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 without any ugly technicalities.
Definition 5.3.1 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 5.3.2 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.
Then G⊇ σ
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 define

I want to show G_{A} 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 contains σ
Since K is given to be a π system, K⊆G_{A}. Property 3 is obvious because if

because A ∩ B_{i} ∈G and the property 3 of G.
It remains to verify Property 2 so let B ∈G_{A}. I need to verify that B^{C} ∈G_{A}. In other words, I need to show that A ∩ B^{C} ∈G. However,

Here is why. Since B ∈G_{A}, A∩B ∈G and since A ∈K⊆G it follows A^{C} ∈G by assumption 2. It follows from assumption 3 the union of the disjoint sets, A^{C} and

I just proved K⊆G_{B}. The other arguments are identical to show G_{B} 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

Therefore, G⊇ σ
Example 5.3.3 Suppose you have