Real Analysis Lecture Notes.pdf

REAL ANALYSIS LECTURE NOTES

Contents

1. Introduction and Preliminaries 12. The Structure of R 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 345.3. Vitali’s 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 Stone–Weierstrass Theorem 638. General Measure and Integration 678.1. 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 Ascoli–Arzela Theorem 113Appendix A. Convergence in Measure 116References 118Index 119

REAL ANALYSIS LECTURE NOTES 1

1. Introduction and Preliminaries

The following notes are from a year of graduate Real Analysis that I took at the Uni-versity of Colorado, Boulder. We used Royden’s text Real Analysis, 3rd edition (see [3]).There are no proofs in what follows, only statements of Propositions, Theorems, andLemmas, as well as many Definitions, Examples and Remarks throughout. I wanted alist of all results that I could easily and quickly navigate and that was extensively cross-referenced to itself and the original source. It proved enormously helpful while doinghomework and studying for exams. Since I find Royden difficult to follow, I originallyintended to eventually fill in all proofs and provide very detailed, careful explanations.Luckily that has largely been done for me in the very helpful book Measure, Integraland Probability by Capinski and Kopp (see [1]). Another good treatment of the subjectis [4]. For an excellent account of the history of the subject (and an easy, delightfulread), see [2]. My primary goal is clarity, which for me involves showing steps, explain-ing reasons, and justifying answers. Some may find these notes too verbose and preferthe terse treatment in [3]. This is a matter of personal preference. Any comments orsuggestions are welcome, as are corrections. Errors are of course mine. We begin witha brief motivational review of Riemann integration, followed by some elementary settheory and structure of the real line.

Jonathan KishBoulder, ColoradoJuly, [email protected]

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, let

mi = infxi−1≤x≤xi

f(x) and Mi = supxi−1≤x≤xi

f(x).

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

Lp(f) =n∑

i=1

mi(xi − xi−1).

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

Up(f) =n∑

i=1

Mi(xi − xi−1).

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 P of [a, b]. Moreover, note that the

set{Lp(f)

}is bounded above and so has a supremum. Similarly, the upper Riemann

integral of f over [a, b], written ∫ b

a

f(x)dx,

is defined by

(3)

∫ b

a

f(x)dx = infP

{Up(f)

}.

We always have that ∫a

b

f(x)dx ≤∫ b

a

f(x)dx.

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

b

f(x)dx =

∫ b

a

f(x)dx =

∫ b

a

f(x)dx.


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 = 01√x

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 by

f(x) =

{0 if x is rational

1 if x is irrational.

For any partition P of [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 =

∫ b

a

f(x)dx.

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

∫lim

n→∞fn(x)

if limn→∞

fn(x) = f(x) exists.

Remark 1.4. We denote the power set of a set X by 2X or P(X). The latter shouldcause no confusion with the notation P that stands for a partition, as the meaning ofthe symbol P will always be clear from the context.

Definition 1.5. Let A be a subset of the underlying set X. The complement of A in

X, A, is defined by

A = {x ∈ X : x /∈ A}.


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 DeMorgan’s Laws). Let C be a collection of subsets ofX. Then ⋂

A∈C

A =⋃A∈C

A and⋃A∈C

A =⋂A∈C

A.

See Exercise 13, p. 16 of [3].

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

A∆B = (A ∼ B) ∪ (B ∼ A)

= (A ∩ B) ∪ (B ∩ A).

Note that A∆B = B∆A 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 A is 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, thenn⋃

i=1

Ai ∈ A.

(iv) If A1, . . . , An ∈ A, thenn⋂

i=1

Ai ∈ A.


(v) If A,B ∈ A, then B ∩ A ∈ A.(vi) If A,B ∈ A, then A∆B ∈ 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 A be the “smallest” algebra of sets of Xcontaining C. See Proposition 1.13.

Proposition 1.13 (Proposition 1, §1.4, p. 17 of [3]). Let X 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, then B also contains the algebra A.

Definition 1.14. The smallest algebra containing C as in Proposition 1.13 is called thealgebra generated by C.

Proposition 1.15 (Proposition 2, §1.4, p. 17 of [3]). Let X 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 6= n.

(ii) For every n ∈ N,n⋃

i=1

Bi =n⋃

i=1

Ai. (these are in the algebra A)

(iii)∞⋃i=1

Bi =∞⋃i=1

Ai. (these infinite unions might not be in A, but as subsets of X

they 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 σ-algebra of subsets of X if, whenever {Ai}∞i=1 ⊂ A,∞⋃i=1

Ai ∈ A. Thus we are

allowing countably infinite unions now, not just finite unions.

Remark 1.18 (Facts about σ-algebras).

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

(ii) Let {Ai}∞i=1 ⊂ A for A a σ-algebra. Then∞⋂i=1

Ai ∈ A.


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

Proposition 1.20 (Proposition 3, §1.4, p. 19 of [3]). Let X be a nonempty set and let Cbe a nonempty collection of subsets of X. Then there is a unique σ-algebra A containingC that is the smallest possible in the sense that if B is any other σ-algebra and C ⊂ B,then A ⊂ B. Compare to Proposition 1.13, and see Exercise 19, p. 19 of [3].

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


2. The Structure of R

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

Remark 2.2. Note that ∅ is open as there is nothing to verify. Moreoever, all of R isopen. Now, if x ∈ (a, b), let δ = min(x − a, x − b). Then (x − δ, x + δ) ⊂ (a, b), whichimplies that (a, b) is open.

Proposition 2.3 (Properties of open sets).

(i) Let C be any collection of open subsets of R. Then⋃O∈C

O is open (Proposition 7,

§2.5, p. 41 of [3]).

(ii) If {Oi}ni=1 is any finite collection of open subsets of R, then

n⋂i=1

Oi is open (Corol-

lary 6, §2.5, p. 41 of [3]).

Remark 2.4. If we have a countably infinite family of open subsets of R, {Oi}∞i=1, then∞⋂i=1

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

On =(− 1

n, 1

n

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

∞⋂n=1

On =∞⋂

n=1

(− 1

n,1

n

)= {0}.

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

Proposition 2.5 (Proposition 8, §2.5, p. 42 of [3]). 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 δ > 0, (x− δ, x+ δ) ∩ E 6= ∅.

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 ∀ δ > 0, (x− δ, x+δ) ∩ E 6= ∅. However, E 6⊂ E in general.


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 [3]).

(iii) Note, however, that in general, A ∩B 6= A ∩B.

(iv) If E ⊂ R, then E = E (Proposition 11, §2.5, p. 43 of [3]).

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

Proposition 2.11 (Properties of closed sets).

(i) Let C be any collection of closed subsets of R. Then⋂

F∈CF is closed (Proposition

13, §2.5, p. 44 of [3]).

(ii) If {Fi}ni=1 is any finite collection of closed subsets of R, then

n⋃i=1

Fi is closed

(Proposition 12, §2.5, p. 44 of [3]).

Compare to Proposition 2.3.

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

then∞⋃i=1

Fi need not be closed. For example, let Fn =[−2 + 1

n, 3− 1

n

]for every n ∈ N,

each of which is closed. Then∞⋃

n=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 [3]). 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 σ-algebra containing all theopen intervals, that is, containing C = {(a, b) : −∞ ≤ a < b ≤ ∞}. B(R) is calledthe σ-algebra of Borel sets of R. B(R) also contains every open subset of R since


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 of R (see Proposition 2.11(iii)).

Definition 2.16. A set A ⊂ R is said to be a Gδ-set if we can write

A =∞⋂i=1

Oi,

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

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

B =∞⋃i=1

Fi,

where each Fi is closed. Note that an Fσ-set need not be closed (cf. Remark 2.12).

Remark 2.18. Note that Gδ-sets and Fσ-sets are all Borel sets. Moreover, we could con-sider (Gδ)σ-sets (the countable union of Gδ-sets) or (Fσ)δ-sets (the countable intersectionof Fσ-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

αn = inf{xn, xn+1, xn+2, . . . } and βn = sup{xn, xn+1, xn+2, . . . }.Thus we have two new sequences {αn}∞n=1 and {βn}∞n=1 where the αn’s are increasing,the βn’s are decreasing, and αn ≤ βn for all n.

Define the lower limit of {xk}∞k=1 by

lim xk = lim inf xk = sup{αn}∞n=1 = limn→∞

αn

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

Similarly, define the upper limit of {xk}∞k=1 by

lim xk = lim sup xk = inf{βn}∞n=1 = limn→∞

βn

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

lim xk = limn→∞

αn ≤ limn→∞

βn = lim xk.


Example 2.20. Consider the sequence

{x1, x2, . . . , x2n−1, x2n, . . . } =

{−1, 3,−1

2,5

2, . . . ,− 1

n, 2 +

1

n, . . .

}.

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


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 of R 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 of R, we want

˜(⋃i≥1

Ai

)=∑i≥1

˜(Ai).

(iv) Translation invariance: If A ⊂ R and r ∈ R, then ˜(A + r) = ˜(A), whereA+ 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 won’t be able toget countable additivity in general (condition (iii)). We can satisfy all four conditionsfor Borel sets, but not for every subset of R.

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

m∗(A) = inf

{∑n≥1

`(In) : all countable collectionsof open intervals {In}n≥1

suchthat A ⊂

⋃n≥1

In

}.

Here we allow∑

n≥1 `(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}n≥1 such that A ⊂

⋃n≥1 In,∑

n≥1 `(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 [3]). Let I be an interval in R. Thenm∗(I) = `(I).


Proposition 3.5 (Countable Subadditivity - Proposition 2, §3.2, p. 57 of [3]). If{An}n≥1 is a countable collection of subsets of R, then

m∗

(⋃n≥1

An

)≤∑n≥1

m∗(An).

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

Corollary 3.7 (Corollary 4, §3.2, p. 58 of [3]). 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}n≥1

in 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) <m∗(A1) +m∗(A2).

Remark 3.13. At this point we want to define a class of subsets M⊆ P(R) containingall the intervals - indeed, containing all the Borel sets: B ⊂M. It will turn out thatM isthe σ-algebra of all Lebesgue measurable subsets of R, and m∗ will be countably additivewhen restricted to M. See Definition 3.14, Theorems 3.21 and 3.24, and Proposition3.26.


Definition 3.14. Let E ⊂ R. We say E is Lebesgue measurable, and write E ∈M,if for every A ⊂ R,

(4) m∗(A) = m∗(A ∩ E) +m∗(A ∩ E).

Remark 3.15. So E ⊂ R is Lebesgue measurable if E splits any A ⊂ R into “two niceparts.” Note that

A = A ∩ R

= A ∩ (E ∪ E)

= (A ∩ E) ∪ (A ∩ E).

Also,

(A ∩ E) ∩ (A ∩ E) = A ∩ (E ∩ E) = A ∩ ∅ = ∅,so A∩E and A∩ E are disjoint and their union is A. By Proposition 3.5, we know that

m∗(A) = m∗((A ∩ E) ∪ (A ∩ E))≤ m∗(A ∩ E) +m∗(A ∩ E).

So in order to show equality in (4) (i.e., that E is Lebesgue measurable), it is enough toshow that, for every A ⊂ R,

(5) m∗(A) ≥ m∗(A ∩ E) +m∗(A ∩ E).

Remark 3.16. We want to show that M is a σ-algebra of subsets of R containing theBorel sets. Thus, among other things, it must be shown that

(i) ∅, R ∈M. (so M is nonempty)

(ii) If E ∈M, then E ∈M.(iii) If {En}n≥1 ⊂M, then

⋃n≥1

En ∈M.

This will establish that M is a σ-algebra of sets.

Proposition 3.17 (Basic Facts About Measurable Sets).

(i) If E ⊂ R and m∗(E) = 0, then E ∈M (see Lemma 6, §3.3, p. 58 of [3]).

(ii) If E ⊂ R, then E ∈M ⇐⇒ E ∈M.(iii) R ∈M.

Lemma 3.18 (Lemma 7, §3.3, p. 59 of [3]). If E1, E2 ∈M, then E1 ∪ E2 ∈M.

Corollary 3.19 (Corollary 8, §3.3, p. 59 of [3]). M is an algebra of subsets of R.

Proof. See Remark 3.16 (i), (ii) and Lemma 3.18. �


Lemma 3.20 (Lemma 9, §3.3, p. 59 of [3]). If A ⊂ R and {Ei}ni=1 is a family of pairwise

disjoint, Lebesgue measurable subsets of R, then

(6) m∗

(A ∩

(n⋃

i=1

Ei

))=

n∑i=1

m∗(A ∩ Ei).

Theorem 3.21 (Theorem 10, §3.3, p. 60 of [3]). The collection M of Lebesgue measur-able subsets of R is a σ-algebra of subsets of R.

Lemma 3.22 (Lemma 11, §3.3, p. 60 of [3]). Fix a ∈ R. Then I = (a,∞) ∈M.

Corollary 3.23. Every open interval is Lebesgue measurable.

Theorem 3.24 (Theorem 12, §3.3, p. 61 of [3]). Every Borel subset of R is Lebesguemeasurable.

Definition 3.25. We now define the Lebesgue measure on sets E ∈ M by m(E) =m∗(E). Thus m is just m∗ restricted to M. Note that m has all the properties of m∗

(see Remark 3.3 and Proposition 3.4) and in addition has the property of countableadditivity (see Proposition 3.26). This is our generalization of length of intervals.

Proposition 3.26 (Countable Additivity - Proposition 13, §3.3, p. 62 of [3]). Let{Ei}i≥1 be a countable sequence of pairwise disjoint Lebesgue measurable subsets of R.Then

(7) m

(⋃i≥1

Ei

)=∑i≥1

m(Ei).

Corollary 3.27. Suppose {Ei}∞i=1 is a sequence of Lebesgue measurable subsets of R.Then

(8) m

(∞⋃i=1

Ei

)= lim

N→∞m

(N⋃

i=1

Ei

).

Corollary 3.28. If {Ei}∞i=1 is an increasing sequence of Lebesgue measurable subsets ofR, then

(9) m

(∞⋃i=1

Ei

)= lim

N→∞m(EN).


Proposition 3.29 (Proposition 14, §3.3, p. 62 of [3]). Let {En}∞n=1 be a decreasingsequence of Lebesgue measurable subsets of R such that m(E1) <∞. Then

(10) m

(∞⋂

n=1

En

)= lim

n→∞m(En).

Remark 3.30. Note that Proposition 3.29 need not be true ifm(E1) = ∞. For instance,let En = [n,∞) for every n ∈ N. These are certainly decreasing since [n,∞) ⊃ [n+1,∞)where m(En) = `(En) = ∞. Thus limn→∞m(En) = ∞. On the other hand,

m

(∞⋂

n=1

En

)= m(∅) = 0

(see Remark 3.3(ii)), which is strictly less than limn→∞m(En). See Exercise 11, p. 64of [3].

Example 3.31. Let C denote the Cantor ternary set (cf. Exercises 37-39, p. 46

of [3]). Let F1 = [0, 1] and let G1 = (13, 2

3). Then F2 = F1 ∩ G1 = [0, 1

3] ∪ [2

3, 1] is

closed. Let G2 = (19, 2

9) ∪ (7

9, 8

9) be the “middle thirds” of the intervals in F2. Then

F3 = F2 ∩ G2 is closed. Continue this process. Fn consists of 2n−1 disjoint, closedintervals each of length 1/3n−1. Gn is defined as above to be the “middle thirds” of the

intervals in Fn. Then Fn+1 = Fn ∩ Gn is still closed and consists of 2n intervals. Notethat F1 ⊃ F2 ⊃ F3 ⊃ · · · is a decreasing sequence of Lebesgue measurable subsets of Rwhere m(F1) = m

([0, 1]

)= 1 <∞. Hence by Proposition 3.29 applied to

C =∞⋂

n=1

Fn,

we have

m(C) = limn→∞

m(Fn) = limn→∞

2n−1 · 1

3n−1= lim

n→∞

(2

3

)n−1

= 0.

Therefore m(C) = 0.

The consequence: recall that C is uncountable. Thus there exists an uncountable setwith Lebesgue measure zero. Furthermore, if A ⊂ C, then 0 ≤ m∗(A) ≤ m(C) = 0, whichimplies that m∗(A) = 0. Hence every subset of C is measurable (see Proposition 3.17(ii)), and there are 2c (the cardinality of P(C)) such subsets, where c is the cardinality ofthe contiuum. Therefore there are more Lebesgue measurable sets than Borelsets.


Proposition 3.32 (Proposition 5, §3.2, p. 58 of [3]).

(i) Let A ⊂ R. Then for every ε > 0, there exists an open set U ⊃ A such thatm(U) ≤ m∗(A) + ε.

(ii) Given A ⊂ R, there exists a Gδ-set G with G ⊃ A and m(G) = m∗(A).


Proposition 3.33 (Proposition 15, §3.3, p. 63 of [3]). Let E ⊂ R. Then the followingfive conditions are equivalent:

(i) E ∈M.

(ii) For every ε > 0, there exists an open set U ⊃ E such that m∗(U ∩ E) < ε.

(iii) For every ε > 0, there exists a closed set F ⊂ E such that m∗(E ∩ F ) < ε.

(iv) There exists a Gδ-set G ⊃ E such that m∗(G ∩ E) = 0.

(v) There exists an Fσ-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 ε > 0, there exists a finite collection {Ui}Ni=1 of open intervals such that

(11) m∗

[(N⋃

i=1

Ui

)∆E

]< ε.


Remark 3.34. Part (vi) of Proposition 3.33 is the precise statement of Littlewood’sFirst Principle: “Every Lebesgue measurable set is almost a finite union of openintervals.” 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 ε(E) denote the family of equivalence classes of E with respect to ∼. For eachequivalence class γ ∈ ε(E), choose xγ ∈ γ ⊂ E = [0, 1] (using the Axiom of Choice). Let

P =⋃

γ∈ε(E)

{xγ}.

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 thatm(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 [3] for the full details.


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 [3]). Let f be an extended real-valued function whose domain E is measurable. Then the following four conditions areequivalent:

(i) ∀α ∈ R {x ∈ E : f(x) > α} ∈ M.(ii) ∀α ∈ R {x ∈ E : f(x) ≥ α} ∈ M.

(iii) ∀α ∈ R {x ∈ E : f(x) < α} ∈ M.(iv) ∀α ∈ R {x ∈ E : f(x) ≤ α} ∈ M.

All of the conditions (i)− (iv) imply

(v) ∀α ∈ R {x ∈ E : f(x) = α} ∈ M (actually this is true ∀α ∈ R∪ {−∞,∞}).Note that conditions (i)− (iv) can be rewritten as

(i′) ∀α ∈ R f−1((α,∞]

)∈M.

(ii′) ∀α ∈ R f−1([α,∞]

)∈M.

(iii′) ∀α ∈ R f−1([−∞, α)

)∈M.

(iv′) ∀α ∈ R f−1([−∞, α]

)∈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 α ∈ R. Consider{x ∈ R : f(x) > α

}= f−1

((α,∞)

).

Note that (α,∞) is open. Since f is continuous, f−1((α,∞)

)is open and hence

f−1((α,∞)

)∈ M (see Proposition 18, §2.6, p. 47 of [3]). 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) χA : R −→ R by

χA(x) =

{1 if x ∈ A0 if x ∈ A

.

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


Lemma 3.41. Let A ⊂ R. Then χA 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 [3]). 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 of R if

m({x ∈ E : P is not true for x}

)= 0.

Proposition 3.45 (Proposition 21, §3.5, p. 69 of [3]). If f and g are defined on E ∈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 [3]). 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}n

i=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) = lim

n→∞fn(x) exists in the extended real numbers for every x ∈ E, then f is

measurable.

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


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

(12) ϕ(x) =n∑

i=1

ciχAi(x)

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

n⋃i=1

Ai = E, we have

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

Remark 3.50 (Converse to Proposition 3.49). If

g(x) =n∑

i=1

ciχAi(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 simplefunction ϕ. Note that ϕ corresponds one set to each element in the range of ϕ.

Remark 3.52. If ϕ1 and ϕ2 are simple functions and c, d ∈ R, then cϕ1+dϕ2, cϕ1−dϕ2

and ϕ1ϕ2 (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

n→∞fn(x) = f(x). That is, ∀ ε > 0, ∃N ∈ N

(depending on ε and on x) such that

|fn(x)− f(x)| < ε

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 ∀ ε > 0, ∃N ∈ N (depending only on ε and not on x) such that

|fn(x)− f(x)| < ε

for every n ≥ N and for every x ∈ E (the same N ∈ N works for all x ∈ E). This meanslim

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

Theorem 3.55. Let f : E −→ [0,∞] be a measurable, nonnegative extended real-valuedfunction defined on E ∈ M. Then there is an increasing sequence {ϕn}∞n=1 of simplefunctions with lim

n→∞ϕn(x) = f(x) for all x ∈ E. If f is bounded, this convergence is

uniform. See Exercise 4, p. 89 of [3].


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 ∈M by Proposition 3.37, P ∩N = ∅ and P ∪N = E. Let

(13) f+ = fχP =

{f(x) if x ∈ P0 if x ∈ N = E ∩ P

(where we define −∞ · 0 = 0) and

(14) f− = −fχN =

{−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 χP and χN 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). Thismotivates the following corollary.

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

n→∞ϕn(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 ∈ M with m(E) < ∞, then given ε > 0 there exists a simplefunction ϕ such that

m({x ∈ E : |f(x)− ϕ(x)| ≥ ε}

)< ε.

That is, if we let

B = {x ∈ E : |f(x)− ϕ(x)| ≥ ε},then for all x ∈ E ∩ B, |f(x) − ϕ(x)| < ε. In other words, ϕ approximates f very well

on E ∩ B. Related to this is Littlewood’s Second Principle: “Every measurable


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 [3]). Let f be a measurable functiondefined on a closed interval [a, b]. Assume f takes on the values ±∞ on a set of measure0. Then given any ε > 0 there exists a step function g and a continuous function h suchthat

m({x ∈ [a, b] : |f(x)− g(x)| ≥ ε}

)< ε

and

m({x ∈ [a, b] : |f(x)− h(x)| ≥ ε}

)< ε.

