152 CHAPTER 6. MEASURES AND MEASURABLE FUNCTIONS

Now consider the assertion of outer regularity. The assertion of outer regularity is nothard to get. Letting E be any set µ (E)< ∞, there exist open intervals covering E denotedby {(ai,bi)}∞

i=1 such that

µ (E)+ ε >∞

∑i=1

bi−ai =∞

∑i=1

µ (ai,bi)≥ µ (V )

where V is the union of the open intervals just mentioned. Thus

µ (E)≤ µ (V )≤ µ (E)+ ε.

This shows outer regularity. If µ (E) = ∞, there is nothing to show.Now consider the assertion of inner regularity 6.16. Suppose I is a closed and bounded

interval and E ⊆ I with E ∈F . By outer regularity, there exists open V containing I∩EC

such thatµ(I∩EC)+ ε > µ (V )

Then since µ is additive on F , it follows that µ(V \(I∩EC

))< ε. Then K ≡ VC ∩ I is a

compact subset of E. This is because V ⊇ I∩EC so VC ⊆ IC ∪E and so

VC ∩ I ⊆(IC ∪E

)∩ I = E ∩ I = E.

Also,E \(VC ∩ I

)= E ∩V =V \EC ⊆V \

(I∩EC) ,

a set of measure less than ε . Therefore,

µ(VC ∩ I

)+ ε ≥ µ

(VC ∩ I

)+µ

(E \(VC ∩ I

))= µ (E) ,

so the desired conclusion holds in the case where E is contained in a compact interval.Now suppose E is arbitrary and let l < µ (E) . Then choosing ε small enough, l + ε <

µ (E) also. Letting En ≡ E ∩ [−n,n] , it follows from Lemma 6.2.4 that for n large enough,µ (En)> l + ε. Now from what was just shown, there exists K ⊆ En such that µ (K)+ ε >µ (En). Hence µ (K)> l. This shows 6.16. ■

Definition 6.8.3 The countable union of closed sets is called an Fσ set and thecountable union of open sets is called a Gδ set. These are Borel sets.

Proposition 6.8.4 For m Lebesgue measure, m([a,b]) = m((a,b)) = b− a. Also m istranslation invariant in the sense that if E is any Lebesgue measurable set, then m(x+E)=m(E).

Proof: Let K consist of the open intervals including R and /0. Then K is a π system.Also m(x+ I) = m(I) is obvious for any I ∈K . Let G denote those Borel sets E such thatfor all x,m(x+E ∩ (−n,n)) = m(E ∩ (−n,n)). Thus K ⊆ G . If Ei are disjoint sets in G ,x+∪i (Ei∩ (−n,n)) = ∪(x+Ei∩ (−n,n)) and so

m(x+∪iEi∩ (−n,n)) = m(∪(x+Ei∩ (−n,n))) = ∑i

m(x+Ei∩ (−n,n))

= ∑i

m(Ei∩ (−n,n)) = m(∪iEi∩ (−n,n))