FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez IV. DIFFERENTIATION and SIGNED MEASURES IV.1. Differentiation of Monotonic

Home , Almost everywhere, Finite measure, Monotonic function, Null set

FUNDAMENTALS OF REAL ANALYSIS by

Do˘ganC¸¨omez

IV. DIFFERENTIATION AND SIGNED MEASURES

IV.1. Diﬀerentiation of Monotonic Functions R Question: Can we calculate E f easily? More explicitly, can we hope for a result like the Fundamental Theorem of Calculus for Riemann Integral? Answer: Yes, but with lots of work and patience. The following fact is well-known:

Proposition. Let f :[a, b] → R be a monotonic function. Then the set of discontinuities of f is at most countable (hence has measure 0).

Exercise. Prove that if E be a countable subset of [a, b], then there exists f :(a, b) → R which is continuous only on (a, b) \ E.

Convention: Throughout this chapter, [a, b] will be an interval with −∞ < a < b < ∞.

Deﬁnition. A collection F of closed and bounded intervals is called a Vitali cover for a set E ⊂ R if for all x ∈ E, for all > 0 there exists I ∈ F such that x ∈ I and `(I) < . Example. Let E ∈ R be a bounded set (say E ⊂ [c, d]). Then the collection F = {[α, β]: c ≤ α < β ≤ d, α, β ∈ Q} is a Vitali Cover for E.

Next theorem is an important one!

Theorem.1 (Vitali Covering Lemma) Let E ⊂ R with m∗(E) < ∞ and let F be a Vitali n cover for E. Then for all > 0 there exists a ﬁnite disjoint sub-collection {Ik}k=1 ⊂ F such that n ! ∗ [ m E \ Ik < . k=1

Comments:

(a) One can always ﬁnd an open set O such that E ⊂ O and m(O) < ∞. (b) Without loss of generality, we can assume that [ X I ∈ F ⇒ I ⊂ O ⇒ m Ik = m(Ik) ≤ m(O) < ∞. | {z } =`(Ik)

Hence `(Ik) → 0 if the collection F is inﬁnite. (c) For any interval I, Ie will be the closed interval having the same midpoint as I and has length 5 times the length of I.

Proof (of Theorem.1) We will consider two cases.

n Case 1: There exists a disjoint sub-collection {Ik}k=1 ⊂ F such that for all I ∈ F, I ∩ Sn ( k=1 Ik) 6= ∅. 1 2 Sn In this case we claim that E ⊂ k=1 Ik (and hence the theorem is proven). If the claim is Sn false, then there exists a point x ∈ E \ k=1 Ik. Then x 6∈ Ik for all 1 ≤ k ≤ n. Since F is a Vitali cover, there exists I ∈ F such that x ∈ I. By assumption

I ∩ Ik0 6= ∅ for some 1 ≤ k0 ≤ k. Without loss of generality we can assume that I ∩ Ik = ∅ for k 6= k0. First 1 of all I 6= Ik0 , hence `(I \ Ik0 ) > 0. Let < 2 `(I \ Ik), then there exists J ∈ F such that x ∈ J and `(J) < . Thus n ! [ J ∩ Ik = ∅. k=1 This contradicts our assumption, proving the claim.

n Case 2: No such ﬁnite family exists. That is, given any ﬁnite disjoint collection {Ik}k=1, there Sn exists I ∈ F such that I ∩ ( k=1 Ik) = ∅. ∞ [In this case we will construct (inductively) a countably disjoint sub-collection {Ik} ∈ F S k=1 such that E \ k Ik is as small as we wish.]

Let I1 be arbitrary. By assumption there exists I (there may be others) such that I ∩ I1 = ∅. 1 Pick I1 ∈ F such that I2 ∩ I1 = ∅ and `(I2) > 2 S1 where S1 = `(I). Now assume we have a n disjoint collection {Ik}k=1 ⊂ F constructed in this manner. Then let

Sn = sup{`(I): I ∈ F,I ∩ Ik = ∅, k = 1, . . . n}.

1 n+1 Pick In+1 ∈ F such that In+1 ∩ Ik = ∅ for all k = 1, . . . n and `(In+1) > 2 Sn. So {Ik}k=1 is a ∞ disjoint family. Hence, by induction we have a countably disjoint family {Ik}k=1 ⊂ F such that each Ik ⊂ O for all k, where O is an open set such that E ⊂ O and m(O) < ∞. Then `(Ik) → 0 must hold. Claim: This collection has the property that for all n ≥ 1

n ∞ [ [ E \ Ik ⊂ Iek. k=1 k=n+1 Sn To prove this claim let n ≥ 1 and let x ∈ E \ k=1 Ik. Then there exists I ∈ F such that x ∈ I and n ! [ I ∩ Ik = ∅. k=1

Then I ∩ Im 6= ∅ for some m ≥ n + 1 must hold (otherwise, if I ∩ Im = ∅ for all m, then 1 1 `(Im) > 2 `(I) for all m, which implies that `(Em) 6→ 0). So now we have `(Im) ≥ 2 Sm. Let cm be the center of Im. So 1 1 |x − c | < `(I ) + `(I) < `(I ) + 2`(I ). m 2 m 2 n m S∞ Hence x must belong to Iem. Therefore x ∈ k=n+1 Iek. This implies

n ∞ [ [ E \ Ik ⊂ Iek. k=1 k=n+1 This now proves the above claim. 3 S∞ P∞ So given > 0, since {Ik} are disjoint, m( k=1 Ik) = k=1 `(Ik) < m(O). Hence, we can pick P∞ n such that k=n+1 `(Ik) < 5 . Then n ! ∞ ! ∞ ∗ [ ∗ [ X m E \ Ik ≤ m Iek ≤ `(Iek) < . k=1 k=n+1 k=n+1

Observe: if f :[a, b] → R is continuous and diﬀerentiable on (a, b), by the Mean Value Theorem, if f 0(x) > α > 0 on (c, d) ⊂ (a, b), then f(d) − f(c) ≥ α(d − c). (∗)

Exercise.1 Let f :[a, b] → R be a continuous function which is diﬀerentiable on (a, b). Let E = {x ∈ (a, b): f 0(x) > α} and let F = {[c, d] ⊂ (a, b) : (*) holds}. Then the collection F is a Vitali Cover for E.

