Lemma 16.1.5. Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then
For any $A \in \cm$ positive and $B \in \cm$ with $B \subset A$, $B$ is positive.
For any $\seq{A_n}\subset \cm$ positive, $\bigcup_{n \in \natp}A_{n}$ is positive.
Proof. (1): For any $C \in \cm$ with $C \subset B$, $C \subset A$, so $\mu(C) \ge 0$. Hence $B$ is positive.
(2): For each $N \in \natp$, let $B_{N} = A_{N} \setminus \sum_{n = 1}^{N-1}A_{n}$. By (1), $B_{N}$ is positive with $\bigcup_{n \in \natp}A_{n} = \bigsqcup_{n \in \natp}B_{n}$. For any $C \subset \bigcup_{n \in \natp}B_{n}$,
\[\mu(C) = \sum_{n \in \natp}\mu(C \cap B_{n}) \ge 0\]
$\square$