If a measure is semifinite, then there are sets of arbitrarily large but finite measure

Let $\mathcal{F}=\{F\subset E: F$ is measurable and $0<\mu(F)<\infty \}$. Since $\mu$ is semifinite, $\mathcal{F}$ is non-empty. Let $s=\sup_{}\{\mu(F):F\in\mathcal{F}\}$. It suffices to show that $s=\infty$.

Choose $\{F_n\}_{n\in\mathbb{N}}\subset\mathcal{F}$, such that $\lim_{n\to\infty}\mu(F_n)=s$. Then $F=\cup_{n\in\mathbb{N}}F_n\subset E$ and $\mu(F)=s$ (see remark below). If $s<\infty$, then $\mu(E\setminus F)=\infty$, and hence there exists $F'\subset E\setminus F$, such that $0<\mu(F')<\infty$. Then $F\cup F'\subset E$ and $s<\mu(F\cup F')<\infty$, i.e. $F\cup F'\in\mathcal{F}$, which contradicts to the definition of $s$.

Remark: For every $k \in \mathbb{N}$, $\cup_{n=0}^kF_n \in \mathcal{F}$, so, $\mu(\cup_{n=0}^kF_n)<s$. So we have $$ \mu(F_k) \leqslant \mu(\cup_{n=0}^kF_n)<s$$ Since $\lim_{k\to\infty}\mu(F_k)=s$, we have $\lim_{k\to\infty}\mu(\cup_{n=0}^kF_n)=s$. Since $\cup_{n=0}^kF_n \nearrow F$, we have $\mu(F)= \lim_{k\to\infty}\mu(\cup_{n=0}^kF_n)=s$.

I don't think 23rd's answer is quite right because the collection $\{F_n\}$ might not be disjoint. Here's my attempt:

Let $\mathcal{F}$ be the collection of all measurable sets $F\subseteq E$ such that $0<\mu(F)<\infty$. This set is nonempty because $\mu$ is semifinite. Let $M=\sup_{F\in\mathcal{F}}\mu(F)$ and choose a sequence $\{G_{n}\}$ in $\mathcal{F}$ such that $\mu(G_{n})\to M$. Let $G=\bigcup_{n=1}^{\infty}G_{n}$.

Suppose that $M<\infty$ and $\mu(G)<\infty$; then $G \in \mathcal{F}$, so $\mu(G)\leq M$. But, since $\mu(G_{n})\to M$, we have $\mu(G)= M$.

Note that $\mu(E\setminus G)=\infty$, because $\mu(E)=\infty$. Choose a measurable set $H\subseteq E\setminus G$ such that $0<\mu(H)<\infty$. Then $G\cup H\in\mathcal{F}$, so $$ M<\mu(G)+\mu(H)=\mu(G\cup H)\le M. $$ This is a contradiction, so either $M=\infty$ or $\mu(G)=\infty$. If $M=\infty$ then, for any $C>0$, there is some $F$ such that $C<\mu(F)<+\infty$. If $\mu(G)=\infty$ then there is some $N$ such that $C<\mu\left(\bigcup_{n=1}^{N}G_{n}\right)<+\infty$.

However i think in 23rd's proof, μ(F)=s is not an obvious nor a travial statement. Here's my proof.

Define $\mathcal{E}= \{ F \subset E : F \in \mathcal{M}, \mu (F) < \infty \}$, i.e., the set of all the finite measurable subsets of E. And $C = \sup \{ \mu (F) : F \in \mathcal{E} \}$, it suffices to show that $C = \infty$.

Select a sequence $\{ E_n \}_{n = 1}^{\infty} \subset \mathcal{E}$ with $\lim_{n \rightarrow \infty}^{} \mu (E_n) = C$. Then let $F_n = \bigcup_{i = 1}^n E_i$ and $F = \bigcup_{n = 1}^{\infty} F_n$.

Due to the continuity from below, we have $\lim_{n \rightarrow_{} \infty} \mu (F_n) = \mu (F) .$ It's easy to see that for all $n, \mu (F_n) \geqslant \mu (E_n)$, thus $\mu (F) = \lim_{n \rightarrow \infty} \mu (F_n) \geqslant \lim_{n \rightarrow \infty} \mu (E_n) = C$.

Here the question is whether F is finite? The answer is yes.

Suppose not, i.e, $\mu (F) = \infty$. Since $\lim_{n \rightarrow \infty} \mu (F_n) = \mu (F) = \infty$, then there exists some N, such that for all $n > N, \mu (F_n) > C$. But $F_n = \bigcup_{i = 1}^n E_i$ is always finite, which means $\forall n, F_n \in \mathcal{E}, \mu (F_n) \leqslant C$.

Thus we must have $\mu (F) < \infty$,so $F \in \mathcal{E} \Longrightarrow \mu (F) \leqslant C$. Combine with $\mu (F) \geqslant C$ we have $\mu (F) = C$.

If $C < \infty$, we can choose a finite measurable set $W \subset E - F$. Then $\mu \left( W \bigcup F \right) > C$, which contradicts the definition of C. So we must have $C = \infty .$