Definition. Let f :[a, b] → R. The upper derivative of f at c ∈ (a, b) is the value (possibly +∞) defined by " # f(x + t) − f(x) Df(c) = lim sup . h→0 0<|t|≤h t Similarly, the lower derivative of f at c ∈ (a, b) is f(x + t) − f(x) Df(c) = lim inf . h→0 0<|t|≤h t If Df(c) = Df(c) and is < ∞, then we say that f is differentiable at c, and call this common value as the derivative of f at c, which is denoted by f 0(c). Exercise.2

(i) Let f :[−1, 1] → R be given as   x if 0 ≤ x ∈ Q f(x) = 0 if 0 ≤ x ∈ R \ Q  x 2 if x < 0. Find Df(0) and Df(0). (ii) Construct f :[a, b] → R continuous such that Df(c) 6= Df(c), for some c ∈ (a, b). (iii) Construct f :[a, b] → R such that Df(c) = ∞ whereas Df(c) = 0, for some c ∈ (a, b).

Fact.1 Let f :[a, b] → R be increasing. Then for all α > 0 1 (a) m∗({x ∈ (a, b): Df(x) ≥ α}) ≤ [f(b) − f(a)] and α (b) m∗({x ∈ (a, b): Df(x) = +∞}) = 0.

Proof. (a) Let α, > 0 be arbitrary. Let

Eα = {x ∈ (a, b): Df(x) ≥ α}. Choose 0 < β < α, and let F = {[c, d] ⊂ (a, b): f(d) − f(c) ≥ β(d − c)}. Then F is a Vitali Cover for Eα. Then by the Vitali Covering Lemma, " n # " n #! n ! ∗ ∗ [ [ ∗ [ m (Eα) = m Eα \ [ck, dk] ∪ [ck, dk] < + m [ck, dk] k=1 k=1 k=1 4

n for some disjoint sub-collection {[ck, dk]}k=1 ⊂ F. Hence n ! n n [ X 1 X + m∗ [c , d ] ≤ + (d − c ) < + [f(d ) − f(c )] k k k k β k k k=1 k=1 k=1 1 1 ≤ + [f(b) − f(a)] ⇒ m∗(E ) < [f(b) − f(a)]. β α α ∞ ∞ For E = {x ∈ (a, b): Df(x) = +∞}, E ⊂ EN for any N ≥ 1. Therefore, 1 m∗(E∞) ≤ m∗(E ) ≤ [f(d) − f(c)] −→ 0 as N → ∞, implying (b). N N

Corollary. If f is monotonic and diﬀerentiable, then f 0 < ∞ almost everywhere.

Question: Which functions are diﬀerentiable? (In particular, are monotonic functions diﬀeren- tiable?)

Theorem.2 (Lebesgue Diﬀerentiation Theorem) Let f :(a, b) → R be monotonic. Then f is diﬀerentiable almost everywhere on (a, b). Proof. We will assume that f is increasing on (a, b). Let

Eα,β = {x ∈ (a, b): Df(x) > β > α > Df(x)}. Note that f fails to be diﬀerentiable on points at which Df(x) < Df(x). Then the family

{Eα,β}α,β∈Q is a countable collection of sets on which f is not differentiable. Hence, if [ E = Eα,β, α,β∈Q α<β ∗ ∗ we need to show that m (E) = 0, which will be the case if we show that m (Eα,β) = 0 for each α and β. So, fix α < β and let F = Eα,β. Given > 0, pick O open such that F ⊂ O ⊂ (a, b) and m(O) < m∗(F ) + . Let F = {[c, d]: f(d) − f(c) < α(d − c)}. Then F is a Vitali Cover for F by Exercise.1. Hence by the Vitali Covering Lemma there exists n a finite collection {[ck, dk]}k=1 ⊂ F such that n ! ∗ [ m F \ [ck, dk] < . k=1 Then n n n ! X X [ [f(dk) − f(ck)] < α (dk − ck) = αm [ck, dk] k=1 k=1 k=1 ≤ αm(O) < α(m∗(F ) + ).

Now focus on the interval (ck, dk). f|(ck,dk) is increasing. By Fact.1, 1 m∗(F ∩ (c , d )) < [f(d ) − f(c )]. k k β k k Thus n X ∗ ∗ β m (F ∩ (ck, dk)) < α(m (F ) + ). k=1 5

Now we have n !! n ! ∗ ∗ [ ∗ [ m (F ) ≤ m F ∩ [ck, dk] + m F \ [ck, dk] k=1 k=1 n X α ≤ m∗(F ∩ (c , d )) + ≤ (m∗(F ) + ). k k β k=1 Hence ! ∗ α ∗ α ∗ α 1 ∗ m (F ) ≤ m (F ) + ⇒ m (F ) < α ⇒ m (F ) < c β β β 1 − β α ∗ where c = β−α . Since is arbitrary, it follows that m (F ) = 0.

Corollary. Let f :[a, b] → R be increasing. Then f 0 is integrable over [a.b] with Z f 0 ≤ f(b) − f(a). [a,b]

Remark. The inequality above can be strict.

