224 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

Note that the above definition immediately implies that if Ei ∈F and the sets Ei arenot necessarily disjoint,

µ(∞⋃

i=1

Ei)≤∞

∑i=1

µ (Ei) .

To see this, let F1 ≡ E1, F2 ≡ E2 \E1, · · · ,Fn ≡ En \∪n−1i=1 Ei, then the sets Fi are disjoint sets

in F and

µ(∞⋃

i=1

Ei) = µ(∞⋃

i=1

Fi) =∞

∑i=1

µ (Fi)≤∞

∑i=1

µ(Ei)

because of the fact that each Ei ⊇ Fi and so

µ (Ei) = µ (Fi)+µ (Ei \Fi)

which implies µ (Ei)≥ µ (Fi) .The following theorem is the basis for most of what is done in the theory of measure

and integration. It is a very simple result which follows directly from the above definition.

Theorem 11.1.5 Let {Em}∞m=1 be measurable sets in a measure space (Ω,F ,µ). Then if

· · ·En ⊆ En+1 ⊆ En+2 ⊆ ·· · ,µ(∪∞

i=1Ei) = limn→∞

µ(En) (11.1.4)

and if · · ·En ⊇ En+1 ⊇ En+2 ⊇ ·· · and µ(E1)< ∞, then

µ(∩∞i=1Ei) = lim

n→∞µ(En). (11.1.5)

Stated more succinctly, Ek ↑ E implies µ (Ek) ↑ µ (E) and Ek ↓ E with µ (E1) < ∞ impliesµ (Ek) ↓ µ (E).

Proof: First note that ∩∞i=1Ei = (∪∞

i=1ECi )

C ∈F so ∩∞i=1Ei is measurable. Also note

that for A and B sets of F , A \B ≡(AC ∪B

)C ∈F . To show 11.1.4, note that 11.1.4 isobviously true if µ(Ek) = ∞ for any k. Therefore, assume µ(Ek)< ∞ for all k. Thus

µ(Ek+1 \Ek)+µ(Ek) = µ(Ek+1)

and soµ(Ek+1 \Ek) = µ(Ek+1)−µ(Ek).

Also,∞⋃

k=1

Ek = E1∪∞⋃

k=1

(Ek+1 \Ek)

and the sets in the above union are disjoint. Hence by 11.1.3,

µ(∪∞i=1Ei) = µ(E1)+

∞

∑k=1

µ(Ek+1 \Ek) = µ(E1)

+∞

∑k=1

µ(Ek+1)−µ(Ek)