove (130). There aretwo cases: f(x) g(x) and g(x) f(x)). Similarly,

f g(x) = minf(x), g(x) = f(x) + g(x) |f(x) g(x)|2

C(X).

Remark 7.34. Not every algebra is a lattice. Consider C[0, 1] and let A denote thealgebra of all polynomials on [0, 1] (see Example 7.30). Take f(x) = x and g(x) = 1 xon [0, 1]. Then neither f g(x) = maxf(x), g(x) nor f g(x) = minf(x), g(x) isin A since neither is a polynomial (graph it

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