Example. Let f : [0, 1] → R be the Cantor-Lebesgue function. Then f is diﬀerentiable almost everywhere on [0, 1] with f 0 = 0 almost everywhere. However, f(1) − f(0) = 1 and Z 0 f = 0 f(1) − f(0). [a,b]

Proof (of Corollary). Deﬁne, for n ≥ 1, a sequence of functions gn :[a, b] → R by 1 f(x + n ) − f(x) gn(x) = 1 . n

Then each gn is a measurable function. Furthermore, since f is diﬀerentiable almost everywhere, 0 0 gn → f a.e. on [a, b]. Hence f is measurable (we will assume that f(x) = f(b) for x ≥ b). Then by Fatou’s Lemma Z Z 0 f ≤ lim inf gn. [a,b] [a,b] Now Z Z Z "Z Z # gn = n f(x + 1/n) − f(x) = n f(x) − f(x) 1 1 [a,b] [a,b] [a,b] [a+ n ,b+ n ] [a,b] " # " # Z Z 1 Z = n f(x) − f(x) = n f(b) − f(x) 1 1 n 1 [b,b+ n ] [a,a+ n ] [a,a+ n ] 1 1 ≤ n f(b) − f(a) = f(b) − f(a), since f(a) ≤ f(x), ∀x ≥ a. n n

Remark. If f :(a, b) → R is increasing, then, following the same line of proof as above, we have Z f 0 ≤ sup f(x) − inf f(x). (a,b) x∈(a,b) x∈(a,b)

Exercises 6

1. Let E ⊂ R be a bounded set (say E ⊂ [a, b]). If V = {[α, β]: a ≤ α < β ≤ b, α, β ∈ Q}, show that V is a Vitali cover for E.

2. Let f :[a, b] → R be a continuous function which is diﬀerentiable on (a, b), and let E = {x ∈ (a, b): f 0(x) > α}. If V = {[c, d] ⊂ (a, b): f(d) − f(c) ≥ α(d − c)}, show that V is a Vitali cover for E.

3. a) Find a function f :[a, b] → R such that Df(c) = ∞, and Df(c) = 0 for some c ∈ (a, b). b) Determine Df(0) and Df(0) for x sin( 1 ) if x 6= 0, f(x) = x 0 if x = 0.

c) Determine Df(x) and Df(x) for the function f : [0, 1] → R given by f(x) = χQ(x).

4. Consider functions f, g :[−1, 1] → R given by x2 cos( 1 ) if x 6= 0, f(x) = x2 0 if x = 0,

x2 cos( 1 ) if x 6= 0, g(x) = x 0 if x = 0.

Find Df(0),Df(0), Dg(0) and Dg(0).

IV.2. Functions of Bounded Variation and Absolute Continuity

In this section and the next, we will improve the assertion of the Corollary above. Since any monotone function is measurable, our aim is to generalize Lebesgue Diﬀerentiation Theorem (and hence the Corollary above) to largest possible class of measurable functions.

n Deﬁnition. Let f :[a, b] → R be a function. Let P = {[xi, xi+1]}i=1 be a partition of [a, b]. Then the quantity n b X Va (f, P) := |f(xi) − f(xi+1)| i=1 is called as the variation of f over [a, b] with respect to P. The quantity b b Ta (f) := sup{Va (f, P): P is a partition of [a, b]} b is called the total variation of f. If Ta (f) < ∞, then f is called a function of bounded variation.

Example. (1) Any increasing (decreasing) function is of bounded variation.

(2) Lipschitz continuous functions (∃c > 0, such that |f(x) − f(y)| ≤ c|x − y|, ∀x, ∈ [a, b]) are of bounded variation. (3) Let f be deﬁned as x sin( 1 ) if x ∈ (0, 1] f(x) = x 0 if x = 0. Then f is not of bounded variation.

Fact.2 Let f :[a, b] → R. Then 7

b b 0 0 (a) Va (f, P) ≤ Va (f, P ) where P is a reﬁnement of P. b c b (b) For all a ≤ c ≤ b, Ta (f) = Ta (f) + Tc (f). x (c) The function x 7→ Ta (f) is a non-decreasing function. x (d) The function x 7→ f(x) + Ta (f) is a non-decreasing function.

y Proof. We will leave (a) and (b) as an exercise. (c) Assume x < y. Then we have Ta (f) = x y y y x Ta (f) + Tc (f) by (b). Thus, by the deﬁnition, Tx (f) ≥ 0. Hence Ta (f) ≥ Ta (f). (d) Assume that x < y. Then

y x y x [f(y) + Ta (f)] − [f(x) + Ta (f)] = [f(y) − f(x)] + [Ta (f) − Ta (f)] y = [f(y) − f(x)] + Tx (f). y b We know this is nonnegative since Tx (f) ≥ |f(y) − f(x)| = Va (f, P) where P = {[x, y]}.

Theorem.3 (Jordan Decomposition of a function) A function f :[a, b] → R is of bounded variation if and only if f = g − h, where g and h are increasing on [a, b].

x x x x Proof. (⇒) Write f(x) = f(x)+Ta (f)−Ta (f). Then, letting g = f(x)+Ta (f) and h = Ta (f), the assertion follows from Fact.2(c) and (d).

(⇐) Let P = {a = x0 < x1 < ··· < xn−1 < xn = b} be a partition of [a, b]. Then n n b X X Va (f, P) = |f(xk) − f(xk−1)| = |[g(xk) − g(xk−1)] − [h(xk) − h(xk−1)]| k=1 k=1 n n X X ≤ |g(xk) − g(xk−1)| + |h(xk) − h(xk−1)| k=1 k=1 = [g(b) − g(a)] + [h(b) − h(a)] < ∞.

Corollary. Any function of bounded variation is diﬀerentiable almost everywhere (and f 0 is integrable).

Proof. By Theorem.3, write f = g − h, where g, h :[a, b] → R are monotonic. Apply Lebesgue Diﬀerentiation Theorem. Question: Is it true that Z f 0 = f(b) − f(a)? [a,b] Note that the Cantor-Lebesgue function is of bounded variation by the previous theorem. Also ϕ0 = 0 almost everywhere. Hence, the Cantor-Lebesgue function, although is of bounded variation, fails to satisfy the Fundamental Theorem of Calculus since Z ϕ(1) − ϕ(0) = 1 6= 0 = f 0. [0,1]

Deﬁnition. A function f :[a, b]:→ R is called absolutely continuous on [a, b] if for all > 0 n there exists δ > 0 such that for any ﬁnite disjoint collection {[ak, bk]}k=1 of subintervals of [a, b] n n X X |bk − ak| < δ ⇒ |f(bk) − f(ak)| < . k=1 k=1

Remark. The Cantor-Lebesgue function is not absolutely continuous. 8

Examples. 1. Lipschitz continuous functions are absolutely continuous on any compact interval. √ 2. x is absolutely continuous on [0, 1].