That is,

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

and

|f(x)− h(x)| < ε except on a set of measure less than ε.

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 [3].

Remark 3.61. The intuitive form of Littlewood’s 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 [3]). Let E be a measurable subset ofR such that m(E) < ∞. Suppose that {fn}∞n=1 is a sequence of measurable real-valuedfunctions defined on E that converges pointwise to some real-valued function f definedon E. Then given ε > 0 and δ > 0 there is a measurable set A ⊂ E with m(A) < δ and

∃N ∈ N such that |fn(x)− f(x)| < ε 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. Theset A depends on both δ and ε so that as you change δ and ε the set A changes.

Corollary 3.64 (Proposition 24, §3.6, p. 73 of [3]). 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.


Proposition 3.65 (Egoroff’s Theorem - See Exercise 30, p. 73 of [3]). Let {fn}∞n=1

be 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 η > 0 there exists a measurable set A ⊂ E with m(A) < η

and {fn}∞n=1 converges uniformly to f on E ∩ A.


4. Lebesgue Integration

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

(17) ϕ(x) =n∑

i=1

ciχAi(x)

where

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

ci0χAi0(x) = 0 ·χAi0

= 0,

so that this term does not contribute to the sum (17). Thus we will henceforth assumethat all the ci’s are not equal to zero.

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

m({x ∈ R : ϕ(x) 6= 0}

)<∞.

This will happen if and only ifn∑

i=1

m(Ai) <∞.

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

ϕ or

∫Rϕ

is defined by

(18)

∫ϕ =

n∑i=1

cim(Ai)

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

(19)

∫E

ϕ =n∑

i=1

cim(Ai ∩ E)

Lemma 4.4. If E ∈M, then ∫E

ϕ =

∫ϕχE.


Lemma 4.5 (Lemma 1, §4.2, p. 78 of [3]). Let ϕ be a simple function that vanishesoutside a set of finite measure. Suppose that we can write

(20) ϕ(x) =m∑

j=1

bjχEj(x)

where the {Ej}mj=1 are measurable, pairwise disjoint subsets of R. Note that (20) might

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

(21)

∫ϕ =

m∑j=1

bjm(Ej).

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

Proposition 4.6 (Proposition 2, §4.2, p. 78 of [3]). Let ϕ and ψ be simple functionsthat both vanish outside a set of finite measure. Then

(i) ∀ a, b ∈ R, ∫(aϕ+ bψ) = a

∫ϕ+ b

∫ψ.

(ii) If ϕ(x) ≥ ψ(x) on R, then ∫ϕ ≥

∫ψ.

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

ψ : ψ is simple and ψ(x) ≥ f(x) ∀x ∈ E}.

Remark 4.8. Note that we could have taken

(24) sup

{∫E

ϕ : ϕ is simple and ϕ(x) ≤ f(x) ∀x ∈ 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).


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) Am(E) ≤∫

E

f ≤ Bm(E).

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

(26) A = inf

{∫E

ψ : ψ is simple and ψ(x) ≥ f(x) ∀x ∈ E},

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

(27) B = sup

{∫E

ϕ : ϕ is simple and ϕ(x) ≤ f(x) ∀x ∈ 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 [3]). Let f be a bounded real-valuedfunction defined on E ∈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 [3]). Let f 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

(∫ b

a

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 [3]). 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.


(i′) ∫E

(f + g) =

∫E

f +

∫E

g.

Hence for every a, b ∈ R,∫E

(af + bg) = a

∫E

f + b

∫E

g.

(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, then

M1m(E) ≤∫

E

f ≤M2m(E).

(v) If A and B are disjoint measurable subsets of E with m(A) <∞ and m(B) <∞,then ∫

A∪B

f =

∫A

f +

∫B

f.

Corollary 4.15 (Corollary to Proposition 4.14 (ii)). If f is a bounded, Lebesguemeasurable function defined on E ∈M with m(E) <∞, then

(30)

∣∣∣∣∫E

f

∣∣∣∣ ≤ ∫E

|f |.

Proposition 4.16 (Bounded Convergence Theorem, or BCT - Proposition 6,§4.2, p. 84 of [3]). Let {fn}∞n=1 be a sequence of Lebesgue measurable functions definedon E ∈M with m(E) <∞. Suppose that there is an M > 0 such that |fn(x)| ≤M forevery n ∈ N and for every x ∈ E. If there exists an f such that f(x) = lim

n→∞fn(x) a.e.

on E, then

(31)

∫E

f = limn→∞

∫E

fn(x).

Note that since f(x) = limn→∞

fn(x), (31) is equivalent to saying that

(32)

∫E

limn→∞

fn(x) = limn→∞

∫E

fn(x).

Remark 4.17. Note that, so far, we can compute Lebesgue integrals for Lebesguemeasurable functions over E ∈M when

(i) f is bounded on E and(ii) m(E) <∞.


But is this really that big of an improvement over the Riemann integral? Consider thefollowing examples:

1.

∫ 1

0

1√xdx. This function is not bounded on [0, 1], but you could compute an

improper Riemann integral.

2.

∫[1,∞)

1

x2. Here, if E = [1,∞), then m(E) = ∞ and we cannot compute the

Lebesgue integral.

In both cases we should be able to compute the Lebesgue integral if it is indeed ageneralization of the Riemann integral. This motivates Definition 4.18.

Definition 4.18. Let f be a nonnegative Lebesgue measurable function defined onE ∈ M (E doesn’t have to be finite here). We define the Lebesgue Integral of fover E by

(33)

∫E

f = sup

{∫E

h : 0 ≤ h ≤ f

},

where h is bounded, Lebesgue measurable and vanishes outside a set of finite measure.Note that it is possible to have ∫

E

f = ∞.

We say that the function f is integrable over E if∫E

f <∞,

that is, if the set {∫E

h : 0 ≤ h ≤ f

}in (33) is bounded above. Note that we always have, for f ≥ 0,

0 ≤∫

E

f ≤ ∞.

Proposition 4.19 (Proposition 8, §4.3, p. 85 of [3]). Let f, g be nonnegative Lebesguemeasurable functions defined on E ∈M and let a ∈ R such that 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.


(iii) If f(x) ≤ g(x) a.e. on E, then∫E

f ≤∫

E

g.

(iv) If E1 and E2 are disjoint measurable subsets of R, then∫E1∪E2

f =

∫E1

f +

∫E2

f.

We use this Proposition to prove Fatou’s Lemma (Lemma 4.20).

Theorem 4.20 (Fatou’s Lemma - Theorem 9, §4.3, p. 86 of [3]). 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)

∫E

f ≤ lim

∫E

fn.

Note that we’re not saying the RHS of (34) isn’t ∞, or that both sides of (34) aren’t ∞.

Theorem 4.21 (Monotone Convergence Theorem, or MCT - Theorem 10, §4.3,p. 87 of [3]). Let {fn}∞n=1 be a sequence of nonnegative, Lebesgue measurable functionsdefined on E ∈M. Suppose also that the fn’s 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”). Then f(x) = limn→∞

fn(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 [3]). 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].


Corollary 4.23 (Proposition 12, §4.3, p. 87 of [3]). Let f be a nonnegative integrablefunction defined on E ∈ M. Let {Ei}∞i=1 be pairwise disjoint measurable subsets of E.Then

(38)

∫∞S

i=1Ei

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 [3].

Proposition 4.27 (Proposition 13, §4.3, p. 88 of [3]). 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

f −∫

E

g.

Proposition 4.28 (Proposition 14, §4.3, p. 88 of [3]). Let f be a nonnegative, integrableextended real-valued function defined on E ∈M. Then for any ε > 0, ∃ δ > 0 such thatwhenever A ⊂ E is measurable with m(A) < δ, we have∫

A

f < ε.

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.


Proposition 4.29 (Proposition 15, §4.4, p. 90 of [3]). Let f and g be integrable functionsover E ∈M. Then

(i) For every c ∈ R, cf is integrable over E and∫E

cf = c

∫E

f.

(ii) f + g is integrable over E and∫E

(f + g) =

∫E

f +

∫E

g.

(iii) If f(x) ≤ g(x) a.e. on E, then∫E

f ≤∫

E

g.

In particular,

(41)

∣∣∣∣∫E

f

∣∣∣∣ ≤ ∫E

|f |.

(iv) If E1 and E2 are disjoint measurable subsets of E, then∫E1∪E2

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 ∀ ε > 0, ∃ δ > 0 such that wheneverA ⊂ E and m(A) < δ,

(42)

∣∣∣∣∫A

f

∣∣∣∣ < ε.

Theorem 4.31 (Lebesgue Dominating Convergence Theorem, or DCT -Theorem 16, §4.4, p. 91 of [3]). 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 thatlim

n→∞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.


fn(x), (43) is equivalent to

(44)

∫E

limn→∞

fn = limn→∞

∫E

fn.


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 [3]). This motivates the following corollary.

Corollary 4.33. Let {fn}∞n=1 be a decreasing sequence of nonnegative, Lebesgue mea-surable functions defined on E ∈M. Suppose further that f1 is integrable over E. Then,letting f = lim

n→∞fn, we have

(45)

∫E

f = limn→∞

∫E

fn.

Note that, since the fn’s are decreasing and nonnegative (i.e., bounded below by 0),lim

n→∞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 [3]). Let {gn}∞n=1 be a sequence of nonnegative,integrable functions defined on E ∈M. Suppose that

(i) limn→∞

gn(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) limn→∞

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

Suppose also that

(iii) |fn(x)| ≤ gn(x) ∀x ∈ E and ∀n ∈ N.

Then, if

(46) limn→∞

∫E

gn =

∫E

g,

then

(47) limn→∞

∫E

fn =

∫E

f.


fn(x), (47) is equivalent to

(48) limn→∞

∫E

fn =

∫E

limn→∞

fn.

Remark 4.35. Note that in Theorem 4.34 each fn is dominated by gn, which mightchange. If all the gn’s are equal, we get the regular DCT (see Theorem 4.31).


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) forevery x ∈ [a, b]. Then

(49)

∫ b

a

f(x)dx = F (b)− F (a) or

∫ b

a

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 thanjust 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 1800’s Karl Weierstrass provided anexample of a function F which is continuous at every x ∈ [a, b] but differentiable at nopoint 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


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

)− 2

xcos(

1x2

)if 0 < x ≤ 1

0 = limx→0+

F (x)−F (0)x−0

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 can’t 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 < 1

2

1 if 12≤ x ≤ 1

and note that

F ′(x) =

{0 if 0 ≤ x < 1

2

0 if 12≤ x ≤ 1

Note also that F ′(x) is not defined at x = 12

since F ′ is not continuous there. Still,F ′(x) = 0 a.e. on [0, 1]. In this case,∫

[0,1]

F ′ =

∫[0,1]

0 = 0,

butF (1)− F (0) = 1,

so we don’t have equality in a Fundamental Theorem of Calculus-like setting.

Remark 5.5. We will be able to define a class of functions on [a, b], called the abso-lutely continuous functions, that avoid problems (i), (ii) and (iii) of Remark 5.1 andsatisfy

(i′) F ′ is defined a.e. on [a, b],(ii′) F ′ is integrable over [a, b], and

(iii′)

∫[a,b]

F ′ = F (b)− F (a).

Our goal, therefore, is a version of the Fundamental Theorem of Calculus that holds fora class of Lebesgue measurable functions.


5.2. Derivatives and Derivates.

Definition 5.6. Recall that if f is differentiable at x ∈ [a, b], we have

limh→0

f(x+ h)− f(x)

h= f ′(x) ∈ R,

the derivative of f at x. Even if f is not differentiable at x, we can define the Diniderivates: D+, read “D upper plus,” is defined as

(51) D+f(x) = limh→0h>0

f(x+ h)− f(x)

h= inf

t>0

{sup

0<h<t

{f(x+ h)− f(x)

h

}}.

To simplify notation, we will rewrite the limits on the lim as h→ 0+ in (51) so that

(52) D+f(x) = limh→0+

f(x+ h)− f(x)

h= inf

t>0

{sup

0<h<t

{f(x+ h)− f(x)

h

}}.

Similarly, D+, read “D lower plus,” is defined as

(53) D+f(x) = limh→0+

f(x+ h)− f(x)

h= sup

t>0

{inf

0<h<t

{f(x+ h)− f(x)

h

}}.

D−, read “D upper minus,” is defined as

(54) D−f(x) = limh→0−

f(x+ h)− f(x)

h= inf

t<0

{sup

t<h<0

{f(x+ h)− f(x)

h

}}.

D−, read “D lower minus,” is defined as

(55) D−f(x) = limh→0−

f(x+ h)− f(x)

h= sup

t<0

{inf

t<h<0

{f(x+ h)− f(x)

h

}}.

Note that (54) and (55) can be rewritten as

(56) D−f(x) = limh→0+

f(x)− f(x+ h)

h= inf

t>0

{sup

0<h<t

{f(x)− f(x+ h)

h

}}and

(57) D−f(x) = limh→0+

f(x)− f(x+ h)

h= sup

t>0

{inf

0<h<t

{f(x)− f(x+ h)

h

}}.

Remark 5.7. By the definition of Dini Derivates (see Definition 5.6), the following twoinequalities are immediate:

(58) D+f(x) ≤ D+f(x) Right-hand derivates


and

(59) D−f(x) ≤ D−f(x) Left-hand derivates.

What if they’re all equal? It turns out that

(60) limh→0+

f(x+ h)− f(x)

hexists and is finite ⇐⇒ D+f(x) = D+f(x)

and

(61) limh→0−

f(x+ h)− f(x)

hexists and is finite ⇐⇒ D−f(x) = D−f(x).

Thus

f ′(x) = limh→0

f(x+ h)− f(x)

hexists and is finite if and

only if D−f(x) = D−f(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+ 1

n

)− f(x)

]is measurable for every

n ∈ N, so if the limit in the right-hand side of (62) exists and is finite a.e., the limitfunction lim

n→∞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.


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

f(t) and limt→x−

f(t) exist. For example, if

f is monotone increasing, then

(63) limt→x+

f(t) = inft>x{f(t)} and lim

t→x−f(t) = sup

t<x{f(t)}.

Interchange these in (63) for f monotone decreasing.

Remark 5.13. If f(x) is monotone on [a, b], f can be discontinuous at at most acountable number of points in [a, b]. Indeed,

(64) f is not continuous at x ⇐⇒ limt→x+

f(t) 6= limt→x−

f(t).

The jump at a discontinuity cannot be too big. Is it possible to have the functionunbounded? We will soon provide an answer to this question.

Proposition 5.14. Let f be monotone. Then f is bounded by Remark 5.11. Let

(65) D = {x ∈ [a, b] : f is not continuous at x}.Then D is countable.


5.3. Vitali’s Covering Lemma.

Definition 5.15. Let E ⊂ R. Let I be a collection of nondegenerate intervals of R.We say that I is a Vitali cover for E, or that I covers E in the sense of Vitali, if∀x ∈ E and ∀ ε > 0, ∃ I ∈ I such that x ∈ I and m(I) = l(I) < ε.

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 ε > 0 is given, let

I =(x− ε

3, x+

ε

3

).

Then x ∈ I and l(I) = 2ε3< ε.

Example 5.17. Let E = [0, 1] and let

I = {(s, t) : 0 ≤ s < t ≤ 1, s, t ∈ Q}.You should convince yourself that this I is a Vitali cover for [0, 1].

Lemma 5.18 (Vitali’s Covering Lemma - Lemma 1, §5.1, p. 98 of [3]). SupposeE ⊂ R with m∗(E) <∞ (E need not be measurable here). Let I be a Vitali cover for E.Then ∀ ε > 0 there exist pairwise disjoint intervals {In}N

n=1 with {In}Nn=1 ⊂ I, Ii∩Ij = ∅

for 1 ≤ i ≤ j ≤ N , i 6= j and such that

(66) m∗

(E ∼

N⋃i=1

Ii

)< ε.

Theorem 5.19 (Theorem 3, §5.1, p. 100 of [3]). Let f 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 derivativesare 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 ′.


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


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(xi−1)∣∣ <∞

(note that (70) is finite since it’s a finite sum). We say that f is of bounded variationover [a, b], written f ∈ BV [a, b], if

(71) supP partitions

of [a,b]

{T (f,P)

}<∞.

If f ∈ BV [a, b], let

(72) T ba(f) = sup

P

{T (f,P)

},

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

Example 5.23. Let

(73) f(x) =

{sin 1

xif 0 < x ≤ 1

0 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 variation

over [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 ofbounded 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, then

f + g ∈ BV [a, b]

and

cf ∈ BV [a, b].


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 ≥ 0

0 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(xi−1)

]+.

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

(81) n(f,P) =n∑

i=1

[f(xi)− f(xi−1)

]−.

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).


This is true for all partitions P of [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 P of [a, b]. Similarly, using (77), we see that

(86) p(f,P) + n(f,P) = T (f,P)

for every partition P of [a, b].

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

(87) P ba(f) = sup

P

{p(f,P)

}.

Similarly, we define the negative variation of f over [a, b] as

(88) N ba(f) = sup

P

{n(f,P)

}.

Remark 5.29. Letting f ∈ BV [a, b], since

(89) p(f,P) ≤ T (f,P) ≤ T ba(f) <∞,

(where the last inequality on the right comes from the assumption that f ∈ BV [a, b]),we see that

(90) 0 ≤ P ba(f) <∞

since T ba(f) is an upper bound on p(f,P) for any partition P of [a, b] by (89), hence

there exists a least upper bound on{p(f,P)

}for all partitions P of [a, b], which we

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

(91) 0 ≤ N ba(f) ≤ T b

a(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 [3]). Let f : [a, b] −→ R be of bounded variationover [a, b]. Then

(92) P ba(f)−N b

a(f) = f(b)− f(a)

and

(93) P ba(f) +N b

a(f) = T ba(f).


Definition 5.31. Let f ∈ BV [a, b]. For a ≤ x ≤ b, consider

T xa (f) : [a, b] −→ [0,∞)

defined by

(94) T xa (f) =

{0 if x = a

T xa (f) if a < x ≤ b

Similarly, define

P xa (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 variationover [a, x] for all x ∈ (a, b]. Thus, by Lemma 5.30 we have, for every x ∈ [a, b],

(95) f(x)− f(a) = P xa (f)−Nx

a (f)

and

(96) T xa (f) = P x

a (f) +Nxa (F ).

If x = a, everything is zero.

Proposition 5.32. If f ∈ BV [a, b], then T xa (f), P x

a (f) and Nxa (f) are monotone in-

creasing functions on [a, b].

Theorem 5.33 (Theorem 5, §5.2, p. 103 of [3]). Let f : [a, b] −→ R. Then f isof 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 [3]). 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 don’t have an analogue of the Fundamental Theoremof Calculus. Recall Exercise 5, §4.3, p. 89 of [3]: If f : [a, b] −→ [0,∞], and f isintegrable 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.


Lemma 5.36 (Lemma 7, §5.3, p. 105 of [3]). Suppose that f : [a, b] −→ [−∞,∞] isintegrable over [a, b]. Then

(98) F (x) =

∫ x

a

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) =

∫ x

a

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 [3]). Suppose that f : [a, b] −→ [−∞,∞] isintegrable over [a, b], and suppose that

(99) F (x) =

∫ x

a

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 [3]). Let f : [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) +

∫ x

a

f(t)dt.

Then F ′(x) = f(x) a.e. on [a, b].

Theorem 5.40 (Theorem 10, §5.3, p. 107 of [3]). Let f be a Lebesgue integrable functionover [a, b] and suppose that

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

∫ x

a

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 ∀ ε > 0, ∃ δ > 0 such that whenever

{(ci, di)

}n

i=1is a pairwise disjoint collection

of intervals in [a, b] with

(102)n∑

i=1

(di − ci) < δ,


then

(103)n∑

i=1

∣∣g(di)− g(ci)∣∣ < ε.

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) =

∫ x

a

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) =

∫ x

a

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). Theconverse 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 [3]). 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 [3] (andbe careful - the Exercise as stated in the book is incorrect. Can you come up with acounterexample?).


Lemma 5.48 (Lemma 13, §5.4, p. 109 of [3]). 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 [3]). Call this function f . Then f : [0, 1] −→ [0, 1], f is monotone increasing, f ′

exists 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 [3]). A function F : [a, b] −→ R can be expressed as anindefinite integral of a Lebesgue integrable function over [a, b] if and only if F is absolutelycontinuous on [a, b].

Corollary 5.51 (Corollary 15, §5.4, p. 110 of [3]). Every absolutely continuous functionF on [a, b] can be written as the indefinite integral of its derivative plus a constant, F (a).


5.5. Convexity.

Definition 5.52. Suppose that ϕ : (a, b) −→ R is a function defined on an open interval(a, b) in R. We say that ϕ is convex if for each x, y ∈ (a, b), x < y and for every λ ∈ [0, 1],

(105) ϕ(λx+ (1− λ)y

)≤ λϕ(x) + (1− λ)ϕ(y).

This relates to “concave up” in the Calc 1 sense of the term (whereas concave means“concave down” in Calc 1 language). Consider the chord joining

(x, ϕ(x)

)and

(y, ϕ(y)

).

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

Lemma 5.53 ((Lemma 16, §5.5, p. 113 of [3]). Let ϕ 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)ϕ(d1)− ϕ(c1)

d1 − c1≤ ϕ(d2)− ϕ(c2)

d2 − c2.

The inequality in (106) is equivalent to

(107) (d2 − c2)ϕ(d1) + (d1 − c1)ϕ(c2) ≤ (d1 − c1)ϕ(d2) + (d2 − c2)ϕ(c1).

