# Use Markov inequality to prove that a positive random variable is null a.s

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

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