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 and countably subadditive.

Why introduce the outer measure? Want to describe a general constructive 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 open rectangles 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∗ on P(X )) Let E ⊂ P(X ) be an algebra of sets and ρ : E → R+ ∪ {0} an set valued function such that ρ(∅) = 0. For A ⊂ X , we deﬁne   X∞  µ∗(A) = inf ρ(E ): A ⊂ ∪∞ E , E ∈ E .  j j=1 j j  j=1

Then µ∗ is an outer measure.

Proof. 1. Non-negative. By its deﬁnition, µ∗(∅) = 0 and µ(A) ≥ 0 for A ⊂ X . ∞ ∞ 2. Monotone. If A ⊂ B and B ⊂ ∪j=1Ej , then A ⊂ ∪j=1Ej and µ∗(A) ≤ µ∗(B). 3. See the next page. 3 Proposition. (Continue... ) n o ∗ P∞ ∞ 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 ² > 0 be given. • For each j = 1, 2, ···, ∃Ejk ∈ E s .t. X∞ ∞ ∗ ∗ −j Aj ⊂ ∪k=1Ejk & µ (Ejk ) ≤ µ (Aj ) + ²2 k=1 and therefore X∞ X∞ X∞ X∞ ∗ ∗ ¡ ∗ −j ¢ ∗ µ (A) ≤ µ (Ejk ) ≤ µ (Aj ) + ²2 ≤ µ (Aj )+² j=1 k=1 j=1 j=1

∗ P∞ ∗ Since ² 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 to be 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 complete measure space (X , M, µ∗) where M is the collection of all measurable sets. • Example. The Lebesgue measure on X = R is an extension of the pre-measure defined by ρ((a, b]) = b − a. 1. Let X = R. Let E be the smallest σ−algebra generated by half-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 all measurable 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 or metric space. The corresponding outer measure of Lebesgue measure µ is

µ∗(A) = inf {ρ(U): A ⊂ U, U open }

where ρ is a pre-measure defined on open sets in X . For example 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 → R such 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 of fingerprints in Seoul city police department. • Motivation: Design an efficient access system to find a target. • We need to define a dissimilarity function stating the distance between the data. The distance d(x, y) between two data x and y must satisfy the above four rules. • Similarity queries. For a given target x∗ ∈ X and ² > 0, arrest all having finger print y ∈ X such that d(y, x∗) < ². 8 Chapter 3. Measurable functions • A function f : Rn → R is Lebesgue measurable if f −1(U) is Lebesgue measurable for every open set U. • Let X be a metric space and let (X , M, µ) be a measure space. A function f : X → R is measurable if f −1(U) ∈ M whenever 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 ). £ ¤c • f −1(E c ) = f −1(E) . • 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 are f + 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}. (f +g)2−f 2−g 2 • fg is measurable because fg = 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 because f +g+|f −g| f +g−|f −g| f ∨ g = 2 , f ∧ g = 2 . 2 Theorem. (3.1.3)

If {fj } is a sequence of measurable functions, then lim 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 infj fj is measurable.

3 3.2 Integration of non-negative functions Let (X , M, µ) be measure space where X is a metric space. If your mathematical background is poor, you regard X as X = R2 and µ as the standard Lebesgue measure, that is, µ(A) = the area of A. Throughout this lecture, E, Ej are a measurable set. • The characteristic function of E denoted by χE is the function deﬁned by ½ 1 if x ∈ E χ (x) = E 0 otherwise

• A simple function is a ﬁnite linear combination of characteristic functions Xn

φ = cj χEj j=1

Hence, Ej = {φ = cj }. 4 Theorem. (3.2.1) Let f : X → R be measurable and f ≥ 0. Then

2X2n−1 −n n φn = k2 χEn,k + 2 χFn % f k=0

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

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

Proof. Straightforward. From the above theorem, we can prove that for any measurable function f there is a sequence of simple functions φn such that φn → f on any set on which f is bounded. 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 The Pn integral of a measurable simple function φ = j=1 cj χEj is deﬁned to be Z Xn φdµ = cj µ(Ej ) j=1

• We use the convention that 0 · ∞ = 0. R • If φ is a simple function, then φ ≥ 0 =⇒ φdµ ≥ 0. • Let Simple be a vector spaceR of measurable simple functions. Then the integralR ¤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 Z Z

ν(A) = φ dµ = φχA dµ A X Then (X , M, ν) is also a measure space.