Definition 5.54. Suppose f : (a, b) −→ R and let x0 ∈ (a, b). If D−f(x0) = D−f(x0)exist, are finite and equal to each other, we say that f is left differentiable at x0 andwrite the common value D−f(x0) = D−f(x0) as f ′L(x0). Thus

(108) limx→x−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 rightdifferentiable at x0 and write the common value D+f(x0) = D+f(x0) as f ′R(x0). Thus

(109) limx→x+

0

f(x)− f(x0)

x− x0

= f ′R(x0).

Remark 5.55. Note that, if perchance m ∈ R is such that ϕ′L(x0) ≤ m ≤ ϕ′R(x0), thenwe always have

(110)ϕ(x)− ϕ(x0)

x− x0

≤ m ≤ ϕ(y)− ϕ(x0)

y − x0

whenever a < x < x0 < y < b. We will need this to prove Jensen’s inequality (seeProposition 5.61).


Proposition 5.56 (Proposition 17, §5.5, p. 113 of [3]). Let ϕ : (a, b) −→ R be convex.Then on every closed subinterval of (a, b) ϕ is absolutely continuous. Moreover, the right-and left-hand derivatives ϕ′L and ϕ′R exist at each point in (a, b) and ϕ′L(x) = ϕ′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, ϕ′L and ϕ′R are monotone increasing functions and ϕ′L(x) ≤ ϕ′R(x) at each pointx ∈ (a, b).

Proposition 5.57 (Proposition 18, §5.5, p. 114 of [3]). Suppose ϕ is continuous on(a, b) and suppose that one derivate, say D+ϕ is finite and monotone increasing on(a, b). Then ϕ is convex on [a, b].

Definition 5.58. If ϕ is convex on (a, b), m ∈ R and x0 ∈ (a, b), we say the line

(111) y = m(x− x0) + ϕ(x0)

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

ϕ(x)− ϕ(x0)

x− x0

≥ m

for x > x0 andϕ(x)− ϕ(x0)

x− x0

≤ m

for x < x0. Thus if ϕ is differentiable and m ∈[ϕ′L(x0), ϕ

′R(x0)

], then (125) is a

supporting line for ϕ. 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 [3]). Let ϕ : (a, b) −→ R be twicedifferentiable on (a, b). Then ϕ is convex on (a, b) if and only if ϕ′′(x) ≥ 0 for everyx ∈ (a, b).

Example 5.60. To illustrate Corollary 5.59, consider

(i) Let ϕ(x) = ex = exp(x) for every x ∈ R. Since ϕ′′(x) = ex > 0 for every x ∈ R,ϕ is convex.

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

Proposition 5.61 (Jensen’s Inequality - Proposition 20, §5.5, p. 115 of [3]). Let ϕbe convex on (−∞,∞) = R and suppose that f : [0, 1] −→ R is Lebesgue integrable over[0, 1]. Suppose that either


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

Then

(112)

∫[0,1]

ϕ(f(t)

)dt ≥ ϕ

(∫[0,1]

f(t)dt

).

Corollary 5.62 (Corollary 21, §5.5, p. 116 of [3]). 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 Jensen’s inequality and its corollary (See Propo-sition 5.61 and Corollary 5.62) are as follows: If ψ is concave and h is Lebesgue integrableover [0, 1], then

(114)

∫[0,1]

ψ(h(t)

)dt ≤ ψ

(∫[0,1]

h(t)dt

).

Otherwise, let h(x) be positive and ψ(x) = lnx. 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 ϕ : (a, b) −→ R beconvex. Suppose that t1, t2, . . . , tn ∈ [0, 1] satisfy

∑ni=1 ti = 1. If x1, x2, . . . , xn ∈ (a, b),

then

ϕ

(n∑

i=1

tixi

)≤

n∑i=1

tiϕ(xi).


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 [3]) and we say in this case that f is in L1(X, m), and write

(116) ‖f‖1 =

∫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 gp(t) = tp. Note that gp is continuous, thus gp

(|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) ‖f‖p =

[∫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 = 01√x

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) = ∞.


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

t→∞lnx]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 thatf 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.


f(x) =

0 if x = 0

1 if x ∈ R\Q1x

if x ∈ Q\{0}.

This function is not bounded since limn→∞

f( 1n) = lim

n→∞11n

= ∞. Note, however, that f is

essentially bounded by 1 since |f | = 1 a.e.


Definition 6.10. We say that f is in L∞(X, m) if f 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), then ‖f‖∞ = 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,∞), ‖f‖p = 0 ⇐⇒ f = 0 (this shows uniqueness ofthe zero vector).

(ii) For every α ∈ R and for all f ∈ Lp(X,m), αf ∈ Lp(X,m) and ‖αf‖p = |α|‖f‖p.(iii) If f1, f2 ∈ Lp(X,m), then f1 + f2 ∈ Lp(X,m) and (Minkowski’s inequality)

‖f1 + f2‖p ≤ ‖f1‖p + ‖f2‖p.

The problem that arises is that if f = 0 a.e., then ‖f‖p = 0 but f is not identically zero.To remedy this situation, if ‖f‖p < ∞ and ‖g‖p < ∞, we say f and g are equivalent,written f ∼ g, if

(120) m({x : f(x) 6= 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 of Lp(X, m) are equivalence classes of functions

(121){[f ] : f : X −→ [−∞,∞] is Lebesgue measurable and ‖f‖p <∞

},

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 of Lp(X,m) instead of [f ], [g].This should cause no confusion.

Proposition 6.14 (See 1, §6.2, p. 120 of [3]). 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 don’tget a norm for p ∈ (0, 1).


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

(122) ‖f‖p =

[∫

X|f(x)|pdx

]1/pif 1 ≤ p <∞

ess sup |f(x)| if p = ∞.

That is, ‖ · ‖p : Lp(X,m) −→ [0,∞) defined in (122) satisfies

(i) ‖f‖p = 0 ⇐⇒ f = 0.(ii) For every α ∈ R and for every f ∈ Lp(X,m), ‖αf‖p = |α|‖f‖p.

(iii) Minkowski’s inequality: If f1, f2 ∈ Lp(X,m), then

‖f1 + f2‖p ≤ ‖f1‖p + ‖f2‖p.

(See 1, §6.2, p. 120 of [3] and Theorem 6.16).

Theorem 6.16 (Minkowski’s Inequality - See 1, §6.2, p. 120 of [3]). This is theproof of Proposition 6.15(iii). If f, g ∈ Lp(X,m) with 1 ≤ p ≤ ∞, then

(123) ‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Furthermore, if 1 < p < ∞, then equality will hold in (123) if and only if ∃α, β ∈ Rsuch that βf = αg.

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 inthe 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. If p ∈ [1,∞], we defined the conjugate exponent q to p as follows:

q =

∞ if p = 1

pp−1

if 1 < p <∞

1 if p = ∞.

Note that 1 < pp−1

< ∞ in the middle case above and limp→1+

pp−1

= ∞. Furthermore,

limp→∞

pp−1

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

1

p+

1

q=

1

p+

1p

p−1

=1

p+p− 1

p= 1.

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


Proposition 6.19 (Young’s Inequality - See Exercise 8, p. 123 of [3]). Let a ≥ 0,b ≥ 0 and p ∈ (1,∞). Let q be the conjugate exponent to p. Then

(124) ab ≤ ap

p+bq

q

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

Theorem 6.20 (Holder’s Inequality - See 4, §6.2, p. 121 of [3]). Let (X, β,m)be([0, 1],M,m

)or (R,M,m). Let p ∈ [1,∞] and let q be the conjugate exponent to

p. Let f ∈ Lp(X,m) and g ∈ Lq(X,m). Then fg is Lebesgue integrable over X (i.e.,fg ∈ L1(X,m)) and

(125) ‖fg‖1 ≤ ‖f‖p‖g‖q,

with equality holding in (125) if and only if ∃ α, β ≥ 0 with β|f(x)|p = α|g(x)|q a.e.

Corollary 6.21 (Cauchy-Schwarz Inequality). If f, g ∈ L2(X,m), then fg ∈L1(X,m) (i.e., fg is Lebesgue integrable over X) and

‖fg‖1 ≤ ‖f‖2‖g‖2.

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 − g‖p.

Minkowski’s 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 ε > 0 there is an N ∈ N such that for all n ≥ N and all m ≥ N ,‖fn − fm‖ < ε. 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.


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. Then∞∑

k=1

(−1)k

kis summable

(the alternating harmonic series).

(ii){yk = (−1)k

k2

}∞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 = 1

10, y3 = 4

100, y4 = 1

1000,

y5 = 510000

, etc. That is, let

yn =nth digit of π

10n.

Then∞∑

k=1

‖yk‖ =∞∑

k=1

|yk| <∞

converges to π in R but∑∞

k=1 yk is not summable in (Q, | · |).


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 is

summable, 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 [3]). 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 ‖fk‖p <∞. It must be shown that there exists an f ∈ Lp(X,m) such that

limn→∞

∥∥∥∥∥f −n∑

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 [3]). Letp ∈ [1,∞]. Then Lp(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 ε > 0 there exists s step function ϕ and a continuous functionψ such that ∫

X

∣∣f(x)− ϕ(x)∣∣dm < ε and

∫X

∣∣f(x)− ψ(x)∣∣dm < ε

with |ϕ(x)| < |f(x)| and |ψ(x)| < |f(x)| (see Proposition 3.60). We want to do the samething in Lp(X,m), but we will need p <∞. We first do the bounded case.

Lemma 6.36 (Lemma 7, §6.4, p. 127 of [3]). Let 1 ≤ p < ∞ and let f ∈ Lp(X,m).Then there exists an M > 0 and a bounded Lebesgue measurable function fM ∈ Lp(X,m)such that |fM(x)| ≤ |f(x)| for every x ∈ X and ‖f − fM‖p < ε for any ε > 0.


Proposition 6.37 (Proposition 8, §6.4, p. 128 of [3]). Let f ∈ Lp[0, 1], p ∈ [1,∞). Givenε > 0 there exists a step function ϕ ∈ Lp[0, 1] and a continuous function ψ ∈ Lp[0, 1]such that

‖f − ϕ‖p < ε and ‖f − ψ‖p < ε.

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 (αf1 +βf2) = αF (f1)+βF (f2) for all α, β ∈ R and for all f1, f2 ∈ Y (linearity)(ii) There exists an M > 0 such that |F (f)| ≤M‖f‖ 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)| ≤ M‖f1 − f2‖ ⇐⇒ |F (f1)− F (f2)| ≤ M‖f1 − 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 [3]). Fix g ∈ L2[0, 1]. Note thatby Holder’s 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] 1

2[∫

[0,1]

|f |2dx] 1

2


by Holder’s inequality again, which implies that |Tg(f)| ≤ ‖g‖2‖f‖2. Thus for [f ] 6= 0 inL2[0, 1],

|Tg(f)|‖f‖2

≤ ‖g‖2 =⇒ ‖Tg‖ ≤ ‖g‖2

since ‖g‖2 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‖ ≤ ‖g‖q.

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) < 0

Note that sgn g(x) is bounded and measurable (it’s 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 [3]). If p and q are conjugate expo-nents in [1,∞], then ‖Tg‖ = ‖g‖q.

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 [3]). Let p ∈ [1,∞). Let g be a real-valued,Lebesgue measurable function on [0, 1] with the following property: ∃M > 0 such thatfor every bounded, Lebesgue measurable function f on [0, 1], fg is integrable with∣∣∣∣∫

[0,1]

fg

∣∣∣∣ ≤M‖f‖p.


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

Theorem 6.47 (Riesz Representation Theorem - Theorem 13, §6.5, p. 132 of[3]). Let F be a bounded linear functional on Lp[0, 1], 1 ≤ p <∞. Let q 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‖ = ‖g‖q.

Corollary 6.48. Let F 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. �


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.(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 = Lp[0, 1], d(f, g) = ‖f − g‖p, p ∈ [1,∞].

(iii) X = Rn, d(x, y) =

[n∑

i=1

(xi − yi)2

]1/2

, 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 ε > 0 there exists an N ∈ Nsuch that d(x0, xn) < ε whenever n ≥ N .

(ii) The sequence {xn}∞n=1 ⊂ X is Cauchy if and only if for every ε > 0 there existsan N ∈ N such that if m > n ≥ N , d(xm, xn) < ε.

(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 δ > 0 such that whenever y ∈ X and d(x, y) < δ, then y ∈ U .

Remark 7.5 (See Proposition 1, §7.2, p. 142 of [3]). Note that ∅ and U = X are bothopen subsets of X.

Example 7.6. Let x ∈ X, fix δ > 0 and define B(x, δ) = {y ∈ X : d(x, y) < δ}. ThenB(x, δ) is open in X (prove this - it doesn’t simply follow from the definition), calledthe open ball at x with radius δ.


7.2. Compactness.

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

⋃α∈I

Uα ⊇ K and each Uα 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 and∞⋃

n=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 {Uα}α∈I of K admits a finite subcover Uα1 , . . . , UαN

withN⋃

j=1

Uαj⊇ K. X is said to be a compact metric space if X is compact in (X, d) itself.

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

1n+1

, nn+1

).

Then∞⋃

n=1

= (0, 1) but {Un}∞n=1 contains no finite subcover since U1 ⊂ U2 ⊂ U3 ⊂ · · · ⊂

(0, 1) and thusN⋃

i=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 Heine–Borel Theorem: a subset K of Rn 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 [3] 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 sequentiallycompact 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

k→∞xnk

= 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 [3]).


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 δ > 0. Fix x0 ∈ X and let K = B(x0, δ) (see Example 7.6). Thenif x, y ∈ K,

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

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

< δ + δ

= 2δ.

Thus B(x0, δ) is bounded.

Definition 7.16. We say that K ⊆ (X, d) is totally bounded if for every ε > 0 there

exist x1, . . . , xn ∈ X such thatn⋃

i=1

B(xi, ε) ⊃ K, that is, we can cover K with a finite

number of ε-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 [3]).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.


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 ε > 0 there exists a δ > 0 such that wheneverx ∈ X and d(x, x0) < δ, then |f(x)− f(x0)| < ε. 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 of R. Inparticular, the range of f 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 α ∈ R and f ∈ C(X), then αf ∈ C(X) where (αf)(x) = α

(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‖ = supx∈X

|f(x)| = maxx∈X

|f(x)|.

Note that this definition makes sense since the supremum exists (the f ’s 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 α ∈ R and f ∈ C(X), then ‖αf‖ = |α|‖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 [3], Assignment 1. We can use this result to prove thatC(X) is complete just like L∞ (see Exercise 11, p. 126 of [3], 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).


7.4. The Stone–Weierstrass Theorem.

Remark 7.25. A natural question to ask is what subspaces are dense in C(X)? Thisis the aim of the Stone–Weierstrass 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 6= x2, there exists an f ∈ S with f(x1) 6= 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 A is a subspace (over R) of C(X). Wesay 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 f∨g and f∧g are in L where f∨g(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, . . . , fn} ⊂ L, it followseasily by induction that f1 ∧ f2 ∧ · · · ∧ fn(x) = min

{f1(x), . . . , fn(x)

}and f1 ∨ f2 ∨ · · · ∨

fn(x) = max{f1(x), . . . , fn(x)

}define functions f1 ∧ f2 ∧ · · · ∧ fn and f1 ∨ f2 ∨ · · · ∨ fn

in L.

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

(130) f ∨ g(x) = max{f(x), g(x)

}=f(x) + g(x) + |f(x)− g(x)|

2∈ C(X)


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) = min{f(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 the

algebra 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) = max

{f(x), g(x)

}nor f ∧ g(x) = min

{f(x), g(x)

}is

in A since neither is a polynomial (graph it – the maximum is the upper “V” and theminimum is the lower “inverted V”).

Theorem 7.35 (Stone–Weierstrass Theorem - Theorem 34, §9.9, p. 212 of [3]).Let (X, d) be a compact metric space. Let A ⊂ C(X) be an algebra of functions suchthat

(i) A separates the points of X and(ii) A contains the constant functions fc(x) = c for every x ∈ X and c ∈ R (these

are continuous).

Then given any f ∈ C(X) and given any ε > 0 there exists a g ∈ A such that |g(x) −f(x)| < ε for every x ∈ X. This tells you that ‖g − f‖ < ε, that is, A is dense inC(X) in the topology on C(X) induced by its supremum norm. We need to establishfour Lemmas and Propositions before we can prove this. See Remark 7.42 for the proof.

Remark 7.36. Note that the algebra A of all polynomials on X = [0, 1] satisfies theconditions of Theorem 7.35 (see Example 7.30). Define f ∈ A by f(x) = x. Thusif x1 6= x2 ∈ [0, 1], then f(x1) = x1 6= x2 = f(x2) and A separates the points ofX. A also contains the constant functions (the zero-degree polynomials). Thus A isdense in C

([0, 1]

)in the supremum norm. This is a special case of the Weierstrass

Approximation Theorem (see Corollary 7.43).

Lemma 7.37 (Lemma 31, §9.9, p. 211 of [3]). Let (X, d) be a compact metric space.Suppose L is a lattice in C(X) such that

(i) L separates the points of C(X) and(ii) If f ∈ L and c ∈ R, then cf ∈ L and c+ f ∈ L.

Then given a, b ∈ R and x, y ∈ X with x 6= y, there exists an f ∈ L such that f(x) = aand f(y) = b.

Lemma 7.38 (Lemma 32, §9.9, p. 212 of [3]). Let L be as in Lemma 7.37. Let a, b ∈ Rwith a ≤ b. Let F ⊆ X be a closed subset of X and suppose p ∈ X ∼ F (which is open).Then there exists an f ∈ L such that f(x) ≥ a for every x ∈ X, f(p) = a and f(x) > b

for every x ∈ F .


Proposition 7.39 (Proposition 29, §9.9, p. 211 of [3]). Let (X, d) be a compact metricspace. Let L be a lattice in C(X) (any lattice at all will do). Suppose that h ∈ C(X)and suppose in addition h satisfies h(x) = inf

f∈Lf(x). Then for every ε > 0 there exists a

g ∈ L such that 0 ≤ g(x)−h(x) < ε for every x ∈ X. Note that we know we can do thispointwise, but the consequence is g does it uniformly for every x ∈ X.

Proposition 7.40 (Proposition 30, §9.9, p. 211 of [3]). Let (X, d) be a compact metricspace. Let L be a lattice in C(X) satisfying conditions (i) and (ii) of Lemma 7.37. Thengiven h ∈ C(X), for every ε > 0 there exists a g ∈ L such that 0 ≤ g(x)− h(x) < ε forevery x ∈ X.

Lemma 7.41 (Lemma 33, §9.9, p. 212 of [3]). Let ε > 0 be fixed. Then there exists apolynomial p in the variable s such that for every s ∈ [−1, 1],

∣∣p(s) − |s|∣∣ < ε, that is,−ε < p(s) − |s| < ε. In other words, absolute value can be uniformly approximated bypolynomials on [−, 1, 1].

Remark 7.42. We are now ready to prove the Stone–Weierstrass Theorem.

Corollary 7.43 (Weierstrass Approximation Theorem - Corollary 35, §9.9, p.213 of [3]). Let f be continuous and real-valued on [a, b], −∞ < a < b < ∞. Then forevery ε > 0 there is a polynomial p such that |f(x)− p(x)| < ε for every x ∈ [a, b].

Example 7.44. Let X = {z ∈ C : |z| = 1}, or X = {(x, y) ∈ R2 : x2 +y2 = 1}. Then(X, d) is a compact metric space by the Heine-Borel theorem (it’s closed and boundedin R2, see Remark 7.11) with metric d(z, w) = |z − w|. Let

T0 = {f ∈ C(X) : f(z) = c, c ∈ R}

and

Tn =

{f ∈ C(X) : f(z) = a0 +

n∑k=1

akRe(zk) +

n∑k=1

bkIm(zk), {ak}nk=0, {bk}n

k=1 ⊂ R

}.

Recall that if z = eit, then Re(z) = cos t, Im(z) = sin t and (cos t + i sin t)n = cosnt +i sinnt for every n ∈ N. We’re really talking about trigonometric polynomials. Let

T =∞⋃

n=0

Tn. Then you can show that T is an algebra of functions that satisfies conditions

(i) and (ii) of Theorem 7.35 in C(X). See Exercise 42, p. 213 of [3], Assignment 3.


Remark 7.45 (See Example 7.44). We can also define Tn ⊂ C[0, 2π] by

Tn =

{a0 +

n∑k=1

ak cos kt+n∑

k=1

bk sin kt : {ak}nk=0, {bk}n

k=1 ⊂ R

}and let

T =∞⋃

n=0

Tn ⊂ C[0, 2π].

Is T dense in C[0, 2π]?


8. General Measure and Integration

8.1. Introduction.

Definition 8.1. Let X be a nonempty set. Recall B ⊆ 2X is called a σ-algebra ofsubsets of X if

(i) ∅ ∈ B.

(ii) If A ∈ B, then X ∼ A = A ∈ B.(iii) If {An}n≥1 ⊂ B, then

⋃n≥1

An ∈ B.

See also Definition 1.17 and Remark 1.18.

Example 8.2 (Examples of σ-algebras). (i) Recall thatM is a σ-algebra, whereM denotes all Lebesgue measurable subsets of R (see Theorem 3.21).

(ii) Recall also that B is a σ-algebra, where here B denotes the Borel subsets of R.(iii) Now, take X a topological space and let BX denote the smallest σ-algebra con-

taining all the open sets in X. This is called the Borel subsets of X.(iv) If X is any nonempty set, then 2X and {∅, X} are examples of σ-algebras.

Definition 8.3. If X is a nonempty set and B is a σ-algebra of subsets of X, we saythe pair (X,B) is a measurable space, and the sets in B are called the measurablesubsets of X.

Example 8.4. (R,M) in Example 8.2(i) is a measurable space.

