Let $\mu$ be a semi-finite (and countably-additive) measure and $\mu(E)=\infty$. Then for any $C>0$, there exists a measurable set $F$ such that $F \subseteq E$ and $C<\mu(F)<\infty$.
By definition of semi-finiteness, I know I am guaranteed that there will exist a measurable-set $F$ satisfying $F \subseteq E$ and $0<\mu(F)<\infty$. I am not sure how can I construct a set satisfying what I am required to prove. Will appreciate some hint/help. Many thanks.
Let $\mathcal{F}=\{\mu(F):\ F\subset E, \mu(F)<\infty\}$ and $K=\sup(\mathcal{F})$. If $K=\infty$, then there is a sequence $F_n\subset\mathcal{F}$ such that $\mu(F_n)\to\infty$. Then one of them will have finite measure larger than $C$.
Assume that $K<\infty$. Let $F_n\in\mathcal{F}$ such that $\mu(F_n)\nearrow K$. Then $\infty>\mu(\bigcup_{n=1}^m F_n)\geq \mu(F_n)\nearrow K$. So, if $\mu(\bigcup_{n=1}^{m}F_n)\to\infty$ we are done again. Otherwise, $\bigcup_{n=1}^{\infty}F_n\in\mathcal{F}$. Then $K\leq\mu(\bigcup_{n=1}^{\infty}F_n)\geq K$ and we get that $\mu(\bigcup_{n=1}^{\infty}F_n)=K$. Then $E\setminus \bigcup_{n=1}^{\infty}F_n$ has infinite measure. By the semi-finiteness of the measure it contains an $F$ of finite and positive measure. Then $K\cup\bigcup_{n=1}^{\infty}F_n\in\mathcal{F}$ and has measure strictly larger than $K$. This is a contradiction. So, this case doesn't occur.
