# 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.

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, $$\mu(\cup_{n=0}^kF_n). So we have $$\mu(F_k) \leqslant \mu(\cup_{n=0}^kF_n) 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 $$\mu(F)= \lim_{k\to\infty}\mu(\cup_{n=0}^kF_n)=s$$.

• 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 :)
– user48547
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 $\lim_{n\to\infty}\mu(F_n)=\mu(F)=s$. 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
• @23rd but why $F=\cup F_n \in\mathcal{F}$?
– user469065
Sep 13 '19 at 18:03

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

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

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 $$C = \infty .$$