Question: Are absolutely continuous functions diﬀerentiable? Answer YES!!

Fact.3 if f :[a, b] → R is absolutely continuous, then its of bounded variation (hence diﬀeren- tiable almost everywhere).

Proof. Let f :[a, b] → R be absolutely continuous, let = 1 and δ > 0 be the appropriate real + b−a number satisfying the deﬁnition. Pick N ∈ Z such that N < δ. Now let

a = x0 < x1 < ··· < xn−1 < xn = b be such that xk − xk−1 < δ, 1 ≤ k ≤ n.

xk+1 xk+1 Then take Txk (f) = sup Vxk (f, P). Now, for any partition P of [xk, xk+1], say P = {xk = t0 < t1 < ··· < tm−1 < tm = xk+1}, m X V xk+1 (f, P) = |f(t ) − f(t )|. xk i i−1 i=1 P Since |ti − ti−1| ≤ |xk − xk−1| < δ and since f is absolutely continuous, , X |f(ti) − f(ti−1)| < 1 = . i=1

n−1 x X So T k+1 (f) ≤ 1. Therefore, T b(f) = T xk+1 (f) = n < ∞. xk a xk k=0

Theorem.4 Let f :[a, b] → R be absolutely continuous. If f 0(x) = 0 almost everywhere on [a, b], then f is constant. Proof. It is enough to prove that f(t) = f(a) for all t ∈ [a, b]. So given > 0, let δ > 0 be chosen to satisfy the absolute continuity of f. Let

E = {x ∈ (a, t): f 0(x) = 0}.

Then E ∈ F and m(E) = t − a. Given any x ∈ E, we can ﬁnd hx > 0 such that h |f(x + h ) − f(x)| < x x t − a 0 (since f (x) = 0). Then the collection of intervals of the form [x, x + hx] forms a Vitali Cover. n So, by Vitali Covering Lemma, there exists a ﬁnite disjoint sub-collection, say {[xi, xi+1]}i=1 such that n ! [ m E \ [xi, xi + hxi ] < δ. i=1 Sn Hence, letting F = E \ i=1[xi, xi + hi] and let a = x0, t = xn+1, and h0 = 0, hi = hxi , we have

F = [a, x1] ∪ [x1 + h1, x2] ∪ [x2 + h2, x3] ∪ · · · ∪ [xn−1 + hn−1, xn] ∪ [xn + hn, t]. 9

|f(t) − f(a)| = |f(a) + f(x1) − f(x1) + f(x1 + h1) − f(x1 + h1) − f(x2) + f(x2) + ...

Exercises 1. Consider function x sin( 1 ) if x 6= 0, f(x) = x 0 if x = 0.

a) Determine if f a function of bounded variation on [0, 1]. b) Show that f uniformly continuous on on [0, 1].

2. Consider functions f, g :[−1, 1] → R given by x2 cos( 1 ) if x 6= 0, f(x) = x2 0 if x = 0,

x2 cos( 1 ) if x 6= 0, g(x) = x 0 if x = 0. Determine if f and g are of bounded variation on [−1, 1].

3. Let f, g : [0, 1] → R be given by x2 sin( 1 ) if x 6= 0, f(x) = x 0 if x = 0,

x sin( 1 ) if x 6= 0, g(x) = x2 0 if x = 0. Determine if f and g are absolutely continuous.

4. For any E ⊂ R, let BV (E) denote the collection of all R-valued functions of bounded variation on E. Show that if f, g ∈ BV (E) and α ∈ R, then f +g, αf ∈ BV (E). Is fg ∈ BV (E)? (Prove or provide a counterexample.) 10

5. Let f ∈ BV ([a, b]). Show that there exists a countable family {Pn}n≥1 of partitions of [a, b] such that b b lim Va (f, Pn) = Ta (f). n

6. Show that for any Lebesgue integrable function f :[a, b] → R, the function g(x) = R [a,x) f(y)dy is absolutely continuous.

7. Let f, g : [0, 1] → R be functions be given by x2| sin( 1 )| if x > 0 √ f(x) = x g(x) = x. 0 if x = 0, Show that

a) f and g are absolutely continuous on [0, 1]. b) f ◦ g is absolutely continuous on [0, 1]. c) g ◦ f is not absolutely continuous on [0, 1].

IV.3. Fundamental Theorem for Lebesgue Integration

Definition. Let f :[a, b] → R be integrable. The indefinite integral of f is defined as Z F (x) = f + c, x ∈ [a, b]. [a,x]

Recall: If f is Riemann integrable with F 0(x) = f(x), then Z x f dx = [F (x) − F (a)]. a Question: How nice is F (x) if f is Lebesgue integrable?

Fact.4 Let f :[a, b] → R be integrable. Then the indeﬁnite integral F of f is uniformly continuous on [a, b] and is of bounded variation (hence diﬀerentiable almost everywhere). R Proof. Observe if y > x, then |F (y) − F (x)| ≤ [x,y] |f|. Since f is integrable, given > 0, there exists δ > 0 such that whenever A ∈ F with m(A) < δ, then Z |f| < . A R So pick x and y such that |x − y| < δ. Then m([x, y]) < δ. So [x,y] |f| < and F is uniformly continuous.

