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

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.

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## 2 Answers

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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OK, thanks for answering. But, I did not understand why you take an $s$ such that, and why it suffices to show that $s = \infty$. I ask these questions to learn about motivations, not only for logically true solutions. If you give some motivation, I would really appreciate that. Thanks for help again :) – oeda Nov 11 '12 at 17:25
@John: $s=\infty$ implies that for any $C>0$, there exists $F\in\mathcal{F}$, such that $\mu(F)>C$. Then the result follows from the definition of $\mathcal{F}$. The motivation is roughly that: (i) since $\mu(E)=\infty$, for each $F\subset E$ with $\mu(F)<\infty$, $\mu(E\setminus F)=\infty$. (ii) since $\mu$ is semifinite, we can choose $F$ in (i) with $\mu(F)>0$, and $\mu(F)$ cannot be bounded from above, because we can consistently repeat the procedure by replacing $E$ with $E\setminus F$. Because my English is not very well, I can hardly explain clearer. Sorry about that. – 23rd Nov 11 '12 at 18:03
I think there are one point that has not been made clear here. The $F_n$'s should be taken such that $F_n\subset F_{n+1}$. Only then we can use the monotone limit rule to bring the limit in to get $\lim_{n\to\infty}\mu(F_n)=\mu(F)=s$. – user84731 Jul 2 '13 at 13:41
@user84731: I think everything is clear in my answer. By definition, $\cup_{k=1}^n F_k\in \mathcal{F}$ and $\mu(F)=\lim_{n\to\infty}\mu(\cup_{k=1}^n F_k)=s$. – 23rd Jul 3 '13 at 2:31

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

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