Pn Proof. Let φ = k=1 ck χEk where Ek ∈ M. Assume A = ∪j Aj where Aj ∈ M are mutually disjoint. Then Z Xn Z Xn Z

ν(A) = φχA dµ = ck χEk χAdµ = ck χEk ∩Adµ k=1 k=1 Xn Xn X∞ = ck µ(Ek ∩ A) = ck µ(Ek ∩ Aj ) k=1 k=1 j=1 X∞ Xn X∞ Z X = ck µ(Ek ∩ Aj ) = φdµ = ν(Aj ) j=1 k=1 j=1 Aj j 7 Deﬁnition. (Lebesgue integral of non-negative measurable function) The integral of a non-negative measurable function f is deﬁned by Z ½Z ¾

f dµ = sup φdµ : φ ≤ f & φ ∈ Simple

Recall that the integral of a measurable simple function Pn φ = j=1 cj χEj is deﬁned to be Z Xn φdµ = cj µ(Ej ) j=1

From the deﬁnition, we obtain Z Z f ≤ g =⇒ fdµ ≤ gdµ

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

If {fn} is a nondecreasing sequence of non-negative measurable functions, then Z Z lim fn dµ = lim fn dµ n n

∃ • Since fn %, limn fn = f and is measurable. Note that it is possible that f (x) = ∞ at some x. R R R • Since fndµ % and fn ≤ f , fn ≤ f and therefore Z Z lim fndµ ≤ f dµ n

R R • It remains to prove limn fndµ ≥ f dµ.

9 Theorem. (3.2.3: Continue ...... MCT) R R If fn %, then limn fn dµ = limn fn dµ R R Continue... Aim to prove lim f dµ ≥ f dµ. R R n n • Since f dµ = sup{ φdµ : φ ≤ f , φ ∈ Simple }, it suﬃces to prove that for any α, 0 < α < 1 and any φ ∈ Simple with φ ≤ f , Z Z lim fndµ ≥ α φ dµ n

• Let En = {fn ≥ αφ}. Then Z Z Z deﬁne fndµ ≥ fndµ ≥ αφ dµ : = α ν(En) En En

• Since ν is a measure andR En % X , R R limn ν(En) = ν(X ) = φdµ. Thus, limn fndµ ≥ α φdµ.

10 Corollary. (3.2.4: φn % f ) + Let Mable be the set of non-negative measurable functions. • If φn ∈ Simple and φn % f for some f ∈ Mable , then Z Z lim φndµ = fdµ n

R + • The map ¤dµ : Mable → R is linear.

+ + + Proof. Let f , g ∈ Mable , φ ∈ Simple % f , and ψ ∈ Simple % g. Then R R • f + gR dµ = limn(φn +R ψn) dµ R =lim (φ + ψ ) dµ = f dµ + g dµ R n n R n R R • αf dµ = limn αφndµ=α limn φn dµ = f dµ.

11 Proposition. (3.2.7: f = 0 almost everywhere) + Let f ∈ Mable . Then Z fdµ = 0 ⇐⇒ f = 0 a.e.

Proof. + • If f ∈ Simple , then the statement is immediate. • RIf f = 0 a.eR. and φ ≤ f , then φ = 0 a.e., and hence f = sup{ φ : φ ≤ f , φ ∈ S+ } = 0. R imple • Conversely, let f dµ = 0. Then Z 1 1 0 = f dµ ≥ µ({f > }), n = 1, 2, ··· n n ∴ f = 0 a.e. Why? µ ¶ 1 1 µ({f > 0}) = µ ∪∞ {f > } ≤ ∪∞ µ({f > }) = 0. n=1 n n=1 n 12 Lemma. (3.2.9: Fatou’s Lemma) + For any sequence fn ∈ Mable , we have Z Z lim inf fn dµ ≤ lim inf fn dµ n n