Let a = x0 < x1 < ··· < xn−1 < xn = b be a partition of [a, b], call it P. Then n n X X Z Z |F (xi) − F (xi−1)| ≤ |f| = |f| < ∞. i=1 i=1 [xi−1,xi] [a,b] b b So Va (f, P) < ∞. Therefore Ta (F ) < ∞; and hence, F is of bounded variation. R Fact.5 Let f :[a, b] → R be integrable. If [a,x] f = 0 for all x ∈ [a, b], then f = 0 almost everywhere. 11 R Proof. Observe that the hypothesis states that E f = 0 for any open or closed E ⊂ [a, b]. Assume that assertion is false. That is there exists F ∈ F, F ⊂ [a, b] such that m(F ) > 0 and f > 0 (or f < 0) on F . Then there exists an open set O ⊃ [a, b] \ F with m(O) > b − a. Then [a, b] \ O is closed and [a, b] \ O ⊂ F with m([a, b] \ O) > 0. Further f > 0 on [a, b] \ O. Hence R [a,b]\O f > 0. This is a contradiction. R 0 Theorem.5 Let f :[a, b] → R be integrable and F (x) = F (a) + [a,x] f. Then F = f almost everywhere on [a, b]. Proof. Assume that f is bounded, say |f| ≤ M. F is of bounded variation (Fact.4); and hence, 0 diﬀerentiable a.e., implying that F (x) exists. Deﬁne gn :[a, b] → R by 1 F (x + n ) − F (x) gn(x) = 1 . n Then

Z Z |gn(x)| = n|F (x + 1/n) − F (x)| = n f ≤ n |f| ≤ M. 1 1 [x,x+ n ] [x,x+ n ] 0 Also gn(x) → F (x) almost everywhere. Hence the Bounded Convergence Theorem now implies Z Z Z 0 lim gn = F = F (b) − F (a) = f, n→∞ [a,b] [a,b] [a,b] where the the last equality follows from the deﬁnition of F and the second equality follows from R 0 the fact that F is Riemann-integrable (since it is continuous by Fact.4). Therefore, [a,b](F − f) = 0; and hence, by Fact.5, F 0 = f almost everywhere.

+ When f is an arbitrary integrable function, do the same for fm = f ∧ M, for any m ∈ Z .

Theorem.6 (F.T.C. for Lebesgue Integral) Let f :[a, b] → R be integrable. The function f is absolutely continuous on [a, b] if and only if f 0 is integrable where Z f(x) = f(a) + f 0 ∀x ∈ [a, b]. [a,x]

Proof. (⇐) Since f 0 is integrable, by Corollary.2 of DCT, given > 0, there exists δ > 0 such R 0 n that whenever A ∈ F with m(A) < δ, we have A |f |dm < . Let {[xi, xi+1]}i=0 be a collection R 0 of sub-intervals of [a, b]. Then, from the assumption we have f(y) − f(x) = [x,y] f for any x, y ∈ [a, b]; and hence, n n Z n Z Z X X 0 X 0 0 |f(xi+1) − f(xi)| = f ≤ |f | ≤ |f |,

i=0 i=0 [xi,xi+1] i=0 [xi,xi+1] E Sn n where E = i=0[xi, xi+1]. Therefore, for any collection {[xi, xi+1]}i=0 such that n X |xi+1 − xi| < δ, i=0 we have n Z X 0 |f(xi+1) − f(xi)| ≤ |f | < . i=0 E Hence, f must be absolutely continuous. 12

(⇒) Assume that f is absolutely continuous. Then, by the Corollary of Theorem.3, f is diﬀerentiable and f 0 is integrable. So let Z g(x) = f 0. [a,x] Then f 0 = g0 almost everywhere by Theorem.5. Now look at the function h := f − g. Note that h is absolutely continuous and h0 = 0 almost everywhere. Thus h must be constant by Theorem.4. So Z Z 0 0 f(x) = h(x) + g(x) = h(a) + g(x) = h(a) + f = f(a) + f . [a,x] [a,x]

Corollary.1 (Lebesgue Decomposition of a function) Let f :[a, b] → R be of bounded variation. Then there exists functions of bounded variation g, h :[a, b] → R such that g is absolutely continuous on [a, b], h0 = 0 almost everywhere on [a, b] and f = g − h almost everywhere.

R 0 Proof. Set g(x) = [a,x] f and h = f − g.

Corollary.2 Every function of bounded variation is the indeﬁnite integral of its derivative. Proof. Exercise. Question. Can we claim that FTC holds in the setting of Corollary.1 and Corollary.2?

R 0 b Exercise. Show that f :[a, b] → R is absolutely continuous if and only if [a,b] |f | = Ta (f).

IV.4 Signed Measures and Radon-Nikod´ymTheorem

We will let (X, A) be a measurable space and (X, A, µ) be a measure space. Recall that if f : X → R+ is (µ−) measurable and f ≥ 0, then the set function Z ν = f dµ, E E ∈ A, deﬁnes a measure on A. Question: What if non-negativity of f is dropped? Then f = f + − f − where f + and f − are non-negative. So Z Z Z ν(E) = f dµ = f + dµ − f − dµ. E E E Note that now ν is not a measure anymore; however, it is still a nice set function since:

• ν(∅) = 0 • ν(E) is well deﬁned (as an extended real number). • If {En}n ⊂ A is a disjoint collection, then ∞ ! Z Z Z [ + − ν En = f = f − f S S S n=1 n En n En n En ∞ Z ∞ Z ∞ Z ∞ X + X − X X = f − f = f = ν(En). n=1 En n=1 En n=1 En n=1

Deﬁnition. Let (X, A) be a measurable space. A set function ν : A → R# such that 13

• ν(∅) = 0 • ν takes at most one of either +∞ or −∞ as a value. • If {En} ⊂ A is a disjoint collection of measurable sets, then

∞ ! ∞ [ X ν En = ν(En) n=1 n=1 S and the series is absolutely convergent if ν( n En) < ∞, is called a signed measure (on A or X). Question: Can we characterize signed measures in terms of simple (ordinary) measures?

