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I am trying to solve following exercise from Folland,

If $\mu$ is a semifinite measure and $\mu(E) = \infty$, for any $C > 0$, $\exists$ $F \subset E$ with $C < \mu(F) < \infty$.

It seems to follow from definition of semifinite measures, which you can find here, but I couldn't prove it.

3 Answers 3

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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 $\mu(F)=s$. 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 $s$.

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    @user84731: I think everything is clear in my answer. By definition, $\cup_{k=1}^n F_k\in \mathcal{F}$ and $\mu(F)=\lim_{n\to\infty}\mu(\cup_{k=1}^n F_k)=s$.2013-07-03
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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 $\mu(G)\ge M$ because $\mu(G_{n})\to M$, and $\mu(E\setminus G)=\infty$ 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 there is some $N$ such that $\mu(G_{N})>C$. If $\mu(G)=\infty$ then there is some $N$ such that $\mu\left(\bigcup_{n=1}^{N}G_{n}\right)>C$.