Caratheodory Theorem

Definition. (2.2.1; Outer measure)

• Let (X ,M, µ) be a measure space.• Recall

(i) X is a set.(ii) M is a σ−algebra, that is, closed under a countable union

and complementations.(iii) µ is a measure on M, non-negative & countably additive .

• A null set is a set N s.t. µ(N) = 0• If σ−algebra M includes all null set, then µ is said to be

complete.

• An outer measure on a non-empty set X is a set function µ∗

defined on P(X ) which is non-negative, monotone andcountably subadditive.

Why introduce the outer measure? Want to describe a generalconstructive procedure for obtaining complete measure.

1

Example of outer measure in X = R2

• X = R2, E=the σ-algebra generated by the set of all openrectangles in R2, and define

ρ(E ) = the area of E , E ∈ E

• (X , E , ρ) is a measure space but it may not be complete.

• This ρ is called pre-measure.

• For A ⊂ X , we define

µ∗(A) = inf{ρ(E ) : A ⊂ E , E ∈ E}.

Then µ∗ is an outer measure.

2

Proposition. (2.2.2: Construction of outer measure µ∗ onP(X ))

Let E ⊂ P(X ) be an algebra of sets and ρ : E → R+ ∪ {0} an setvalued function such that ρ(∅) = 0. For A ⊂ X, we define

µ∗(A) = inf

∞∑

j=1

ρ(Ej) : A ⊂ ∪∞j=1Ej , Ej ∈ E .

Then µ∗ is an outer measure.

Proof.

1. Non-negative. By its definition, µ∗(∅) = 0 and µ(A) ≥ 0 forA ⊂ X .

2. Monotone. If A ⊂ B and B ⊂ ∪∞j=1Ej , then A ⊂ ∪∞j=1Ej andµ∗(A) ≤ µ∗(B).

3. See the next page.

3

Proposition. (Continue... )

µ∗(A) = inf{∑∞

j=1 ρ(Ej) : A ⊂ ∪∞j=1Ej , Ej ∈ E}

: outer measure.

Continue...

3. It remains to prove countable subadditivity.Let A = ∪∞j=1Aj . Let ε > 0 be given.• For each j = 1, 2, · · ·, ∃Ejk ∈ E s .t.

Aj ⊂ ∪∞k=1Ejk &∞∑

k=1

µ∗(Ejk) ≤ µ∗(Aj) + ε2−j

and therefore

µ∗(A) ≤∞∑

j=1

∞∑

k=1

µ∗(Ejk) ≤∞∑

j=1

(µ∗(Aj) + ε2−j

) ≤∞∑

j=1

µ∗(Aj)+ε

Since ε is arbitrary small, µ∗(A) ≤ ∑∞j=1µ

∗(Aj).

4

Definition. (2.2.3: µ∗-measurable by Caratheodory)

Let µ∗ be an outer measure on a set X . A subset A ⊂ X is said tobe µ∗-measurable if

∀E ⊂ X , µ∗(E ) = µ∗(E ∩ A) + µ∗(E \ A)

• This is SUPER CLEVER defintion!

• This definition provides a method of constructing a completemeasure space (X ,M, µ∗) where M is the collection of allmeasurable sets.

• Example. The Lebesgue measure on X = R is an extensionof the pre-measure defined by ρ((a, b]) = b − a.

1. Let X = R. Let E be the smallest σ−algebra generated byhalf-open intervals (a, b]. Then (R, E , ρ) is an measure space.

2. Define the outer measure µ∗ as in Prop 2.2.2.3. Denote by M the collection of all measurable sets.4. Then (R,M, µ∗) is a complete measure space.

5

Theorem. (2.2.4: Caratheodory extension theorem)

Let µ∗ is an outer measure on X . Let M be the collection of allmeasurable sets. Then M is σ−algebra and the restriction of µ∗

to M is a complete measure.

Proof.

• Prove that M is σ−algebra. Easy.

• Prove that (X ,M, µ∗) is a measure space. Easy.

• Prove that µ∗ is complete measure.Proof. If µ∗(A) = 0, then for any E ⊂ X

µ∗(E ) ≤ µ∗(E ∩ A) + µ∗(E \ A) ≤ µ∗(A) + µ∗(E ) = µ∗(E )

Hence, µ∗(E ) = µ∗(E ∩ A) + µ∗(E \ A) for any E ⊂ X .Hence, A ∈M.

6

Lebesgue-Stieltjes measure

• Example: Lebesgue-Stieltjes measure on X = R.• Let E be the algebra containing half open intervals (a, b].• Define ρF ((a, b]) = F (b)− F (a) where F : R→ R is a

monotone increasing function.• ρF is a pre-measure measure on E but ρF is not complete.• Let µ∗ be the outer measure defined as before.• Denote by M the collection of all measurable sets.• Denote by µ = µ∗|M the restriction of µ∗ on M.• This µ is called a Lebesgue-Stieltjes measure generated by F .

• Example: Lebesgue-Stieltjes measure on X = Rn ormetric space. The corresponding outer measure of Lebesguemeasure µ is

µ∗(A) = inf {ρ(U) : A ⊂ U, U open }

where ρ is a pre-measure defined on open sets in X . Forexample in X = R2, ρ(U) = the volume of U.

7

Definition. (Metric Space (X , d) equipped with d =distance)

A metric space (M, d) is a set M and a function d : M ×M → Rsuch that

1. d(x , y) ≥ 0 for all x , y ∈ X.

2. d(x , y) = 0 iff x = y.

3. d(x , y) = d(y , x) for all x , y ∈ X.

4. d(x , y) ≤ d(x , z) + d(z , y) for all x , y ∈ X.

