248 CHAPTER 9. MEASURES AND MEASURABLE FUNCTIONS

The problem is, I don’t know F ∈S and so it is not clear that µ (F1 \F) = µ (F1)−µ (F).However, µ (F1 \F)+µ (F)≥ µ (F1) and so µ (F1 \F)≥ µ (F1)−µ (F). Hence

limn→∞

(µ (F1)−µ (Fn)) = µ (F1 \F)≥ µ (F1)−µ (F)

which implies limn→∞ µ (Fn) ≤ µ (F) . But since F ⊆ Fn, µ (F) ≤ limn→∞ µ (Fn) and thisestablishes 9.10. Note that it was assumed µ (F1)< ∞ because µ (F1) was subtracted fromboth sides.

It remains to show S is closed under countable unions. Recall that if A ∈ S , thenAC ∈S and S is closed under finite unions. Let Ai ∈S , A = ∪∞

i=1Ai, Bn = ∪ni=1Ai. Then

µ(S) = µ(S∩Bn)+µ(S\Bn) (9.12)= (µ⌊S)(Bn)+(µ⌊S)(BC

n ).

By Lemma 9.5.3 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 9.10and 9.9 to the outer measure µ⌊S in 9.12 and let n→ ∞. Thus Bn ↑ A, BC

n ↓ AC and thisyields µ(S) = (µ⌊S)(A)+(µ⌊S)(AC) = µ(S∩A)+µ(S\A).

Therefore A ∈S and this proves Parts 9.8, 9.9, and 9.10.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. ■

Corollary 9.5.5 Completeness is the same as saying that if (E \E ′)∪(E ′ \E)⊆N ∈Fand µ (N) = 0, then if E ∈F , it follows that E ′ ∈F also.

Proof: If the new condition holds, then suppose G⊆ F where µ (F) = 0,F ∈F . Then= /0︷ ︸︸ ︷

(G\F)∪ (F \G)⊆ F and µ (F) is given to equal 0. Therefore, G ∈F .Now suppose the earlier version of completeness and let(

E \E ′)∪(E ′ \E

)⊆ N ∈F

where µ (N) = 0 and E ∈F . Then we know (E \E ′) ,(E ′ \E) ∈F and all have measurezero. It follows E \ (E \E ′) = E ∩E ′ ∈F . Hence

E ′ =(E ∩E ′

)∪(E ′ \E

)∈F ■

9.6 Measurable Sets Include Borel Sets?If you have an outer measure, it determines a measure. This section gives a very convenientcriterion which allows you to conclude right away that the measure is a Borel measure.

Theorem 9.6.1 Let µ be an outer measure on the subsets of (X ,d), a metric space.If µ(A∪B) = µ(A)+µ(B) whenever dist(A,B)> 0, then the σ algebra of measurable setsS contains the Borel sets.

248 CHAPTER 9. MEASURES AND MEASURABLE FUNCTIONSThe problem is, I don’t know F € .Y and so it is not clear that u (F, \F) =u (Fi) -L(F).However, u (Fi \F)+u(F) > (Fi) and so pw (Fi \F) > uw (Fi) —U(F). Hencelim (ut (Fi) — u (Fa) = H(i \F) > (Fi) —H(F)n-oowhich implies limp... HM (Fn) < U(F). But since F C Fy, w(F) < limps. UM (F,) and thisestablishes 9.10. Note that it was assumed LU (F,) < c¢ because pt (F) was subtracted fromboth sides.It remains to show .Y is closed under countable unions. Recall that if A € .Y, thenAC € SF and -Y is closed under finite unions. Let A; € Y, A = U2, Ai, Bn = UL, Aj. ThenM(S) = B(SOBn)+M(S\Bn) (9.12)= (u|S)(Bn) + (HS) (Br )-By Lemma 9.5.3 B, is (u|S) measurable and so is BC. I want to show (S$) > u(S\A) +(SNA). If u(S) =, there is nothing to prove. Assume p1(S) < co. Then apply Parts 9.10and 9.9 to the outer measure |S in 9.12 and let n + oo. Thus B, + A, BS | A© and thisyields 11(S) = (u[S)(A) + (uLS)(AS) = M(SMA) + W(S\ A).Therefore A € .¥ and this proves Parts 9.8, 9.9, and 9.10.It only remains to verify the assertion about completeness. Letting G and F be asdescribed above, let S C Q. I need to verify u (S) > uw (SAG) + pu (S\ G). However,U(SAG)+u(S\G) < w(SOF)+u(S\F)+u(F\G)= u(SNF)+u(S\F)=H(S)because by assumption, uw (F \\G) <u(F)=0. EfCorollary 9.5.5 Completeness is the same as saying that if (E\ E')U(E'\E) CNE Fand (N) = 0, then if E € F, it follows that E' € F also.Proof: If the new condition holds, then suppose G C F where (F) =0,F € ¥. Then=0_—_—, . .(G\ F)U(F\G) CF and u (F) is given to equal 0. Therefore, G € ¥.Now suppose the earlier version of completeness and let(E\E')U(E'\E) CNEFwhere U (N) = 0 and E € ¥. Then we know (E£ \ E’),(E’\ E) € F and all have measurezero. It follows E \ (E\ E') =ENE' € ¥. HenceE'=(ENE')U(E'\E)e FO9.6 Measurable Sets Include Borel Sets?If you have an outer measure, it determines a measure. This section gives a very convenientcriterion which allows you to conclude right away that the measure is a Borel measure.Theorem 9.6.1 Lez Lt be an outer measure on the subsets of (X,d), a metric space.If u(AUB) = w(A) + UW (B) whenever dist(A, B) > 0, then the o algebra of measurable setsS contains the Borel sets.