324 CHAPTER 14. THE LEBESGUE INTEGRAL

14.2 Dynkin’s LemmaRather than attempt to show S is a σ algebra, I will show that S contains a σ algebra. Thisis fairly easy because of a very elegant lemma due to Dynkin which is part of the abstracttheory of measures and integrals. This lemma is more interesting than the assertion that Scontains a σ algebra.

Lemma 14.2.1 Let C be a set whose elements are σ algebras each containing somesubset K of the set of all subsets. Then ∩C is a σ algebra which contains K .

Proof: /0,Ω are in ∩C because these are each in each σ algebra of C . If Ei ∈ ∩C , thenif F ∈ C it follows that ∪∞

i=1Ei ∈F and so, since F is arbitrary, this shows this union isin ∩C . If E ∈ ∩C , then EC ∈F for each F ∈ ∩C and so, as before, EC ∈ ∩C . Thus ∩Cis a σ algebra.

Definition 14.2.2 Let Ω be a set and let K be a collection of subsets of Ω. ThenK is called a π system if /0,Ω∈K and whenever A,B∈K , it follows A∩B∈K . σ (K )will denote the intersection of all σ algebras containing K . The set of all subsets of Ω

is one such σ algebra which contains K . Thus σ (K ) is the smallest σ algebra whichcontains K .

The following is the fundamental lemma which shows these π systems are useful. Thisis due to Dynkin. Note that the open intervals in R constitute a π system.

Lemma 14.2.3 Let K be a π system of subsets of Ω, a set. Also let G be a collectionof subsets of Ω which satisfies the following three properties.

1. K ⊆ G

2. If A ∈ G , then AC ∈ G

3. If {Ai}∞

i=1 is a sequence of disjoint sets from G then ∪∞i=1Ai ∈ G .

Then G ⊇ σ (K ) , where σ (K ) is the smallest σ algebra which contains K .

Proof: First note that if

H ≡ {G : 1 - 3 all hold for G }

then ∩H yields a collection of sets which also satisfies 1 - 3. Therefore, I will assume inthe argument that G is the smallest collection satisfying 1 - 3. Let A ∈K and define

GA ≡ {B ∈ G : A∩B ∈ G } .

I want to show GA satisfies 1 - 3 because then it must equal G since G is the smallestcollection of subsets of Ω which satisfies 1 - 3. This will give the conclusion that forA ∈K and B ∈ G , A∩B ∈ G . This information will then be used to show that if A,B ∈ Gthen A∩B ∈ G . From this it will follow very easily that G is a σ algebra which will implyit contains σ (K ). Now here are the details of the argument.

Since K is given to be a π system, K ⊆ G A. Property 3 is obvious because if {Bi}is a sequence of disjoint sets in GA, then A∩∪∞

i=1Bi = ∪∞i=1A∩Bi ∈ G because A∩Bi ∈ G

and the property 3 of G .