# (How) follows P(X=0)=0 if E(1/X) finite?

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?

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

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 ${\rm P}(X=0)=0$.