Proof. Z Z Z

lim inf fn dµ = sup inf fj dµ ≥ sup inf fj dµ n k≥1 j≥k k≥1 j≥k

Since gk = infj≥k fj %, Z Z Z

sup inf fj dµ = lim inf fj dµ = lim inf fj dµ k≥1 j≥k k→∞ j≥k k→∞j≥k

13 L1(X , dµ): Complete metric space

Deﬁnition. (Integrability)

Let (X , M, µ) be measure space.R A function f : X → R is 1 integrable if f ∈ Mable & |f |dµ < ∞. We denote by L (X , dµ) the class of all integrable functions. For f ∈ L1(X , dµ), we deﬁne Z Z Z 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 Z kf k = |f |dµ (its metric : d(f , g) = kf − gk )

To answer this question, we need to study several convergence theorems. 14 Proposition. L1(X , dµ) is a normed space equipped with the norm Z kf k = |f |dµ

This means that L1(X , dµ) is a vector space satisfying 1. kf k ≥ 0, ∀f ∈ L1 2. kf k = 0 iﬀ f = 0 a.e.. 3. kλf k = |λ|kf k, ∀f ∈ L1 and every scaler λ. 4. kf + gk ≤ kf k + kgk, ∀f , g ∈ L1 . The proof is the straightforward.

15 Theorem. ( Lebesgue Dominate Convergence Theorem ) 1 1 Assume {fn} ⊂ L such fn → f a.e. and ∃g ∈ L so that |fn| ≤ g a.e. for all n. Then Z Z 1 f ∈ L & f dµ = lim fn dµ n

Proof. Since g + fn ≥ 0, Z Z Fatou0s Lemma lim inf (g + fn) dµ ≤ lim inf (g + fn) dµ n n R R Hence, f dµ ≤ limn inf fn dµ. Applying the same argument to the sequence g − fn ≥ 0, we obtain Z Z Z − f dµ ≤ lim inf (−fn) dµ = −lim sup fn dµ n n

16 Example. (Gaussian function) The fundamental solution of the heat equation in 1-D is

1 2 G(x, t) = √ e−x /4t . 4πt Then Z G(x, t) dx = 1 for all t > 0 & lim G(x, t) = 0 a.e. R t→0+

• Let fn(x) = G(x, 1/n). Then fn → f = 0 a.e. and Z Z f dµ = 0 6= 1 = lim fn dµ n

This is the reason why LDC requires the assumption that {fn} is dominated by a ﬁxed L1-function g. 17 Corollary. (3.3.2) R 1 P 1 Let {fj } ⊂ L s.t. j |fj | dµ < ∞. Then ∃ f ∈ L such that

Xn Z X Z lim fj = f a.e. & f dµ = fj dµ n→∞ j=1 j

P∞ (> Denote f = j=1 fj .)

Pn P∞ 1. Let gn = j=1 |fj | and g = j=1 |fj |. 2. Since gn % g, it follows from the monotone convergence theorem that Z Z X Z gdµ = lim gn dµ = |fj | dµ < ∞ n j 1 Hence,¯ g ∈ L ¯and g < ∞ a.e. ¯Pn ¯ 1 3. Since ¯ j=1 fj ¯ < g a.e and g ∈ L , the result follows by the Dominate Convergence Theorem. 18 1 Theorem. (3.3.3: Simple is dense in L )

1 1 1 • Simple is dense in L , i.e., every element in L is a L −limit of a sequence of elements in Simple . 1 • C0(R) is dense in L (R, M, µ), where µ is any Borel measure on R. Here, the deﬁnition of C0(R) is C0(R) := {f ∈ C(R): ∃N s.t. f (x) = 0 for |x| > N}.

1 Proof of the ﬁrst statement: Simple is dense in L . • Let f ∈ L1. By Thm 3.2.1,

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

• By LDCT(LebesgueR Dominate Convergence Theorem), kφn − f k = |φn − f | dµ → 0. This completes the proof.

19 1 Proof of the second statement: C0(R) is dense in L (R, M, µ). 1 1. Since Simple is dense in L , it suﬃces to prove that any 1 φ ∈ Simple ∩ L can be approximated by a sequence {fn} ⊂ C0(R).

