11.7. REGULAR MEASURES IN A METRIC SPACE 263

1. For each E ∈ F , there is an Fσ set F and a Gδ set G such that F ⊆ E ⊆ G andµ (G\F) = 0.

2. Also if f ≥ 0 is F measurable, then there exists g≤ f such that g is Borel measurableand g = f a.e. and h≥ f such that h is Borel measurable and h = f a.e.

3. If E ∈ F is a bounded set contained in a ball B(x0,r) = V , then there exists asequence of continuous functions in Cc (V ) {hn} having values in [0,1] and a setof measure zero N such that for x /∈ N,hn (x)→XE (x) . Also

∫|hn−XE |dµ → 0.

Letting Ñ be a Gδ set of measure zero containing N,hnXÑC →XF where F ⊆ E andµ (E \F) = 0.

4. If f ∈ L1 (X ,F ,µ) , there exists g ∈Cc (X) , such that∫

X | f −g|dµ < ε. There alsoexists a sequence of functions in Cc (X) {gn} which converges pointwise to f .

Proof: 1. Let Rn ≡ B(x0,n) ,R0 = /0. If E is Lebesgue measurable, let En ≡ E ∩(Rn \Rn−1) . Thus these En are disjoint and their union is E. By outer regularity, thereexists open Un ⊇ En such that µ (Un \En) < ε/2n. Now if U ≡ ∪nUn, it follows thatµ (U \E) ≤ ∑

∞n=1

ε

2n = ε . Let Vn be open, containing E and µ (Vn \E) < 12n ,Vn ⊇ Vn+1.

Let G ≡ ∩nVn. This is a Gδ set containing E and µ (G\E) ≤ µ (Vn \E) < 12n and so

µ (G\E)= 0. By inner regularity, there is Fn an Fσ set contained in En with µ (En \Fn)= 0.Then let F ≡ ∪nFn. This is an Fσ set and µ (E \F)≤ ∑n µ (En \Fn) = 0. Thus F ⊆ E ⊆ Gand µ (G\F)≤ µ (G\E)+µ (E \F) = 0.

2. If f is measurable and nonnegative, there is an increasing sequence of simple func-tions sn such that limn→∞ sn (x) = f (x) . Say ∑

mnk=1 cn

kXEnk(x) . Let mp

(En

k \Fnk

)= 0 where

Fnk is an Fσ set. Replace En

k with Fnk and let s̃n be the resulting simple function. Let

g(x) ≡ limn→∞ s̃n (x) . Then g is Borel measurable and g ≤ f and g = f except for a setof measure zero, the union of the sets where sn is not equal to s̃n. As to the other claim,let hn (x) ≡ ∑

∞k=1 XAkn (x)

k2n where Akn is a Gδ set containing f−1

(( k−1

2n , k2n ])

for whichµ(Akn \ f−1

(( k−1

2n , k2n ]))≡ µ (Dkn) = 0. If N = ∪k,n Dkn, then N is a set of measure zero.

On NC, hn (x)→ f (x) . Let h(x) = liminfn→∞ hn (x).3. Let Kn ⊆ E ⊆Vn with Kn compact and Vn open such that Vn ⊆ B(x0,r) and

µ (Vn \Kn)< 2−(n+1).

Then from Lemma 11.7.2, there exists 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

Letting An≡[|hn−XE |>

( 23

)n], the set of x which is in infinitely many An is N≡∩n∪k≥n

Ak and so

µ (∩n∪k≥n Ak)≤ µ (∪k≥nAk)≤∞

∑k=n

(34

)k

=

(34

)n( 11/4

)