634 CHAPTER 32. MEASURES AND INTEGRALS

Proof: Let G1 = F1 and if G1, · · · ,Gn have been chosen disjoint, let

Gn+1 ≡ Fn+1 \∪ni=1Gi

Thus the Gi are disjoint. In addition, these are all measurable sets. Now

µ (Gn+1)+µ (Fn+1 ∩ (∪ni=1Gi)) = µ (Fn+1)

and so µ (Gn)≤ µ (Fn). Therefore,

µ (∪∞i=1Gi) = µ (∪∞

i=1Fi) = ∑i

µ (Gi)≤ ∑i

µ (Fi) .

Now consider the increasing sequence of Fn ∈ F . If F ⊆ G and these are sets of F

µ (G) = µ (F)+µ (G\F)

so µ (G)≥ µ (F). AlsoF = ∪∞

i=1 (Fi+1 \Fi)+F1

Then

µ (F) =∞

∑i=1

µ (Fi+1 \Fi)+µ (F1)

Now µ (Fi+1 \Fi)+µ (Fi) = µ (Fi+1). If any µ (Fi) = ∞, there is nothing to prove. Assumethen that these are all finite. Then

µ (Fi+1 \Fi) = µ (Fi+1)−µ (Fi)

and so

µ (F) =∞

∑i=1

µ (Fi+1)−µ (Fi)+µ (F1)

= limn→∞

n

∑i=1

µ (Fi+1)−µ (Fi)+µ (F1) = limn→∞

µ (Fn+1)

Next suppose µ (F1)< ∞ and {Fn} is a decreasing sequence. Then

F1 \Fn

is increasing to F1 \F and so by the first part,

µ (F1)−µ (F) = µ (F1 \F) = limn→∞

µ (F1 \Fn) = limn→∞

(µ (F1)−µ (Fn))

This is justified because µ (F1 \Fn)+µ (Fn) = µ (F1) and all numbers are finite by assump-tion. Hence

µ (F) = limn→∞

µ (Fn) .

Next is a discussion of the notion of a measurable function.

Notation 32.2.4 In whatever context, f−1 (S) ≡ {ω : f (ω) ∈ S}. It is called the inverseimage of S and everything in the theory of the Lebesgue integral is formulated in terms ofinverse images. For a real valued f , f−1 (λ ,∞) may be written as [ f > λ ].