2. If φ = χ(0,1), then a sequence of continuous functions   1 if 0 ≤ x ≤ 1  0 if 0 < −1/n f (x) := → φ = χ in L1-sense. n  0 if x > 1 + 1/n (0,1)  linear otherwise

Indeed, kfn − φk = 1/n → 0. 3. Similarly, if A is a ﬁnite union of bounded open intervals, then φ = χA can be approximated by a sequence {fn} ⊂ C0(R).

20 Continue....

4. Let E be a Borel measurable set with µ(E) = kχE k < ∞. That is,   X  µ(E) = inf µ(I ): E ⊂ ∪I , I = (a , b ) < ∞  j j j j j  j

5. Hence, for any ² > 0, ∃ a ﬁnite union of open intervals N A = ∪j=1Ij such that

kχE − χAk = µ(E4A) < ²

where E4A = (E \ A) ∪ (A \ E).

6. Since ² > 0 is arbitrary, χE can be approximated by a sequence {fn} ⊂ C0(R).

21 R R 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.

−n n 1. Let Pn = {j2 : j = 1, ··· , 2 }, a partition of [a, b]. −n −n 2. Denote En,j := (j2 , (j + 1)2 ] and

mn,j := infx∈En,j f (x) Mn,j := supx∈En,j f (x) P2n P2n φn = j=1 mn,j χEn,j ψn = j=1 Mn,j χEn,j

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

4. Hence, ∃ψ = limn ψn and ∃ψ = limn ψn. R 1 R 1 5. By def’n, L(Pn, f ) = 0 φn(x)dx ≤ U(Pn, f ) = 0 ψn(x)dx

22 Continue.... 6. By deﬁnition of the Lebesgue integral for simple functions, Z Z L(Pn, f ) = φndµ & U(Pn, f ) = ψndµ

7. From Riemann integrability of f , R 1 infn U(Pn, f ) = supn L(Pn, f ) = 0 f (x) dx 8. By LDCT, Z Z φdµ = lim φn dµ = lim U(Pn, f ) n n

= inf U(Pn, f ) = sup L(Pn, f ) = lim L(Pn, f ) n n Z Zn = lim ψn dµ = ψdµ n

R R 1 R 9. Hence, φdµ = 0 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 is diﬀerentiable with respect to t and that ¯ ¯ ¯ ¯ ¯ ∂ ¯ 1 g(x) := sup ¯ f (x, t)¯ ∈ L (X , dµ) t∈[a,b] ∂t R Then F (t) = f (x, t)dµ is diﬀerentiable on a ≤ t ≤ b and Z Z ∂ ∂ f (x, t)dµ = f (x, t)dµ ∂t ∂t

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

f (x, tn) − f (x, t) hn(x) = , tn → t tn − t

(∵ |hn| ≤ g from the mean value theorem.) 24 3.5 Pointwise, Uniform, Norm convergence, & Convergence in measure Four important examples. R 1 • Consider φn ∈ L (R, dµ) s.t. kφnk = R |φn|dµ = 1 9 0 but φn → 0 in some sense. • 1 If ♥ φn = n χ(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 k ♣ ψk,j = χ(j2−k ,(j+1)2−k ) for j = 0, ··· , 2 − 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), ··· −k 3. kψk,j − 0k = 2 → 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

∀² > 0, lim µ({|fn − f | ≥ ²}) = 0 n→∞

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

∀² > 0, lim µ({|fn − fm| ≥ ²}) = 0 n,m→∞

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

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.

• ∃ fnk s.t. fnk → f a.e. • f is uniquely determined a.e.

Proof.

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

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

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

1. For x ∈ X \ ZN and j > i, we have

Xj−1 |gN+i (x) − gN+j (x)| ≤ |gN+k (x) − gN+k+1(x)| k=i Xj−1 ≤ 2−N−k ≤ 2−N−i+1 k=i

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 and f (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,...).

• Prove that fn converges to f in measure. Proof. For any ² > 0, we have ² ² µ({|f − f | ≥ ²}) = µ({|f − g | ≥ }) + µ({|g − f | ≥ }) n k 2 k n 2 → 0 as k, n → ∞

5 Corollary. (3.5.3) 1 If fn → f in L , then there is a subsequence fnk such that fnk → f a.e. Proof. For any ² > 0, Z 1 µ({|f − f | > ²}) ≤ |f − f | dµ → 0. n ² n

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

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

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 ² > 0, there exist F ⊂ E so that µ(E \ F ) < ² and fn → f uniformly on F .

