Kenneth Kuttler

346 CHAPTER 11. REGULAR MEASURES

If the function F is not absolutely continuous, then there exists ε > 0 and open setsEn consisting of unions of finitely many non-overlapping open intervals such that if En =∪mn

i=1 (xni ,y

ni ) , then ∑

mi=1 |yn

i − xni |= m(En)< 2−n but∫

[a,b]XEn (x)Dmµ (x)dm = µ (En) =

∑i=1

µ (xni ,y

ni ) =

∑i=1

(F (yni )−F (xn

i ))≥ ε (11.22)

However, XEn (x)→ 0 a.e. because ∑n m(En) < ∞ and so, by the Borel Cantelli lemma,there is a set of measure zero N such that for x /∈ N, x is in only finitely many of the En. Inparticular, XEn (x) = 0 for all n large enough if x /∈ N. Then by the dominated convergencetheorem, the inequality 11.22 cannot be valid for all n because the limit of the integral onthe left equals 0. This is a contradiction. Hence F must be absolutely continuous after all.⇐Next suppose the function F is absolutely continuous. Suppose m(E) = 0. Does it

follow that µ (E) = 0? Let ε > 0 be given. Let δ correspond to ε/2 in the definition ofabsolute continuity. Let E ⊆ V where V is an open set such that m(V ) < δ . By Theorem3.11.8, V =∪i (ai,bi) where these open intervals are disjoint. It follows that for each n, ε

2 >

∑ni=1 F (bi)−F (ai) = µ

(∪n

i=1 (ai,bi)). Then letting n→ ∞,ε > ε

2 ≥ µ (∪∞i=1 (ai,bi)) =

µ (V )≥ µ (E). Since ε > 0 is arbitrary, it follows that µ (E) = 0 and so µ ≪ m. ■An example which shows that increasing and continuous is not enough, see Problem 5

on Page 269.

Corollary 11.14.4 F is increasing on [a,b] and absolutely continuous if and only ifF ′ (x) exists for a.e. x and F ′ is in L1 ([a,b] ,m) and for every x,y such that a≤ x≤ y≤ b

F (y)−F (x) =∫ y

xF ′ (t)dm

Proof:⇒Suppose first that F is absolutely continuous. Then by Theorem 11.13.2, forµ defined above, µ (E) =

∫E Dmµ (x)dm for all E Borel. In particular,

F (y)−F (x) =∫(x,y)

Dmµ (t)dm(t) (11.23)

Since Dmµ is in L1 ([a,b] ,m) , it follows that almost every point is a Lebesgue point and sofor such Lebesgue points x,∣∣∣∣F (x+h)−F (x)

h−Dmµ (x)

∣∣∣∣= ∣∣∣∣1h∫[x,x+h]

(Dmµ (t)−Dmµ (x))dm(t)∣∣∣∣

≤ 2∣∣∣∣ 12h

∫[x−h,x+h]

|Dmµ (t)−Dmµ (x)|dm(t)∣∣∣∣

which converges to 0 as h→ 0 since x is a Lebesgue point. Similarly, at each Lebesguepoint, limh→0

F(x)−F(x−h)h = Dmµ (x) Thus F is differentiable at each Lebesgue point and

the derivative equals Dmµ at these points. Now 11.23 yields the desired result that thefunction can be recovered from integrating its derivative.⇐Next suppose F (y)−F (x) =

∫ yx F ′ (t)dm where F ′ (t) exists a.e. and F ′ is in L1.

Then if {Ii}i are nonoverlapping intervals,∫∪iIi F ′ (t)dm = m(∪iF (Ii)) < ε if m(∪iIi) is

small enough, as an application of the dominated convergence theorem or as in the firstpart of Theorem 11.14.3. ■

The importance of the intervals being non overlapping is discussed in the followingproposition. I think it is also interesting that it implies F is Lipschitz. In this proposition,F is defined on some interval, possibly R.

346 CHAPTER 11. REGULAR MEASURESIf the function F is not absolutely continuous, then there exists € > 0 and open setsE,, consisting of unions of finitely many non-overlapping open intervals such that if E, =Une (xy), then 27", |y! —x?| = m(E,) < 2~" butI, Ke, (x) Dm (x)dm = UW (En) = ye (x7, y?) = y (F (vy?) -—F(x#}))>e (11.22)However, 2, (x) > 0 a.e. because )’,,m(E;,) < 0 and so, by the Borel Cantelli lemma,there is a set of measure zero N such that for x ¢ N, x is in only finitely many of the E,,. Inparticular, 2%, (x) = 0 for all n large enough if x ¢ N. Then by the dominated convergencetheorem, the inequality 11.22 cannot be valid for all n because the limit of the integral onthe left equals 0. This is a contradiction. Hence F must be absolutely continuous after all.<=Next suppose the function F is absolutely continuous. Suppose m(E) = 0. Does itfollow that u (EZ) = 0? Let € > 0 be given. Let 6 correspond to €/2 in the definition ofabsolute continuity. Let E C V where V is an open set such that m(V) < 6. By Theorem3.11.8, V =U; (a;,b;) where these open intervals are disjoint. It follows that for each n, 5 >Lt, F (bi) — F (ai) = w (UL, (ai,b;)). Then letting n > &,€ > § > w (UZ, (a;,b;)) =L(V) >u(E). Since € > 0 is arbitrary, it follows that u (EZ) =0 and sou <m.An example which shows that increasing and continuous is not enough, see Problem 5on Page 269.Corollary 11.14.4 F is increasing on [a,b] and absolutely continuous if and only ifF' (x) exists for a.e. x and F' is in L' (|a,b] ,m) and for every x,y such thata<x<y<bF(y)-F@)= [P'()dmProof: =Suppose first that F is absolutely continuous. Then by Theorem 11.13.2, forLt defined above, 1 (E) = J; Dm (x) dm for all E Borel. In particular,F(y)-F@)= ( Pmt (1) dm (t) (11.23)uySince Di» is in L' ([a,b] ,m) , it follows that almost every point is a Lebesgue point and sofor such Lebesgue points x,Pen Fe)PFO pywe(s)|=[F [Date t) Dot (x) dn)hl< PF]. —<7) 5h I. en Dnt) — Pm) dentwhich converges to 0 as h — 0 since x is a Lebesgue point. Similarly, at each Lebesguepoint, limp_,o Fe) Fert) = Dy, (x) Thus F is differentiable at each Lebesgue point andthe derivative equals D,,j at these points. Now 11.23 yields the desired result that thefunction can be recovered from integrating its derivative.<Next suppose F (y) — F (x) = [2 F’(t)dm where F’ (t) exists a.e. and F’ is in L!.Then if {/;}; are nonoverlapping intervals, f\,,, F’ (t)dm = m(UjF (I;)) < € if m(Ujli) issmall enough, as an application of the dominated convergence theorem or as in the firstpart of Theorem 11.14.3. HfThe importance of the intervals being non overlapping is discussed in the followingproposition. I think it is also interesting that it implies F is Lipschitz. In this proposition,F is defined on some interval, possibly R.