322 CHAPTER 14. THE LEBESGUE INTEGRAL

Example 14.1.2 As a simple example, let Ω =N and let F =P (N) the set of all subsets,and let µ (E) be the number of elements of E. You might verify that this is a measure space.

Observation 14.1.3 If (Ω,F ) is a measurable space and Ei ∈F , then ∩∞i=1Ei ∈F .

This is because Ei ∈F and by DeMorgan’s laws,

∩∞i=1Ei =

(∪∞

i=1ECi)C ∈F since each EC

i ∈F

Measures have the following fundamental property.

Lemma 14.1.4 If µ is a measure and Fi ∈ F , then µ (∪∞i=1Fi) ≤ ∑

∞i=1 µ (Fi). Also if

Fn ∈F and Fn ⊆ Fn+1 for all n, then if F = ∪nFn,

µ (F) = limn→∞

µ (Fn)

Symbolically, if Fn ↑ F, then µ (Fn) ↑ µ (F). If Fn ⊇ Fn+1 for all n, then if µ (F1) < ∞ andF = ∩nFn, then

µ (F) = limn→∞

µ (Fn)

Symbolically, if µ (F1)< ∞ and Fn ↓ F, then µ (Fn) ↓ µ (F).

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)