14.3 Semifinite Measures
Definition 14.3.1 (Semifinite). Let $(X, \cm, \mu)$ be a measure space, then the following are equivalent:
For any $E \in \cm$ with $\mu(E) = \infty$, there exists $F \subset E$ with $0 < \mu(F) < \infty$.
For any $E \in \cm$,
\[\mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}\]
If the above holds, then $\mu$ is a semifinite measure.
Proof. $(1) \Rightarrow (2)$: If $\mu(E) < \infty$, then the result holds directly. Assume that $\mu(E) = \infty$, and suppose for contradiction that
Let $\seq{F_n}\subset \cm$ such that $F \subset E$, $\mu(F) < \infty$, and $\mu(F_{n}) \upto M$. Let $F = \bigcup_{n \in \nat}F_{n}$, then $\mu(F) = M$, and $\mu(E \setminus F) = \infty$. By (1), there exists $G \in \cm$ with $G \subset E \setminus F$ and $0 < \mu(G) < \infty$. Thus $F \cup G \subset E$ with $M < \mu(F) + \mu(G) = \mu(F \cup G) < \infty$. This contradicts the fact that $M$ is the supremum.$\square$
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 $n \in \natp$,