Deﬁnition. Let (X, A) be a measurable space and ν a signed measure on A. A set A ∈ A is called ν-positive (ν-negative, ν-null) if for every measurable E ⊂ A, ν(E) ≥ 0 (ν(E) ≤ 0, ν(E) = 0, resp.). Remark. (1) If A ⊂ B and |ν(B)| < ∞, then |ν(A)| < ∞.

(2) Every ν-null set has ν-measure zero; however a set with ν-measure zero may not be ν-null. (Exercise: Find an example of such a set.)

Fact.6 Let ν be a signed measure on (X, A). Then

(a) Any measurable subset of a ν-positive set is ν-positive. (b) Any countable union of ν-positive sets is a ν-positive set.

Proof. Exercise.

Fact.7 (Continuity of signed measures) Let ν be a signed measure on (X, A) and {Ak}k ⊂ A be a collection of subsets of X.

∞ (a) If Ak ⊂ Ak+1 for all k ≥ 1, then ν(∪1 Ak) = limk ν(Ak). ∞ (b) If Ak ⊃ Ak+1 for all k ≥ 1, then ν(∩1 Ak) = limk ν(Ak).

Proof. Exercise. [Hint: Mimic the proof of Theorem.3 in Chapter II.)

Next, we will show that given a measurable space (X, A) and a signed measure on it, one can decompose X into a disjoint union of a positive and a negative set w.r.t. the signed measure. For, we need the following statement. Lemma. If ν is a signed measure on (X, A) and E ∈ A such that 0 < ν(E) < ∞, then there exists a measurable set A ⊂ E which is positive with ν(A) > 0. Proof. See p: 344 of Royden.

Theorem.7 (Hahn Decomposition) If ν is a signed measure on (X, A), then there exist a positive set P and a negative set N such that P ∪ N = X and P ∩ N = ∅. Proof. By the deﬁnition of signed measure we can assume that ν does not assume +∞ (for otherwise, consider −ν). Deﬁne λ = sup{ν(E): E is a positive set}. 14

Since ∅ is a positive set, λ ≥ 0. Let {Ak} be a countable collection of positive sets such that λ = limk ν(Ak). If P = ∪kAk, then P is a positive set with ν(P ) = λ. Since ν does not assume +∞, it d follows that λ < ∞. Next, let N = X \ P ; we will prove that N is a negative set. For, assume that N is not a negative set. Hence, ∃B ⊂ N with positive measure. Then, by Lemma above, there is a positive set A ⊂ B with positive measure. It follows that P ∪ A is a positive set and ν(P ∪ A) = ν(P ) + ν(A) > λ, which is a contradiction.

Remark. Hahn decomposition is not unique; however, it is unique up to a set of measure zero.

Observe that, given a measure space (X, A, µ), one can ﬁnd functions f, g : X → R with R A = supp(f),B = supp(g), where A ∩ B = ∅, and A ∪ B = X. Hence, if ν1(.) = fdµ and R ν2(.) = gdµ, then A is ν2-null and B is ν1-null; i.e., ν1 and ν2 live on disjoint sets. More generally,

Deﬁnition. Let (X, A) be a measurable space. The signed measures ν1 and ν2 on A are called singular (or ν1 (ν2) is singular with respect to ν2 (ν1)) if there exists E1,E2 ∈ A such that E1 ∩ E2 = ∅,E1 ∪ E2 = X and Ei is νi-null, i = 1, 2.

Notation. ν1 ⊥ ν2.

Theorem.8 (Jordan Decomposition of a measure) If ν is a signed measure on a measurable space (X, A), then there exist unique positive measures ν+ and ν−such that ν = ν+ − ν− and ν+ ⊥ ν−.

Proof. The existence follows from Hahn decomposition by letting ν+(E) = ν(E ∩ P ) and ν−(E) = −ν(E ∩ N∩) for any E ∈ A. Then ν = ν+ − ν− and ν+ ⊥ ν−. For uniqueness, if ν = µ+ − µ− and µ+ ⊥ µ− is another decomposition, then pick A, B ∈ A with A ∩ B = ∅,A ∪ B = X and µ+(B) = 0, µ−(A) = 0. Then this gives another Hahn decomposition for ν, and consequently, P 4A will be ν-null. Hence, for any E ∈ A, µ+(E) = µ+(E ∩ A) = ν(E ∩ A) = ν(E ∩ P ) = ν+(E), + + − − i.e., µ = ν . Similarly, µ = ν . Observe the similarity between the decomposition in Theorem.8 and the Jordan decomposition of a function of bounded variation (into difference of two monotonic functions). Hence, the following definition is appropriate. Definition. The measures ν+ and ν− are called the positive and negative variations of ν; ν = ν+ − ν− is called the Jordan decomposition of ν, and for the measure |ν| defined by |ν| = ν+ + ν−, kνk := |ν|(X) is called the total variation of ν. R Example. Let f : R → R be Lebesgue integrable function, and let ν(E) = E fdm for any E ∈ F. Then ν is a signed measure on (R, F). Let P = {x ∈ R : f(x) ≥ 0},N = {x ∈ R : f(x) < 0} and define, for E ∈ F, Z Z ν+(E) = fdm and ν−(E) = − fdm. P ∩E N∩E Then {P,N} is a Hahn decomposition of R w.r.t. ν, and ν = ν+ − ν− is Jordan decomposition of ν.

Pn n Exercises. 1) |ν|(X) = sup{ k=1 |ν(Ek)| : {Ek}1 ⊂ A disjoint collection}. 15

2) E ∈ A is ν-null if and only if |ν|(E) = 0. 3) ν ⊥ µ if and only if |ν| ⊥ µ if and only if ν+ ⊥ µ and ν− ⊥ µ. 4) If ν omits the value ∞, then ν+(X) = ν(P ); hence ν+ is a ﬁnite measure and ν is bounded above by ν+(X). The same is also valid if ν omits the value −∞ (with appropriate changes).