Example [Fingerprint Recognition] Let X be a data set offingerprints in Seoul city police department.

• Motivation: Design an efficient access system to find a target.

• We need to define a dissimilarity function stating thedistance between the data. The distance d(x , y) between twodata x and y must satisfy the above four rules.

• Similarity queries. For a given target x∗ ∈ X and ε > 0,arrest all having finger print y ∈ X such that d(y , x∗) < ε.

8

Chapter 3. Measurable functions

• A function f : Rn → R is Lebesgue measurable if f −1(U) isLebesgue measurable for every open set U.

• Let X be a metric space and let (X ,M, µ) be a measurespace. A function f : X → R is measurable if f −1(U) ∈Mwhenever U is an open or closed interval, or open ray(a,∞). It is a simple exercise to show the followings:• f −1(E ∪ F ) = f −1(E ) ∪ f −1(F ).• f −1(E ∩ F ) = f −1(E ) ∩ f −1(F ).• f −1(E c) =

[f −1(E )

]c.

• In particular, f : X → R is measurable if{x ∈ X : f (x) > a} ∈ M for all a ∈ R.

• Given two function f and g we define

f ∨ g = max{f , g} f ∧ g = min{f , g}f + = f ∨ 0 f − = (−f ) ∨ 0

1

Proposition. (3.1.2)

If f and g are measurable, then so aref + g , fg , f ∨ g , f ∧ g , f +, f −, and |f |.Proof. We will denote {f > a} := {x ∈ X : f (x) > a}• f + g is measurable because∀a ∈ R, {f + g > a} = ∪t∈Q ({f > t} ∩ {g > a− t}).Q := the set of rational numbers.

• f 2 is measurable since {f 2 > a} = X if a < 0 and∀a ≥ 0, {f 2 > a} = {f >

√a} ∪ {f < −√a}.

• fg is measurable because fg = (f +g)2−f 2−g2

2 .

• f + is measurable because {f + > a} = X if a < 0 &{f + > a} = {f > a} if a ≥ 0.

• |f | is measurable because |f | = f + + f −.

• f ∨ g , f ∧ g are measurable becausef ∨ g = f +g+|f−g |

2 , f ∧ g = f +g−|f−g |2 .

2

Theorem. (3.1.3)

If {fj} is a sequence of measurable functions, thenlim supj fj , lim supj fj are measurable.

Proof. Denote φ := lim supj fj .

1. Recall φ := lim supj fj = limn→∞ gn where gn = supj≥n fj .

2. {gn > a} = ∪j≥n{fj > a}. Hence, gn is measurable.

3. Since gn ↘ , lim supj fj = infn≥0 gn.

4. Hence, {φ > a} = ∩∞n=1{gn > a}.5. Therefore φ is measurable.

6. A similar proof shows that lim inf j fj is measurable.

3

3.2 Integration of non-negative functions

Let (X ,M, µ) be measure space where X is a metric space. Ifyour mathematical background is poor, you regard X asX = R2 and µ as the standard Lebesgue measure, that is,µ(A) = the area of A. Throughout this lecture, E , Ej are ameasurable set.• The characteristic function of E denoted by χE is the

function defined by

χE (x) =

{1 if x ∈ E0 otherwise

• A simple function is a finite linear combination ofcharacteristic functions

φ =n∑

j=1

cjχEj

Hence, Ej = {φ = cj}.4

Theorem. (3.2.1)

Let f : X → R be measurable and f ≥ 0. Then

φn =22n−1∑

k=0

k2−nχEn,k+ 2nχFn ↗ f

where En,k = f −1((k2−n, (k + 1)2−n]) , Fn = f −1((2n,∞]).Moreover, each φn satisfies

φn ≤ φn+1 & 0 ≤ f (x)− φn(x) ≤ 2−n for x ∈ X \ Fn

Proof. Straightforward.From the above theorem, we can prove that for anymeasurable function f there is a sequence of simplefunctions φn such that φn → f on any set on which f isbounded.Why? f = f + − f − where f = max{f , 0} and f − = max{−f , 0}.

5

Let (X ,M, µ) be a measure space.

• Definition of Lebesgue Integral for simple functions Theintegral of a measurable simple function φ =

∑nj=1 cjχEj

isdefined to be ∫

φdµ =n∑

j=1

cj µ(Ej)

• We use the convention that 0 · ∞ = 0.

• If φ is a simple function, then φ ≥ 0 =⇒ ∫φdµ ≥ 0.

• Let Simple be a vector space of measurable simplefunctions. Then the integral

∫¤dµ can be viewed as a linear

functional on Simple , that is,∫

¤ dµ : Simple → R is linear.

6

Lemma. (3.2.2)

Let (X ,M, µ) be a measure space. Given a non-negative,measurable simple function φ and A ∈M, define

ν(A) =

∫

Aφ dµ =

∫

XφχA dµ

Then (X ,M, ν) is also a measure space.

Proof. Let φ =∑n

k=1 ckχEkwhere Ek ∈M. Assume A = ∪jAj where

Aj ∈M are mutually disjoint. Then

ν(A) =

∫φχA dµ =

n∑

k=1

∫ckχEk

χAdµ =n∑

k=1

∫ckχEk∩Adµ

=n∑

k=1

ckµ(Ek ∩ A) =n∑

k=1

∞∑

j=1

ckµ(Ek ∩ Aj)

=∞∑

j=1

n∑

k=1

ckµ(Ek ∩ Aj) =∞∑

j=1

∫

Aj

φdµ =∑

j

ν(Aj)

7

Definition. (Lebesgue integral of non-negative measurablefunction)

The integral of a non-negative measurable function f is defined by

∫f dµ = sup

{∫φdµ : φ ≤ f & φ ∈ Simple

}

Recall that the integral of a measurable simple functionφ =

∑nj=1 cjχEj

is defined to be

∫φdµ =

n∑

j=1

cj µ(Ej)

From the definition, we obtain

f ≤ g =⇒∫

fdµ ≤∫

gdµ

8

Theorem. (3.2.3: MCT(Monotone Convergence Thm))

If {fn} is a nondecreasing sequence of non-negative measurablefunctions, then

∫limn

fn dµ = limn

∫fn dµ

• Since fn ↗, limn fn = ∃f and is measurable. Note that it is possible

that f (x) = ∞ at some x .

