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Let $(\Omega, \mathcal{A}, \mu)$ be a measure space. Let $f:\Omega \to [0, \infty]$ $\mathcal{A}-$measurable such that $\int_{\Omega}f~d\mu< \infty$. Prove that the set $$\Omega_+^f= \lbrace x \in \Omega ~|~ f(x)>0 \rbrace$$ is $\sigma-$finite.

I know that i need to show that there exists $(E_n)_{n \in \mathbb{N}} \subset \mathcal{A}$ with $\bigcup_{n=1}^{\infty}=\Omega_+^f$ and $\mu(E_n)<\infty$ for all $n \in \mathbb{N}$. But i don't know how to continue?

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  • $\begingroup$ I'm thinking about the connection between $\int_{\Omega}f~d\mu< \infty$ and $\mu$ to show $\mu(E_n)< \infty$ for all $n$? $\endgroup$
    – user487904
    Commented Nov 15, 2017 at 21:56

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For $n\in\Bbb N$ let $A(n)=f^{-1}[1/n,n].$ Then $1\leq nf(x) $ for $x\in A(n)$ so $\mu(A(n))\leq n\int_{A(n)}fd\mu \leq n\int_{\Omega}fd\mu<\infty$.

Now $f^{-1}(0,\infty)=\cup_{n\in \Bbb N}A(n).$ And with $B= f^{-1}\{\infty\}$ we have $\mu(B)=0$, otherwise $\int_{\Omega}fd\mu\geq \int_B fd\mu=\infty. $

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  • $\begingroup$ +1 Is there a reason why $A_n = f^{-1}[\frac{1}{n}, \infty)]$ wouldn't work as well? $\endgroup$ Commented Nov 15, 2017 at 22:25
  • $\begingroup$ @HennoBrandsma. Of course it's just as good.. $\endgroup$ Commented Nov 15, 2017 at 22:29

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