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I found this one

Finite almost surely implies integrable?

which shows that $X$ finite a.s. does not imply $X$ integrable...

But how can I show that the converse is true?

Many thanks in advance.

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    $\begingroup$ If $P[X=\infty]>0$ then $E[|X|]=E[|X|\cdot 1_{X<\infty}] + E[\infty\cdot 1_{X=\infty}]\geq \infty\cdot P[X=\infty]=\infty$. $\endgroup$
    – user539964
    Commented Mar 15, 2018 at 17:41

1 Answer 1

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Let $S_{n}=\{\omega\in \Omega: |X(\omega)|\geq n\}$ and $S=\displaystyle\bigcap_{n=1}S_{n}$, then $P(S)\leq P(S_{n})\leq n^{-1}E[|X|]\rightarrow 0$, so $P(S)=0$. Then $P[|X|<\infty]=1-P(S)=1$. Since $P[|X|<\infty] \leq P[X<\infty]$, now we have $P[X<\infty] = 1$.

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  • $\begingroup$ So you say, basically, "use Markov's inequality"? $\endgroup$
    – dEmigOd
    Commented Mar 15, 2018 at 19:35
  • $\begingroup$ Yes, Markov's inequality takes place for that $P(S_{n})\leq n^{-1}E[|X|]$. $\endgroup$
    – user284331
    Commented Mar 15, 2018 at 19:37

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