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You have $E(|1/X|) < \infty$, then how exactly follows $P(X=0)=0$?

First I would put $ E(|1/x|) \geq \int_{ \{ \omega: X(\omega)=0 \} } |\frac{1}{X}| P(d\omega)$

The integral would be zero for $P(X=0)=0$ and $\infty$ in all other cases, is this a right way and how should one detail it?

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    $\begingroup$ $\frac 1{|X|}$ is almost surely bounded and $P(|X|\neq 0) =\lim_{n\to +\infty}P\left(\frac 1{|X|}\leq n\right)$. $\endgroup$
    – Davide Giraudo
    Aug 2, 2011 at 9:36

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In what follows, suppose that $|1/0|$ is defined to be equal to $+\infty$.

First, if ${\rm P}(X=0) > 0$, then $|1/X|$ might not even be defined as a random variable, since it is not real-valued. (By a common definition, a random variable is real-valued.)

However, suppose that we allow $|1/X|$ to take the value $+\infty$. Note that $$ \bigg|\frac{1}{X}\bigg| \ge \infty {\mathbf 1}(X = 0), $$ where ${\mathbf 1}$ is the indicator function. Hence $$ {\rm E}\bigg|\frac{1}{X}\bigg| \ge {\rm E}[\infty {\mathbf 1}(X = 0)] = \infty {\rm P}(X = 0). $$ So the condition ${\rm E}|1/X| < \infty$ implies that ${\rm P}(X = 0)=0$.

EDIT: With regard to your approach, indeed $$ {\rm E}\bigg|\frac{1}{X}\bigg| \ge \int_{\{ \omega :X(\omega ) = 0\} } {\bigg|\frac{1}{{X(\omega )}}\bigg|{\rm P}(d\omega )} . $$ Now, using the definition $|1/0|=+\infty$, we have $$ \int_{\{ \omega :X(\omega ) = 0\} } {\bigg|\frac{1}{{X(\omega )}}\bigg|{\rm P}(d\omega )} = \int_{\{ \omega :X(\omega ) = 0\} } {\infty {\rm P}(d\omega )} = \infty {\rm P}(X = 0), $$ and the conclusion is as before.

Remark: Note that $\infty {\rm P}(X=0)$ is defined to be equal to $0$ when ${\rm P}(X=0)=0$.

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