9.5. MEASURES FROM OUTER MEASURES 247

It is required to show that µ (S) = µ (S\ (A\B))+ µ (S∩ (A\B)) . First consider S \(A\B) . From the picture, it equals(

S∩AC ∩BC)∪ (S∩A∩B)∪(S∩AC ∩B

)Therefore, µ (S)≤ µ (S\ (A\B))+µ (S∩ (A\B))

≤ µ(S∩AC ∩BC)+µ (S∩A∩B)+µ

(S∩AC ∩B

)+µ (S∩ (A\B))

= µ(S∩AC ∩BC)+µ (S∩A∩B)+µ

(S∩AC ∩B

)+µ

(S∩A∩BC)

= µ(S∩AC ∩BC)+µ

(S∩A∩BC)+µ (S∩A∩B)+µ

(S∩AC ∩B

)= µ

(S∩BC)+µ (S∩B) = µ (S)

and so this shows that A\B ∈S whenever A,B ∈S .Since Ω ∈S , this shows that A ∈S if and only if AC ∈S . Now if A,B ∈S , A∪B =

(AC ∩ BC)C = (AC \ B)C ∈ S . By induction, if A1, · · · ,An ∈ S , then so is ∪ni=1Ai. If

A,B ∈S , with A∩B = /0,

µ(A∪B) = µ((A∪B)∩A)+µ((A∪B)\A) = µ(A)+µ(B).

By induction, if Ai∩A j = /0 and Ai ∈S ,

µ(∪ni=1Ai) =

n

∑i=1

µ(Ai). (9.11)

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

∞i=1 µ(Ai) ≥ µ(A) ≥ µ(∪n

i=1Ai) =

∑ni=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 9.8.

Consider part 9.9. Without loss of generality µ (Fk)< ∞ for all k since otherwise thereis nothing to show. Suppose {Fk} is an increasing sequence of sets of S . Then lettingF0 ≡ /0, {Fk+1 \Fk}∞

k=0 is a sequence of disjoint sets of S since it was shown above thatthe difference of two sets of S is in S . Also note that from 9.11

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, letting F ≡ ∪∞k=1Fk which also equals ∪∞

k=1 (Fk+1 \Fk) , it follows from part 9.8just 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 9.10, let the Fn be as given there. Then, since (F1 \Fn) increases to(F1 \F), 9.9 implies

limn→∞

(µ (F1)−µ (Fn)) = limn→∞

µ (F1 \Fn) = µ (F1 \F) .