640 CHAPTER 32. MEASURES AND INTEGRALS

Proof: First note that /0 and Ω are obviously in S . Now suppose A,B ∈S . I will showA\B ≡ A∩BC is in S . To do so, consider the following picture.

S⋂

AC⋂BC

S⋂

AC⋂B

S⋂

A⋂

BS⋂

A⋂

BC

A

B

S

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). (32.9)

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).