• Since∫

fndµ ↗ and fn ≤ f ,∫

fn ≤∫

f and therefore

limn

∫fndµ ≤

∫f dµ

• It remains to prove limn

∫fndµ ≥ ∫

f dµ.

9

Theorem. (3.2.3: Continue ......MCT)

If fn ↗, then∫

limn fn dµ = limn

∫fn dµ

Continue... Aim to prove limn

∫fndµ ≥ ∫

f dµ.

• Since∫

f dµ = sup{∫ φdµ : φ ≤ f , φ ∈ Simple}, it sufficesto prove that for any α, 0 < α < 1 and any φ ∈ Simple withφ ≤ f ,

limn

∫fndµ ≥ α

∫φ dµ

• Let En = {fn ≥ αφ}. Then

∫fndµ ≥

∫

En

fndµ ≥∫

En

αφ dµ :define= α ν(En)

• Since ν is a measure and En ↗ X ,limn ν(En) = ν(X ) =

∫φdµ. Thus, limn

∫fndµ ≥ α

∫φdµ.

10

Corollary. (3.2.4: φn ↗ f )

Let M+able be the set of non-negative measurable functions.

• If φn ∈ Simple and φn ↗ f for some f ∈Mable , then

limn

∫φndµ =

∫fdµ

• The map∫

¤dµ : M+able → R is linear.

Proof. Let f , g ∈M+able , φ ∈ S+

imple ↗ f , and ψ ∈ S+imple ↗ g .

Then

• ∫f + g dµ =

∫limn(φn + ψn) dµ

=limn

∫(φn + ψn) dµ =

∫f dµ +

∫g dµ

• ∫αf dµ =

∫limn αφndµ=α limn

∫φn dµ =

∫f dµ.

11

Proposition. (3.2.7: f = 0 almost everywhere)

Let f ∈M+able . Then

∫fdµ = 0 ⇐⇒ f = 0 a.e.

Proof.• If f ∈ S+

imple , then the statement is immediate.• If f = 0 a.e. and φ ≤ f , then φ = 0 a.e., and hence∫

f = sup{∫ φ : φ ≤ f , φ ∈ S+imple} = 0.

• Conversely, let∫

f dµ = 0. Then

0 =

∫f dµ ≥ 1

nµ({f >

1

n}), n = 1, 2, · · ·

∴ f = 0 a.e. Why?

µ({f > 0}) = µ

(∪∞n=1{f >

1

n})≤ ∪∞n=1µ({f >

1

n}) = 0.

12

Lemma. (3.2.9: Fatou’s Lemma)

For any sequence fn ∈M+able , we have

∫limn

inf fn dµ ≤ limn

inf

∫fn dµ

Proof.

limn

inf

∫fn dµ = sup

k≥1infj≥k

∫fj dµ ≥ sup

k≥1

∫infj≥k

fj dµ

Since gk = inf j≥k fj ↗,

supk≥1

∫infj≥k

fj dµ = limk→∞

∫infj≥k

fj dµ =

∫lim

k→∞infj≥k

fj dµ

13

L1(X , dµ): Complete metric space

Definition. (Integrability)

Let (X ,M, µ) be measure space. A function f : X → R isintegrable if f ∈Mable &

∫ |f |dµ < ∞. We denote by L1(X , dµ)the class of all integrable functions. For f ∈ L1(X , dµ), we define

∫f dµ =

∫f + dµ −

∫f − dµ

Question: Prove that L1(X , dµ) is a complete normed space(or Banach space) when it is equipped with the norm

‖f ‖ =

∫|f |dµ (its metric : d(f , g) = ‖f − g‖ )

To answer this question, we need to study several convergencetheorems.

14

Proposition.

L1(X , dµ) is a normed space equipped with the norm

‖f ‖ =

∫|f |dµ

This means that L1(X , dµ) is a vector space satisfying

1. ‖f ‖ ≥ 0, ∀f ∈ L1

2. ‖f ‖ = 0 iff f = 0 a.e..

3. ‖λf ‖ = |λ|‖f ‖, ∀f ∈ L1 and every scaler λ.

4. ‖f + g‖ ≤ ‖f ‖+ ‖g‖, ∀f , g ∈ L1

.

The proof is the straightforward.

15

Theorem. ( Lebesgue Dominate Convergence Theorem )

Assume {fn} ⊂ L1 such fn → f a.e. and ∃g ∈ L1 so that|fn| ≤ g a.e. for all n. Then

f ∈ L1 &

∫f dµ = lim

n

∫fn dµ

Proof. Since g + fn ≥ 0,

∫limn

inf (g + fn) dµFatou′s Lemma≤ lim

ninf

∫(g + fn) dµ

Hence,∫

f dµ ≤ limn inf∫

fn dµ.Applying the same argument to the sequence g − fn ≥ 0, we obtain

−∫

f dµ ≤ limn

inf

∫(−fn) dµ = −lim

nsup

∫fn dµ

16

Example. (Gaussian function)

The fundamental solution of the heat equation in 1-D is

G (x , t) =1√4πt

e−x2/4t .

