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

Solution 1:

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 $\mu(F)= \lim_{k\to\infty}\mu(\cup_{n=0}^kF_n)=s$.

Solution 2:

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 $C<\mu\left(\bigcup_{n=1}^{N}G_{n}\right)<+\infty$.