Semifinite Measure Prove:
Let $(X,\mathcal{M},\mu)$ be a measure space.
If $\mu$ is a semifinite measure, then for any
$C >0$ and $E \in \mathcal{M}$ with $\mathcal{M}=\infty$,
there must exist $F \in \mathcal{M}$, s.t.
$F \subseteq E$ and $C < \mu(F) < \infty$.
My attempt: Let $E \in \mathcal{M}$, then
$\exists \; F_1 \subseteq E \; ,$
s.t. $F_1 \in \mathcal{M}$ and $0 < \mu(F_1) < \infty$.
Then $\mu(E \setminus F_1) = \infty$, so
$\exists \; F_2 \subseteq E \setminus F_1 \; ,$
s.t. $F_2 \in \mathcal{M}$ and $0 < \mu(F_2) < \infty$.
Repeating this process, we can get a squence of
pairwisely disjoint $F_i$, with $F_i \subseteq E$
and $0 < \mu(F_i) < \infty$, $i \in \mathbb{N}^+$.
But we could not say $C < \sum_{i=1}^{\infty}\mu(F_i)
= \mu(\bigsqcup_{i=1}^{\infty} F_i)$, for instance,
when $\mu(F_i)=2^{-i}$, $C=2$, this inequality fails.
 A: Suppose for a contradiction that there exists $C \in(0, \infty)$ such that every measurable subset $F \subseteq E$ satisfies $\mu(F) \leq C $ or $ \mu(F)=\infty$.
Set
$$M:=\sup \{\mu(F) \mid F \subseteq E~{\rm is ~measurable ~and }~ \mu(F)<\infty\} ,$$ and note that $ 0 \leq M \leq C $. For each $ n \in \mathbb{N} $ there exists a measurable subset $ E_{n} \subseteq E $ such that
$$ M-n^{-1} \leq \mu\left(E_{n}\right)<\infty .$$
Set $ F_{n}:=\cup_{i=1}^{n} E_{i} $ for each $ n \in \mathbb{N} $ and define $ F:=\cup_{n \in \mathbb{N}} F_{n} $. Note that $ M-n^{-1} \leq \mu\left(E_{n}\right) \leq \mu(F) $ and also $ \mu\left(F_{n}\right) \leq \sum_{i=1}^{n} \mu\left(E_{n}\right)<\infty $ for all $ n \in \mathbb{N} $, so
$$ M \leq \mu(F)=\lim _{n \rightarrow \infty} \mu\left(F_{n}\right) \leq M .$$
This shows that $ \mu(F)=M $, so $ \mu(E \backslash F)=\infty $. Since $ \mu $ is semifinite, there exists a measurable subset $ A \subseteq E \backslash F $ such that $ 0<\mu(A)<\infty $. This contradicts the definition of $ M $, because $ A \cup F \subseteq E $ but
$$ \mu(F)<\mu(A)+\mu(F)=\mu(A \cup F)<\infty .$$
Therefore, for any $ C \in(0, \infty) $ there exists a measurable subset $ F \subseteq E $ such that $ C<\mu(F)<\infty $.