Proof. Since fn → f a.e., there exist Z ⊂ E so that µ(Z) = 0 and fn(x) → f (x) for x ∈ E \ Z. It suﬃces to prove the theorem for the case when Z = ∅. ∞ 1 1. Let Em,n = ∩j=m{|fj − f | < n }. 2. Then limm→∞ µ(Em,n) = E for n = 1, 2, ··· . (why?) −n 3. Hence, ∃ mn s.t. µ(E \ Emn,n) ≤ ²2 . ∞ 4. Let F = ∩n=1Emn,n. Then µ(E \ F ) < ². (why?) 1 5. Moreover, if j > mn, then supx∈F |fj (x) − f (x)| < n . Hence, fn → f uniformly on F .

7 Chapter 4. Product spaces

Throughout this chapter, we assume that (Xj , Mj , µj ), j = 1, 2 is two 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 let R = {E1 × E2 : Ej ∈ Mj }. A product measure space (X , M, µ) is constructed as follows: • Define the pre-measure µ0 on R by 0 µ (E1 × E2) = µ1(E1)µ2(E2) • By Carathéodory’s theorem, we obtain a complete measure µ on X whose σ-algebra of measurable sets contain the product algebra 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 Z Z 1 1 f (x, y)dµ2(y) ∈ L (dµ1), f (x, y)dµ1(x) ∈ L (dµ2) X2 X1 and Z Z ·Z ¸ Z ·Z ¸

fdµ = f (x, y)dµ2(y) dµ1(x) = f (x, y)dµ1(x) dµ2(y) X X1 X2 X2 X1

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 functions Throughout this subsection, we consider a bounded function f :[a, b] → R. We will study a necessary and sufficient condition that f 0 exist almost everywhere and Z y f (y) − f (x) = f 0(x)dµ, µ((x, y)) = |y − x|. x

• If f is Cantor function, then f 0 = 0 almost everywhere but Z y 1 = f (1) − f (0) 6= 0 = f 0(x)dµ x • Lebesgue’s Theorem 5.1.1: Every monotonic function f :[a, b] → R is differentiable almost everywhere. • Recall that the derivative of f at x exists if the following all four numbers are the same finite value: f (x + h) − f (x) f (x + h) − f (x) lim inf ± , lim sup ± h→0 h h→0 h 3 Definition. (Bounded Variations) Let f :[a, b] → R be a function. The total variation of f on [a, x] is defined to be Xn Tf (a, x) = sup sup |f (xj ) − f (xj−1)|, n Pn j=1 where Pn = {a = x0 < x1 < ··· < xn = x}. The class of functions of bounded variation on [a, b] is denoted by BV [a, b].

It is well known that the space BV [a, b] is a Banach space with norm kf kvar = 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 by showing 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 ∀² > 0, there exist δ such that whenever a sequenceP of non-overlapping subintervals (xj , yj ) ⊂ [a, b] satisfies j (yj − xj ) < δ, then X |f (yj ) − f (xj )| < ² j

Note that the Cantor function is not absolute continuous. Theorem. (5.1.6: Absolute continuity) If f 0 exist almost everywhere, f 0 ∈ L1(dµ), and

Z x f (x) = f 0(x) dµ, x ∈ (a, b] a then f is absolute continuous. 6 Proof.

PWe want to prove thatP for a given ² > 0, there exist δ s.t. j (yj − xj ) < δ ⇒ j |f (yj ) − f (xj )| < ². 0 ² 1. If |f | is bounded, then we choose δ = 0 and kf k∞ Z X X yj X 0 0 |f (yj )−f (xj )| ≤ |f |dµ ≤ C (yj −xj ) < kf k∞δ = ² j j xj j

2. If f 0 ∈ L1(dµ) but not bounded, then we decompose Z ² f 0 = g + h where g is bounded and |h|dµ < 2 This is possible because the bounded functions are dense in L1(dµ). The results follows by choosing δ = ² . 2kgk∞

7 Theorem. (5.1.7: absolute + singular ) Let f be continuous and non-decreasing. Then f can be decomposed into the sum of an absolute continuous function and a singular function, both monotone.

