238 CHAPTER 10. REGULAR MEASURES
3. Let Kn ⊆ E ⊆ Vn with Kn compact and Vn open such that Vn ⊆ B(x0,r) and alsothat µ (Vn \Kn)< 2−(n+1). Then from Lemma 3.12.4, there is hn with Kn ≺ hn ≺Vn. Then∫|hn−XE |dµ < 2−n and so
µ
(|hn−XE |>
(23
)n)<
((32
)n ∫[|hn−XE |>( 2
3 )n]|hn−XE |dµ
)≤(
34
)n
By Lemma 8.2.5 there is a set of measure zero N such that if x /∈N, it is in only finitely manyof the sets
[|hn−XE |>
( 23
)n]. Thus on NC, eventually, for all k large enough, |hk−XE | ≤( 2
3
)kso hk (x)→XE (x) off N. The assertion about convergence of the integrals follows
from the dominated convergence theorem and the fact that each hn is nonnegative, boundedby 1, (Kn ≺ hn ≺ Vn) and is 0 off some ball. In the last claim, it only remains to verifythat hnXÑC converges to an indicator function because each hnXÑC is Borel measurable.(Ñ ⊇N and Ñ is a Borel set and µ
(Ñ \N
)= 0) Thus its limit will also be Borel measurable.
However, hnXÑC converges to 1 on E ∩ ÑC,0 on EC ∩ ÑC and 0 on Ñ. Thus E ∩ ÑC = Fand hnXÑC (x)→XF where F ⊆ E and µ (E \F)≤ µ
(Ñ)= 0.
4. It suffices to assume f ≥ 0 because you can consider the positive and negative partsof the real and imaginary parts of f and reduce to this case. Let fn (x)≡XB(x0,n) (x) f (x) .Then by the dominated convergence theorem, if n is large enough,
∫| f − fn|dµ < ε. There
is a nonnegative simple function s ≤ fn such that∫| fn− s|dµ < ε. This follows from
picking k large enough in an increasing sequence of simple functions {sk} converging to fnand the dominated convergence theorem. Say s(x) = ∑
mk=1 ckXEk (x) . Then let Kk ⊆ Ek ⊆
Vk where Kk,Vk are compact and open respectively and ∑mk=1 ckµ (Vk \Kk)< ε . By Lemma
3.12.4, there exists hk with Kk ≺ hk ≺Vk. Then∫ ∣∣∣∣∣ m
∑k=1
ckXEk (x)−m
∑k=1
ckhk (x)
∣∣∣∣∣dµ ≤ ∑k
ck
∫ ∣∣XEk (x)−hk (x)∣∣dx
< 2∑k
ckµ (Vk \Kk)< 2ε
Let g≡ ∑mk=1 ckhk (x) . Thus
∫|s−g|dµ ≤ 2ε. Then∫
| f −g|dµ ≤∫| f − fn|dµ +
∫| fn− s|dµ +
∫|s−g|dµ < 4ε
Since ε is arbitrary, this proves the first part of 4. For the second part, let gn ∈Cc (X) suchthat
∫| f −gn|dµ < 2−n. Let An ≡
{x : | f −gn|>
( 23
)n}. Then
µ (An)≤(
32
)n ∫An
| f −gn|dµ ≤(
34
)n
.
Thus, if N is all x in infinitely many An, then by the Borel Cantelli lemma, µ (N) = 0 andif x /∈ N, then x is in only finitely many An and so for all n large enough, | f (x)−gn (x)| ≤( 2
3
)n. ■
10.2 Fundamental Theorem of CalculusIn this section the Vitali covering theorem, Proposition 4.5.3 will be used to give a gener-alization of the fundamental theorem of calculus. Let f be in L1 (Rp) where the measure isLebesgue measure as discussed above.