Theorem 14.1.6 (Dynkin’s Uniqueness Theorem). Let $(X, \cm)$ be a measurable space, $\mathcal{P}\subset \cm$ be a $\pi$-syste, and $\mu, \nu: \cm \to [0, \infty]$ be measures. If
$\sigma(\mathcal{P}) = \cm$.
$\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
There exists $\seq{E_n}\subset \mathcal{P}$ such that $E_{n} \upto X$ and $\mu(E_{n}) < \infty$ for all $n \in \natp$.
then $\mu = \nu$.
Proof. Let $F \in \mathcal{P}$ with $\mu(F) < \infty$ and
then $\alg(F) \supset \mathcal{P}$ by (b), and
For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
\begin{align*}\mu((E' \setminus E) \cap F)&= \mu(E' \cap F) - \mu(E \cap F) \\&= \nu(E' \cap F) - \nu(E \cap F) = \nu((E' \setminus E) \cap F)\end{align*}For any $\seq{E_n}\subset \alg$ with $E_{n} \subset E_{n+1}$ for all $n \in \natp$ and $F \in \mathcal{P}$,
\[\mu\paren{\bigcup_{n \in \nat}E_n \cap F}= \limv{n}\mu(E_{n} \cap F) = \limv{n}\nu(E_{n} \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}\]by continuity from below (Proposition 14.1.5).
so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin’s $\pi$-$\lambda$ theorem (Theorem 13.2.5), $\alg(F) = \cm$.
Let $\seq{E_n}$ as in assumption (c), then $\mu(E_{n} \cap F) = \mu(E_{n} \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (Proposition 14.1.5),