Kenneth Kuttler

948 CHAPTER 26. INTEGRALS AND DERIVATIVES

and each [x′,x′+h′] is contained in an interval (xi−hi,xi). Then by the Vitali cover-ing theorem again, Corollary 13.4.6 there are countably many disjoint closed intervals{[

x′j,x′j +h′j

]}∞

j=1whose union includes all of Npq except for a set of measure zero such

that each of these is contained in some (xi−hi,xi) described earlier. Then as before,

qm(Npq) ≤ q∑j

h′j ≤∑j

f(x′j +h′j

)− f

(x′j)≤∑

if (xi)− f (xi−hi)

≤ p∑i

hi ≤ pm(V )≤ p(m(Npq)+ ε)

Then as before, this shows that qm(Npq) ≤ pm(Npq) and so m(Npq) = 0. Then taking theunion of all such for p,q ∈Q yields D+ f (x) = D− f (x) for a.e. x. Taking the union of allthese sets of measure zero and considering points not in this union, it follows that f ′ (x)exists for a.e. x. Thus f ′ (t)≥ 0 and is a limit of measurable even continuous functions fora.e. x so f ′ is clearly measurable. The issue is whether f (y)− f (x) =

∫X[x,y] (t) f ′ (t)dm.

Up to now, the only thing used has been that f is increasing.Let h > 0. ∫ x

f (t)− f (t−h)h

dt =1h

∫ x

af (t)dt− 1

∫ x

af (t−h)dt

=1h

∫ x

af (t)dt− 1

∫ x−h

a−hf (t)dt

=1h

∫ x

x−hf (t)dt− 1

∫ a

a−hf (t)dt

=1h

∫ x

x−hf (t)dt− f (a)

Therefore, by continuity of f it follows from Fatou’s lemma that∫ x

aD− f (t)dt =

∫ x

af ′ (t)dt ≤ lim inf

h→0+

∫ x

f (t)− f (t−h)h

dt = f (x)− f (a)

and this shows that f ′ is in L1. This part only used the fact that f is increasing and contin-uous. That f is Lipschitz has not been used.

If it were known that there is a dominating function for t→ f (t)− f (t−h)h , then you could

simply apply the dominated convergence theorem in the above inequality instead of Fatou’slemma and get the desired result. But from Lipschitz continuity, you have∣∣∣∣ f (t)− f (t−h)

∣∣∣∣≤ K

and so one can indeed apply the dominated convergence theorem and conclude that∫ x

af ′ (t)dt = f (x)− f (a)

The last claim follows right away from consideration of intervals since the restriction of aLipschitz function is Lipschitz.

948 CHAPTER 26. INTEGRALS AND DERIVATIVESand each [x’,x’+h’] is contained in an interval (x;—/;,x;). Then by the Vitali cover-ing theorem again, Corollary 13.4.6 there are countably many disjoint closed intervals{ [x + n, \ _, Whose union includes all of Ng except for a set of measure zero such, iS jethat each of these is contained in some (x; — hj,x;) described earlier. Then as before,am(Npo) <a SE (+H) —F 0) < EL) —f (hy)J J i< phi < pm(V) < p(m(Npg) +€)IAThen as before, this shows that gm(Npq) < pm(Npq) and so m(Npq) = 0. Then taking theunion of all such for p,g € Q yields D, f (x) = D7 f (x) for a.e. x. Taking the union of allthese sets of measure zero and considering points not in this union, it follows that f’ (x)exists for a.e. x. Thus f” (t) > 0 and is a limit of measurable even continuous functions fora.e. x so f” is clearly measurable. The issue is whether f(y) — f (x) = J ix (t) f’ (t)dmUp to now, the only thing used has been that f is increasing.Leth > 0.[spite = 5 froe-j[ne-n= ult i)dr—7 "P(ton= 5 “ pyatf f(ax—h1 f roan 4)Therefore, by continuity of f it follows from Fatou’s lemma that[e-t@ar= [of Oar sim inf [ FO FCW i p(y) — F(a)h-0+and this shows that f’ is in L'. This part only used the fact that f is increasing and contin-uous. That f is Lipschitz has not been used.fO=f(t-h)If it were known that there is a dominating function for + —““~+~—, then you couldsimply apply the dominated convergence theorem in the above inequality instead of Fatou’slemma and get the desired result. But from Lipschitz continuity, you havef=) <xh Ssand so one can indeed apply the dominated convergence theorem and conclude that[ foa=reo-r@The last claim follows right away from consideration of intervals since the restriction of aLipschitz function is Lipschitz. Jj