642 CHAPTER 32. MEASURES AND INTEGRALS

By Lemma 32.4.3 Bn is (µ⌊S) measurable and so is BCn . I want to show µ(S)≥ µ(S\A)+

µ(S∩A). If µ(S) = ∞, there is nothing to prove. Assume µ(S)< ∞. Then apply Parts 32.8and 32.7 to the outer measure µ⌊S in 32.10 and let n → ∞. Thus

Bn ↑ A, BCn ↓ AC

and this yields µ(S) = (µ⌊S)(A)+(µ⌊S)(AC) = µ(S∩A)+µ(S\A).Therefore A ∈ S and this proves Parts 32.6, 32.7, and 32.8.It only remains to verify the assertion about completeness. Letting G and F be as

described above, let S ⊆ Ω. I need to verify

µ (S)≥ µ (S∩G)+µ (S\G)

However,

µ (S∩G)+µ (S\G) ≤ µ (S∩F)+µ (S\F)+µ (F \G)

= µ (S∩F)+µ (S\F) = µ (S)

because by assumption, µ (F \G)≤ µ (F) = 0.The measure m which results from the outer measure of Theorem 33.1.1 is called

Lebesgue measure. The following is a general result about completion of a measure space.This is coming up, but first is another general result about completion of a measure space.

Proposition 32.4.5 Let (Ω,F ,µ) be a measure space. Also let µ̂ be the outer measuredefined by

µ̂ (F)≡ inf{µ (E) : E ⊇ F and E ∈ F}

Then µ̂ is an outer measure which is a measure on F̂ , the set of µ̂ measurable sets. Alsoµ̂ (E) = µ (E) for E ∈ F and F ⊆ F̂ . If (Ω,F ,µ) is already complete, then no new setsare obtained from this process and F = F̂ .

Proof: The first part of this follows from Proposition 32.3.2. It only remains to verifythat F ⊆ F̂ . Let S be a set and let E ∈ F , ES ⊇ S,ES ∈ F . Then

µ (ES) = µ (ES \E)+µ (ES ∩E)

due to the fact that µ is a measure. As usual, if µ̂ (S) = ∞, it is obvious that µ̂ (S) ≥µ̂ (S\E)+ µ̂ (S∩E) . Therefore, assume this is not ∞. Then let µ̂ (S) > µ (ES)− ε. Thenfrom the above,

ε + µ̂ (S)≥ µ (ES \E)+µ (ES ∩E)≥ µ (S\E)+µ (S∩E)

Since ε is arbitrary, this shows that E ∈ F̂ . Thus F ⊆ F̂ .Why are these two σ algebras equal if (Ω,F ,µ) is complete? Suppose now that

(Ω,F ,µ) is complete. Let F ∈ F̂ . Then there exists E ⊇ F such that µ (E) = µ̂ (F) . Thisis obvious if µ̂ (F) = ∞. Otherwise, let En ⊇ F, µ̂ (F)+ 1

n > µ (En) . Just let E = ∩nEn.Now µ̂ (E \F) = 0. Now also, there exists a set of F called W such that µ (W ) = 0 andW ⊇ E \F. Thus E \F ⊆W, a set of measure zero. Hence by completeness of (Ω,F ,µ) ,it must be the case that E \F = E ∩FC = G ∈ F . Then taking complements of both sides,EC ∪F = GC ∈ F . Now take intersections with E. F ∈ E ∩GC ∈ F .