Proposition 21.2.3.label Let $(X, \cm)$ be a measurable, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measurethen:
- (1)
For any $\seq{A_n}\subset \cm$ with $A_{n} \subset A_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}A_n}= \limv{n}\mu(A_{n})$.
- (2)
For any $\seq{A_n}\subset \cm$ with $A_{n} \supset A_{n+1}$ for all $n \in \nat$, $\mu(\bigcap_{n \in \natp}A_{n}) = \limv{n}\mu(A_{n})$.
Proof. (1): Let $A_{0} = \emptyset$. For each $n \in \natp$, let $B_{n} = A_{n} \setminus B_{n-1}$, then $\bigcup_{n \in \natp}A_{n} = \bigsqcup_{n \in \natp}B_{n}$ and
\[\limv{n}\mu(A_{n}) = \sum_{n = 1}^{\infty} \mu(B_{n}) = \mu\paren{\bigcup_{n \in \natp}A_n}\]
(2): By (1),
\begin{align*}\limv{n}\mu(A_{n})&= \mu(A_{1}) - \limv{n}\mu(A_{1} \setminus A_{n}) \\&= \mu(A_{1}) - \mu\paren{A_1 \setminus \bigcup_{n \in \natp}A_n}= \mu\paren{\bigcap_{n \in \natp}A_n}\end{align*}
$\square$
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