Proof. 1. f 0 exist almost everywhere by Lebesgue’s theorem. 2. Since f is continuous, R x f (t+h)−f (t) 1 R x+h 1 R a+h a h dµ = h x f (t)dµ − h a f (t)dµ → f (x) − f (a) as h → 0. f (t+h)−f (t) 0 3. Since h → f (t) a.e. as h → 0, by Fatou’s lemma, Z Z x x f (t + h) − f (t) f 0(t)dµ ≤ lim inf dµ = f (x) − f (a) a a h R 0 0 1 x 0 4. Since f ≥ 0, f ∈ L (dµ). Set g = a f (t)dt and h = f − g. Then g is absolute continuous and h0 = 0a.e.. 8 Lebesgue-Radon-Nikodym Theorem The goal is to understand the main structure of the following theorem 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 that ν ¿ µ (ν is absolute continuous w.r.t.µ). Then Z ∃ f ∈ L1(X , dµ) s.t. ν(E) = f dµ E 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 to understand several concepts. Throughout this section, X is a metric space. Deﬁnition.

• A real valued measure ν, defined on Borel subset of X , is called a signed measure if it can be written in the form ν = µ1 − µ2 where µ1 and µ2 are positive Borel measures, at least one of which is finite. • Define the total variation of ν, denoted by |ν|, to be   X  |ν|(E) = sup |ν(E )|  j  j

where the sup is taken over all disjoint collection {Ej } such that E = ∪j Ej . 2 Theorem. (5.2.1: Jordan decomposition) If ν is a signed measure, then its total variation |ν| is a positive, + |ν|+ν − |ν|−ν countably additive measure. Moreover, ν = 2 and ν = 2 are 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 aP 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 signed measure 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. + −j ν(Ej ) > ν (X ) − 2 . c 1. For A ⊂ Ej ,

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

− + + −j 2. ν (Ej ) = −ν(Ej ) + ν (Ej ) ≤ −ν(Ej ) + ν (X ) < 2 c 3. Let lim supj Ej = P & lim infj Ej = 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 ∀ ² > 0 ∃ δ > 0 s.t. µ(E) < δ ⇒ |ν(E)| < ²

Proof. By Jordan decomposition, it suﬃces to prove the result when ν = ν+. The suﬃciency (⇐) is immediate; so we have only to establish its necessity. 1. To derive a contradiction, suppose ν ¿ µ but that ² − δ condition fails. −j 2. Then ∃ ² > 0 and ∃ {Ej } s.t. ν(Ej ) > ² while µ(Ej ) < 2 .

3. Let E = lim supj Ej = ∩k≥1 ∪j≥k Ej . 4. By Fatou’s lemma, ν(E) ≥ lim supj ν(Ej ) ≥ ². 5. By monotone convergence theorem, µ(E) = limk µ(∪j≥k Ej ) ≤ P∞ P∞ −j −k+1 limk j=k µ(Ej ) ≤ limk j=k 2 = limk 2 = 0. 6. From 4 and 5, ν is not absolutely continuous w.r.t. µ. 5 Theorem. (5.3.3) 1 Let f ∈ RL (X , dµ), where µ is a positive measure. Deﬁne ν(E) = E f dµ for E ∈ M. Then ¯Z ¯ ¯ ¯ ¯ ¯ ∀ ² > 0 ∃ δ > 0 s.t. µ(E) < δ ⇒ ¯ f dµ¯ < ² E

Proof. R 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 ½ Z ¾ F = f ∈ L1(dµ): ν(E) ≥ f dµ for all E ∈ M E Then ∃ unique g ∈ F s.t. Z Z gdµ = sup f dµ f ∈F

Proof. The sup makes sense since F= 6 ∅ by Lemma 5.3.4. R 1. LetR M = supf ∈F f dµ and let {fj } be a sequence in F s.t. fj dµ → M. 2. Let gnR = max{f1, f2, ··· , fn}. Then gn % and M = limj gj dµ by monotone convergence thm.