Definition 8.5. Let (X,B) be a measurable space. Let µ : B −→ [0,∞] be defined onall measurable subsets of X. We say µ is a measure on (X,B) if

(i) µ(∅) = 0.(ii) If {Ai}i≥1 is a countable collection of pairwise disjoint subsets of B, then

µ

(⋃i≥1

Ai

)=∑i≥1

µ(Ai),

where ∞ + r = r + ∞ = ∞ for any r ∈ R. This property is referred to ascountable additivity.

The triple (X,B, µ) is called a measure space.


Example 8.6 (Examples of Measure Spaces).

(i) (R,M,m) where m denotes Lebesgue measure on M.(ii) Let X = N and B = 2N. Define ν : B −→ [0,∞] by ν(E) = #E, the number of

elements in E ⊂ N. This function ν is called the counting measure and takeson values in N ∪ {0,∞}.

Proposition 8.7 (Proposition 1, §11.1, p. 254 of [3]). Let (X,B, µ) be a measure space.Let A,B ∈ B with A ⊆ B. Then µ(A) ≤ µ(B). This property is called monotonicityof µ.

Remark 8.8. Note that the monotonicity of µ in Proposition 8.7 follows automati-cally from countable additivity, so it’s helpful to have this as an axiom. Compare toProposition 3.26.

Proposition 8.9 (Proposition 2, §11.1, p. 255 of [3]). Let (X,B, µ) be a measurespace, and let {En}∞n=1 ⊂ B be a decreasing sequence of measurable subsets of X, i.e.,E1 ⊇ E2 ⊇ · · · , with µ(E1) <∞. Then

µ

(∞⋂

n=1

En

)= lim

n→∞µ(En).

Proof. The proof is exactly the same as that of Proposition 3.29. �

Proposition 8.10. If (X,B, µ) is a measure space and {En}∞n=1 is an increasing se-quence of measurable subsets of X, i.e., E1 ⊆ E2 ⊆ E3 ⊆ · · · , then

µ

(∞⋃

n=1

En

)= lim

n→∞µ(En).

Proof. The proof is the same as in the Lebesgue measure case. See Corollary 3.28. �

Proposition 8.11 (Countable Subadditivity - Proposition 3, §11.1, p. 255 of [3]).Suppose that (X,B, µ) is a measure space, and let {En}n≥1 ⊂ B be a countable family ofmeasurable subsets of X (the En’s are not necessarily pairwise disjoint anymore). Then

µ

(⋃n≥1

En

)≤∑n≥1

µ(En).

Remark 8.12. Note that we no longer have translation invariance like we did in theLebesgue measure situation (unless we impose further conditions).


Definition 8.13. Let (X,B, µ) be a measure space. We say that (X,B, µ) is finiteis µ(X) < ∞. We say that (X,B, µ) is σ-finite if there exists {Xn}n≥1 ⊆ B with⋃n≥1

Xn = X and µ(Xn) <∞ for all n.

Example 8.14.

(i) (R,M,m) is σ-finite. Let Xn = [−n, n]. Then m(Xn) = 2n < ∞ and⋃

n≥1

Xn =

R.(ii) Let X be a nonempty set, take B = {∅, X} and define µ by µ(∅) = 0 and

µ(X) = ∞. Then (X,B, µ) is not σ-finite.

Definition 8.15. (X,B, µ) is called complete if whenever E ∈ B and µ(E) = 0, thenfor any A ⊂ E, A ∈ B.

Example 8.16.

(i) We showed that (R,M,m) is complete (see ??).(ii) (R,BR,m) is not complete (BR denotes the Borel subsets of R). Recall, however,

that if A ∈ M, we could write A = B ∪ Z with B a Borel set, m(Z) = 0 andZ ⊂ B′ ∈ BR with m(B′) = 0 (see ??).

(iii) Let X = {0, 1, 2}, B ={∅, {0, 1}, {2}, {0, 1, 2} = X

}. You should verify that B

is a σ-algebra of subsets of X. Define µ : B −→ [0,∞] by

µ(∅) = 0, µ({0, 1}

)= 0, µ

({2})

= 1, µ(X) = 1,

You should also check that (X,B, µ) is a measure space. Note that (X,B, µ) isnot complete, however, since {1} ⊂ {0, 1} and {0} ⊂ {0, 1}, {0, 1} ∈ B withµ({0, 1}

)= 0 yet {0}, {1} /∈ B.

Proposition 8.17 (Proposition 4, §11.1, p. 257 of [3]). Let (X,B, µ) be a measurespace. Then there is a complete measure space (X,B0, µ0) such that

(i) B ⊂ B0.(ii) For every E ∈ B, µ0(E) = µ(E).

(iii) If C ⊂ B0, C can be written as C = A∪B where A ∈ B and B ⊂ Z where Z ∈ Band µ(Z) = 0 (compare to Example 8.16(ii)). Then define µ0(C) = µ0(A∪B) =µ(A).


8.2. Measurable Functions.

Proposition 8.18 (Proposition 5, §11.2, p. 259 of [3]). Let (X,B) be a measurablespace, and let f be an extended real-valued function defined on X. Then the followingstatements are equivalent:

(i) ∀α ∈ R {x ∈ E : f(x) > α} ∈ B.(ii) ∀α ∈ R {x ∈ E : f(x) ≥ α} ∈ B.

(iii) ∀α ∈ R {x ∈ E : f(x) < α} ∈ B.(iv) ∀α ∈ R {x ∈ E : f(x) ≤ α} ∈ B.

Proof. The proof is the same as in the Lebesgue measure case. See Proposition 3.37. �

Definition 8.19. Let (X,B, µ) be a measure space. Let f : X −→ [−∞,∞]. Wesay that f is measurable if any one (and hence all) of the equivalent conditions inProposition 8.18 are satisfied.

Example 8.20. Characteristic functions. Let E ⊂ X, E ∈ B. Define the characteris-tic function

χE(x) =

{1 if x ∈ E0 if x /∈ E

Recall that χE is a measurable function with respect to (X,B, µ). See Lemma 3.41.

Definition 8.21 (See Definition 3.44). Recall that we say a property P on holds µ-a.e.on (X,B, µ) if

{x ∈ X : P does not hold }has µ-measure zero.

Proposition 8.22 (Proposition 8, §11.2, p. 260 of [3]). Let f, g : X −→ [−∞,∞].Suppose f is measurable and f(x) = g(x) µ-a.e. Then g is measurable.

Proof. Essentially the same as in the Lebesgue measure case. See Proposition 3.45. �

Example 8.23. Let X = {0, 1, 2}, B ={∅, {0, 1}, {2}, {0, 1, 2}

}, and let µ be as defined

in Example 8.16(iii). Let

f(x) = χ{2} =

{0 if x = 0, 1

1 if x = 2.


Note that f is measurable (see Example 8.20). Let

g(x) =

−1 if x = 0

2 if x = 1

1 if x = 2

Then f(x) = g(x) µ-a.e. but g is not measurable since{x ∈ {0, 1, 2} : g(x) < −1

2

}= {0} /∈ B.

Remark 8.24. Example 8.23 illustrates one reason we want complete measure spaces –in a complete measure space, we can change a function on a set of measure zero and thefunction stays measurable. The altered function doesn’t go “bad,” and we can identifyit with the original function.

Theorem 8.25 (Theorem 6, §11.2, p. 259 of [3]). Let (X,B, µ) be a complete measurespace. Let c ∈ R and suppose that f and g are both real-valued, measurable functions.Then

(i) c+ f ,(ii) c · f ,

(iii) f + g,(iv) f · g,(v) f ∨ g = max{f, g},

(vi) f ∧ v = min{f, g}, and(vii) |f |

are all measurable. Moreover, if {fn}∞n=1 is a sequence of measurable functions, then

(viii) sup{fn}∞n=1,(ix) inf{fn}∞n=1,(x) lim {fn}, and

(xi) lim {fn}are all measurable.

Proof. Essentially the same as in the Lebesgue measure case. See Proposition 3.43 andTheorem 3.46. �

Corollary 8.26. Let (X,B, µ) be a measure space and suppose that {Ei}ni=1 ⊂ B and

{αi}ni=1 ⊂ R. Then the simple function

n∑i=1

αiχEi

is measurable.


Proposition 8.27 (Proposition 7, §11.2, p. 260 of [3]). Let (X,B, µ) be a measure space.Let f : X −→ [0,∞) be a nonnegative, real-valued measurable function. Then thereexists an increasing sequence of simple functions {ϕn}∞n=1 such that lim

n→∞ϕn(x) = f(x)

for every x ∈ X. If f is bounded, the convergence is uniform. Moreover, if (X,B, µ) isσ-finite, it is possible to choose the ϕn such that

µ({x ∈ X : ϕn(x) 6= 0}

)<∞

for every n ∈ N. See Theorem 3.55.


8.3. Integration.

Remark 8.28 (Arithmetic in [0,∞]). We make the following conventions: If c ∈[0,∞], then c + ∞ = ∞ + c = ∞. If c > 0, then c · ∞ = ∞ · c = ∞. Finally0 · ∞ = ∞ · 0 = 0. Why do we make the last convention? Consider 0 ·χR = 0. Then

0 =

∫R

0 ·χR dm(x) = 0 ·m(R) = 0 · ∞,

so if we want this to work out the way we expect, we must make the convention that0 · ∞ = ∞ · 0 = 0.

Definition 8.29 (Integrating Nonnegative Simple Functions). Let (X,Bµ) bea measure space. Let E ∈ B. If {Ei}n

i=1 ⊂ B is a family of measurable subsets of X and{αi}n

i=1 ⊂ [0,∞] are nonnegative real numbers, and if we consider

ϕ =n∑

i=1

αiχEi,

then we define ∫E

ϕdµ =n∑

i=1

αi µ(E ∩ Ei).

It is standard convention to write∫ϕdµ for

∫X

ϕdµ =n∑

i=1

αi µ(Ei).

Example 8.30 (See Example 8.16(iii) and Example 8.23). Let X = {0, 1, 2}, B ={∅, {0, 1}, {2}, {0, 1, 2}

}. Take ϕ = 2χ{0,1} + 3χ{2}. As before µ(∅) = µ

({0, 1}

)= 0,

µ({2})

= µ({0, 1, 2}

)= 1. Then∫

{0,1}ϕdµ = 2µ

({0, 1} ∩ {0, 1}

)+ 3µ

({2} ∩ {0, 1}

)= 2µ

({0, 1}

)+ 3µ(∅) = 0

and∫{2}ϕdµ = 2µ

({2} ∩ {0, 1}

)+ 3µ

({2} ∩ {2}

)= 2µ(∅) + 3µ

({2})

= 2 · 0 + 3 · 1 = 3.

Also ∫{0,1,2}

ϕdµ = 3.

Remark 8.31. One can show (as in the Lebesgue measure case) that

∫E

ϕdµ is inde-

pendent of the representation chosen for ϕ. For example, we could have written ϕ fromExample 8.30 as ϕ = 2χ{0,1,2} +χ{2}.


Remark 8.32 (Properties of Integrals of Simple Functions). Let (X,B, µ)be a measure space with E ∈ B. Let ϕ and τ be simple, nonnegative functions.

(i) If c ∈ [0,∞), then ∫E

cϕ dµ = c

∫E

ϕdµ.

(ii) If a, b ∈ [0,∞), then aϕ+ bτ is simple and nonnegative, and∫E

(aϕ+ bτ)dµ = a

∫E

ϕdµ+ b

∫E

τ dµ.

(iii) If E,F ∈ B with E ∩ F = ∅, then∫E∪F

ϕdµ =

∫E

ϕdµ+

∫F

ϕdµ.

Compare to Proposition 4.6.

Definition 8.33. Let (X,B, µ) be a measure space and let f be a nonnegative measur-able functions on X. Let E ∈ B. We define the integral of f over E by(131)∫

E

f dµ = sup

{∫E

ϕdµ : ϕ is simple, nonnegative and 0 ≤ ϕ(x) ≤ f(x) ∀x ∈ E}.

The set in (131) might not be bounded above so we might get +∞ for the integral,which is fine. Note that since f ≥ 0,

∫Ef dµ ∈ [0,∞] since the set in (131) is bounded

below by zero.

Remark 8.34. Recall in the Lebesgue measure case that for f measurable and zerooutside a set of finite measure, we defined∫

e

f dx = inf

{∫E

ψ dx : ψ is nonnegative, simple and ψ(x) ≥ f(x)

}.

We could have taken nonnegative simple functions ϕ(x) ≤ f(x) and used the supremuminstead. See Remark 4.8 and Proposition 4.12.

Remark 8.35 (Basic Facts About Integrals of Nonnegative Measurable

Functions). Let (X,B, µ) be a measure space, let f be a nonnegative, measurablefunction on (X,B, µ), and let E ∈ B.

(i) If c ∈ [0,∞) so that c · f is nonnegative and measurable on (X,B, µ), then∫E

c · f dµ = c

∫E

f dµ.

(ii) If E,F ∈ B with E ∩ F = ∅, then∫E∪F

f dµ =

∫E

f dµ+

∫F

f dµ.


(iii) If E ∈ B and µ(E) = 0, then ∫E

f dµ = 0.

(iv) If f and g are nonnegative, measurable functions on (X,B, µ) and if f(x) ≤ g(x)for every x ∈ X (or for µ-almost all x ∈ X), then∫

E

f dµ ≤∫

E

g dµ.

Lemma 8.36 (Fatou’s Lemma - Theorem 11, §11.3, p. 264 of [3]). Let (X,B, µ) be ameasure space, and {fn}∞n=1 and f are nonnegative, simple functions defined on (X,B, µ)such that lim

n→∞fn(x) = f(x) µ-a.e. on X. Then for every E ∈ B,∫

E

f dµ ≤ lim

∫E

fn dµ.

Proof. Note that we don’t have BCT to prove it this time. See Theorem 4.20. �

Theorem 8.37 (Monotone Convergence Theorem, or MCT - Theorem 12,§11.3, p. 265 of [3]). Let {fn}∞n=1 be a sequence of nonnegative, increasing measurablefunctions on the measure space (X,B, µ). Suppose that lim

n→∞fn(x) = f(x) µ-a.e. on X.

Then for every E ∈ B,

limn→∞

∫E

fn dµ =

∫E

f dµ.

Proof. The proof depends on Fatou’s Lemma the same way as in the Lebesgue measurecase. See Theorem 4.21. �

Corollary 8.38 (Proposition 13, §11.3, p. 266 of [3]). Let (X,B, µ) be a measure space,and let f, g be nonnegative measurable functions on X. Let a, b ∈ [0,∞). Then af + bg

is nonnegative and measurable, and for any E ∈ B,∫E

(af + bg)dµ = a

∫E

f dµ+ b

∫E

g dµ.

Corollary 8.39 (Corollary 14, §11.3, p. 266 of [3]). Let (X,B, µ) be a measure space,and let {fn}∞n=1 be a sequence of nonnegative, measurable functions defined on X. Thenfor every E ∈ B, ∫

E

∞∑n=1

fn(x) dµ =∞∑

n=1

∫E

fn(x) dµ,

where we could have +∞ on both sides.

Proof. Same as before. See Corollary 4.22. �


Definition 8.40. Let (X,B, µ) be a measure space, and let f : X → [0,∞] be a

measurable function. We say that f is integrable over E ∈ B if

∫E

f dµ <∞.

Definition 8.41. Suppose that f : X → [−∞,∞] is measurable. Write

f+ = max{f, 0} = f ∨ 0

f− = max{−f, 0} = −f ∨ 0

Note that if we write

P = {x ∈ X : f(x) ≥ 0} and N = {x ∈ X : f(x) < 0},

then

f+ = fχP and f− = fχN .

f+ and f− are called the positive part and negative part of f , respectively. Notethat f+ and f− are nonnegative and measurable, and

f = f+ − f− and |f | = f+ + f−

on X. Compare to Definition 3.56.

Definition 8.42. Let (X,B, µ) be a measure space, and let f : X → [−∞,∞] bemeasurable. We say that f is integrable over E ∈ B if both f+ and f− are integrableover E. In this case we define∫

E

f dµ =

∫E

f+ dµ−∫

E

f− dµ.

Remark 8.43. If f : X → [−∞,∞] is measurable, then f is integrable over E ∈ B ifand only if |f | integrable over E.

Example 8.44 (See Example 8.6(ii)). Let X = N, B = 2N and let ν be countingmeasure, that is,

ν(E) =

{#E if E 6= ∅0 if E = ∅

where ν(E) could equal +∞ if E has infinite cardinality. Note that ν is σ-finite since

ν({n}

)= 1 <∞ for every n ∈ N and N = X =

∞⋃n=1

{n} (note that Q with the counting

measure is also a σ-finite measure space). Now, f : N −→ [−∞,∞] corresponds to asequence {an}∞n=1 with an = f(n) ∈ [−∞,∞]. Note that any function is measurablesince all subsets of N are measurable by definition. Thus if f = {an}∞n=1 is a function, it


is measurable and will be integrable over N iff |f | = {|an|}∞n=1 is integrable over N. Weuse MCT to determine this. Let

|f |k =k∑

n=1

|an|χ{n}, |f |k : N −→ [0,∞].

The sequence corresponding to |f |k is{|a1|, |a2|, . . . , |ak|, 0, 0, . . .

}.

Thus ∫N|f |k dν =

∫N

k∑n=1

|an|χ{n} dν =k∑

n=1

|an|ν({n}

)=

k∑n=1

|an|.

Note that the{|f |k}∞

k=1is an increasing sequence of nonnegative functions on N with

limk→∞

|f |k(i) = |f |(i) = |ai|. By MCT,∫N|f | dν = lim

k→∞

∫N|f |k dν = lim

k→∞

k∑n=1

|an| =∞∑

n=1

|an|,

which may converge or diverge. Hence f is integrable if and only if{f(n) = an

}∞n=1

is

such that∞∑

n=1

|an| <∞ and ∫N|f | dν =

∞∑n=1

|an|.

Moreover, ∫Nf dν =

∞∑n=1

an,

which converges if the series is absolutely convergent. Therefore integration on N withthe counting measure ν is basically just adding and subtracting. For example,

{1n

}∞n=1

is not integrable in (N, 2N, ν) but{

1n2

}∞n=1

is integrable.

Proposition 8.45 (Proposition 15, §11.3, p. 267 of [3]). Let f, g be integrable overE ∈ B, where (X,B, µ) is a measure space. Let c1, c2 ∈ R. Then

(i) c1f + c2g is integrable over E with∫E

(c1f + c2g)dµ = c1

∫E

f dµ+ c2

∫E

g dµ.

(ii) If h : X → [−∞,∞] is measurable and |h(x)| ≤ |f(x)| µ-a.e. on E, then h isintegrable over E.

(iii) If f(x) ≥ g(x) µ-a.e. on E, then

∫E

f dµ ≥∫

E

g dµ.

Proof. Same as in the Lebesgue measure case. Prove (iii) first, then use it to prove (ii).See Proposition 4.29. �


Theorem 8.46 (Dominating Convergence Theorem, or DCT - Theorem 16,§11.3, p. 267 of [3]). Let (X,B, µ) be a measure space and suppose that g is a nonnegativefunction which is integrable over E ∈ B. Suppose {fn}∞n=1 is a sequence of measurablefunctions on X with |fn(x)| ≤ g(x) for every n ∈ N and for every x ∈ E (we can alsodo this µ-a.e. on E but it doesn’t affect the integral). Suppose f is measurable andlim

n→∞fn(x) = f(x) µ-a.e. on X (we don’t need f measurable if X is complete). Then

limn→∞

∫E

fn dµ =

∫E

f dµ <∞.

Proof. Same as in the Lebesgue measure case. See Theorem 4.31. �

Theorem 8.47 (Generalized Dominating Convergence Theorem, or GDCT- Proposition 18, §11.4, p. 270 of [3]). Let (X,B, µ) be a measure space and {µn}∞n=1

a sequence of measures on B that converge setwise to a measure µ. Let {fn}∞n=1 and{gn}∞n=1 be two sequences of measurable functions such that lim

n→∞fn(x) = f(x) and

limn→∞

gn(x) = g(x) for µ-a.e. x ∈ X. Suppose that, for any E ∈ B, |fn(x)| ≤ g(x)

for all x ∈ E and that

limn→∞

∫E

gn dµn =

∫E

g dµ <∞

for every n ∈ N. Then

limn→∞

∫E

fn dµn =

∫E

f dµ.

Proof. Apply Theorem 8.46 to the sequences {gn + fn}∞n=1 and {gn − fn}∞n=1. See alsoTheorem 4.34. �

Remark 8.48. We’re missing translation invariance. Recall that∫Rf(x+ t)dx =

∫Rf(x)dx

for Lebesgue measure. This is no longer true for general measure.


8.4. Lp Spaces.

Definition 8.49. Let p ∈ [1,∞) and f : X → [−∞,∞] be measurable. We say that fis in Lp(µ) if ∫

X

|f(x)|p dµ <∞.

In this case we define

‖f‖p =

[∫X

|f(x)|p dµ] 1

p

.

Note that ‖f‖p = 0 if and only if f(x) = 0 µ-a.e. on X.

Definition 8.50. If p = ∞ and f : X → [−∞,∞] is measurable, we say M is anessential bound for |f | if |f(x)| ≤M µ-a.e.

Definition 8.51. The measurable function f is said to be in L∞(µ) if |f | has anessential bound. We define

‖f‖∞ = inf{M : M is an essential bound for |f |}.

Remark 8.52. Just as before (see Proposition 6.11), it can be shown that ‖f‖∞ is anessential bound for |f |, and it is the least such bound. Again, ‖f‖∞ = 0 if and only iff = 0 µ-a.e. As before,

Lp(µ) ={[f ] : f is measurable on X, ‖f‖p <∞

},

where f1 ∼ f2 if f1(x) = f2(x) µ-a.e.