Then∫

RG (x , t) dx = 1 for all t > 0 & lim

t→0+G (x , t) = 0 a.e.

• Let fn(x) = G (x , 1/n). Then fn → f = 0 a.e. and

∫f dµ = 0 6= 1 = lim

n

∫fn dµ

This is the reason why LDC requires the assumption that {fn}is dominated by a fixed L1-function g .

17

Corollary. (3.3.2)

Let {fj} ⊂ L1 s.t.∑

j

∫ |fj | dµ < ∞. Then ∃ f ∈ L1 such that

limn→∞

n∑

j=1

fj = f a.e. &

∫f dµ =

∑

j

∫fj dµ

(> Denote f =∑∞

j=1 fj .)

1. Let gn =∑n

j=1 |fj | and g =∑∞

j=1 |fj |.2. Since gn ↗ g , it follows from the monotone convergence

theorem that∫gdµ = lim

n

∫gn dµ =

∑

j

∫|fj | dµ < ∞

Hence, g ∈ L1 and g < ∞ a.e.

3. Since∣∣∣∑n

j=1 fj

∣∣∣ < g a.e and g ∈ L1, the result follows by the

Dominate Convergence Theorem.18

Theorem. (3.3.3: Simple is dense in L1)

• Simple is dense in L1, i.e., every element in L1 is a L1−limit ofa sequence of elements in Simple .

• C0(R) is dense in L1(R,M, µ), where µ is any Borel measureon R. Here, the definition of C0(R) isC0(R) := {f ∈ C (R) : ∃N s.t. f (x) = 0 for |x | > N}.

Proof of the first statement: Simple is dense in L1.

• Let f ∈ L1. By Thm 3.2.1,

∃ φn ∈ Simple s.t. φn → f a.e. & |φn| < |f | a.e.

• By LDCT(Lebesgue Dominate Convergence Theorem),‖φn − f ‖ =

∫ |φn − f | dµ → 0. This completes the proof.

19

Proof of the second statement: C0(R) is dense in L1(R,M, µ).

1. Since Simple is dense in L1, it suffices to prove that anyφ ∈ Simple ∩ L1 can be approximated by a sequence{fn} ⊂ C0(R).

2. If φ = χ(0,1), then a sequence of continuous functions

fn(x) :=

1 if 0 ≤ x ≤ 10 if 0 < −1/n0 if x > 1 + 1/n

linear otherwise

→ φ = χ(0,1) in L1-sense.

Indeed, ‖fn − φ‖ = 1/n → 0.

3. Similarly, if A is a finite union of bounded open intervals, thenφ = χA can be approximated by a sequence {fn} ⊂ C0(R).

20

Continue....

4. Let E be a Borel measurable set with µ(E ) = ‖χE‖ < ∞.That is,

µ(E ) = inf

∑

j

µ(Ij) : E ⊂ ∪Ij , Ij = (aj , bj)

< ∞

5. Hence, for any ε > 0, ∃ a finite union of open intervalsA = ∪N

j=1Ij such that

‖χE − χA‖ = µ(E4A) < ε

where E4A = (E \ A) ∪ (A \ E ).

6. Since ε > 0 is arbitrary, χE can be approximated by asequence {fn} ⊂ C0(R).

21

Theorem. (Riemann∫

v.s. Lebesgue∫

)

Let f : [0, 1] → R be a bounded and Riemann integrable. Then

• f ∈ L1([0, 1], dµ) where µ((a, b]) = b − a.

• Lebesgue and Riemann integrals agrees.

1. Let Pn = {j2−n : j = 1, · · · , 2n}, a partition of [a, b].

2. Denote En,j := (j2−n, (j + 1)2−n] and

mn,j := infx∈En,jf (x) Mn,j := supx∈En,j

f (x)

φn =∑2n

j=1 mn,j χEn,jψn =

∑2n

j=1 Mn,jχEn,j

3. Therefore φn ≤ φn+1 ≤ f ≤ ψn+1 ≤ ψn.

4. Hence, ∃ψ = limn ψn and ∃ψ = limn ψn.

5. By def’n, L(Pn, f ) =∫ 10 φn(x)dx ≤ U(Pn, f ) =

∫ 10 ψn(x)dx

22

Continue....

6. By definition of the Lebesgue integral for simple functions,

L(Pn, f ) =

∫φndµ & U(Pn, f ) =

∫ψndµ

7. From Riemann integrability of f ,infn U(Pn, f ) = supn L(Pn, f ) =

∫ 10 f (x) dx

8. By LDCT,

∫φdµ = lim

n

∫φn dµ = lim

nU(Pn, f )

= infn

U(Pn, f ) = supn

L(Pn, f ) = limn

L(Pn, f )

= limn

∫ψn dµ =

∫ψdµ

9. Hence,∫

φdµ =∫ 10 fdx =

∫φdµ

10. Therefore, ψ = f = φ a.e..23

Theorem. (3.4.3)

Let f (x , t) : X × [a, b] → R be a mapping. Suppose that f isdifferentiable with respect to t and that

g(x) := supt∈[a,b]

∣∣∣∣∂

∂tf (x , t)

∣∣∣∣ ∈ L1(X , dµ)

Then F (t) =∫

f (x , t)dµ is differentiable on a ≤ t ≤ b and

∂

∂t

∫f (x , t)dµ =

∫∂

∂tf (x , t)dµ

For each t ∈ (a, b), we can apply LDCT to the sequence

hn(x) =f (x , tn)− f (x , t)

tn − t, tn → t

(∵ |hn| ≤ g from the mean value theorem.)24

3.5 Pointwise, Uniform, Norm convergence, &Convergence in measure

Four important examples.

