216 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

each interval being contained in V ∩ (−r,r). Thus you have

m(Nrab) = m(∪iNr

ab∩ (ai,bi)) .

Next show there exist disjoint intervals (a j,b j) such that each of these is containedin some (ai,bi), the (a j,b j) are disjoint, f (b j)− f (a j)≥ bm(a j,b j) , and

∑j

m(Nrab∩ (a j,b j)) = m(Nr

ab)

. Then you have the following thanks to the fact that f is increasing.

a(m(Nrab)+ ε) > am(V )≥ a∑

i(bi−ai)> ∑

if (bi)− f (ai)

≥ ∑j

f (b j)− f (a j)≥ b∑j

b j−a j

≥ b∑j

m(Nrab∩ (a j,b j)) = bm(Nr

ab)

and since ε > 0,am(Nr

ab)≥ bm(Nrab)

showing that m(Nr

ab

)= 0. This is for any r and so m(Nab) = 0. Thus the derivative

from the right exists for a.e. x by taking the complement of the union of the Nab fora,b nonnegative rational numbers. Now do the same thing to show that the derivativefrom the left exists a.e. and finally, show that D− f (x) = D+ f (x) for almost a.e. x.Off the union of these three exceptional sets of measure zero all the derivates are thesame and so the derivative of f exists a.e. In other words, an increasing function hasa derivative a.e.

22. This problem is on Eggoroff’s theorem. This was presented earlier in the book. Theidea is for you to review this by going through a proof. Suppose you have a measurespace (Ω,F ,µ) where µ (Ω) < ∞. Also suppose that { fk} is a sequence of mea-surable, complex valued functions which converge to f pointwise. Then Eggoroff’stheorem says that for any ε > 0 there is a set N with µ (N) < ε and convergence isuniform on NC.

(a) Define Emk ≡ ∪∞r=m{

ω : | f (ω)− fr (ω)|> 1k

}. Show Emk ⊇ E(m+1)k for all m

and that ∩mEmk = /0

(b) Show that there exists m(k) such that µ(Em(k)k

)< ε2−k.

(c) Let N ≡ ∪∞k=1Em(k)k. Explain why µ (N) < ε and that for all ω /∈ NC, if r >

m(k) , then | f (ω)− fr (ω)| ≤ 1k . Thus uniform convergence takes place on NC.

23. Suppose you have a sequence { fn} which converges uniformly on each of finitelymany sets A1, · · · ,An. Why does the sequence converge uniformly on ∪n

i=1Ai?

24. ↑Now suppose you have µ is a finite Radon measure on Rp. For example, you couldhave Lebesgue measure. Suppose you have f has nonnegative real values for all xand is measurable. Then Lusin’s theorem says that for every ε > 0, there exists anopen set V with measure less than ε and a continuous function defined on Rp suchthat f (x) = g(x) for all x /∈V. That is, off an open set of small measure, the functionis equal to a continuous function.