Proposition 16.1.4. Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then:
For any $\seq{E_n}\subset \cm$ and $E \in \cm$ with $E_{n} \upto E$, $\mu\paren{\bigcup_{n \in \natp}E_n}= \limv{n}\mu(E_{n})$.
For any $\seq{E_n}\subset \cm$ and $E \in \cm$ with $E_{n} \downto E$ and $|\mu(E_{1})| < \infty$, $\mu\paren{\bigcap_{n \in \natp}E_n}= \limv{n}\mu(E_{n})$.
Proof. (1): For each $N \in \natp$, let $F_{N} = E_{N} \setminus \bigcup_{n = 1}^{N-1}E_{n}$, then $E_{N} = \bigsqcup_{n = 1}^{N} F_{n}$. By (M2),
\[\mu(E) = \limv{N}\sum_{n = 1}^{N} \mu(F_{N}) = \limv{N}\mu(E_{N})\]
(2): By (1),
\[\mu(E) = \mu(E_{1}) - \mu(E_{1} \setminus E) = \limv{n}[\mu(E_{1}) - \mu(E_{1} \setminus E_{n})] = \limv{n}\mu(E_{n})\]
$\square$