• Consider φn ∈ L1(R, dµ) s.t. ‖φn‖ =∫R |φn|dµ = 19 0 but

φn → 0 in some sense.• If ♥ φn = 1

nχ(0,n), then φn → 0 uniformly.• If ♦ φn = χ(n,n+1), then φn → 0 pointwise.• If ♠ φn = nχ(0,1/n), then φn → 0 a.e.

• For each k = 0, 1, 2, · · · , define a sequence♣ ψk,j = χ(j2−k ,(j+1)2−k) for j = 0, · · · , 2k − 1.

1. Denote φ1 = ψ0,0, φ1 = ψ1,0, φ2 = ψ1,1, φ3 = ψ2,0, φ4 =ψ2,1, φ3 = ψ2,2, φ3 = ψ2,3, φ4 = ψ3,0.

2. Then φ1 = χ(0,1), φ2 = χ(0,2−1), φ3 = χ(2−1,1), · · ·3. ‖ψk,j − 0‖ = 2−k → 0, while φn(x) 9 0 for any x .

1

Definition. (3.5.1: fn → f (meas))

• {fn} is said to converge in measure to f if

∀ε > 0, limn→∞µ({|fn − f | ≥ ε}) = 0

• {fn} is said to be Cauchy sequence in measure if

∀ε > 0, limn,m→∞µ({|fn − fm| ≥ ε}) = 0

• The sequences ♠ ♣,♥ converge to 0 in measure. That is,φn = 1

nχ(0,n), χ(0,1/n), χ(j2−k ,(j+1)2−k ) → 0 [meas].

• The sequence ♦ φn = χ(n,n+1) does not converge to 0 inmeasure.

2

Theorem. (3.5.2: fn: Cauchy in meas ⇒ fn →∃ f in meas)

Suppose {fn} is a Cauchy seq in measure. Then

• ∃ f ∈Mable s.t. fn → f in measure.

• ∃ fnks.t. fnk

→ f a.e.

• f is uniquely determined a.e.

Proof.

1. Choose a subsequence nk such that gk = fnk,

µ(Ek) ≤ 2−k , Ek = {|gk − gk+1| ≥ 2−k}

2. Let Zk := ∪∞j=k+1Ej . Then µ(Zk) ≤ 2−k .

3. Let Z = ∩∞k=1Zk . Then µ(Z ) = 0.

4. Prove that limk→∞ gk(x) = ∃f (x) for all x ∈ X \ Z

5. Prove that gk converges to f uniformly on X \ ZN

(N=1,2,...).

3

• Prove that limk→∞ gk(x) = ∃f (x) for all x ∈ X \ ZProof.

1. For x ∈ X \ ZN and j > i , we have

|gN+i (x)− gN+j(x)| ≤j−1∑

k=i

|gN+k(x)− gN+k+1(x)|

≤j−1∑

k=i

2−N−k ≤ 2−N−i+1

2. ∴ gk(x) is Cauchy sequence if x ∈ X \ ZN , N = 1, 2, · · · .3. ∴ limk→∞ gk(x) exists if x ∈ X \ Z .

4. Define f : X → R by f (x) = limk→∞ gk(x) for x ∈ X \ Z andf (x) = 0 for x ∈ Z .

5. Since µ(Z ) = 0, gk → f a.e..

4

• Prove that gk converges to f uniformly on X \ZN (N=1,2,...).

Proof. For x ∈ X \ ZN , we have|gN+k(x)− f (x)| = limj |gN+k(x)− gN+j(x)| ≤ 2−N−k+1,that is,

supX\ZN

|gN+k − f | ≤ 2−N−k+1

• Prove that fn converges to f in measure.

Proof. For any ε > 0, we have

µ({|f − fn| ≥ ε}) = µ({|f − gk | ≥ ε

2}) + µ({|gk − fn| ≥ ε

2})

→ 0 as k, n →∞

5

Corollary. (3.5.3)

If fn → f in L1, then there is a subsequence fnksuch that

fnk→ f a.e.

Proof. For any ε > 0,

µ({|fn − f | > ε}) ≤ 1

ε

∫|fn − f | dµ → 0.

Hence, fn → f in measure, and from the proof of Theorem 3.5.2we can choose a subsequence nk such that gk = fnk

,

µ(Ek) ≤ 2−k , Ek = {|gk − gk+1| ≥ 2−k}.

From Theorem 3.5.2, fnk= gk → f a.e..

6

Theorem. (3.5.4: Egorov’s Theorem)

Let µ(E ) < ∞ and let fn → f a.e.. Then, for any ε > 0, thereexist F ⊂ E so that µ(E \ F ) < ε and fn → f uniformly on F .

Proof. Since fn → f a.e., there exist Z ⊂ E so that µ(Z ) = 0 andfn(x) → f (x) for x ∈ E \ Z . It suffices to prove the theorem forthe case when Z = ∅.

1. Let Em,n = ∩∞j=m{|fj − f | < 1n}.

2. Then limm→∞ µ(Em,n) = E for n = 1, 2, · · · . (why?)

3. Hence, ∃ mn s.t. µ(E \ Emn,n) ≤ ε2−n.

4. Let F = ∩∞n=1Emn,n. Then µ(E \ F ) < ε. (why?)

5. Moreover, if j > mn, then supx∈F |fj(x)− f (x)| < 1n . Hence,

fn → f uniformly on F .

