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

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FUNDAMENTALS of REAL ANALYSIS by Do˘Gan C¸Ömez IV. DIFFERENTIATION and SIGNED MEASURES IV.1. Differentiation of Monotonic 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) n E. Convention: Throughout this chapter, [a; b] will be an interval with −∞ < a < b < 1. Definition. A collection F of closed and bounded intervals is called a Vitali cover for a set E ⊂ R if for all x 2 E, for all > 0 there exists I 2 F such that x 2 I and `(I) < . Example. Let E 2 R be a bounded set (say E ⊂ [c; d]). Then the collection F = f[α; β]: c ≤ α < β ≤ d; α; β 2 Qg is a Vitali Cover for E. Next theorem is an important one! Theorem.1 (Vitali Covering Lemma) Let E ⊂ R with m∗(E) < 1 and let F be a Vitali n cover for E. Then for all > 0 there exists a finite disjoint sub-collection fIkgk=1 ⊂ F such that n ! ∗ [ m E n Ik < . k=1 Comments: (a) One can always find an open set O such that E ⊂ O and m(O) < 1. (b) Without loss of generality, we can assume that [ X I 2 F ) I ⊂ O ) m Ik = m(Ik) ≤ m(O) < 1: | {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 fIkgk=1 ⊂ F such that for all I 2 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 2 E n k=1 Ik. Then x 62 Ik for all 1 ≤ k ≤ n. Since F is a Vitali cover, there exists I 2 F such that x 2 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 n Ik0 ) > 0. Let < 2 `(I n Ik), then there exists J 2 F such that x 2 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 fIkgk=1; there Sn exists I 2 F such that I \ ( k=1 Ik) = ;. 1 [In this case we will construct (inductively) a countably disjoint sub-collection fIkg 2 F S k=1 such that E n 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 2 F such that I2 \ I1 = ; and `(I2) > 2 S1 where S1 = `(I). Now assume we have a n disjoint collection fIkgk=1 ⊂ F constructed in this manner. Then let Sn = supf`(I): I 2 F;I \ Ik = ;; k = 1; : : : ng: 1 n+1 Pick In+1 2 F such that In+1 \ Ik = ; for all k = 1; : : : n and `(In+1) > 2 Sn. So fIkgk=1 is a 1 disjoint family. Hence, by induction we have a countably disjoint family fIkgk=1 ⊂ F such that each Ik ⊂ O for all k; where O is an open set such that E ⊂ O and m(O) < 1. Then `(Ik) ! 0 must hold. Claim: This collection has the property that for all n ≥ 1 n 1 [ [ E n Ik ⊂ Iek: k=1 k=n+1 Sn To prove this claim let n ≥ 1 and let x 2 E n k=1 Ik. Then there exists I 2 F such that x 2 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 jx − c j < `(I ) + `(I) < `(I ) + 2`(I ): m 2 m 2 n m S1 Hence x must belong to Iem. Therefore x 2 k=n+1 Iek. This implies n 1 [ [ E n Ik ⊂ Iek: k=1 k=n+1 This now proves the above claim. 3 S1 P1 So given > 0, since fIkg are disjoint, m( k=1 Ik) = k=1 `(Ik) < m(O). Hence, we can pick P1 n such that k=n+1 `(Ik) < 5 . Then n ! 1 ! 1 ∗ [ ∗ [ X m E n 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 = fx 2 (a; b): f 0(x) > αg and let F = f[c; d] ⊂ (a; b) : (*) holdsg: Then the collection F is a Vitali Cover for E. Definition. Let f :[a; b] ! R. The upper derivative of f at c 2 (a; b) is the value (possibly +1) defined by " # f(x + t) − f(x) Df(c) = lim sup : h!0 0<jt|≤h t Similarly, the lower derivative of f at c 2 (a; b) is f(x + t) − f(x) Df(c) = lim inf : h!0 0<jt|≤h t If Df(c) = Df(c) and is < 1, 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 8 < x if 0 ≤ x 2 Q f(x) = 0 if 0 ≤ x 2 R n 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 2 (a; b): (iii) Construct f :[a; b] ! R such that Df(c) = 1 whereas Df(c) = 0; for some c 2 (a; b): Fact.1 Let f :[a; b] ! R be increasing. Then for all α > 0 1 (a) m∗(fx 2 (a; b): Df(x) ≥ αg) ≤ [f(b) − f(a)] and α (b) m∗(fx 2 (a; b): Df(x) = +1g) = 0: Proof. (a) Let α; > 0 be arbitrary. Let Eα = fx 2 (a; b): Df(x) ≥ αg: Choose 0 < β < α, and let F = f[c; d] ⊂ (a; b): f(d) − f(c) ≥ β(d − c)g: Then F is a Vitali Cover for Eα. Then by the Vitali Covering Lemma, " n # " n #! n ! ∗ ∗ [ [ ∗ [ m (Eα) = m Eα n [ck; dk] [ [ck; dk] < + m [ck; dk] k=1 k=1 k=1 4 n for some disjoint sub-collection f[ck; dk]gk=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 1 1 For E = fx 2 (a; b): Df(x) = +1g, E ⊂ EN for any N ≥ 1. Therefore, 1 m∗(E1) ≤ m∗(E ) ≤ [f(d) − f(c)] −! 0 as N ! 1; implying (b): N N Corollary. If f is monotonic and differentiable, then f 0 < 1 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 = fx 2 (a; b): Df(x) > β > α > Df(x)g: Note that f fails to be differentiable on points at which Df(x) < Df(x). Then the family fEα,βgα,β2Q is a countable collection of sets on which f is not differentiable. Hence, if [ E = Eα,β; α,β2Q 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 = f[c; d]: f(d) − f(c) < α(d − c)g: Then F is a Vitali Cover for F by Exercise.1. Hence by the Vitali Covering Lemma there exists n a finite collection f[ck; dk]gk=1 ⊂ F such that n ! ∗ [ m F n [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).
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