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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.

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

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    $\begingroup$ 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 :) $\endgroup$
    – user48547
    Nov 11 '12 at 17:25
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    $\begingroup$ @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. $\endgroup$
    – 23rd
    Nov 11 '12 at 18:03
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    $\begingroup$ 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 $\lim_{n\to\infty}\mu(F_n)=\mu(F)=s$. $\endgroup$
    – user84731
    Jul 2 '13 at 13:41
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    $\begingroup$ @user84731: I think everything is clear in my answer. By definition, $\cup_{k=1}^n F_k\in \mathcal{F}$ and $\mu(F)=\lim_{n\to\infty}\mu(\cup_{k=1}^n F_k)=s$. $\endgroup$
    – 23rd
    Jul 3 '13 at 2:31
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    $\begingroup$ @23rd but why $F=\cup F_n \in\mathcal{F}$? $\endgroup$
    – user469065
    Sep 13 '19 at 18:03
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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 $G \in \mathcal{F}$, 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 $C<\mu\left(\bigcup_{n=1}^{N}G_{n}\right)<+\infty$.

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  • $\begingroup$ Why $G=\cup_{n=1}^{\infty} G_{n} \subset E$? $\endgroup$
    – user469065
    Sep 13 '19 at 16:18
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    $\begingroup$ You wouldn’t need the sets to be disjoint. I can’t see any case where he uses such a condition. $\endgroup$
    – user443408
    Jan 1 '20 at 22:29
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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 $C = \infty .$

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