7

Chapter 4. Product spaces

Throughout this chapter, we assume that (Xj ,Mj , µj), j = 1, 2 istwo σ−finite measure spaces. Recall that the measure µj is called -finite, if X is the

countable union of measurable sets of finite measure. Let X = X1 × X2 and letR = {E1 × E2 : Ej ∈Mj}.A product measure space (X ,M, µ) is constructed as follows:

• Define the pre-measure µ′ on R byµ′(E1 × E2) = µ1(E1)µ2(E2)

• By Caratheodory’s theorem, we obtain a complete measure µon X whose σ-algebra of measurable sets contain the productalgebra M1 ⊗M2 := σ-algebra generated by R.

• Since µj is σ−finite, so is µ.

1

Theorem. (4.1.1: Fubini )

Assume f ∈ L1(dµ). Then

∫

X2

f (x , y)dµ2(y) ∈ L1(dµ1),

∫

X1

f (x , y)dµ1(x) ∈ L1(dµ2)

and∫

X

fdµ =

∫

X1

[∫

X2

f (x , y)dµ2(y)

]dµ1(x) =

∫

X2

[∫

X1

f (x , y)dµ1(x)

]dµ2(y)

The strategy of the proof.

1. Begin by proving the result for f (x , y) = χE1×E2 . It is trivial!

2. Then, prove it for f ∈ Simple .

3. Finally, extend it for general f ∈ L1(dµ).

2

Differentiation Theory

5.1 Differentiation Theory of functionsThroughout this subsection, we consider a bounded functionf : [a, b] → R. We will study a necessary and sufficientcondition that f ′ exist almost everywhere and

f (y)− f (x) =

∫ y

xf ′(x)dµ, µ((x , y)) = |y − x |.

• If f is Cantor function, then f ′ = 0 almost everywhere but

1 = f (1)− f (0) 6= 0 =

∫ y

xf ′(x)dµ

• Lebesgue’s Theorem 5.1.1: Every monotonic functionf : [a, b] → R is differentiable almost everywhere.

• Recall that the derivative of f at x exists if the following allfour numbers are the same finite value:

lim infh→0±f (x + h)− f (x)

h, lim suph→0±

f (x + h)− f (x)

h3

Definition. (Bounded Variations)

Let f : [a, b] → R be a function. The total variation of f on [a, x ]is defined to be

Tf (a, x) = supn

supPn

n∑

j=1

|f (xj)− f (xj−1)|,

where Pn = {a = x0 < x1 < · · · < xn = x}.The class of functions of bounded variation on [a, b] is denoted byBV [a, b].

It is well known that the space BV [a, b] is a Banach space withnorm ‖f ‖var = Tf (a, b).

♣ If f (x) = A sin nx , then Tf (0, π) = An.

4

Theorem. (5.1.3: Jordan Decomposition)

Every f ∈ BV [a, b] can be written as two non-decreasing functions.

Proof. Let T (x) = Tf (a, x). The theorem will be proved byshowing T − f and T are non-decreasing since f = T − (T − f ).

1. Let x < y . Tf (x , y) = T (y)− T (x)

2. From the definition,

T (x) = Tf (a, x) is a montone non-decreasing function of x

since, for x < y , T (y) = T (x) + T (x , y) ≥ T (x).

3. Clearly, |f (y)− f (x)| ≤ Tf (x , y) = T (y)− T (x).

4. Hence, f (y)− f (x) ≤ T (y)− T (x).

5. Hence, T (x)− f (x) ≤ T (y)− f (y).

5

Definition. (5.1.5: Absolute continuous)

A function f ∈ BV [a, b] is absolute continuous iff ∀ε > 0, thereexist δ such that whenever a sequence of non-overlappingsubintervals (xj , yj) ⊂ [a, b] satisfies

∑j(yj − xj) < δ, then

∑

j

|f (yj)− f (xj)| < ε

Note that the Cantor function is not absolute continuous.

Theorem. (5.1.6: Absolute continuity)

If f ′ exist almost everywhere, f ′ ∈ L1(dµ), and

f (x) =

∫ x

af ′(x) dµ, x ∈ (a, b]

then f is absolute continuous.

6

Proof.

We want to prove that for a given ε > 0, there exist δ s.t.∑j(yj − xj) < δ ⇒ ∑

j |f (yj)− f (xj)| < ε.

1. If |f ′| is bounded, then we choose δ = ε‖f ′‖∞ and

∑

j

|f (yj)−f (xj)| ≤∑

j

∫ yj

xj

|f ′|dµ ≤ C∑

j

(yj−xj) < ‖f ′‖∞δ = ε

2. If f ′ ∈ L1(dµ) but not bounded, then we decompose

f ′ = g + h where g is bounded and

∫|h|dµ <

ε

2

This is possible because the bounded functions are dense inL1(dµ). The results follows by choosing δ = ε

2‖g‖∞ .

7

Theorem. (5.1.7: absolute + singular )

Let f be continuous and non-decreasing. Then f can bedecomposed into the sum of an absolute continuous functionand a singular function, both monotone.

Proof.

1. f ′ exist almost everywhere by Lebesgue’s theorem.

2. Since f is continuous,∫ xa

f (t+h)−f (t)h dµ = 1

h

∫ x+hx f (t)dµ− 1

h

∫ a+ha f (t)dµ →

f (x)− f (a) as h → 0.

3. Since f (t+h)−f (t)h → f ′(t) a.e. as h → 0, by Fatou’s lemma,

∫ x

af ′(t)dµ ≤ lim inf

∫ x

a

f (t + h)− f (t)

hdµ = f (x)− f (a)

4. Since f ′ ≥ 0, f ′ ∈ L1(dµ). Set g =∫ xa f ′(t)dt and h = f − g .

Then g is absolute continuous and h′ = 0a.e..

8

Lebesgue-Radon-Nikodym Theorem

The goal is to understand the main structure of the followingtheorem which is a remarkable achievement.

