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I am trying to prove every $\sigma$-finite measure is semifinite. This is what I have tried:

Definition of $\sigma$-finiteness: Let $(X,\mathcal{M},\mu)$ is a measure space. Then, $ \mu$ is $\sigma$-finite if $X = \bigcup_{i=1}^{\infty}E_i$ where $E_i \in \mathcal{M}$ and $\mu(E_i) < \infty$ for all $ j \in N$. (Real Analysis: Modern Techniques and Their Applications 2nd Edition by Foland).

Definition of semifiniteness: $\mu $ is simifinite if for each $E \in \mathcal{M}$ with $\mu(E) = \infty$ $\exists$ $F \subset E$ and $F \in \mathcal{M}$ and $0 < \mu(F) < \infty$.

So, take $A$ s.t. $\mu(A) = \infty$. We know $X \cap A = A$. Then, $A = A \cap \bigcup E_j$ hence $A = \bigcup E_j \cap A$. By subadditivity,

$$\infty = \mu(A) = \mu\left(\bigcup E_j \cap A\right) \leq \sum_1^{\infty} \mu(E_j \cap A) $$

OK, I am here. But I do not understand how to continue, or even this is a right approach. Thanks.

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    $\begingroup$ I think you meant $F\subseteq E$ in the definition of semifinite. Also there are something off in the definition of $\sigma$-finite (e.g. "There are $E_i$ such that...). $\endgroup$
    – Asaf Karagila
    Nov 11, 2012 at 14:30
  • $\begingroup$ I think your method is a good one. Assume all the sets $E_{j} \cap A$ have measure 0, then A must have measure 0, so at least one is positive, and we already know they aren't infinite because they are the special sets from the definition of sigma-finite. $\endgroup$ Aug 1, 2021 at 20:34

2 Answers 2

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We can find $N$ such that $\mu\left(A\cap E_N\right)>0$ (otherwise, we would have for each $n$ that $\mu\left(A\cap\bigcup_{j=1}^nE_j\right)=0$ and $\mu\left(A\right)=\lim_{n\to +\infty}\mu\left(A\cap\bigcup_{j=1}^nE_j\right)$), and we have $\mu\left(A\cap E_N\right)\leqslant \mu\left( E_N\right)<+\infty$. Furthermore, $A \cap E_N\subset A$, hence the choice $F:=A\cap E_N$ does the job. This proves that $\mu$ is semi-finite.

The converse is not true: counting measure on the subsets of $[0,1]$ is semi-finite but not $\sigma$-finite.

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  • $\begingroup$ Hi, is there any reason you use the union of $E_j$ instead of $E_j$ by themselves? We can find N such that $u(A \cap E_N)>0$ as otherwise since $A = \cup_{j} (E_j \cap A)$, the measure of A would be 0 by subadditivity, contradiction? $\endgroup$
    – user172377
    Nov 12, 2017 at 20:56
  • $\begingroup$ I have the same question as the above user. $\endgroup$
    – user389056
    Jan 3, 2018 at 22:44
  • $\begingroup$ @user172377 Indeed, this is not needed and shortens the proof. Thanĸs. $\endgroup$ Jan 3, 2018 at 23:37
  • $\begingroup$ @user389056 I have edited. Thanĸs for reminding. $\endgroup$ Jan 3, 2018 at 23:37
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Is it possible to think in this direction: From the definition of $\sigma-$ finite measure the $E_i$'s are finite measure since $\nu(E_i)<\infty\:\forall\,i$. Now we know from the definition that $X=\bigcup_{I=1}^{\infty}E_i$ hence for arbitrary E u pick from the sequence of $E_i, E\subset X$. We only need to show that $\nu(X)=\infty$. $\nu(X)=\nu\left(\bigcup_{I=1}^{\infty}E_i\right)$ $=\sum_{i=1}^{\infty}\nu(E_i)=\infty$.

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