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I would like to show that if the expectation of a positive random variable $X$ (integrable) is $0$ then $X$ is null almost surely.

I thought on a proof and would like to know if it is correct, it uses the Markov inequality.

We have for all $\varepsilon>0$

$$ \mathbb{P}(X>\varepsilon) = 0 $$

This implies that $\mathbb{P}(X> 0) = 0$. Indeed, suppose it is not the case, then there exists a set in which $X(\omega)>0$ with nonzero probability. This means that we can find an $\eta>0$ such that $\mathbb{P}(X>\eta)\neq 0$. But this is not possible using Markov inequality with $\varepsilon=\eta$.

Thus we conclude that $\mathbb{P}(X> 0) = 0$ and since $X$ is non negative we get

$$ \mathbb{P}(X =0) = 1 - \mathbb{P}(X>0) = 1 $$

Is this correct and if not what can I improve please ?

Thank you a lot

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    $\begingroup$ $P(X>0)=\lim_kP(X>1/k)=0$ due to continuity of measures $\endgroup$
    – Snoop
    Sep 18 at 18:13
  • $\begingroup$ @Snoop Thank you for this remark it is more efficient indeed ! But what about my argument is it false ? $\endgroup$
    – coboy
    Sep 18 at 18:22

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