Theorem 8.53 (Theorem 25, §11.7, p. 282 of [3]). Let (X,B, µ) be a measure space.Let p ∈ [1,∞] and let q be the conjugate exponent to p (see Definition 6.18). Iff1, f2 ∈ Lp(µ), then f1 + f2 ∈ Lp(µ) and

(i) ‖f1 + f2‖p ≤ ‖f1‖p + ‖f2‖p (Minkowski’s inequality)(ii) If f ∈ Lp(µ) and g ∈ Lq(µ), then fg is integrable, that is, fg ∈ L1(µ), and∫

X

|fg|dµ = ‖fg‖1 ≤ ‖f‖p‖g‖q (Holder’s inequality)

(iii) Lp(µ) is a normed vector space which is a Banach space, that is, Lp(µ) is completein the metric induced by its norm ‖ · ‖p. More specifically ρp : Lp(µ)× Lp(µ) →[0,∞) defined by ρp(f, g) = ‖f − g‖p is a metric.


Example 8.54. Let (X,B, µ) = (N, 2N, ν). What does Lp(ν) look like? For 1 ≤ p <∞,

Lp(ν) =

{f : N → [−∞,∞] :

∞∑n=1

|f(n)|p <∞

}

`p =

{Real-valued sequences {an = f(n)}∞n=1 :

∞∑n=1

|an|p <∞

}For p = ∞,

`∞ ={Real-valued sequences which are bounded, i.e., {an}∞n=1 : sup |an| <∞

}.

Moreover,

∥∥{an}∞n=1

∥∥p

=

(

∞∑n=1

|an|p) 1

p

if 1 ≤ p <∞

sup |an| if p = ∞

Remark 8.55. We also have Minkowski’s inequality for sequence spaces, which iswhat we call the `p. If {an}∞n=1 and {bn}∞n=1 are in `p, where p ∈ [1,∞] is fixed, then{an + bn}∞n=1 ∈ `p and∥∥{an + bn}∞n=1

∥∥p≤∥∥{an}∞n=1

∥∥p+∥∥{bn}∞n=1

∥∥p.

Remark 8.56. We also have Holder’s inequality for sequence spaces. If p, q ∈ [1,∞]are conjugate exponents and {an}∞n=1 ∈ `p and {bn}∞n=1 ∈ `q, then {anbn}∞n=1 ∈ `1 (is anabsolutely convergent series) and∣∣∣∣∣

∞∑n=1

anbn

∣∣∣∣∣ ≤∞∑

n=1

|anbn| ≤∥∥{an}∞n=1

∥∥p·∥∥{bn}∞n=1

∥∥q

This follows directly from Holder’s inequality for general Lp spaces.

Remark 8.57. Fix p ∈ [1,∞]. Then

`p =

{{an}∞n=1 :

∞∑n=1

|an|p <∞}

if p ∈ [1,∞)

{{an}∞n=1 : sup |an| <∞

}if p = ∞

is a Banach space, i.e., is a normed vector space which is complete in the metric inducedby its norm.


Definition 8.58. Let (X,B, µ) be a measure space. Let F : Lp(µ) → R. We saythat F is a bounded linear functional on Lp(µ) if for all a, b ∈ R and for everyf1, f2 ∈ Lp(µ), F (af1 + bf2) = aF (f1) + bF (f2). In addition, F is a bounded linearfunctional if

‖F‖ = supf∈Lp(µ)

[f ] 6=0

|F (f)|‖f‖p

<∞.

If F is bounded, the finite quantity ‖F‖ is called the norm of F .

Example 8.59. Consider Lp(µ) for p ∈ [1,∞]. Fix a g ∈ Lq(µ), where q is the conjugateexponent to p. Define Fg : Lp(µ) → R by

Fg(f) =

∫X

gf dµ.

Note that this integral is defined by Holder’s inequality. We also have by Holder’sinequality that

|Fg(f)| =∣∣∣∣∫

X

f(x)g(x) dµ

∣∣∣∣ ≤ ‖f‖p · ‖g‖q,

hence if [f ] 6= 0, then|Fg(f)|‖f‖p

≤ ‖g‖q.

Therefore Fg is a bounded linear functional (linearity follows from the integral) and‖Fg‖ ≤ ‖g‖q. It is possible to show that, in fact, ‖Fg‖ = ‖g‖q. (see Proposition 6.44).

Theorem 8.60 (Riesz Representation Theorem - Theorem 29, §11.7, p. 284 of[3]). Let (X,B, µ) be a σ-finite measure space. Fix p ∈ [1,∞) and let q be the conjugateexponent to p. If F is a bounded linear functional on Lp(µ), then there exists a uniqueg ∈ Lq(µ) such that F = Fg, that is,

F (f) =

∫X

fg dµ

for every f ∈ Lp(µ). See Theorem 6.47.

Remark 8.61. If p = ∞, Theorem 8.60 does not hold. The bounded linear functionalson L∞(µ) can be more complicated.


8.5. Signed Measures.

Remark 8.62. Let (X,B) be a measurable space. Recall a measure µ on (X,B) is aset-valued function µ : B −→ [0,∞] such that

(i) µ(∅) = 0 and(ii) If {En}n≥1 ⊂ B is a countable collection of pairwise disjoint measurable subsets

of X, then µ

( ⋃n≥1

En

)=∑n≥1

µ(En).

Could µ take on possibly negative values? What would this mean for condition (ii)above? In other words, how could condition (II) hold when we don’t know what−∞+∞is? Do we need other restrictions to avoid this possibility? We now extend the conceptof measure with ν : B −→ R ∪ {−∞,∞} to obtain signed measures.

Definition 8.63. Let (X,B) be a measurable space. A signed measure ν on (X,B)is a function ν : B −→ [−∞,∞] such that

(i) ν takes on at most one of the values ∞ or −∞.(ii) ν(∅) = 0.

(iii) If {En}n≥1 ⊂ B is a countable family of pairwise disjoint measurable subsets of

X, then ν

( ⋃n≥1

En

)=∑n≥1

ν(En) where the sum on the right-hand side either

converges absolutely or properly diverges (goes to ∞ or −∞).

Remark 8.64. Note that condition (i) in Definition 8.63 remedies the potential prob-lem mentioned in Remark 8.62. Moreover, condition (iii) eliminates possibilities such

as∞∑

n=1

(−1)n

nand

∞∑n=1

(−1)n, which does not properly diverge. Note also that ordinary

measures are a special case of signed measures since they satisfy all three conditions inDefinition 8.63.

Example 8.65 (Examples of signed measures).

(i) We can construct signed measures out of ordinary measures. Let (X,B, µ) be aregular measure space. Let g ∈ L1(µ) so that g : X −→ [−∞,∞] and

∫X|g| dµ <

∞ (g is allowed to take on negative values). Define νg : B −→ R by

νg(E) =

∫E

g dµ.

We might get a negative value for this integral, but it doesn’t matter. Note thatνg takes on only real values, hence condition (i) in Definition 8.63 is satisfied.Note also that

νg(∅) =

∫∅g dµ = 0


so (ii) is satisfied in Definition 8.63. As for (iii), if {En}n≥1 is a countablecollection of pairwise disjoint measurable sets, then

νg

(⋃n≥1

En

)=

∫S

n≥1

En

g dµ =

∫X

gχ Sn≥1

Endµ

=

∫X

g∑n≥1

χEndµ

=∑n≥1

∫En

g dµ(132)

=∑n≥1

νg(En).

If the number of sets in the collection is finite, then step (132) is justified simplyby interchanging the sum and integral signs. If the number of sets in the col-lection is countably infinite, then we can use DCT to interchange the sum andintegral signs (prove this – see Theorem 8.46).

(ii) (X,B, µ) =([0, 1],M,m). Let g(x) = x − 1

2, 0 ≤ x ≤ 1. By part (i), νg(E) =∫

Eg dµ is a signed measure. For example, note that

νg

([0,

1

2

])=

∫[0, 1

2

] (x− 1

2

)dx =

x2

2− x

2

] 12

0

=1

8− 1

4= −1

4.

In this example, if g(x) ≥ 0 for every x ∈ X, then the measure νg is an ordinarymeasure since then νg(E) =

∫Eg(x) dµ ≥ 0. We will come back to this idea later

(see Example 8.67 and Remark 8.68).

(iii) Consider (R,M). Let

g(x) =

− 1x2 if −∞ < x < −1

1 if − 1 ≤ x ≤ 1

1x

if 1 ≤ x <∞Note that g(x) is integrable for −∞ < x ≤ 1 but is not integrable for x ≥ 1. Tomakes sure that νg makes sense as defined in part (i), we define

νg(E) =

∫E∩(−∞,1]

g dµ︸︷︷︸finite

+

∫E

g dµ︸︷︷︸either +∞ or finite


Definition 8.66. Let ν be a signed measure on a measurable space (X,B). Let A ∈ B.

(i) We say that A is a positive set for ν if ν(B) ≥ 0 whenever B ⊂ A and B ∈ B.In particular, since A ⊆ A, ν(A) ≥ 0 if A is positive. Note that ν restricted toA and its subsets is an ordinary measure.

(ii) We say that A is a negative set for ν if ν(B) ≤ 0 whenever B ⊂ A andB ∈ B. In particular, ν(A) ≤ 0 if A is negative for ν since A ⊆ A. Note thatν restricted to A and its subsets in this case is minus an ordinary measure. Inother words, negative sets are sets on which we take on minus the value of anordinary measure.

Example 8.67. Consider νg for g(x) as defined in Example 8.65(iii). Let

P = {x ∈ R : g(x) ≥ 0} = [−1,∞).

Note that P is measurable since g(x) is measurable. Furthermore, P is a positive set forνg since for B ⊂ P with B ∈ B,

νg(B) =

∫B

g(x) dx ≥ 0

since g(x) ≥ 0 on B. Similarly, define

N = {x ∈ R : g(x) ≤ 0} = (−∞,−1].

Hence if B′ ⊂ N and B′ ∈ B, then

νg(B′) =

∫B′g(x) dx ≤ 0

since g(x) ≤ 0 on B′.

Remark 8.68. The ideas in Example 8.67 work in general for measures of the form νg

where g ∈ L1(µ), (X,B, µ) is an ordinary measure space and νg : B −→ R is defined by

νg(E) =

∫E

g dµ.

Let

P = {x ∈ X : g(x) > 0} ∈ BN = {x ∈ X : g(x) < 0} ∈ BZ = {x ∈ X : g(x) = 0} ∈ B.

P is a positive set for νg by the same argument as in Example 8.67. Similarly, N is anegative set for νg. Note that Z is both positive and negative, however, for if B ⊂ Zand B ∈ B, then

νg(B) =

∫B

g dµ =

∫B

0 dµ = 0,

so νg(B) = 0 ≤ 0 and νg(B) = 0 ≥ 0. Thus Z is both a positive and a negative set forνg. Sets with this property are called null sets.


Definition 8.69. Let ν be a signed measure on the measurable space (X,B). We sayA ∈ B is a null set for ν if A is both a positive set and a negative set for ν.

Remark 8.70. Note that if A is a null set for ν, then ν(A) ≥ 0 (since A is positive) andν(A) ≤ 0 (since A is negative). Thus ν(A) = 0 for any null set. The converse need notbe true: If A ∈ B and ν(A) = 0, A need not be a null set for ν (see Example 8.71). Fora set A to be a null set of ν, we need that for every subset B ⊂ A that is measurable,ν(B) = 0.

Example 8.71. See Example 8.65(ii). Let (X,B) =([0, 1],M

). Let g(x) = x− 1

2. Let

A = [0, 1] and define νg(E) =

∫E

g dm(x) for E ⊂ [0, 1], E ∈M. Then

νg(A) =

∫[0,1]

(x− 1

2

)dm(x) =

∫ 1

0

(x− 1

2

)dx =

x2

2− x

2

]1

0

= 0.

But A is not a null set for νg (neither is it a positive or a negative set for νg). Look at[0, 1

2

]⊂ [0, 1], which is measurable, but

νg

([0,

1

2

])= −1

4< 0.

Note also that

νg

([1

2, 1

])=

1

4> 0

where[

12, 1]⊂ [0, 1] (look at the graph).

Remark 8.72 (Be Careful!). Signed measures lack the monotone property in general.If B ⊂ A, A,B ∈ B, it need not be true that ν(B) ≤ ν(A). Consider the previousExample 8.71:

[12, 1]⊂ [0, 1] but

νg

([1

2, 1

])=

1

4> 0 = νg

([0, 1]

).

Lemma 8.73 (Lemma 19, §11.5, p. 272 of [3]). Let ν be a signed measure on themeasurable space (X,B).

(i) Let A be a positive set for ν and suppose that A′ ⊂ A and A′ ∈ B. Then A′ is apositive set for ν.

(ii) Let {An}n≥1 be a countable collection of positive sets for ν. Then⋃

n≥1

An is also

a positive set for ν.

Note that a similar result holds for negative sets.


Lemma 8.74 (Lemma 20, §11.5, p. 272 of [3]). Let ν be a signed measure on themeasurable space (X,B). Suppose there exists some E ∈ B with 0 < ν(E) < ∞ (thisdoesn’t mean E is positive). Then there exists an A ⊂ E, A ∈ B such that A is positivefor ν and ν(A) > 0.

Theorem 8.75 (Hahn Decomposition Theorem - Proposition 21, §11.5, p. 273 of[3]). Let ν be a signed measure on the measurable space (X,B). Then there exist setsA,B ∈ B such that A∪B = X, A∩B = ∅, A is a positive set for ν and B is a negativeset for ν. The pair (A,B) is called a Hahn decomposition for (X,B, ν).

Remark 8.76. Note that we’re not saying the Hahn decomposition is unique, but it isunique up to null sets. See Exercise 27, p. 275 of [3] and also Exercise 3, Assignment 3.

Definition 8.77. Let ν be a signed measure on the measurable space (X,B). Let(A,B) be any Hahn decomposition for ν. Define set functions ν+ : B −→ [0,∞] andν− : B −→ [0,∞] on B as

ν+(E) = ν(E ∩ A) and ν−(E) = −ν(E ∩B)

for any E ∈ B. Note that these functions do not depend on the Hahn decomposition byRemark 8.76. In fact, ν+ and ν− define ordinary measures on (X,B) (convince yourselfof this). Furthermore, since ν is a signed measure, at most one of the measures ν+ andν− can take on the value +∞ (we won’t get ∞−∞ if we try and subtract them, forinstance). Now, for any E ∈ B,

(133) E = E ∩X = E ∩ (A ∪B) = (E ∩ A) ∪ (E ∩B)

In addition, the union in (133) is disjoint since

(E ∩ A) ∩ (E ∩B) = E ∩ (A ∩B) = E ∩ ∅ = ∅since (A,B) is a Hahn decomposition for ν. Thus

ν(E) = ν((E ∩ A) ∪ (E ∩B)

)= ν(E ∩ A) + ν(E ∩B)

= ν+(E)− ν−(E)

by the definitions of ν+ and ν−. This decomposition of the signed measure ν into thedifference of two ordinary measures that are “supported” on disjoint sets is called theJordan decomposition of ν (see Definition 8.78 for the definition of “supported”).This decomposition is unique (see Exercise 28, p. 275 of [3]).

Definition 8.78. Let (X,B) be a measurable space, and let ν1 and ν2 be ordinarymeasures on (X,B). We say that ν1 and ν2 are mutually singular if there exist setsA,B ∈ B with A ∩ B = ∅, A ∪ B = X and ν2(A) = ν1(B) = 0. In other words, ν1


is supported entirely on A and ν2 is supported entirely on B with A ∩ B = ∅. (ν1

“supported” on A means that ν1(E) = 0 for any E ∈ B with A ∩ E = ∅). We usuallywrite ν1 ⊥ ν2 if ν1 and ν2 are mutually singular (obviously ν1 and ν2 have to be definedon the same measurable space).

Remark 8.79. Note that if (X,B, ν) is a signed measure space, the measures ν+ andν− are mutually singular since if (A,B) is any Hahn decomposition for (X,B, ν), then

ν+(B) = ν(B ∩ A) = ν(∅) = 0 = ν(A ∩B) = ν−(A).

Remark 8.80. Note also that we can interchange the order of ν1 and ν2 in Definition8.78 so that ν1 ⊥ ν2 ⇐⇒ ν2 ⊥ ν1.

Proposition 8.81 (Proposition 22, §11.5, p. 274 of [3]). Let ν be a signed measureon the measurable space (X,B). Then there exists a pair of mutually singular ordinarymeasures ν+ and ν− with ν = ν+−ν− on B. Moreover there is only one such pair (hencewe may call it “the” Jordan decomposition of ν).

Definition 8.82. The measure ν+ is called the positive variation of ν and ν− iscalled the negative variation of ν (see Definition 8.77 and Proposition 8.81). Wewill see this is similar to work we did on absolutely continuous functions. Now, define|ν| : B −→ [0,∞] by |ν|(E) = ν+ + ν− for any E ∈ B. This is an ordinary measure on(X,B) called the total variation of ν.

Example 8.83. Let (X,B, µ) be an ordinary measure space. Take g integrable (not

necessarily nonnegative). Define in the usual way νg : B → R by νv(E) =

∫E

g dµ for

any E ∈ B. Recall νg is a signed measure on (X,B) (see Remark 8.68). Let

A = {x ∈ X : g(x) ≥ 0}B = {x ∈ X : g(x) < 0}.

Note that A ∩B = ∅, A ∪B = X. Let E ∈ B with E ⊂ A. Then

νg(E) =

∫E

g(x) dµ ≥ 0

since g(x) ≥ 0 on E ⊂ A. Thus A is a positive set for νg. Similarly, if E ′ ⊂ B, E ′ ∈ B,then νg(E

′) ≤ 0 so that B is a negative set for νg. It follows that (A,B) is a Hahn


decomposition for (X,B, νg). Now, let E ∈ B. Then

(νg)+(E) = νg(E ∩ A)

=

∫E∩A

g dµ

=

∫E∩A

g+ dµ since g+ = g on A

=

∫E∩A

g+ dµ+

∫E∼A

g+ dµ since g+ = 0 on E ∼ A

=

∫E

g+ dµ.

Thus

(νg)+(E) =

∫E

g+ dµ =⇒ ν(g+) = (νg)+.

Similarly,

(νg)−(E) = −νg(E ∩B)

= −∫

E∩B

g dµ

=

∫E∩B

g− dµ since −g− = g on A

=

∫E∩B

g− dµ+

∫E∼B

g− dµ since g− = 0 on E ∼ B

=

∫E

g− dµ.

Thus

(νg)− = ν(g−).

Finally,

|νg|(E) = (νg)+(E) + (νg)−(E)

=

∫E

g+ dµ+

∫E

g− dµ

=

∫E

(g+ + g−) dµ

=

∫E

|g| dµ

= ν|g|(E)(134)

using the same notation. Thus

|νg| = ν|g|.

Go back and look at T xa (g).


Definition 8.84. Let (X,B) be a measurable space. Let µ, ν be ordinary measures on(X,B). We say that ν is absolutely continuous with respect to µ, and write ν � µif for every E ∈ B such that µ(E) = 0, ν(E) = 0 as well.

Example 8.85. Let g : X → [0,∞] be integrable on X with respect to µ. Define

νg : B → [0,∞) by νg(E) =

∫E

g dµ. Then νg � µ since if E ∈ B with µ(E) = 0, then

νg(E) is the supremum of simple functions ϕ ≤ f over E, which is always 0.

Definition 8.86. For signed measures µ and ν on the measurable space (X,B), we saythat ν is absolutely continuous with respect to µ if |ν| � |µ|. So null sets of µ arenull sets of ν since |ν| = ν+ + ν− and |µ| = µ+ + µ−. Thus |µ|(Z) = 0 ⇐⇒ Z is a nullset for µ since µ+(Z) and µ−(Z) are both zero.

Definition 8.87. If µ and ν are signed measures on a measurable space (X,B), we saythat µ is mutually singular with respect to ν, denoted µ ⊥ ν, if |µ| ⊥ |ν|.

Remark 8.88. Unlike mutually singular (µ ⊥ ν ⇐⇒ ν ⊥ µ, see Remark 8.80),absolute continuity of measures is not a symmetric property. That is, ν � µ does notimply µ � ν in general. Consider (X,B) = ([0, 1],M[0,1]). Let µ be m, Lebesgue

measure. Let ν(E) =

∫E

χ[ 12,1] dm for every E ∈ M[0,1]. Then ν � m by Example 8.85

with g = χ[ 12,1]. But m 6� ν since

ν

([0,

1

2

))=

∫[0, 1

2)

χ[ 12,1] dm =

∫[0,1]

χ[0, 12)χ[ 1

2,1] dm = 0

since χ[0, 12)χ[ 1

2,1] = χ[0, 1

2)∩[ 1

2,1] = χ∅ = 0 whereas m

([0, 1

2))

= 12.


8.6. Radon-Nikodym Theorem.

Proposition 8.89 (Proposition 10, §11.1, p. 261 of [3]). Let (X,B, µ) be a completemeasure space. Let D be a countable dense subset of R and suppose there exists a mapfrom D to B, α 7→ Bα, such that for each fixed α ∈ D, whenever β ∈ D and β > α,µ(Bα ∼ Bβ) = 0. Then there exists a measurable function f : X → [−∞,∞] such thatwhenever x ∈ Bα, f(x) ≤ α and f(x) ≥ α for all x ∈ X ∼ Bα.

Example 8.90. Consider (R,M,m) and let D = Q. Let Bα = {x ∈ R : x ≤ α}.Then X ∼ Bα = R ∼ Bα = {x ∈ R : x > α}. For β > α, Bα ∼ Bβ = ∅ and m(∅) = 0,so the conditions of Proposition 8.89 are satisfied. What is the function f? Note thatf(x) = x works.

