190 CHAPTER 8. MEASURES AND MEASURABLE FUNCTIONS

such that F ⊇ E and µ̄ (F) = µ (F) = µ̄ (E) . The same conclusion holds if (Ω,F ,µ) is aσ finite measure space meaning that Ω = ∪∞

k=1Ωk where the Ωk ∈F and are disjoint withµ (Ωk)< ∞.

Proof: µ̄ (E) ≡ inf{µ (F) : F ⊇ E,F ∈F} . Let Fn ∈ F , Fn ⊇ E and µ̄ (E) + 1n >

µ (Fn) . By taking intersections, we can also assume that Fn ⊇ Fn+1. Let F ≡ ∩nFn. Thenusing Proposition 8.4.2 we have µ (F) = µ̄ (F) = µ̄ (E) . In σ finite case, it was justshown that there exists Fk ⊆ Ωk,Fk ∈ F such that µ̄ (Fk) = µ (Fk) = µ̄ (E ∩Ωk). Thenlet F ≡ ∪kFk. µ̄ (E) = ∑

∞k=1 µ̄ (E ∩Ωk) = ∑

∞k=1 µ (Fk) = µ (F) = µ̄ (F) ■

As a corollary, we can say something about functions.

Corollary 8.5.7 Let (Ω,F ,µ) be σ finite and let (Ω,S , µ̄) be the completion justdiscussed. Then if f ≥ 0 and S measurable, there exists h ≥ f such that h = f for µ̄ a.e.and h is F measurable.

Proof: From Theorem 8.1.6 there is a sequence of S measurable simple nonnegativefunctions sn (ω) = ∑

mnk=1 cn

kXEnk(ω) which converges pointwise to f . From Proposition

8.5.6, sn (ω) = ŝn (ω) where ŝn (ω) ≡ ∑mnk=1 cn

kXÊnk(ω) with Ên

k ⊇ Enk , µ̄

(Ên

k \Enk

)= 0.

Then letting h(ω) ≡ limsupn→∞ ŝn (ω), it follows that h(ω) is F measurable, h(ω) =f (ω) µ̄ a.e., and h(ω)≥ f (ω). ■

8.6 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 8.6.1 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.

Proof: It suffices to show that closed sets are in S , the σ -algebra of measurable sets,because then the open sets are also in S and consequently S contains the Borel sets. LetK be closed and let S be a subset of Ω. Is µ(S)≥ µ(S∩K)+µ(S\K)? It suffices to assumeµ(S) < ∞. Let Kn ≡

{x : dist(x,K)≤ 1

n

}. By Lemma 3.12.1 on Page 83, x→ dist(x,K)

is continuous and so Kn is a closed set having K as a subset. That in KCn is at a positive

distance from K. By the assumption of the theorem,

µ(S)≥ µ((S∩K)∪ (S\Kn)) = µ(S∩K)+µ(S\Kn) (8.11)

Nowµ(S\Kn)≤ µ(S\K)≤ µ(S\Kn)+µ((Kn \K)∩S). (8.12)

If limn→∞ µ((Kn \K)∩ S) = 0 then the theorem will be proved because this limit alongwith 8.12 implies limn→∞ µ (S\Kn) = µ (S\K) and then taking a limit in 8.11, µ(S) ≥µ(S∩K)+µ(S\K) as desired. Therefore, it suffices to establish this limit.

Since K is closed, a point, x /∈ K must be at a positive distance from K and so

Kn \K = ∪∞k=nKk \Kk+1.

Therefore

µ(S∩ (Kn \K))≤∞

∑k=n

µ(S∩ (Kk \Kk+1)). (8.13)