6.6. MEASURES FROM OUTER MEASURES 147

Theorem 6.6.4 Let Ω be a set and let µ be an outer measure on P (Ω). The col-lection of µ measurable sets S , forms a σ algebra and

If Fi ∈S, Fi∩Fj = /0, then µ(∪∞i=1Fi) =

∞

∑i=1

µ(Fi). (6.6)

If · · ·Fn ⊆ Fn+1 ⊆ ·· · , then if F = ∪∞n=1Fn and Fn ∈S , it follows that

µ(F) = limn→∞

µ(Fn). (6.7)

If · · ·Fn ⊇ Fn+1 ⊇ ·· · , and if F = ∩∞n=1Fn for Fn ∈S then if µ(F1)< ∞,

µ(F) = limn→∞

µ(Fn). (6.8)

This measure space is also complete which means that if µ (F) = 0 for some F ∈S thenif G⊆ F, it follows G ∈S also.

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)