Theorem 8.91 (Radon-Nikodym Theorem - Theorem 23, §11.6, p. 276 of [3]).Let (X,B) be a measurable space, and let µ, ν be measures on (X,B) such that (X,B, µ)is σ-finite, complete, and ν � µ. Then there exists a nonnegative, measurable functionf : X → [0,∞] such that for every E ∈ B,

ν(E) =

∫E

f dµ,

that is, ν = νf in the notation from Remark 8.68. Moreover, f is unique in the sensethat if there exists another such function g : X → [0,∞] (measurable, nonnegative withν = νg), then f = g µ-a.e. on X. [f ] is called the Radon-Nikodym derivative of νwith respect to µ, written

[f ] =dν

dµ.

Think f dµ = dν to try and remember this.

Remark 8.92. We need (X,B, µ) to be σ-finite to get the previous result. Take X ={1}, B =

{∅, {1}

}. Let µ(∅) = 0 and µ({1}) = +∞. This is not σ-finite. Define

ν : B → [0,∞) such that ν(∅) = 0 and ν({1}) = 1. Then ν � µ since µ(∅) = 0 = ν(∅),but there does not exist a function f : X → [0,∞] such that ν(E) =

∫E

f dµ. If there

were such a function,

1 = ν({1}) =

∫{1}f dµ =

{∞ if f is positive

0 if f = 0,

which is ridiculous. On the other hand, ν is finite and hence σ-finite and µ� ν. Apply-

ing the Radon-Nikodym theorem to µ we find a g : X → [0,∞] such that µ(E) =

∫E

g dν,

g = dµdν

and g({1}) = ∞.


Remark 8.93. The following theorem investigates how to compare two measures onthe same measurable space.

Theorem 8.94 (Lebesgue Decomposition Theorem - Proposition 24, §11.6, p.278 of [3]). Let (X,B) be a measurable space, and let µ and ν be σ-finite (ordinary)measures on (X,B). Then there exist σ-finite measures ν0 and ν1 on (X,B) such thatν = ν0+ν1, ν0 ⊥ µ and ν1 � µ. This is called the Lebesgue decomposition of ν withrespect to µ. This decomposition is unique in that if there is another decompositionν = ν ′0 + ν ′1 with ν ′0 ⊥ µ and ν ′1 � µ, then ν0 = ν ′0 and ν1 = ν ′1.


9. Outer Measure and Extensions of Measures

9.1. Outer Measure.

Remark 9.1. Suppose f : [0, 1] −→ R is a continuous, monotone increasing functionwith f(0) = 0. Consider C ⊂ 2[0,1] where C =

{(a, b] : 0 ≤ a < b ≤ 1

}. Note

that C ⊂ B[0,1], the Borel subsets of [0, 1]. Define a set function µf : C −→ [0,∞)

by µf

(a, b]

)= f(b) − f(a) ≥ 0. It turns out that µf can be extended to a measure

µf : B −→ [0,∞] on [0, 1]. If f is an absolutely continuous, monotone increasingfunction on [0, 1] with f(0) = 0, consider the corresponding function µf ,

µf

((a, b]

)= f(b)− f(a) =

∫ b

a

f ′(t) dµ(t)

where recall f ′ exists µ-a.e. on [0, 1], f ′ ≥ 0, and f ′ ∈ L1[0, 1] by our characterization ofabsolutely continuous functions (see Section 5.4). In this case µf obviously extends toE ⊂ B by

µf (E) =

∫E

f ′(t) dµ(t)

and µf is absolutely continuous with respect to Lebesgue measure, i.e., µf � m on [0, 1](see Example 8.85).

Example 9.2. Recall the Cantor ternary set, C =∞⋂

n=1

Fn, where Fn is closed in [0, 1],

Fn = [0, 1] ∼ Gn and G1 =(

13, 2

3

), G2 =

(19, 2

9

)∪(

79, 8

9

), . . . , Gn =

2n−1⋃k=1

(ak

3n ,bk

3n

). On[

ak

3n ,bk

3n

], let f(x) = ak

3n , 1 ≤ k ≤ 2n−1. The function f so obtained, called the Cantorternary function, is monotone increasing and continuous (prove this). Consider µf :C −→ [0,∞) (where C =

{(a, b] : 0 ≤ a < b ≤ 1

}) defined by µf

((a, b]

)= f(b)− f(a).

Then

µf

((ak

3n,bk3n

])= f

(ak

3n

)− f

(bk3n

)=ak

3n− ak

3n= 0.

Thus µf (Gn) = 0 for every n ∈ N if we extend this to a measure on B[0,1] since

µf (Gn) = µf

(2n−1⋃k=1

(ak

3n,bk3n

))=

2n−1∑k=1

µf

((ak

3n,bk3n

))≤

2n−1∑k=1

µf

((ak

3n,bk3n

])= 0

where we have used the fact that the(

ak

3n ,bk

3n

)are all disjoint. Now, µf

([0, 1]

)= 0 so

that µf (Fn) = µf

([0, 1] ∼ Gn

)= µf

([0, 1]

)− µf (Gn) = 1− 0 = 1 for every n ∈ N. Thus

µf (C) = µf

(∞⋂

n=1

Fn

)= lim

n→∞µf (Fn) = 1

and

µf

([0, 1] ∼ C

)= µf

(∞⋃

n=1

Gn

)= 0.


How does µf compare to m, Lebesgue measure? Recall m(C) = 0 and m([0, 1] ∼ C

)=

m

(∞⋃

n=1

Gn

)= 1. Therefore m and µf and mutually singular, µf ⊥ m. This µf gives

information about subsets of C whereas m misses it since it’s always zero.

Remark 9.3. If f is any arbitrary monotone increasing continuous function on [0, 1]with f(0) = 0, we can write f as f0 + f1, where f0 and f1 are both monotone increasing,f ′0 = 0 m−a.e., f1 is absolutely continuous and µf = µf0 + µf1 with µf0 ⊥ m andµf1 � m. In other words, µf0 + µf1 is the Lebesgue decomposition for µf (which isunique) with respect to Lebesgue measure m on [0, 1].

Remark 9.4. If f is a continuous, monotone increasing function on [0, 1] with f(0) = 0and C =

{(a, b] : 0 ≤ a < b ≤ 1

}, the big claim is that µf : C → [0,∞) defined by

µf

((a, b]

)= f(b)− f(a) extends to a measure on the Borel subsets of [0, 1].

Definition 9.5. Let X be a nonempty set, let A be a family of subsets of X, i.e.,A ⊂ 2X . Recall that A is called an algebra of subsets of X if

(i) ∅ ∈ A(ii) If A ∈ A,, then X ∼ A ∈ A

(iii) If A1, A2 ∈ A, then A1 ∪ A2 ∈ A.

Remark 9.6. Suppose we have a set function µ : A → [0,∞] that satisfies

(i) µ(∅) = 0 and(ii) If {Ai}i≥1 ⊂ A is a countable family of pairwise disjoint subsets of X,

⋃i≥1

Ai ∈ A

(which is guaranteed if the family is finite), and

µ

(⋃i≥1

Ai

)=∑i≥1

µ(Ai).

Let B be the smallest σ-algebra containing A. The big question is: does there exista measure µ : B → [0,∞] on (X,B) such that µ

∣∣A = µ? This is the extension

problem. It turns out the answer to this question is yes, but µ need not be uniqueunless µ is σ-finite restricted to A. This construction, known as the CaratheodoryExtension Procedure, requires the notion of outer measure. The process will looklike the construction of Lebesgue measure.

Definition 9.7. If X is a nonempty set, we say a function µ∗ : 2X → [0,∞] is an outermeasure if

(i) µ∗(∅) = 0


(ii) If A ⊂ B ⊂ X, then µ∗(A) ≤ µ∗(B) (monotonicity)(iii) If {Ai}i≥1 is a countable family of subsets of X, then

µ∗

(⋃i≥1

Ai

)≤∑i≥1

µ∗(Ai) (countable subadditivity)

Example 9.8. Lebesgue outer measure m∗ on R is an outer measure. See Definition3.2.

Remark 9.9. Suppose A is an algebra of subsets of X and µ : A → [0,∞] is a measure

on the algebra, i.e., µ(∅) = 0 and µ∗(⋃

i≥1

Ai

)=∑i≥1

µ∗(Ai) if {Ai}i≥1 are pairwise

disjoint and⋃i≥1

Ai ∈ A. How do we extend µ to an outer measure µ∗ on all of X? See

the following Definition.

Definition 9.10. Given the situation in Remark 9.9, define µ∗ : 2X → [0,∞] by

µ∗(E) = inf

{∑i≥1

µ(Ai) : {Ai}i≥1 ⊂ A and E ⊂⋃i≥1

Ai

}.

This is our outer measure.

Remark 9.11. We will show that µ∗ satisfies (i), (ii) and (iii) of Definition 9.7 and thatµ∗ determines a σ-algebra B of measurable subsets of X given by

E ∈ B ⇐⇒ ∀B ⊂ X, µ∗(B) = µ∗(B ∩ E) + µ∗(B ∼ E),

that is, E splits up B into two disjoint pieces on which µ∗ is additive. Furthermore, if werestrict µ∗ to B, we obtain a complete measure space (X,B, µ∗

∣∣B = µ) with µ(A) = µ(A)

for every A ∈ A (we’ll need A ⊂ B). This is called the Caratheodory ExtensionProcedure.


9.2. The Extension Theorem.

Lemma 9.12 (Lemma 2, §12.2, p. 292 of [3]). Let µ be a measure on the algebra A ofsubsets of a nonempty set X. Suppose that A ∈ A and {Ai}i≥1 ⊂ A satisfies A ⊂

⋃i≥1

Ai.

Then

µ(A) ≤∑i≥1

µ(Ai).

Consequently µ(A) ≤ µ∗(A) be definition of µ∗ as an infimum.

Corollary 9.13 (Corollary 3, §12.2, p. 292 of [3]). For every A ∈ A,

µ∗(A) = µ(A).

Proof. Note that all we need is µ∗(A) ≤ µ(A) since we already have that µ(A) ≤ µ∗(A)by Lemma 9.12. �

Lemma 9.14 (Lemma 4, §12.2, p. 292 of [3]). Let µ be a measure on an algebra A ofsubsets of a nonempty set X. Define µ∗ in the standard way (see Definition 9.10). Thenµ∗ satisfies the properties of an outer measure. See Definition 9.7.

Theorem 9.15 (Caratheodory - Theorem 1, §12.1, p. 289 of [3]). Let µ be a measureon an algebra A of subsets of a nonempty set X. Let

B = {E ⊂ X : µ∗(B) = µ∗(B ∩ E) + µ∗(B ∼ E) ∀B ⊂ X}.

Then B is a σ-algebra of subsets of X.

Proof. Same as in the Lebesgue measure case. See Theorem 3.21. �

Lemma 9.16 (Lemma 5, §12.2, p. 293 of [3]). Let A be an algebra of subsets of anonempty set X, let µ be a measure on A and let µ∗ and B be the outer measure on X

and the σ-algebra of measurable sets with respect to µ∗ discussed in Theorem 9.15. ThenA ⊂ B.

Remark 9.17. So far we have X a nonempty set, A an algebra of subsets of X,µ : A → [0,∞] a measure on A, µ∗ an outer measure defined on every subset of Xdetermined by µ, B a σ-algebra of measurable subsets of X with respect to µ∗, andA ⊂ B with µ∗(A) = µ(A) for every A ∈ A. One can show that µ∗

∣∣B gives a measure on

the σ-algebra B (do this), which we denote by µ (don’t confuse this with the conjugate).So (X,B, µ) is a measure space which extends the measure µ on A.


But what about uniqueness? When the measure µ on A is σ-finite, we will show thatthis extension of µ is unique. Recall a measure µ is σ-finite if we can write X =

⋃n≥1

Xn,

Xn ∈ A, and µ(Xn) < ∞ for all n. Without loss of generality we can assume the Xn

are pairwise disjoint. Hence σ-finite is as good as finite. Let S(A) denote the smallestσ-algebra of subsets of X that contains A:

S(A) =⋂

M a σ-algebraM⊃A

M.

Note that A ⊂ S(A) ⊂ B but we might not have equality. We will show that if µ isanother measure on S(A) such that µ(A) = µ(A) for every A ∈ A (in the case that µ isσ-finite of course), then µ(B) = µ(B) for all B ∈ S(A).

Proposition 9.18 (Proposition 6, §12.2, p. 293 of [3]). Let µ be a measure on an algebraA of subsets of a nonempty set X. Let

Aσ =

{⋃i≥1

Ai : Ai ∈ A

}⊂ S(A).

Let

Aσδ = (Aσ)δ =

{⋂n≥1

Bn : Bn ∈ Aσ

}⊂ S(A).

Let E ⊂ X be any set. Then for every ε > 0 there exists an A ∈ Aσ with E ⊂ Aand µ∗(A) ≤ µ∗(E) + ε. Furthermore, there exists a B ∈ Aσδ such that E ⊂ B andµ∗(B) = µ∗(E). This is reminiscent of Littlewood’s First Principle. See Remark3.34.

Remark 9.19. If E is in addition measurable and if µ(E) < ∞, then for the B inthe Proposition 9.18 we can write µ∗(B) = µ∗(B ∩ E) + µ∗(B ∼ E). But E ⊂ B,which implies that µ∗(B ∩ E) = µ∗(E) so that µ∗(B) = µ∗(E) + µ∗(B ∼ E) and hence0 = µ∗(B ∼ E) since µ∗(B) = µ∗(E). Therefore if E is measurable, µ(E) < ∞, thenthere exists a Z = B ∼ E with µ(Z) = 0 and E ∪ Z ∈ Aσδ.

Proposition 9.20 (Proposition 7, §12.2, p. 294 of [3]). Let µ be a σ-finite measuredefined on an algebra A of subsets of a nonempty set X. Let µ∗ be the outer measuredetermined by µ. Then E ⊂ X is measurable with respect to µ∗ if and only if E can bewritten E = B ∼ Z with B ∈ Aσδ and Z ⊂ X with µ∗(Z) = 0. Furthermore, any such

Z is contained in Z ⊂ Aσδ with µ(Z) = 0.


Theorem 9.21 (Caratheodory Extension Theorem - Theorem 8, §12.2, p. 295of [3]). Let µ be a measure on an algebra A of subsets of a nonempty set X. Let µ∗ andµ be the outer measure on X and the measure on B constructed previously. Then µ isan extension of µ to a σ-algebra B containing A. Moreover, (X,B, µ) is complete. If µis σ-finite (or finite) so is µ and µ is the unique measure on S(A) that is an extensionof µ.

Definition 9.22. Let X be a nonempty set and suppose C ⊂ 2X . We say C is asemialgebra of subsets of X if

(i) C1 ∩ C2 ∈ C whenever C1, C2 ∈ C(ii) For C ∈ C, X ∼ C =

n⋃i=1

Ci, where each Ci ∈ C and Ci ∩ Cj = ∅ whenever i 6= j.

Note that the complement X ∼ C doesn’t have to be in C.


C = {∅}∪{R}∪{(−∞, β] : β ∈ R}∪{(α,∞) : α ∈ R}∪{(α, β] : −∞ < α < β <∞}.Is this a semialgebra? Note that

(α1, β1] ∩ (α2, β2] =

{∅ if max{α1, α2} ≥ min{β1, β2}(max{α1, α2},min{β1, β2}

]if min{β1, β2} > max{α1, α2},

so that condition (i) in Definition 9.22 is satisfied. For condition (ii), note that, forinstance,

R ∼ (α, β] = (−∞, α] ∪ (β,∞)

and these sets are disjoint and each in the semialgebra. Similarly, if we let X = [a, b],

C = {∅} ∪{[a, b]

}∪{{a}∪

{(α, β] : a ≤ α < β ≤ b

}is a semialgebra (prove this).

Remark 9.24. If we have a semialgebra C of subsets of X, C determines an algebraA(C), the smallest algebra containing C:

A(C) = {∅} ∪ {X} ∪

{n⋃

i=1

Ci : n ∈ N, Ci ∈ C

}.

The natural question arises: Given a semialgebra C of subsets of a nonempty set X,when does a mapping µ : C → [0,∞] extend to a measure on A(C)? See the followingProposition.


Proposition 9.25 (Proposition 9, §12.2, p. 297 of [3]). Let C be a semialgebra of subsetsof X. Let µ : C → [0,∞] satisfy

(i) µ(∅) = 0

(ii) If C ∈ C and C can be written as the finite disjoint union C =n⋃

i=1

Ci with Ci ∈ C,

then

µ(C) =n∑

i=1

µ(Ci) (finite additivity)

(iii) If C ∈ C and C ⊂∞⋃i=1

Ci where Ci ∈ C and Ci ∩ Cj = ∅ whenever i 6= j, then

µ(C) ≤∞∑i=1

µ(Ci).

Then µ can be extended to a measure on A(C).

Theorem 9.26. Let F : R −→ [0,∞) be right continuous and monotone increasing.Suppose in addition that lim

x→−∞F (x) = 0. Let C be the semialgebra on R given in Example

9.23. Define νF : C → [0,∞] by

νF

((α, β]

)= F (β)− F (α) νF (∅) = 0 νF (R) = lim

x→∞F (x)

νF

((−∞, β]

)= F (β) νF

(α,∞)

)= lim

x→∞F (x)− F (α)

Then νF satisfies the conditions of Proposition 9.25 and hence determines a measure onA(C).

Remark 9.27. Now, by the Caratheodory Extension Theorem (see Theorem 9.21),νF can be uniquely extended to a measure on the smallest σ-algebra containing C,which is the Borel sets. So such a function determines a Borel measure. Conversely,Borel measures determine functions of this type (these measures have to have certainproperties though, in fact they have to be Baire measures, the subject to which we nowturn).


9.3. Baire Measures on R.

Definition 9.28. Let µ be a measure on (R,B), where B is the σ-algebra of Borel sets.We say µ is a Baire measure if µ(E) <∞ whenever E ∈ B and E is bounded.

Definition 9.29. Let µ be a finite Baire measure on (R,B) (that is, µ(R) < ∞, aprobability measure). Define Fµ : R → [0,∞) by Fµ(x) = µ

((−∞, x]

). Fµ is called a

cumulative distribution function associated to µ.

Example 9.30. Let f(t) = e−γt2 , where we’ve chosen γ > 0 so that

∫Rf(t) dt = 1.

Define µ(E) =

∫E

f(t) dt. Then

Fµ(x) = µ((−∞, x]

)=

∫ x

−∞f(t) dt.

What can we say about Fµ? Well, F ′µ = f and it turns out that Fµ is right continuous.

Lemma 9.31 (Lemma 10, §12.3, p. 300 of [3]). Let µ be a finite Baire measure on(R,B). Let Fµ be the cumulative distribution function associated to µ. Then Fµ ismonotone increasing, right continuous (i.e., lim

x→b+Fµ(x) = Fµ(b) for all b ∈ R) and

limx→−∞

Fµ(x) = 0. Furthermore, Fµ is continuous at a ∈ R if and only if µ({a})

= 0.

Lemma 9.32 (Lemma 11, §12.3, p. 301 of [3]). If F is a bounded, monotone increasing

function on R, and if (a, b] ⊂∞⋃i=1

(ai, bi], then

F (b)− F (a) ≤∞∑i=1

F (bi)− F (ai).

Proposition 9.33 (Proposition 12, §12.3, p. 301 of [3]). Let F : R → [0,∞) be a mono-tone increasing, right-continuous function such that lim

x→−∞F (x) = 0 and lim

x→+∞F (x) =

M <∞. Then there is a unique finite Baire measure µF on (R,B) such that µF

((a, b]

)=

F (b)−F (a) whenever a, b ∈ R and a < b. In fact, F is the cumulative distribution func-tion associated to µ.

Example 9.34. Given X a real random variable and given an expectation E(X ≤ α)for α ∈ R, this determines a function FE : R → [0, 1]. Our theory tells us that E,FE

determines a measure, a probability measure.


Definition 9.35. Let ϕ be a bounded Borel measurable function on R and let F bea monotone increasing function satisfying the conditions of Proposition 9.33. Then wedefine the Lebesgue–Stieltjes integral of ϕ with respect to F by∫

RϕdF =

∫RϕdµF ,

where µF is the finite Baire measure constructed in Proposition 9.33. If ϕ is unboundedbut Borel, we say that ϕ is integrable with respect to F if ϕ ∈ L1(µF ), that is, if∫

R|ϕ| dµF <∞.


9.4. Product Measures.

Remark 9.36. Let (X,A, µ) and (Y,B, ν) be two measure spaces. We want to define ameasure on the Cartesian product space X × Y = {(x, y) : x ∈ X, y ∈ Y } in order tointegrate certain “appropriate” functions (to be determined later).

Example 9.37. Consider an example from Calculus 3. Let X = Y = [0, 1], A = B =M, µ = ν = m (Lebesgue measure on [0, 1]) and define

T = {(x, y) : x, y ∈ [0, 1], 0 ≤ y ≤ x ≤ 1}.

Then ∫∫T

f dA =

∫ 1

0

∫ x

0

[f(x, y) dy

]dx =

∫ 1

0

∫ 1

y

[f(x, y) dx

]dy.

When f is continuous (or piecewise continuous), we can do this iterated integration.How can we generalize this? We will define a semialgebra of subsets of X × Y , find ameasure and extend it to a complete measure on a σ-algebra.

Definition 9.38. Let (X,A, µ) and (Y,B, ν) be measure spaces. Let

R = {A×B : A ∈ A, B ∈ B}

where

A×B = {(a, b) : a ∈ A ⊂ X, b ∈ B ⊂ Y }.R is called the family of measurable rectangles of X × Y . This will turn out tobe our semialgebra (see Proposition 9.40).

Remark 9.39. Be aware that, even in the case where X = Y = R and µ = ν = m,A×B need not be a geometric rectangle. Consider

A =

[0,

1

3

]∪(

2

3, 1

]⊂ R = X and B =

[0,

1

2

]⊂ R = Y.