Let f = χP − χN , where X = P ∪ N is the Hahn decomposition for ν, and let µ = |ν|, then R ν(E) = E fdµ for any E ∈ A. Generalizing this idea, one deﬁnes integration w.r.t. a signed measure ν by Z Z Z fdν = fdν+ − fdν− for any ν+ and ν−-integrable function f.

In regards to differentiation w.r.t. a signed measure, we will follow similar steps as in the differentiation w.r.t. a measure we did above. Definition. Let ν be a signed measure and µ be a measure on A. ν is called absolutely continuous w.r.t. µ, denoted by ν << µ, if ν(E) = 0 for all E ∈ A with µ(E) = 0.

Example. Let µ be a measure on a measurable space (X, A) and f be an integrable function. R Then the signed measure ν deﬁned by ν(E) = E fdµ, for all E ∈ A, is absolutely continuous w.r.t. µ.

Exercises. 1. ν << µ if and only if |ν| << µ if and only if ν+ << µ and ν− << µ.

2. If λ1 ⊥ µ and λ2 ⊥ µ, then λ1 + λ2 ⊥ µ.

3. If ρ1 << µ and ρ2 << µ, then ρ1 + ρ2 << µ.

Theorem.9 Let ν be a finite signed measure and µ be a measure on (X, A). Then ν << µ if and only if for every > 0 there exists δ > 0 such that |ν(E)| < whenever µ(E) < δ. Proof. By the exercise above, it is enough to prove the statement for ν = |ν| is positive. The direction (⇐) is obvious from the definitions. So, we’ll prove the converse by contrapositive. For, if −δ condition is not satisfied, then there exists an > 0 such that for all n ∈ N ∃En ∈ A with 1 ∞ ∞ P −n µ(En) < 2n and ν(En) ≥ . Letting Fk = ∪n=k and F = ∩k=1, we have µ(Fk) < n≥k 2 = 1−k 2 ; hence, µ(F ) = 0. On the other hand, ν(Fk) ≥ for all k ≥ 1; hence, ν(F ) = limk ν(Fk) ≥ . But this contradicts with the fact that ν << µ.

Remarks. 1. Compare this theorem with Theorem.12 in Section III.3. 2. Finiteness of ν is essential. For otherwise, if X = (0, 1), A = F, µ = m, and ν is deﬁned R 1 by ν(A) = A x dm, then the assertion of Theorem.9 does not hold. Another counterexample P 1 is given by X = N, A = P(N), µ is deﬁned by µ(E) = n∈E 2n , ∀E ⊂ N, and ν =counting measure. R Notation. If ν(E) = E fdµ, then this relationship can also be expressed by dν = fdµ.

Lemma. Let ν and µ be ﬁnite measures on (X, A). Then, either ν ⊥ µ, or there exists > 0 and E ∈ A such that µ(E) > 0 and E is a positive set for the measure ν − µ. Proof. Exercise. [Hint: Use Hahn decomposition for ν − n−1µ.] 16

Theorem.10 (Radon-Nikodým) Let (X, A, µ) be a finite measure space and ν : A → R be a finite measure with ν << µ. Then there exists a non-negative A-measurable function, unique µ-a.e., such that Z ν(E) = fdµ, ∀E ∈ A. E

Proof. (Uniqueness) If there exists non-negative A-measurable functions f1 and f2 such that R R ν(E) = E fidµ, ∀E ∈ A, i = 1, 2, then 0 = E(f1 − f2)dµ ∀E ∈ A. Hence, f1 = f2 µ-a.e. on X. (Existence) If ν(E) = 0 ∀E ∈ A, the assertion holds with f ≡ 0. Hence, assume that ν does not vanish on all of A. Define Z G = {g : X → [0, ∞]: g ≤ ν(E), ∀E ∈ A}. E Since 0 ∈ G, we see that G= 6 ∅. Also, if h, g ∈ G, then by letting A = {x : h(x) > g(x)}, for any E ∈ A, it follows that Z Z Z (h ∨ g)dµ = hdµ + gdµ ≤ ν(E ∩ A) + ν(E \ A) = ν(E) E E∩A E\A R implying that h ∨ g ∈ G. Now, define α = supg∈G gdµ. Since α ≤ ν(X) < ∞, we can pick R {gn} ⊂ G such that gn → α. Let fn = max{g1, . . . , gn} , then fn ∈ G for all n and fn ↑ f R R pointwise, for some non-negative A-measurable function f. Since fndµ ≥ gndµ, it follows R R that fn → α; and hence, by MCT, f ∈ G with fdµ = α (in particular, f < ∞ µ-a.e.). R Next, define a new measure η by η(E) = ν(E) − E fdµ for all E ∈ A. The, by the Lemma above, we have two cases: (i) η ⊥ µ, or (ii) ∃ > 0 & A ∈ A such that (η − µ)(A) ≥ 0 & µ(A) > 0. In the first case, since η << µ by construction and ν << µ, it follows that η ≡ 0. Hence, R ν(E) = E fdµ for all E ∈ A, proving the assertion. R R In the second case, we have ν(A) − A fdµ − µ(A) ≥ 0. Thus, ν(A) ≥ A fdµ + µ(A) = R ˆ ˆ ˆ R ˆ R A(f + f)dµ, where f = χA. Since µ(A) > 0, we have f 6= 0 and A fdµ > 0. Thus, X (f + ˆ R f)dµ > X fdµ = α, contradiction. Hence (ii) is not possible, proving the theorem. Corollary. a) Theorem.10 is valid if both µ and ν are σ-finite. b) Theorem.10 is valid if and ν is a σ-finite signed measure and µ is a σ-finite measure.

Proof. a) Let X = ∪iXi and deﬁne µi(E) = µ(Xi ∩ E,E ∈ A, and νi(E) = ν(Xi ∩ E,E ∈ A. P Apply Radon-Nikod´ymTheorem to {µi, νi} to obtain fi. Let f = i fi. b) Let ν = ν+ − ν− and apply (a) to {µ, ν+} and {µ, ν−} to obtain f + and f −, respectively. Let f = f + − f −.

dν Notation. The function f in Radon-Nikod´ymTheorem and its Corollary is denoted by dµ and is called as the Radon-Nikod´ymderivative of ν with respect to µ.

