 Solution

Dynkin lemma, also called the pi-lambda theorem, states that if a Dynkin system $\mathcal{D}$ contains a pi-system $\mathcal{P}$, then

Let $\lambda (\mathcal{P})$ be the smallest Dynkin system containing $\mathcal{P}$. It is enough to prove that $\lambda (\mathcal{P})$ is a sigma-algebra, because then

First fix $A\in \mathcal{P}$ and defineThis is a Dynkin system. It contains $\Omega$ because $A\in \lambda (\mathcal{P})$; closure under relative complements follows fromand closure under disjoint unions follows by distributing $A\cap -$ over the union. Since $\mathcal{P}$ is closed under intersections, $\mathcal{P}\subseteq {\mathcal{G}}_A$. Minimality therefore givesThus $A\cap B\in \lambda (\mathcal{P})$ whenever $A\in \mathcal{P}$ and $B\in \lambda (\mathcal{P})$.

Now fix $B\in \lambda (\mathcal{P})$ and defineThe same argument shows that ${\mathcal{H}}_B$ is a Dynkin system, and the preceding paragraph shows that it contains $\mathcal{P}$. Hence ${\mathcal{H}}_B=\lambda (\mathcal{P})$. We have proved that $\lambda (\mathcal{P})$ is itself a pi-system.

A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore $\lambda (\mathcal{P})=\sigma (\mathcal{P})$, proving the lemma.