Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

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 $s$.

share|improve this answer
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 :) –  Deniz Nov 11 '12 at 17:25
I wish, I could too :) If you learn completely new things, most of the times, even the simplest cases may seem complicated. This is true for me, at least. Then, these notations and thinking style are new to me. I want to learn motivations by solving exercises. So, anyway, thanks. –  Deniz Nov 11 '12 at 17:52
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.