Theorem. (5.3.6: Radon-Nikodym, Riesz representation)

Let µ and ν be σ−finite measures on a set X , and suppose thatν ¿ µ (ν is absolute continuous w.r.t.µ). Then

∃ f ∈ L1(X , dµ) s.t. ν(E ) =

∫

Ef dµ

for all EM for which ν(E ) < ∞

• Definition : ν ¿ µ iff µ(E ) = 0 ⇒ ν(E ) = 0 for all E ∈M.

• Definition : µ and ν are mutually singular, writing µ ⊥ ν, iff

∃ E , F ∈M s.t. X = E ∪F , E ∩F = ∅, µ(E ) = 0 = ν(F ).

1

To prove Radon-Nikodym thm, we need tounderstand several concepts.

Throughout this section, X is a metric space.

Definition.

• A real valued measure ν, defined on Borel subset of X , iscalled a signed measure if it can be written in the formν = µ1 − µ2 where µ1 and µ2 are positive Borel measures, atleast one of which is finite.

• Define the total variation of ν, denoted by |ν|, to be

|ν|(E ) = sup

∑

j

|ν(Ej)|

where the sup is taken over all disjoint collection {Ej} suchthat E = ∪jEj .

2

Theorem. (5.2.1: Jordan decomposition)

If ν is a signed measure, then its total variation |ν| is a positive,

countably additive measure. Moreover, ν+ = |ν|+ν2 and ν− = |ν|−ν

2are positive measures. We thus have the decompositions:

ν = ν+ − ν− (Jordan decomposition) & |ν| = ν+ + ν−

Proof. It is easy to prove the followings:

1. |ν|(∅) = 0 and |ν| is monotone and finitely additive.

2. If {Ej} is a countable partition of E , then|ν|(∪Ej) =

∑ |ν|(Ej).

3. ν+ and ν− are positive measures.

3

Theorem. ( 5.2.3: Hahn Decomposition )

Let ν = ν+ − ν− be the Jordan decomposition of the signedmeasure ν. Then, ν+⊥ ν−, and the exist P, N ∈M s.t.

P ∪ N = X , P ∩ N = ∅, ν+(N) = ν−(P) = 0

Moreover, ν(E ) ≥ 0, ∀ E ⊂ P, while ν(E ) ≤ 0, ∀ E ⊂ N.

Proof. Assume that ν+(X ) < ∞ and let {Ej} be a sequence s.t.ν(Ej) > ν+(X )− 2−j .

1. For A ⊂ E cj ,

ν(A) ≤ ν+(E cj ) = ν+(X )− ν+(Ej) ≤ ν+(X )− ν(Ej) ≤ 2−j .

2. ν−(Ej) = −ν(Ej) + ν+(Ej) ≤ −ν(Ej) + ν+(X ) < 2−j

3. Let lim supj Ej = P & lim inf j E cj = N

4. Then ν+(N) = 0 = ν−(P).

4

Theorem. (5.3.2)

Let ν be finite and µ a positive measure on (X ,M). Then

ν ¿ µ iff ∀ ε > 0 ∃ δ > 0 s.t. µ(E ) < δ ⇒ |ν(E )| < ε

Proof. By Jordan decomposition, it suffices to prove the resultwhen ν = ν+. The sufficiency (⇐) is immediate; so we have onlyto establish its necessity.

1. To derive a contradiction, suppose ν ¿ µ but that ε− δcondition fails.

2. Then ∃ ε > 0 and ∃ {Ej} s.t. ν(Ej) > ε while µ(Ej) < 2−j .

3. Let E = lim supj Ej = ∩k≥1 ∪j≥k Ej .

4. By Fatou’s lemma, ν(E ) ≥ lim supj ν(Ej) ≥ ε.

5. By monotone convergence theorem, µ(E ) = limk µ(∪j≥kEj) ≤limk

∑∞j=k µ(Ej) ≤ limk

∑∞j=k 2−j = limk 2−k+1 = 0.

6. From 4 and 5, ν is not absolutely continuous w.r.t. µ.5

Theorem. (5.3.3)

Let f ∈ L1(X , dµ), where µ is a positive measure. Defineν(E ) =

∫E f dµ for E ∈M. Then

∀ ε > 0 ∃ δ > 0 s.t. µ(E ) < δ ⇒∣∣∣∣∫

Ef dµ

∣∣∣∣ < ε

Proof.

1. Define ν(E ) =∫E f dµ for E ∈M.

2. Then ν ¿ µ.

3. The result follows from 2 and 5.3.2

6

Lemma. (5.3.5)

Let ν be finite and positive, and put

F =

{f ∈ L1(dµ) : ν(E ) ≥

∫

Ef dµ for all E ∈M

}

Then ∃ unique g ∈ F s.t.

∫gdµ = sup

f ∈F

∫f dµ

Proof. The sup makes sense since F 6= ∅ by Lemma 5.3.4.

1. Let M = supf ∈F∫

f dµ and let {fj} be a sequence in F s.t.∫fjdµ → M.

2. Let gn = max{f1, f2, · · · , fn}. Then gn ↗ andM =

∫limj gj dµ by monotone convergence thm.

7

Continue the proof

4. To prove gn ∈ F ,∫E gn dµ ≤ ν(E ).

Proof.• Let E1 = E ∩ {gn = f1} and

Ej = E ∩ {gn = fj} \ ∪j−1k=1Ek , j = 2, · · · , n.

• Then E = ∪nj=1Ej and Ej are disjoint. Hence,

∫

E

gn dµ =n∑

j=1

∫

Ej

fj dµ ≤n∑

j=1

ν(Ej) = ν(E )

4. From 4 and monotone convergence theorem, g = limj gj ∈ F .

8

Theorem. ( 5.3.6: Radon-Nikodym )

Let µ and σ be σ−finite measures on a set X , and suppose thatν ¿ µ. Then ∃ f ∈ L1(X , dµ) s.t.

