7.4 Dynkin’s Lemma
Dynkin’s lemma is a very useful result. It is used quite a bit in books on probability. It will be
used here to obtain n dimensional Lebesgue measure, also to establish an important result on
regularity in the next section.
Definition 7.4.1 Let Ω be a set and let K be a collection of subsets of Ω. Then
K is called a π system if ∅,Ω ∈K and whenever A,B ∈K, it follows A ∩ B ∈K.
The following is the fundamental lemma which shows these π systems are useful. This is
due to Dynkin.
Lemma 7.4.2 Let K be a π system of subsets of Ω, a set. Also let G be a collection
of subsets of Ω which satisfies the following three properties.
- If A ∈G, then AC ∈G
i=1∞ is a sequence of disjoint sets from G then ∪i=1∞Ai ∈G.
Then G⊇ σ
, where σ
is the smallest σ algebra which contains K.
Proof: First note that if
then ∩ℋ yields a collection of sets which also satisfies 1 - 3. Therefore, I will assume
in the argument that G is the smallest collection satisfying 1 - 3. Let A ∈K and
I want to show GA satisfies 1 - 3 because then it must equal G since G is the smallest
collection of subsets of Ω which satisfies 1 - 3. This will give the conclusion that for A ∈K
and B ∈G, A ∩ B ∈G. This information will then be used to show that if A,B ∈G then
A ∩ B ∈G. From this it will follow very easily that G is a σ algebra which will imply it
. 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
sequence of disjoint sets in
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,
Here is why. Since B ∈GA, A∩B ∈G and since A ∈K⊆G it follows AC ∈G by assumption
2. It follows from assumption 3 the union of the disjoint sets, AC and
then from 2
the complement of their union is in G
. Thus GA
and this implies
is the smallest such, that GA ⊇G
. However, GA
is constructed as a subset of G
This proves that for every B ∈G
and A ∈K
, A ∩ B ∈G
. Now pick B ∈G
I just proved K⊆GB. The other arguments are identical to show GB satisfies 1 - 3 and is
therefore equal to G. This shows that whenever A,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 under
countable unions because then it follows G is a σ algebra. Let
. Then let A1′
because finite intersections of sets of G
are in G
. Since the Ai′
are disjoint, it follows
Therefore, G⊇ σ