Definition 14.3.2 (Semifinite Part). Let $(X, \cm, \mu)$ be a measure space and
then $\mu_{0}$ is a semifinite measure, and the semifinite part of $\mu$.
Proof. Firstly, for any $F \in \cm$ with $\mu(F) < \infty$, $\mu(F) = \mu_{0}(F)$.
Let $\seq{E_n}\subset \cm$ be pairwise disjoint. For any $F \in \cm$ with $F \subset \bigsqcup_{n \in \natp}E_{n}$ and $\mu(F) < \infty$,
Thus $\mu_{0}\paren{\bigcup_{n \in \natp}E_n}\le \sum_{n \in \natp}\mu_{0}(E_{n})$.
On the other hand, let $n \in \natp$ and $\seqf{F_k}\subset \cm$ such that $F_{k} \subset E_{k}$ for each $1 \le k \le n$ and $\mu(F_{k}) < \infty$, then $F = \bigsqcup_{k = 1}^{n} F_{k} \subset \bigsqcup_{k \in \natp}E_{k}$. Thus
As this holds for all such $\seqf{F_k}$,
Since this holds for all $n \in \natp$,