Then A×B is the union of two geometric rectangles, which is not a rectangle itself.

Proposition 9.40. Let (X,A, µ) and (Y,B, ν) be measure spaces. Let R be the collectionof measurable rectangles of X × Y . Then R is a semialgebra of subsets of X × Y .

Definition 9.41. We now define a function λ = µ× ν : R −→ [0,∞] given by

λ(A×B) = µ(A) · ν(B).


In case one of µ(A) or ν(B) is infinite, recall that zero overrides ∞ in our arithmetic(see Remark 8.28) so that

(135) ∞ · r = r · ∞ =

{∞ if 0 < r ≤ ∞0 if r = 0

Remark 9.42. Our aim: We want to show that λ defined in Definition 9.41 satisfiesconditions (i) and (ii) of Proposition 9.25. It will then follow that λ extends to a measure,denoted by λ again, on the algebra A(R), the smallest algebra containing R. Then byTheorem 9.21 (Caratheodory), we can extend λ from A(R) to a complete measure,denoted λ, on the σ-algebra S to obtain a complete measure space (X × Y,S, λ), whereR ⊂ S and λ(A × B) = λ(A × B) = µ(A) · ν(B) for every A × B ∈ R. As far asuniqueness is concerned, if (X,A, µ) and (Y,B, ν) are σ-finite, so is the product measurespace (X × Y,S, λ). Moreover, λ is the unique extension of λ to the smallest σ-algebraS containing the measurable rectangles R.

Lemma 9.43 (Lemma 14, §12.4, p. 304 of [3]). Let (X,A, µ) and (Y,B, ν) be measurespaces, let R be the family of measurable rectangles of X×Y , and suppose {Ai×Bi}i≥1 isa countable collection of pairwise disjoint measurable rectangles in R with

⊔i≥1

Ai×Bi =

A×B ∈ R (here⊔

siginifies the disjoint union). Then

λ(A×B) =∑i≥1

λ(Ai ×Bi).

Remark 9.44. If (Ai ×Bi) ∩ (Aj ×Bj) = ∅, it need not be true that Ai ∩Aj = ∅. Forexample, consider X = Y = [0, 1], µ = ν = m, A1 = A3 =

[0, 1

2

), B1 = B2 =

[0, 1

2

),

A2 = A4 =[

12, 1], and B3 = B4 =

[12, 1]. Then

A1 ×B1 =[0, 1

2

)×[0, 1

2

)= R1

A2 ×B2 =[

12, 1]×[0, 1

2

)= R2

A3 ×B3 =[0, 1

2

)×[

12, 1]

= R3

A4 ×B4 =[

12, 1]×[

12, 1]

= R4

these are pairwise disjoint

but A1 = A3, not A1 ∩A3 = ∅. Thus some care is needed when dealing with measurablerectangles.

Example 9.45. Suppose that (X,A, µ) = (Y,B, ν) = (R,M,m). Then (R,M,m) isσ-finite and we can construct (R × R,S, λ = m ×m). This measure is invariant undertranslation of vectors and is called the Lebesgue measure on R2, where S containsall open subsets of R2 and hence all Borel subsets of R2. The same thing can be doneon X = Y = [0, 1] to obtain the Lebesgue measure on the unit square.


Remark 9.46. Returning to the general case, consider E ⊂ X × Y , E ∈ S withλ(E) <∞. Then

(136) λ(E) =

∫X×Y

χE(x, y) dλ.

Is it true that iterated integration works here? We do it in Calculus 3, as remarkedabove (see Example 9.37), so it had better. In other words, is

λ(E) =

∫X

[∫Y

χE(x, y) dν(y)

]dµ(x)

alright to do? And do we get the same real number as in (136)? Similarly, does

λ(E) =

∫Y

[∫X

χE(x, y) dµ(x)

]dν(y)?

If this is true, we move on to simple functions and then integrable functions. But weneed to ask, is χE(x, y) measurable? And once we integrate with respect to either µ(x)or ν(y), is the resulting function integrable? Is the resulting function even measurable?We will answer these questions by first introducing the notion of cross sections. Seethe following Definition.

Definition 9.47. Fix E ⊂ X × Y and fix x inX. Define Ex ⊂ Y by

Ex = {y ∈ Y : (x, y) ∈ E}.

We note that Ex could be empty. Similarly, for y ∈ Y fixed, define Ey ⊂ X by

Ey = {x ∈ X : (x, y) ∈ E}.

Again, Ey could be empty, but it’s still in X in this case. Ex and Ey are called crosssections to E.

Remark 9.48. If E ⊂ X × Y and x ∈ X is fixed, then χEx(y) = χE(x, y) for every

y ∈ Y . Moreover, if y ∈ Y is fixed, then χEy(x) = χE(x, y) for every x ∈ X.

Proof. χEx(y) = 1 ⇐⇒ y ∈ Ex ⇐⇒ (x, y) ∈ E ⇐⇒ χE(x, y) = 1. Similarly for

χEx(y) = 0. The proof of the second assertion is identical. �

Proposition 9.49 (Elementary Facts About Cross Sections). Let Ei ⊂ X ×Yfor i ≥ 1. Then

(i) E1 ⊂ E2 =⇒ (E1)x ⊂ (E2)x ⊂ Y for every x ∈ X and (E1)y ⊂ (E2)y ⊂ X forevery y ∈ Y .

(ii)

[⋃i≥1

Ei

]x

=⋃i≥1

(E1)x ⊂ Y for every x ∈ X and

[⋃i≥1

Ei

]y

=⋃i≥1

(E1)y ⊂ X for

every y ∈ Y .


(iii)

[⋂i≥1

Ei

]x

=⋂i≥1

(E1)x ⊂ Y for every x ∈ X and

[⋂i≥1

Ei

]y

=⋂i≥1

(E1)y ⊂ X for

every y ∈ Y .

Thus cross sections preserve containment, unions and intersections.

Definition 9.50. Let R be the semialgebra of measurable rectangles of X × Y , andlet A(R) be the algebra of sets generated by R. Let A(R)σ be the collection of allcountable unions of elements of A(R). Note that A(R)σ = Rσ, the collection of allcountable unions of elements from R (prove this). Similarly, let Rσδ= (Rσ)δ denote thecollection of all countable intersections of elements from Rσ. Compare to Definitions2.16 and 2.17

Lemma 9.51 (Lemma 15, §12.4, p. 305 of [3]). Let E ∈ Rσδ. Suppose that x ∈ X sothat Ex ⊂ Y . Then Ex ∈ B, that is, Ex is a measurable subset of Y .

Remark 9.52. Lemma 9.51 tells us that if E ∈ Rσδ, then for every x ∈ X, Ex ∈ B sothat χEx

is a measurable function defined on Y . Thus

(137) λ(E) =

∫X×Y

χE(x, y) dλ =

∫X×Y

χEx(y) dλ.

In other words, in this case the integration in (137) is alright, but we still don’t know ifwe can do this by iterated integration. This is what the following Lemma is for.

Lemma 9.53 (Lemma 16, §12.4, p. 305 of [3]). Let E ∈ Rσδ with λ(E) < ∞. Defineg : X −→ [0,∞] by g(x) = ν(Ex) (recall Ex ⊂ Y ). Then g is measurable on X, g isintegrable and ∫

X

g(x) dµ(x) = λ(E).

Note that ∫X

g(x) dµ(x) =

∫X

ν(Ex) dµ(x) =

∫X

[∫Y

χEx(y) dν(y)

]dµ(x)

=

∫X

[∫Y

χE(x, y) dν(y)

]dµ(x)

= λ(E)

=

∫X×Y

χE(x, y) dλ.

Hence once we establish this lemma, we know that iterated integration of characteristicfunctions of sets of this type (E ∈ Rσδ) is fine. We can then interchange x and y toobtain iterated integration in the other order.


Lemma 9.54 (Lemma 17, §12.4, p. 306 of [3]). Let (X,A, µ) and (Y,B, ν) be completemeasure spaces. Form the complete product measure space (X × Y,S, λ). Suppose thatE ∈ S and λ(E) = 0. Then for µ-almost all x ∈ X, Ex ∈ B with ν(Ex) = 0. Thussetting g(x) = ν(Ex), g is a measurable function defined µ-a.e. on X with

∫Xg(x) dµ =

λ(E) = 0.

Proposition 9.55 (Proposition 18, §12.4, p. 307 of [3]). Let E ⊂ X × Y , E ∈ S andλ(E) <∞. Then for µ-almost all x ∈ X, Ex ∈ B. Furthermore, defining

g(x) =

{ν(Ex) if Ex ∈ B0 otherwise,

g is measurable and g ∈ L1(µ) with∫X

g(x) dµ(x) = λ(E).

Proposition 9.56. If ϕ : X × Y −→ R is simple with

λ({(x, y) ∈ X × Y : ϕ(x, y) 6= 0}

)<∞,

then for µ-almost all x ∈ X, ϕx : Y −→ [0,∞] defined by ϕx(y) = ϕ(x, y) is in L1(ν).Moreover, defining g : X −→ [0,∞] by

g(x) =

∫Y

ϕx(y) dν(y),

g ∈ L1(µ) and ∫X

g dµ(x) =

∫X×Y

ϕ(x, y) dλ.

Theorem 9.57 (Fubini’s Theorem) - Theorem 19, §12.4, p. 307 of [3]).Let (X,A, µ) and (Y,B, ν) be complete measure spaces. Suppose that f : X × Y −→[−∞,∞] is integrable with respect to the product measure λ. Then

(i) For µ-almost all x ∈ X, the extended real-valued function fx(y) = f(x, y) is aν-integrable function of Y .

(i′) For ν-almost all y ∈ Y , the extended real-valued function f y(x) = f(x, y) is aµ-integrable function of X.

(ii) For µ-almost all x ∈ X,

∫Y

fx(y) dν(y) =

∫Y

f(x, y) dν(y) is finite, and defining

g(x) =

∫Y

fx(y) dν(y), g ∈ L1(µ).

(ii′) For ν-almost all y ∈ Y ,

∫X

f y(x) dµ(x) =

∫X

f(x, y) dµ(x) is finite, and defining

h(y) =

∫X

f y(x) dµ(x), h ∈ L1(ν).


(iii) Iterated integrals work and we get the same real number:∫X

[∫Y

fx(y) dν(y)

]dµ(x) =

∫X×Y

f(x, y)dλ =

∫Y

[∫X

f y(x) dµ(x)

]dν(y).

Remark 9.58. In order to apply Fubini’s Theorem, we need to know our function isintegrable with respect to λ. This is hard to determine in general. Sometimes we canuse Fubini’s Theorem to show a function is not integrable by computing both iteratedintegrals and getting a different answer for each (see Example 9.59). If one wants toprove that a function is integrable, we have Tonelli’s Theorem (see Theorem 9.60).

Example 9.59. Let (X,A, µ) =([0, 1],M,m

)and (Y,B, ν) =

([0, 1],M, ν

)where ν is

counting measure: For E ⊂ Y , ν(E) is the number of elements in E. Note that mostE’s will have ν-measure ∞. Note also that (Y,B, ν) is not σ-finite (if it were, this wouldimply that [0, 1] is countable, which it is not). Form

([0, 1]× [0, 1],S, λ

). Let f = χ∆,

where ∆ = {(t, t) : 0 ≤ t ≤ 1} (so ∆ is the diagonal of the unit square starting at theorigin). We will show that

(138)

∫X

[∫Y

(χ∆)x(y) dν(y)

]dm(x) 6=

∫Y

[∫X

(χ∆)y(x) dm(x)

]dν(y).

First of all, to evaluate the LHS of (138), fix x0 ∈ [0, 1]. Then∫Y

(χ∆)x0(y) dν(y) = ν(∆x0) = ν({y ∈ [0, 1] : (x0, y) ∈ ∆}

)= ν

({y = x0}

)= 1.

This is true for every x0 ∈ X, and thus

(139)

∫X

[∫Y

(χ∆)x(y) dν(y)

]dm(x) =

∫[0,1]

1 dm(x) = 1.

Now, to evaluate the RHS of (138), fix y0 ∈ [0, 1]. Then∫X

(χ∆)y0(x) dm(x) = m({x ∈ [0, 1] : (x, y0) ∈ ∆}

)= m

({x = y0}

)= 0.

Since this is true for every y ∈ Y , we have

(140)

∫Y

[∫X

(χ∆)y(x) dm(x)

]dν(y) =

∫Y

0 dν(y) = 0.

Combining (139) and (140), since 1 6= 0, χ∆ cannot be integrable lest we contradict

Fubini’s Theorem, which we just proved. Hence λ(∆) = ∞.


Theorem 9.60 (Tonelli’s Theorem- Theorem 20, §12.4, p. 309 of [3]). Let (X,A, µ)and (Y,B, ν) be complete and σ-finite measure spaces. Form (X×Y,S, λ), which is alsocomplete and σ-finite. Let f : X × Y −→ [0,∞] be a measurable, nonnegative extendedreal-valued function. Then

(i) For µ-almost all x ∈ X, fx is a measurable function on Y .(i′) For ν-almost all y ∈ Y , f y is a measurable function on X.

(ii) g(x) =

∫Y

fx(y) dν(y) is a measurable function on X.

(ii′) h(y) =

∫X

f y(x) dµ(x) is a measurable function on Y .

(iii)

(141)

∫X

[∫Y

fx(y) dν(y)

]dµ(x) =

∫X×Y

f(x, y) dλ =

∫Y

[∫X

f y(x) dµ(x)

]dν(y).

Thus in the nonnegative, σ-finite case you can do an iterated integral (either the LHS orthe RHS of (141)) to determine whether the function f is integrable with respect to λ.

Remark 9.61. Given a measurable function f : X×Y −→ [−∞,∞], where (X×Y,S, λ)is σ-finite, in order to determine whether or not f is integrable, apply Tonelli’s Theoremto |f(x, y)|, which is nonnegative and measurable. Recall that f is integrable if and onlyif |f | is integrable.

Example 9.62 (Applications of Tonelli’s Theorem).

(i) We first prove that ∫ ∞

0

e−x2

dx =

√π

2.

Suppose that∫

[0,∞]e−x2

dx = c, where c could be ∞. Then∫

[0,∞]e−y2

dy = c by a

change of variables. Thus(∫[0,∞]

e−y2

dy

)(∫[0,∞]

e−x2

dx

)= c2.

Since([0,∞],M,m

)is σ-finite, so is

([0,∞] × [0,∞],S,m × m

), so Tonelli’s

Theorem gives(∫[0,∞]

e−y2

dy

)(∫[0,∞]

e−x2

dx

)=

∫[0,∞]×[0,∞]

(e−x2)(

e−y2)dλ

=

∫[0,∞]×[0,∞]

e−(x2+y2)dλ(142)

where λ = m ×m. Now, use the change of variables x = r cos θ and y = r sin θso that r2 = x2 + y2 and θ = arctan( y

x). Then

{(x, y) : 0 < x, y <∞} =⇒{

(r, θ) : 0 < r <∞, 0 < θ <π

2

}


and (142) becomes

=

∫(0,∞)×(0, π

2)

e−r2

r drdθ

=

∫ ∞

0

∫ π2

0

e−r2

r drdθ by Tonelli’s Theorem again

=

∫ ∞

0

π

2e−r2

r dr

=π

2limt→∞

∫ t

0

e−r2

r dr

=π

2limt→∞

−e−r2

2

∣∣∣∣t0

=π

2limt→∞

1− e−t2

2

=π

2· 1

2

=π

4.(143)

Thus c2 = π4

=⇒ c =√

π2

as claimed.

(ii) We show that

f(x, y) =

{x

1−y2 if y 6= 1

0 if y = 1

is not integrable over([0, 1] × [0, 1],S, λ) where λ = m × m. This function is

nonnegative, and it is measurable since it is continuous on an increasing sequenceof rectangles going to y = 1 (Fill in the details. Let

fn(x, y) =x

1− y2χ[0,1]×[0,1−1/n],


which is measurable for every n ∈ N). Thus limn→∞

fn(x, y) = f(x, y) and∫[0,1]×[0,1]

x

1− y2dλ =

(∫[0,1]

1

1− y2dy

)(∫[0,1]

x dx

)=

∫[0,1]

1

1− y2dy · x

2

2

∣∣∣∣10

=1

2

∫[0,1]

1

1− y2dy

=1

2

[∫[0,1]

(1

2· 1

1− y+

1

2· 1

1 + y

)dy

]=

1

4

∫[0,1]

1

1− ydy +

1

4

∫[0,1]

1

1 + ydy

=1

4lim

t→1−− ln

∣∣1− y∣∣∣∣∣∣t

0

+1

4ln∣∣1 + y

∣∣∣∣∣∣10

=1

4lim

t→1−

(− ln |1− t|

)+

ln 2

4= +∞.

Thus f(x, y) is not integrable over([0, 1]× [0, 1],S, λ

)by Tonelli’s Theorem, as

claimed.


9.5. Integral Operators.

Remark 9.63. Recall that a normed linear space is a Banach space if it is complete inthe metric induced by its norm, d‖·‖(v, w) = ‖v − w‖. Recall also that if (X,B, µ) is ameasure space, then Lp(X,B, µ) is a Banach space for 1 ≤ p ≤ ∞. Let (X1, ‖ · ‖1) and(X2, ‖ · ‖2) be Banach spaces.

(i) We say that T : X1 −→ X2 is linear if T (a1x1 + a2x2) = a1T (x1) + a2T (x2) forevery x1, x2 ∈ X1 and for every a1, a2 ∈ R.

(ii) If T : X1 −→ X2 is linear, we say that T is bounded if there exists an M > 0such that ‖T (x)‖ ≤M‖x‖ for every x ∈ X1. In this case we define

‖T‖ = inf{M > 0 : ‖T (x)‖ ≤M‖x‖ for every x ∈ X1}.

Example 9.64. Let X = Y = L1(R). Fix g ∈ L1(R). For f ∈ L1(R) define

Tg(f)(x) =

∫Rg(y)f(x− y) dm(y)

if this integral exists and is finite. We claim that Tg(f)(x) is defined for almost allx ∈ R, Tg(f) is a measurable function of x and Tg(f) ∈ L1(R).

(i) First we claim that F : R×R −→ [−∞,∞] defined by F (x, y) = g(y)f(x− y) ismeasurable in (R × R,S,m×m). To see this, show that each separate piece ismeasurable. Surely g(y) is measurable, and to show that f(x− y) is measurableapproximate it by continuous functions. We leave the details as an exercise.

(ii) Now we use Tonelli’s Theorem on (R × R,S,m×m) applied to |F |, which isnonnegative and measurable. Then∫

R×R

∣∣F (x, y)∣∣ dm×m =

∫R

[∫R

∣∣F (x, y)∣∣ dm(x)

]dm(y) by Tonelli

=

∫R

[∫R|g(y)| |f(x− y)| dm(x)

]dm(y)

=

∫R|g(y)|

[∫R|f(x− y)| dm(x)

]dm(y) since g does not depend on x

=

∫R|g(y)|

[∫R|f(x)| dm(x)

]dm(y) since m is translation invariant(144)

=

(∫R|g(y)| dm(y)

)(∫R|f(x)| dm(x)

)since f does not depend on y

= ‖g‖1‖f‖1

<∞

(see Exercise ?? for step (144) above). Thus F is Lebesgue integrable over(R× R,S,m×m).


(ii) Next, by Fubini’s Theorem applied to F (x, y), we see that Tg(f)(x) =∫

R g(y)f(x−y) dm(y) is defined for m-almost all x ∈ R. Moreover Tg(f) is measurable andTg(f) is integrable over R, i.e.,∫

R

∣∣Tg(f)(x)∣∣ dm(x) <∞.

Thus Tg(f) ∈ L1(R).

(iii) Lastly, Tg : L1(R) −→ L1(R) is linear, i.e., Tg(f1 + f2) = tg(f1) + Tg(f2) andTg(αf1) = αTg(f1) for all f1, f2 ∈ L1(R) and all α ∈ R simply by the linearity ofthe integral. Is it bounded? Fix an f ∈ L1(R). Then

‖Tg(f)‖L1(R) =

∫R

∣∣Tg(f)(x)∣∣ dm(x)

=

∫R

∣∣∣∣∫Rg(y)f(x− y) dm(y)

∣∣∣∣ dm(x)

≤∫

R

[∫R

∣∣g(y)f(x− y) dm(y)

]dm(x)

=

∫R

[∫R

∣∣g(y)f(x− y)∣∣ dm(x)

]dm(y) by Tonelli’s Theorem

= ‖g‖1‖f‖1

by the same steps as in (i) above. Thus ‖Tg(f)‖1 ≤ ‖g‖1‖f‖1 for every f ∈ L1(R),which means that Tg(f) is a bounded linear transformation from L1(R) to itselfwith ‖Tg‖ ≤ ‖g‖1. It is standard convention to write Tg(f) = g ∗ f , called theconvolution of g and f .

Proposition 9.65. Let X1 = X2 = L2([0, 1],m

). Suppose that k : [0, 1] × [0, 1] −→

[−∞,∞] is in L2([0, 1]× [0, 1],S,m×m

). Define K : L2

([0, 1],m

)−→ L2

([0, 1],m

)by

K(f)(x) =

∫[0,1]

k(x, y)f(y) dm(y).

Then K(f) is defined for m-almost all x ∈ [0, 1], K(f) ∈ L2([0, 1],m

)and K is a

bounded linear operator with ‖K‖L2[0,1] ≤ ‖k‖L2([0,1]×[0,1],m×m)‖f‖L2[0,1].

Remark 9.66. The process involved in Proposition 9.65 is called integrating againstthe kernel, where one function space is mapped to another by means of integration.This can be thought of as a generalization of matrix multiplication, where K is the“matrix” and f is the “vector.” The natural question is can every transformation berepresented in this way? The function K(x, y) in Proposition 9.65 is called a kernel.