7 Continue the proof R 4. To prove gn ∈ F, E gn dµ ≤ ν(E). Proof.

• Let E1 = E ∩ {gn = f1} and j−1 Ej = E ∩ {gn = fj } \ ∪k=1Ek , j = 2, ··· , n. n • Then E = ∪j=1Ej and Ej are disjoint. Hence,

Z Xn Z Xn gn dµ = fj dµ ≤ ν(Ej ) = ν(E) E j=1 Ej j=1

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 that ν ¿ µ. Then ∃ f ∈ L1(X , dµ) s.t. Z ν(E) = f dµ for all ν(E) < ∞ (E ∈ M) E

Proof. 1. From Jordan decomposition, the proof can be reduced to the case in which µ, ν are positive and ﬁnite. 2. Let f be the maximal function generated by Lemma 5.3.5. R 3. Then ν1(E) = ν(E) − E fdµ satisﬁes ν1 ≥ 0 and ν1 ¿ µ. 0 0 4. If ν1 6= 0, ∃ ² > 0 & E s.t. ν1(E ) > 0 and 00 00 00 0 ν1(E ) ≥ ² µ(E ) for E ⊂ E (Why?) R Hence, ν(E) = E f + ²χE 0 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 Banach space with norm k · k = k · kp. • A bounded linear functional on B is a mapping ` : B → R such that |`(f )| ≤ Ckf k for all f ∈ B. • We denote by B∗ the space of linear functionals on B. n o |`(f )| • ∗ Deﬁne k`kB = k`k∗ = kf k : f ∈ B & kf k 6= 0

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

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

• kλ`k∗ = |λ| k`k∗ & k`1 + `2k∗ = k`1k∗ + k`2k∗.

9 p Dual & Isometry: L p−1 = (Lp)∗

p Theorem. (6.2.2: Isometric injection : L p−1 ,→ (Lp)∗) p q Let 1 < p < ∞ and q = p−1 . For g ∈ L , deﬁne `g : B → R by Z p `g (f ) = g f dµ, for f ∈ L = B.

∗ Then `g ∈ B and k`g k∗ = kgkq. This means that the injection q p ∗ q L ,→ (L ) deﬁned by g 7→ `g is an isometric injection of L into (Lp)∗.

Proof. 1. Clearly, `g is linear on B. 2. According to H¨olderinequality, p |`g (f )| = kfgk1 ≤ kf kpkgk p for all f ∈ L = B p−1

3. From the deﬁnition of k · k∗ and 2, k`g k∗ ≤ kgkq. 10 p Proof of isometric injection : L p−1 ,→ (Lp)∗ 4. If g 6= 0 and

|g|p−1sgn(g) g f = q−1 where sgn(g) = kgkq |g| then `g (f ) = kgkq

5. From 4, k`g k∗ ≥ kgkq

6. From 3 and 5, k`g k∗ = kgkq q p ∗ 7. Hence, g 7→ `g is an isometric injection of L into (L ) . ♣ ♣ ♣ ♣ Question : What about (Lp)∗ ,→ Lq? YES! See the next Theorem 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 with µ(X ) < ∞. For 1 < p < ∞ and ` ∈ (Lp(X , dµ))∗, there exist p g ∈ L p−1 such that Z `(f ) = f g dµ for all f ∈ Lp

• SUPER important! This idea with p = 2 provides the key concept 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 p R g ∈ L p−1 such that `(f ) = f g dµ for all f ∈ Lp = B

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

• Denote B²(x) = {y ∈ X : |y − x| < ²}. Due to absolute continuity of ♠ , we can deﬁne Radon-Nikodym derivative in some sense `(χ ) g(x) := lim B²(x) , x ∈ X ²→0 µ(B²(x)) R • Hence, `(χ ) = g dµ for any measurable set E. E R E • Hence, `(φ) = g φ dµ for any φ ∈ Simple . p • The result follows from the facts that Simple is dense in L 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) its + distribution 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 continuously increasing function η : R+ → R+, Z Z

η(|f |)dµ = − η(α) dλf (α).

In particular, for 1 ≤ p < ∞,

Z Z ∞ Z ∞ p p p−1 |f | dµ = − α dλf (α) = p α λf (α) dα 0 0