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Let $(X,\mathcal{M})$ be a measurable space. A measure $\mu: \mathcal{M} \to [0,\infty]$ is $\sigma$-finite if $X=\cup_i X_i$ where $X_i\in \mathcal{M}$ and $\mu(X_i)<\infty$ for each $i\ge 1$. A measure $\mu$ is said to be semi-finite if for any set $E\in \mathcal{M}$ with $\mu(E)=\infty$, one can find $F\subset E, F\in \mathcal{M}$ such that $0<\mu(F)<\infty$.

I read one result that and its proof

every $\sigma$-finite measure is semifinite.

The proof just takes $E=\cup_i (X_i\cap E)$ and takes $F:=X_i\cap E$, which satisfies our condition.

I am working on a similar question that if $\mu$ is $\sigma$-finite with $\mu(X)=\infty$, then for any $\epsilon>0$, there is a set $F\in \mathcal{M}$ so that $\epsilon<\mu(F)<\infty$.

I try to use a similar idea of above proof: Fix $\epsilon>0$. Let $E\in \mathcal{M}$ with $\mu(E)=\epsilon$. Consider the cover $E=\cup_i(E\cap X_i)$ and $X_i$ is as in definition of $\sigma$-finite. Let $E_j:=E\cap X_j$. Then $\mu(E_j)\le \mu(X_j)<\infty$.

However, we will get $0<\mu(E_j)<\epsilon$ for some $j$ with $E_j\in \mathcal{M}$...

Hermi
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    As it stands, your statement : If $\mu$ is σ-finite, then for any $ϵ>0$, there is a set $F∈M$ so that $ϵ<\mu (F)<∞$ is false. Take $X={1,2}, \mu({1})=\mu({2})=1, \mu(X)=2$ and $ϵ>1$. – Kavi Rama Murthy Jan 09 '24 at 09:19
  • @geetha290krm Oh, sorry, I miss one conditions $\mu(X)=\infty$. – Hermi Jan 09 '24 at 16:03

1 Answers1

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A useful trick is to let $Y_n = \bigcup_{i=1}^n X_i$, so that $Y_n \in \mathcal{M}$, $\mu(Y_n) < \infty$, $X = \bigcup_{n=1}^\infty Y_n$, and $Y_1 \subseteq Y_2 \subseteq \dots$. (Or to put it another way, in the definition of $\sigma$-finite, you can always assume without loss of generality that the $X_i$ are increasing.)

Now, using continuity from below, what can you say about $\mu(Y_n)$?

Nate Eldredge
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  • So do you mean that from continuity from below, $\lim_{n\to \infty} \mu(Y_n)=\mu(\lim_{n\to\infty} Y_n)=\mu(X)=\infty$. So for any $\epsilon>0$, there is a $N>0$ so that for any $n>N$, $\mu(Y_n)>\epsilon$? – Hermi Jan 10 '24 at 02:47
  • @Hermi: Yes, exactly. – Nate Eldredge Jan 10 '24 at 05:58
  • Thank you! But I just read this question:https://math.stackexchange.com/questions/234930/if-a-measure-is-semifinite-then-there-are-sets-of-arbitrarily-large-but-finite. Why do they use more complicated way? Say define $\mathcal{F}:={F\subset E: E \mbox{ is measurable and } 0<\mu(F)<\infty }$ and $C:=\sup \mu(F)$. They show that $C=\infty$... – Hermi Jan 10 '24 at 06:40
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    @Hermi: That question assumes only that the measure is semifinite. Your question assumes $\sigma$-finite which makes things much easier. – Nate Eldredge Jan 10 '24 at 07:03