> [!theoremb] Theorem > > Let $(X, \cf)$ be a [[Measure Space|measurable space]] and $\mathcal{P} \subset \cm$ be a $\pi$-system with $\sigma(\mathcal{P}) = \cf$. Let $\mu_1, \mu_2$ be measures on $\cf$, and suppose that > 1. $\mu_1(E) = \mu_2(E)$ for all $E \in \mathcal{P}$. > 2. There exists a sequence $\seq{X_n} \subset \mathcal{P}$ such that $X_n \upto X$ and $\mu_1(X_n) < \infty$ for all $n \in \nat$. > > Then $\mu_1 = \mu_2$. > > *Proof*. Let $G \in \mathcal{P}$ with $\mu_1(G) < \infty$, and let > $ > \alg = \bracs{E \in \cf: \mu_1(E \cap G) = \mu_2(E \cap G)} > $ > then $\alg \supset \mathcal{P}$ with $\Omega \in \alg$, and we claim that $\alg$ is a [[Lambda-System|lambda-system]]. > > Let $\seq{E_n}$ be an increasing sequence of sets in $\alg$, then by continuity from below, $\bigcup_{n \in \nat}E_n \in \alg$. Let $E, F \in \alg$ with $F \supset E$, then > $ > \begin{align*} > \mu_1(G \cap (F \setminus E)) &= \mu_1(G \cap F) - \mu_1(G \cap E) \\ > &= \mu_2(G \cap F) - \mu_2(G \cap E) \\ > &= \mu_2(G \cap (F \setminus E)) > \end{align*} > $ > and $F \setminus E \in \alg$. Then $\alg$ is a $\lambda$-system. By [[Dynkin's Pi-System Lemma]], $\alg = \cf$. > > Let $E \in \cf$, and $\seq{F_n} \subset \mathcal{P}$ such that $\mu_1(F_n) = \mu_2(F_n) < \infty$ for all $n$, and $F_n \upto \Omega$. Now by continuity from below on both measures, > $ > \mu_1(E) = \limv{n}\mu_1(E \cap F_n) = \limv{n}\mu_2(E \cap F_n) = \mu_2(E) > $ >