Theorem.11 (Lebesgue Decomposition of a measure) Let ν be a σ-finite signed measure and µ be a σ-finite measure on a measurable space (X, A). Then there exists µ-a.e. unique σ-finite signed measures ν0 and ν1 on A such that ν0 ⊥ µ, ν1 << µ and ν = ν0 + ν1. Proof. WLOG we can assume that both ν and µ are finite. First, let ν be a measure, and let λ = µ + ν, then µ << λ. Hence, by Radon-NikodýmTheorem, there exists a λ-a.e. unique 17 R A-measurable function f such that µ(E) = E fdλ for all E ∈ A. Observe that f is also µ-a.e. and ν-a.e. A-measurable function. Thus, Z Z µ(E) = fdµ + fdν, ∀ E ∈ A. (∗) E E

Let X1 = {x : f(x) > 0},X0 = {x : f(x) = 0}, and deﬁne ν0 = ν|X0 and ν1 = ν|X1 . Then we have:

(i) ν0 and ν1 are measures on A. (ii) ν = ν0 + ν1.

Also, by construction, X = X0 ∪ X1,X0 ∩ X1 = ∅, and ν0(X1) = ν1(X0) = 0. Since by (*) R R µ(X0) = fdµ + fdν, it follows that ν0 ⊥ µ. X0 X0 R If A ∈ A such that µ(A) = 0, then A fdµ = 0. So, by (*), we have Z Z Z Z Z Z 0 = fdµ = fdµ + fdν = fdν = fdν + fdν. A A A A A∩X0 A∩X1 R Since f = 0 on A ∩ X0 and f > 0 on A ∩ X1, we have fdν = 0, so ν(A ∩ X1) = 0. Thus A∩X1 ν1(A) = 0, implying that ν1 << µ. If ν is a signed measure, then ν = ν+ − ν− where ν+ and ν− are measures. Then apply + − ± ± ± ± the above to ν and ν , respectively, to obtain the decomposition ν = ν0 + ν1 with ν0 ⊥ ± + − + − µ, ν1 << µ. Then, ν = ν0 + ν1, where ν0 = ν0 + ν0 and ν1 = ν1 + ν1 .

Remark. It is essential in Theorem.11 that ν and µ are σ-ﬁnite. [Exercise: Provide counterex- amples when σ-ﬁniteness in Theorem.11 is dropped.]

Exercises

1. Let µ be a probability measure on (R, B). Deﬁne a measure ν : B → R by ν(E) = 1 if 0 ∈ E and ν(E) = 0 otherwise (ν is called the point mass at 0). Find the Jordan decomposition of the signed measure λ = µ − ν.

2. Let (X, A, µ) be a measure space and ν1 and ν1 be signed measures on A. Prove the following.

a) If ν1 ⊥ µ and ν2 ⊥ µ, then ν1 + ν2 ⊥ µ. b) If ν1 << µ and ν2 << µ, then ν1 + ν2 << µ. c) If ν1 << µ, then |ν1| << µ; and conversely. d) If ν1 << µ and ν2 ⊥ µ, then ν1 ⊥ ν2. e) If ν1 << µ and ν1 ⊥ µ, then ν1 ≡ 0.

n n 3. Let (X, A, µ) be a measure space, {Ek}k=1 ⊂ A, and {ck}k=1 be a collection of real numbers. Pn dν For E ∈ A, deﬁne ν(E) = k=1 ck.µ(E ∩ Ek). Show that ν << µ and ﬁnd dµ .

4. Let {µn} be a sequence of measures on a measurable space (X, A) such that there is a constant C with µn(X) ≤ C for all n ≥ 1. Deﬁne ν : A → R by ∞ X µn(E) ν(E) = ,E ∈ A. 2n n=1

Show that ν is a measure on A and that µn << ν for each n ≥ 1. 18

5. Consider the measure space ([0, 1], F|[0,1], m) and let ν be the counting measure on F|[0,1]. Show that

a) m << ν, and R b) there is no function f : [0, 1] → R for which m(E) = E fdν for all E ∈ F|[0,1].

6. Let {an}n be a ﬁxed sequence of real numbers and {pn}n be a sequence of positive real numbers. Deﬁne a set function ν : F → R by X ν(E) = pn,E ∈ F.

an∈E a) Show that ν is a σ-ﬁnite measure on F. b) Find the Lebesgue decomposition of ν with respect to the Lebesgue measure on F.

7. Let f : R → R be given by f(x) = x2 − 6x + 5, and deﬁne a set function ν : B → R by Z ν(E) = f(x)dm, E ∈ B. E a) Show that ν is a σ-ﬁnite measure on B. b) Find the Hahn decomposition of R with respect to ν. c) Find the Jordan decomposition of ν. d) Find the Lebesgue decomposition of ν with respect to the Lebesgue measure on F.

8. Give an example in which the assertion of the Radon-Nikod`ymtheorem fails (that is, λ << µ R but there is no measurable function f such that λ(A) = A fdµ for all measurable set A). 9. Let (X, A, µ) be a measure space and g be a non-negative measurable function on X. Let a measure λ : A → R be deﬁned as Z λ(A) = gdµ, for all A ∈ A. A Prove that if f : X → R is a measurable function then Z Z fdλ = fgdµ, X X in the sense that if one integral exists, so does the other, and the two integrals are equal.

10. Consider the measure space ([0, 1], F|[0,1], m|[0,1]). Deﬁne a set function λ on all subintervals [a, b] ⊂ [0, 1] by λ([a, b]) = φ(b) − φ(a), where φ is the Cantor-Lebesgue function on [0, 1]. If µ is the measure obtained by restricting the outer measure λ∗ to λ∗-measurable subsets of [0, 1], prove that µ ⊥ m|[0,1].