32.4. MEASURES FROM OUTER MEASURES 641

Since this holds for all n, you can take the limit as n → ∞ and conclude,

∞

∑i=1

µ(Ai) = µ(A)

which establishes 32.6.Consider part 32.7. Without loss of generality µ (Fk) < ∞ for all k since otherwise

there is nothing to show. Suppose {Fk} is an increasing sequence of sets of S . Thenletting F0 ≡ /0, {Fk+1 \Fk}∞

k=0 is a sequence of disjoint sets of S since it was shown abovethat the difference of two sets of S is in S . Also note that from 32.9

µ (Fk+1 \Fk)+µ (Fk) = µ (Fk+1)

and so if µ (Fk)< ∞, then

µ (Fk+1 \Fk) = µ (Fk+1)−µ (Fk) .

Therefore, lettingF ≡ ∪∞

k=1Fk

which also equals∪∞

k=1 (Fk+1 \Fk) ,

it follows from part 32.6 just shown that

µ (F) =∞

∑k=0

µ (Fk+1 \Fk) = limn→∞

n

∑k=0

µ (Fk+1 \Fk)

= limn→∞

n

∑k=0

µ (Fk+1)−µ (Fk) = limn→∞

µ (Fn+1) .

In order to establish 32.8, let the Fn be as given there. Then, since (F1 \Fn) increases to(F1 \F), 32.7 implies

limn→∞

(µ (F1)−µ (Fn)) = µ (F1 \F) .

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 implieslimn→∞

µ (Fn)≤ µ (F) .

But since F ⊆ Fn,µ (F)≤ lim

n→∞µ (Fn)

and this establishes 32.8. Note that it was assumed µ (F1) < ∞ because µ (F1) was sub-tracted from both 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) (32.10)= (µ⌊S)(Bn)+(µ⌊S)(BC

n ).