Note that∫ 1

0K(x, y)f(y) dy is a function of x. Is it continuous? Is it in Lp[0, 1]? Consider

the following examples:


(i) Let V = L1[0, 1] and let N denote the indefinite integral transformation, i.e.,N(f) =

∫ x

0f(y) dy. Let K(x, y) be the characteristic function of the lower trian-

gle T = {(x, y) : 0 ≤ y ≤ x}. Then∫χT (x, y)f(y) dy =

∫ x

0

f(y) dy.

(ii) The identity operator, I(f) = f for all f ∈ L1[0, 1], does not have a kernel. Tosee this, suppose to the contrary that it does have a kernel. Then

(145) I(f) = f(x) =

∫ 1

0

K(x, y)f(y) dy

for every f ∈ L1[0, 1]. Multiplying (145) on both sides by g(x) and integratingyields

(146)

∫ 1

0

f(x)g(x) dx =

∫∫K(x, y)f(y)g(x) dydx,

which holds for every f, g ∈ L1[0, 1]. Now, make f and g characteristic functions,f = χE and g = χF . Then (146) becomes

(147)

∫χE∩Fdx =

∫E×F

K(x, y) dλ.

Consider the case when E and F are disjoint – the left hand side of (147) is zero.Thus ∫

E×F

K(x, y) dλ = 0.

Thus K integrates to zero over all such rectangles R = E × F where E ∩ F = ∅.Hence R does not touch the diagonal: E ∩ F = ∅ ⇐⇒ E × F misses thediagonal. Geometrically, K integrates to zero over any rectangle that doesn’ttouch the diagonal. Recall that rectangles generate the σ-algebra for productmeasures. So

∫EK(x, y) dλ = 0 for any measurable E in the lower triangle,

which implies that k = 0 a.e. on the lower triangle. Similarly, K = 0 a.e. on theupper triangle. The diagonal is a set of measure zero, so K = 0 a.e. on the unitsquare. Where is the contradiction? By Fubini’s Theorem, for almost all x it’strue that for almost all y, K(x, y) = 0. Thus∫ 1

0

K(x, y)f(y) dy =

∫ 1

0

0 · f(y) dy = 0

for almost all x. But this is supposed to be f(x), which is a contradiction.


10. The Ascoli–Arzela Theorem

Remark 10.1. Let X be a compact metric space. Often times we’re given {fn}∞n=1 ⊂C(X), and we would like conditions under which we can find a subsequence {fnk

}∞k=1 of{fn}∞n=1 such that {fnk

}∞k=1 converges uniformly to some f ∈ C(X), i.e., converges in thesup norm on C(X): ‖fnk

−f‖ → 0 as k →∞ (see Remark 7.22). Basically this amountsto showing that {fnk

}∞k=1 is a Cauchy sequence since C(X) is a complete normed linearspace. We will determine precisely when this is possible.

Definition 10.2. Let F be a subset of C(X), the real-valued continuous functions on thecompact metric space X. We say F is equicontinuous if given ε > 0 and x ∈ X, thereexists a δ(ε, x) (depending on both ε and x) such that whenever y ∈ X and d(x, y) < δ,then |f(x)−f(y)| < ε for all f ∈ F . Note that F equicontinuous implies that F ⊂ C(X)since every f ∈ F is continuous at x for all x ∈ X. The difference here is that the sameδ works for every f in the family.

Remark 10.3. When X is compact it is possible to show that the δ in Definition 10.2can be chosen independent of x, that is, so it only depends on ε. Prove this, or seeProposition 10.14.

Example 10.4. Let X = [0, 1] and fix M > 0. Let

F = {f ∈ C[0, 1] : f is differentiable on [0, 1] and |f ′(x)| ≤M}.Then F is equicontinuous since fixing ε > 0, take δ = ε

M. If x, y ∈ [0, 1] and |x − y| <

δ = εM

, then by the Mean Value Theorem there is some c between x and y such that

|f(x)− f(y)| = |f ′(c)(x− y)|= |f ′(c)||x− y|≤M |x− y|

< M · εM

= ε.

Definition 10.5. Let F ⊂ C(X) where X is a compact metric space. Recall that forevery f ∈ C(X), there exists an Mf > 0 such that |f(x)| ≤Mf for every x ∈ X (this isbecause X is compact). We say the family F is bounded if there exists an M > 0 suchthat |f(x)| ≤M for every x ∈ X and for every f ∈ F .

Example 10.6. Let X = [0, 2π] and F = {cosnx : n ∈ N}. Then | cosnx| ≤ 1 forevery x ∈ [0, 2π] and for every n ∈ N. Thus F is bounded.


Lemma 10.7 (Lemma 37, §7.10, p. 167 of [3]). Let D be a countable set (in the futureD will be a countable dense subset of a compact metric space X). Let {fn}∞n=1 be asequence of functions, each of them mapping D −→ R. Suppose that for every x ∈ Dthe sequence {fn(x)}∞n=1 is bounded in R. Then there exists a subsequence {fnk

}∞k=1 of{fn}∞n=1 such that {fnk

(x)}∞k=1 converges for every x ∈ D.

Lemma 10.8 (Lemma 38, §7.10, p. 167 of [3]). Let X be a compact metric space andsuppose {fn}∞n=1 is an equicontinuous family of real-valued functions on X. Supposethere exists a countable dense set D ⊂ X such that lim

n→∞fn(x) exists for every x ∈ D.

Then for every y ∈ X, limn→∞

fn(y) exists, and writing f(y) = limn→∞

fn(y), f : X −→ R is

continuous.

Lemma 10.9 (Lemma 39, §7.10, p. 168 of [3]). Let X be a compact metric space.Suppose that {fn}∞n=1 is an equicontinuous family of real-valued functions on X such thatlim

n→∞fn(x) = f(x) for every x ∈ X and for some function f (note that f is continuous

by Lemma 10.8). Then fn → f uniformly on X, that is, limn→∞

‖fn − f‖ = 0 where ‖ · ‖is the sup norm from Remark 7.22.

Theorem 10.10 (Ascoli–Arzela Theorem - Theorem 40, §7.10, p. 169 of [3]). LetF be an equicontinuous family of real-valued functions on the compact metric space X.Let {fn}∞n=1 be a sequence of functions from F such that for every x ∈ X, {fn(x)}∞n=1 isa bounded subset of R. Then there exists a subsequence {fnk

}∞k=1 of {fn}∞n=1 such that{fnk

}∞k=1 converges uniformly on X to a continuous function f .

Remark 10.11. Which subsets of C(X) are compact? The following Corollary investi-gates this question.

Corollary 10.12 (Corollary 41, §7.10, p. 169 of [3]). Let F be an equicontinuous familyof real-valued functions in C(X) that is closed in the supremum norm ‖ · ‖. Supposethat for every x ∈ X, {f(x) : f ∈ F} is a bounded subset of R. Then F is a compactsubset of the metric space C(X) (under the metric determined by its supremum norm,d‖·‖(f, g) = ‖f − g‖ = sup

x∈X|f(x)− g(x)|.

Example 10.13. Let F ⊂ C([a, b]

)(recall [a, b] is compact). Suppose there exists an

M1 > 0 such that |f(x)| ≤ M1 for every x ∈ [a, b]. Suppose also that each f ∈ F isdifferentiable and there exists an M2 > 0 such that |f ′(x)| ≤ M2 for every x ∈ [a, b].Suppose further that F is closed. Then by Corollary 10.12 F is compact in C

([a, b]

).


Proposition 10.14. Let (X, d) be a compact metric space and let F be an equicontinuousfamily of real-valued functions on X. Fix some ε > 0. Then there exists a δ, dependingonly on ε and not on x, such that whenever x, y ∈ X with d(x, y) < δ, |f(x)− f(y)| < εfor every f ∈ F .

Proposition 10.15. Let (X, d) be a compact metric space. Let F be an equicontinuousfamily of real-valued functions defined on X. Suppose that for every x ∈ X, {f(x) :f ∈ F} is bounded, i.e. there exists an Mx > 0 such that |f(x)| ≤Mx for every x ∈ X.Then there exists an M > 0 such that |f(x)| ≤M for every f ∈ F and for every x ∈ X.

Remark 10.16. What Proposition 10.15 is really saying is that pointwise boundednessof the family F implies uniform boundedness of the family on a compact metric space.Hence we’re not saying anything stronger if we assume our equicontinuous family offunctions on a compact metric space is uniformly bounded.

Remark 10.17. The following theorem is sometimes also called the Ascoli–ArzelaTheorem. Let (X, d) be a compact metric space. Recall once again that C(X) isa complete normed linear space, i.e., a Banach space, in the metric induced by itssupremum norm (see Remark 7.22).

Theorem 10.18. Let (X, d) be a compact metric space and let F ⊂ C(X). Then Fitself is a compact subset of C(X) if and only if F is equicontinuous, closed in the normand bounded in the norm, i.e., there exists an M > 0 such that |f(x)| ≤ M for everyf ∈ F and for all x ∈ X.

Example 10.19. Let X = [0, 1]. Consider

B(0, 1) = {f ∈ C[0, 1] : ‖f‖ ≤ 1},where 0 denotes the zero function. This set is certainly closed and bounded. Is itcompact? This is the same as is it equicontinuous? No. For each n ∈ N, let fn(x) = xn.Then ‖fn‖ = 1 since fn(1) = 1. But {fn(x)}∞n=1 has no convergent subsequence. Supposethat {fnk

(x) = xnk}∞k=1 converges uniformly on [0, 1]. Then, letting f(x) = limk→∞

fnk(x),

this should be continuous on [0, 1] (why?). But

f(x) =

{0 if 0 ≤ x < 1

1 if x = 1.

Thus f is not continuous at x = 1, a contradiction. It follows that {fn}∞n=1 has noconvergent subsequence, so it’s not compact by Theorem 10.18. Note that the derivativesare not uniformly bounded.


Appendix A. Convergence in Measure

Definition A.1. Let {fn(x)}∞n=1 be a sequence of measurable functions defined on E ∈M. Suppose that f is a measurable function defined on E ∈ M. We say that thesequence {fn(x)}∞n=1 converges in measure to the limit function f if for every ε > 0there exists an N ∈ N such that whenever n ≥ N ,

m({x ∈ E : |fn(x)− f(x)| ≥ ε}

)< ε.

Remark A.2. Convergence in measure is a weaker notion than pointwise convergence.If m(E) < ∞ and fn converges pointwise to f , then {fn(x)}∞n=1 converges to f inmeasure on E (use Egoroff’s Theorem). On the other hand, if {fn(x)}∞n=1 converges tof in measure on E with m(E) < ∞, it need not be true that fn → f pointwise on E.See the following Example.

Example A.3. Let E = [0, 1]. For n ∈ N, find k ∈ N ∪ {0} such that 2k ≤ n < 2k+1,and let j = 2k+1 − n, 0 ≤ j < 2k. Define fn on [0, 1] by

fn(x) =

{1 if j

2k ≤ x ≤ j+12k

0 otherwise

We claim that fn → 0 in measure. Given n, k, j,

m({x ∈ [0, 1] : fn(x) 6= 0}

)= m

([j

2k,j + 1

2k

])=

1

2k,

but n < 2k+1 = 2(2k) =⇒ n2< 2k =⇒ 2

n> 1

2k , hence as n→∞,

m({x ∈ [0, 1] : fn(x) 6= 0}

)=

1

2k<

2

n→ 0.

Thus {fn} → 0 in measure. Yet the fn’s do not converge pointwise for any x ∈ [0, 1].If x ∈ [0, 1] is given, as are k, j such that 0 ≤ j < 2k, then j

2k ≤ x ≤ j+12k . Let k → ∞.

Taking n = 2k + j, f2k+j(x) = 1. But then for either j − 1 or j − 1 (call it j′), we wouldhave f2k+j′(x) = 0. So as n→∞, fn(x) goes between the values of 0 and 1, most of thetime 0, but once in a while equal to 1. Thus for fixed x, the sequence {fn(x)}∞n=1 doesnot converge.

Remark A.4. In Example A.3, the sequence {fn}∞n=1 has a subsequence {fnk}∞k=1

that converges to 0 almost everywhere on [0, 1]. Which subsequence? Take fnk(x) =

f2k+2k−1(x). Then for n = 2k + 2k − 1, j = 2k − 1 and

f2k+2k−1(x) =

{0 if 0 ≤ x ≤ 2k−1

2k = 1− 12k

1 if 2k−12k ≤ x ≤ 2k

2k = 1


Now,

limk→∞

fnk(x) =

{0 if 0 ≤ x < 1

1 if x = 1

Since m({1})

= 0, we get that fnk→ 0 a.e. on [0, 1]. This is not the only subsequence

that converges, and this is true in general.

Proposition A.5 (Proposition 18, §4.5, p. 95 of [3]). Let E ∈ M and let {fn}∞n=1 bea sequence of measurable functions. Let f be measurable and suppose that {fn} → f inmeasure. Then {fn} has a subsequence {fnk

} such that fnk(x) → f(x) a.e. on E.

Corollary A.6 (Corollary 19, §4.5, p. 96 of [3]). Let {fn} be a sequence of measurablefunctions defined on E ∈ M with m(E) < ∞. Suppose f is a measurable functiondefined on E. Then {fn} → f in measure if and only if every subsequence {fnk

} = {gk}itself has a subsequence {fnkj

} = {gkj} such that fnkj

= gkjconverges to f pointwise a.e.

on E.

Remark A.7 (Proposition 20, §4.5, p. 96 of [3]). Fatou’s Lemma and the convergencetheorems (MCT, DCT and GDCT) remain true if “convergence a.e.” is replaced by“convergence in measure.” See Exercise 21, p. 96 of [3].


References

1. M. Capinski and E. Kopp, Measure, Integral and Probability, 2nd edition, Springer Verlag, 2004.2. T. Hawkins, Lebesgue’s Theory of Integration: It’s Origins and Developments, Chelsea Publishing

Co., 1975.3. H.L. Royden, Real Analysis, 3rd edition, Prentice Hall, 1988.4. R. L. Wheeden and A. Zygmund, Measure and Integral: An Introduction to Real Analysis, Marcel

Dekker, Inc., 1977.

Index

‖f‖∞, 51(X,B), 672X , 3A∆B, 4B ∼ A, 4B(x, δ), 59BV [a, b], 39C(X), 62

is a normed vector space, 62D+, 34D−, 34D+, 34D−, 34Ex, 103Ey, 103Fµ, 99Fσ-set, 9Fg, 81Gδ-set, 9L, 63L1(X, m), 49L∞(X, m), 51L∞(µ), 79Lp(X, m), 49, 51

is a Banach space for 1 ≤ p ≤ ∞, 55is a metric space, 53

Lp(µis a Banach space, 79

Lp(µ), 79Lp(ν), 80M∞, 51N , 20, 76, 84N b

a(f), 41Nx

a (f), 42P , 20, 76, 84P b

a(f), 41P x

a (f), 42T (f,P), 39T b

a(f), 39T x

a (f), 42Tg, 56, 57, 110Z, 84A, 63A(C), 97A(R), 104Aσ, 96Aσδ, 96‖F‖, 56, 81‖T‖, 110‖f‖∞, 79‖f‖p, 79‖f‖1, 49

‖f‖p, 49A, 3B, 8, 67B(R), 8C, 97E, 7M, 13P, 3P(X), 3R, 104S(A), 96ww{an}∞n=1

wwp, 80⊔

, 102`∞, 80`p, 80

is a Banach space, 80dνdµ , 90lim , 9lim , 9µ ⊥ ν, 89µ-a.e., 70ν � µ, 89ν, 68ν1 ⊥ ν2, 87νF , 98νg, 82σ-algebra

generated by, 6of subsets, 5, 67

σ-finite, 96σ-finite measure space, 69sgn g(x), 57f ∨ g, 63f ∧ g, 63f+, 20, 76f−, 20, 76g ∗ f , 111m, 14m∗, 11n(f,P), 40n+, 86n−, 86p(f,P), 40r+, 40r−, 40

a.e., 18µ-, 70

absolutely continuous, 33, 43measures, 89signed measures, 89

absolutely summable, 54119


algebragenerated by, 5of functions in C(X), 63of subsets, 4, 93

almost everywhere, 18arithmetic in [0,∞], 73Ascoli–Arzela Theorem, 114, 115

Baire measure, 99Banach space, 52, 55, 79, 80, 110

over R, 54BCT, see also Bounded Convergence TheoremBolzano–Weierstrass Theorem, 60Borel sets

σ-algebra of, 8, 67bounded, 110

family of functions, 113metric space, 61subset of a metric space, 61

Bounded Convergence Theorem, 26bounded linear functional, 56

are the continuous linear functionals, 56norm of, 56on Lp(µ), 81

bounded variation, 39

canonical representationof a simple function, 19

Cantor ternary function, 92Cantor ternary set, 15, 92Caratheodory Extension Procedure, 93Caratheodory Extension Theorem, 97Caratheodory xtension Procedure, 94Cauchy sequence, 53, 59Cauchy-Schwarz inequality, 53characteristic function, 17, 70closed set, 8closure, 7collection of subsets, 4compact, 60compact metric space, 60complement, 3

of open and closed sets, 8complete

measure space, 69metric space, 59

complete metric space, 53concave, 46

up, 46conjugate exponent, 79conjugate exponent q to p, 52continuity

of the Lebesgue integral, 29continuous function, 62convergent sequence, 59

convergesin measure, 116

convex function, 46convolution, 111countable additivity, 11, 67

of m, 14countable subadditivity, 68

of m∗, 12of outer measure, 94

counting measure, 68cross section, 103cumulative distribution function, 99

DCT, see also Lebesgue DominatingConvergence Theorem, see alsoDominating Convergence Theorem

DeMorgan’s Laws, generalized, 4dense, 64derivative, 34differentiable, 34Dini derivates, 34Dominating Convergence Theorem, 78

Egoroff’s Theorem, 22equicontinuous, 113essential bound, 79essential bound for f , 50essentially bounded on X, 50extension problem, 93

Fatou’s Lemma, 28, 75finite additivity

of a function on a semialgebra, 98finite measure space, 69Fubini’s Theorem, 105Fundamental Theorem of Calculus, 32

for Riemann integrals, 3

GDCT, see also Lebesgue DominatingConvergence Theorem, see alsoGeneralized Dominating ConvergenceTheorem

Generalized Dominating ConvergenceTheorem, 78

Generalized Lebesgue DominatingConvergence Theorem, 31

Holder’s inequality, 52, 53, 79for sequence spaces, 80

Hahn decomposition, 86Hahn Decomposition Theorem, 86Heine–Borel Theorem, 60

indicator function, 17integrable, 100

Lebesgue, 27


over E, 76Riemann, 2

integralof a measurable function, 76of a nonnegative simple function, 73of a nonnegative, measurable function, 74

Jensen’s inequality, 47Jordan decomposition

of a signed measure, 86

kernel, 111kernel, integrating against, 111

latticein C(X), 63

Lebesgue decomposition, 91Lebesgue Decomposition Theorem, 91Lebesgue Dominating Convergence Theorem,

30Lebesgue integral

continuity of, 29of a bounded, measurable, real-valued

function, 24of a measurable function, 29of a nonnegative, measurable function, 27of a simple function, 23

Lebesgue measurablefunction, 17set, 13

Lebesgue measure, 14on R2, 102on the unit square, 102

Lebesgue outer measure, 11Lebesgue–Stieltjes integral, 100left differentiable, 46linear, 110Littlewood’s First Principle, 16, 96Littlewood’s Second Principle, 20Littlewood’s Third Principle, 21lower envelope of f over E, 25lower limit, 9

MCT, see also Monotone ConvergenceTheorem

measurablefunction, 70space, 67subsets, 67

measurable rectangles, family of, 101measure, 67

on an algebra of sets, 94measure space, 67

σ-finite, 69complete, 69finite, 69

metric, 59metric space, 59

bounded, 61compact, 60complete, 53, 59

Minkowski’s inequality, 51, 52, 79and the triangle inequality, 53for sequence spaces, 80

Monotone Convergence Theorem, 28, 75monotone decreasing function, 35monotone increasing function, 35monotone property

of m∗, 11monotonicity, 68

of outer measure, 94mutually singular, 86

signed measures, 89

negative part of a function, 20, 76negative set

for a signed measure, 84negative variation

of a signed measure, 87over [a, b], 41with respect to P, 40

nonmeasurable sets, 16norm

of a bounded linear functional, 56, 81normed vector space, 51, 52, 79, 80

is a Banach space, 55null set

for a signed measure, 85

open ball at x with radius δ, 59open cover, 60open set, 7

in a metric space (X, d), 59outer measure, 93, 94

partition, 3point of closure, 7pointwise convergence

of a sequence of functions, 19positive part of a function, 20, 76positive set

for a signed measure, 84positive variation


power set, 3probability measure, 99

Radon-Nikodym derivative, 90Radon-Nikodym Theorem, 90relative complement, 4


Riemann integrable, 2Riemann integral, 2–3

lower, 2upper, 2

Riemann sumlower, 2upper, 2

Riesz Representation Theorem, 57, 58, 81Riesz-Fischer Theorem, 55right differentiable, 46

semialgebraof subsets, 97

separates the points of X, 63sequence spaces, 80sequentially compact, 60signed measure, 82simple function, 18Stone–Weierstrass Theorem, 63, 64

the proof, 65summable, 54

absolutely, 54supported, 87supporting line, 47symmetric difference, 4

Tonelli’s Theorem, 107total variation


totally boundedsubset of a metric space, 61

translation invariance, 11, 78triangle inequality, 62

uniform convergenceof a sequence of functions, 19

upper envelope of f over E, 25upper limit, 9

vanishes outside a set of finite measure, 23Vitali cover, 37

Weierstrass Approximation Theorem, 64, 65

Young’s inequality, 53

zero outside a set of finite measure, 23

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