272 CHAPTER 12. THE CONSTRUCTION OF MEASURES

Now let A = ∪∞i=1Ai where Ai∩A j = /0 for i ̸= j.

∞

∑i=1

µ(Ai)≥ µ(A)≥ µ(∪ni=1Ai) =

n

∑i=1

µ(Ai).

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

∞

∑i=1

µ(Ai) = µ(A)

which establishes 12.1.3. Part 12.1.4 follows from part 12.1.3 just as in the proof of Theo-rem 11.1.5 on Page 224. That is, letting F0 ≡ /0, use part 12.1.3 to write

µ (F) = µ (∪∞k=1 (Fk \Fk−1)) =

∞

∑k=1

µ (Fk \Fk−1)

= limn→∞

n

∑k=1

(µ (Fk)−µ (Fk−1)) = limn→∞

µ (Fn) .

In order to establish 12.1.5, let the Fn be as given there. Then from what was just shown,

µ (F1 \Fn)+µ (Fn) = µ (F1)

Then, since (F1 \Fn) increases to (F1 \F), 12.1.4 implies

limn→∞

(µ (F1 \Fn)) = limn→∞

(µ (F1)−µ (Fn)) = µ (F1 \F) .

Now I don’t know whether F ∈S and so all that can be said is that

µ (F1 \F)+µ (F)≥ µ (F1)

but this impliesµ (F1 \F)≥ µ (F1)−µ (F) .

Hencelimn→∞

(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 12.1.5. 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) (12.1.6)= (µ⌊S)(Bn)+(µ⌊S)(BC

n ).