ν(E ) =

∫

Ef dµ for all ν(E ) < ∞ (E ∈M)

Proof.

1. From Jordan decomposition, the proof can be reduced to thecase in which µ, ν are positive and finite.

2. Let f be the maximal function generated by Lemma 5.3.5.

3. Then ν1(E ) = ν(E )− ∫E fdµ satisfies ν1 ≥ 0 and ν1 ¿ µ.

4. If ν1 6= 0, ∃ ε > 0 & E ′ s.t. ν1(E′) > 0 and

ν1(E′′) ≥ ε µ(E ′′) for E ′′ ⊂ E ′ (Why?)

Hence, ν(E ) =∫E f + εχE ′ dµ for E ∈M, contradicting the

maximality of f .9

6.2 Duality and Radon-Nikodym Theorem

Definition.

Let B = Lp(X , dµ), 1 ≤ p < ∞. We knew that B is a Banachspace with norm ‖ · ‖ = ‖ · ‖p.

• A bounded linear functional on B is a mapping ` : B → Rsuch that |`(f )| ≤ C‖f ‖ for all f ∈ B.

• We denote by B∗ the space of linear functionals on B.

• Define ‖`‖B∗ = ‖`‖∗ ={ |`(f )|‖f ‖ : f ∈ B & ‖f ‖ 6= 0

}

It is easy to prove that B∗ is a normed vector space, that is,

• ‖`‖∗ ≥ 0 and ‖`‖ = 0 ⇔ ` = 0. Here, ` = 0 means that`(f ) = 0 for all f ∈ B.

• ‖λ`‖∗ = |λ| ‖`‖∗ & ‖`1 + `2‖∗ = ‖`1‖∗ + ‖`2‖∗.

9

Dual & Isometry: Lp

p−1 = (Lp)∗

Theorem. (6.2.2: Isometric injection : Lp

p−1 ↪→ (Lp)∗)

Let 1 < p < ∞ and q = pp−1 . For g ∈ Lq, define `g : B → R by

`g (f ) =

∫g f dµ, for f ∈ Lp = B.

Then `g ∈ B∗ and ‖`g‖∗ = ‖g‖q. This means that the injectionLq ↪→ (Lp)∗ defined by g 7→ `g is an isometric injection of Lq

into (Lp)∗.

Proof.

1. Clearly, `g is linear on B.

2. According to Holder inequality,

|`g (f )| = ‖fg‖1 ≤ ‖f ‖p‖g‖ pp−1

for all f ∈ Lp = B3. From the definition of ‖ · ‖∗ and 2, ‖`g‖∗ ≤ ‖g‖q.

10

Proof of isometric injection : Lp

p−1 ↪→ (Lp)∗

4. If g 6= 0 and

f =|g |p−1sgn(g)

‖g‖q−1q

where sgn(g) =g

|g |

then`g (f ) = ‖g‖q

5. From 4, ‖`g‖∗ ≥ ‖g‖q

6. From 3 and 5, ‖`g‖∗ = ‖g‖q

7. Hence, g 7→ `g is an isometric injection of Lq into (Lp)∗.

♣ ♣ ♣ ♣Question : What about (Lp)∗ ↪→ Lq? YES! See the nextTheorem 6.2.3.

11

Theorem. (6.2.3: Representation combining Radon-Nikodym)

Let X be a metric space and let µ be a Borel measure withµ(X ) < ∞. For 1 < p < ∞ and ` ∈ (Lp(X , dµ))∗, there exist

g ∈ Lp

p−1 such that

`(f ) =

∫f g dµ for all f ∈ Lp

• SUPER important! This idea with p = 2 provides the keyconcept of Lax-Milgram (uniqueness and existence of PDE,Finite Element Method, and so on).

• To understand this theorem completely, we need to study Hahn-decomposition, signed measure, absolute

continuity, Lebesgue differentiation, etc. However, it is a disaster that many recent mathematicians do not

know its usefulness and basic structure even after they mastered Real analysis. This is the main reason why

I introduce the theorem without knowledge of them here to provide a rough insight.

12

Theorem. (6.2.3: Continue...)

♣♦♥ For 1 < p < ∞ and ` ∈ B∗ = (Lp(X , dµ))∗, there exist

g ∈ Lp

p−1 such that `(f ) =∫

f g dµ for all f ∈ Lp = B

Sketch of the proof. Let us understand its structure roughly.

• Denote Bε(x) = {y ∈ X : |y − x | < ε}. Due to absolute continuity of ♠ ,we can define Radon-Nikodym derivative in some sense

g(x) := limε→0

`(χBε(x))

µ(Bε(x)), x ∈ X

• Hence, `(χE ) =∫E g dµ for any measurable set E .

• Hence, `(φ) =∫

g φ dµ for any φ ∈ Simple .

• The result follows from the facts that Simple is dense in Lp

and ` : Lp → R is bounded and linear.

13

6.3 Distribution functions

Given a real valued function f on a measure space (X ,M, µ) itsdistribution function λf : R+ := [0,∞) → R by

λf (α) = µ({x : |f (x)| > α})

Theorem. (6.3.2: Prove it for Simple . EASY! )

Suppose that λf (α) is finite for all α ∈ R+. For any continuouslyincreasing function η : R+ → R+,

∫η(|f |)dµ = −

∫η(α) dλf (α).

In particular, for 1 ≤ p < ∞,

∫|f |pdµ = −

∫ ∞

0αp dλf (α) = p

∫ ∞

0αp−1 λf (α) dα

14