6.3. DYNKIN’S LEMMA 143

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 ⊆ GA. Property 3 is obvious because if {Bi} isa sequence of disjoint sets in GA, then

A∩∪∞i=1Bi = ∪∞

i=1A∩Bi ∈ G

because A∩Bi ∈ G and the property 3 of G .It remains to verify Property 2 so let B ∈ GA. I need to verify that BC ∈ GA. In other

words, I need to show that A∩BC ∈ G . However,(AC ∪ (A∩B)

)C= A∩

(AC ∪B

)= A∩BC

and so

A∩BC =

∈GAC ∪

 ∈G︷ ︸︸ ︷A∩B

C

∈ G

Here is why. Since B ∈ GA, A∩B ∈ G and since A ∈K ⊆ G it follows AC ∈ G by as-sumption 2. It follows from assumption 3 the union of the disjoint sets, AC and (A∩B) isin G and then from 2 the complement of their union is in G . Thus GA satisfies 1 - 3 andthis implies since G is the smallest such, that GA ⊇ G . However, GA is constructed as asubset of G . This proves that for every B ∈ G and A ∈K , A∩B ∈ G . Now pick B ∈ Gand consider GB ≡ {A ∈ G : A∩B ∈ G } . I just proved K ⊆ GB. The other arguments areidentical to show GB satisfies 1 - 3 and is therefore equal to G . This shows that wheneverA,B ∈ G it follows A∩B ∈ G .

This implies G is a σ algebra. To show this, all that is left is to verify G is closed undercountable unions because then it follows G is a σ algebra. Let {Ai} ⊆ G . Then let A′1 = A1and

A′n+1 ≡ An+1 \ (∪ni=1Ai) = An+1∩

(∩n

i=1ACi)= ∩n

i=1(An+1∩AC

i)∈ G

because finite intersections of sets of G are in G . Since the A′i are disjoint, it follows

∪∞i=1Ai = ∪∞

i=1A′i ∈ G

Therefore, G ⊇ σ (K ). ■

Example 6.3.3 Suppose you have (U,F ) and (V,S ) , two measurable spaces. Let K ⊆U×V consist of all sets of the form A×B where A ∈F and B ∈S . This is easily seen tobe a π system. When this is done, σ (K ) is denoted as F ×S .

An important example of a σ algebra is the Borel sets.

Definition 6.3.4 The Borel sets on Rp, denoted by B (Rp) consists of the smallestσ algebra containing the open sets.

Don’t ever try to describe a generic Borel set. Always work with the definition that it isthe smallest σ algebra containing the open sets. Attempts to give an explicit description ofa “typical” Borel set tend to lead nowhere because there are so many things which can bedone.You can take countable unions and complements and then countable intersections of

6.3. DYNKIN’S LEMMA 143I want to show % satisfies 1 - 3 because then it must equal Y since Y is the smallestcollection of subsets of Q which satisfies 1 - 3. This will give the conclusion that forAE # andBEY,ANBEG. This information will then be used to show that if A,B EYthen ANB € Y. From this it will follow very easily that Y is a o algebra which will implyit contains o (.4). Now here are the details of the argument.Since .% is given to be a z system, % C G4. Property 3 is obvious because if {B;} isa sequence of disjoint sets in %, thenANU 1B; = UZ,ANB; € Ybecause AM B; € Y and the property 3 of Y.It remains to verify Property 2 so let B € Y. I need to verify that BC € Y%. In otherwords, I need to show that AN BC € Y. However, (ACU (A nB))© =AN (ACUB) =ANBCand soeG cg ~c_ [4cANB’ =| AV’ Ul ANB eGHere is why. Since B € %, ANB €G and since A € % CG it follows A© € Y by as-sumption 2. It follows from assumption 3 the union of the disjoint sets, AC and (ANB) isin Y and then from 2 the complement of their union is in Y. Thus % satisfies 1 - 3 andthis implies since Y is the smallest such, that Y > Y. However, 4, is constructed as asubset of Y. This proves that for every BE Y andA € #,ANBE®Y. Now pick BEYand consider % = {A EY :ANBE GY}. I just proved % C Gg. The other arguments areidentical to show Gg satisfies 1 - 3 and is therefore equal to Y. This shows that wheneverA,B €@ it follows ANBEY.This implies Y is a o algebra. To show this, all that is left is to verify Y is closed undercountable unions because then it follows Y is a o algebra. Let {A;} C Y. Then let A) =AandAne = Ans \ (UE 1Ai) = Ang 9 (NAF) = a1 (Ans+1 Af) egbecause finite intersections of sets of Y are in Y. Since the A) are disjoint, it followsTherefore, Y Do (.%). HfExample 6.3.3 Suppose you have (U,F) and (V,.%) , two measurable spaces. Let H CU x V consist of all sets of the form A x B where A € F and B € .. This is easily seen tobe am system. When this is done, 0 (.X) is denoted as F x S.An important example of a o algebra is the Borel sets.Definition 6.3.4 The Borel sets on R?, denoted by B(R?) consists of the smallest0 algebra containing the open sets.Don’t ever try to describe a generic Borel set. Always work with the definition that it isthe smallest o algebra containing the open sets. Attempts to give an explicit description ofa “typical” Borel set tend to lead nowhere because there are so many things which can bedone. You can take countable unions and complements and then countable intersections of