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