Outcome of unmeasurable probability I consider a standard normal random variable $X$ and a Vitali set $V$.
$P(X\in V)$ can not be computed as $V$ is not measurable.
Now I consider the outcome of the following experiment $E_S$ : $N_S$ is the number of experiments $X_i$ (all independent and equals to $X$, and $i\in\mathbb N$) such that $X_i\in S$.


*

*If $P(X\in S)=0$ then $P(N_S=0)=1$

*If $P(X\in S)>0$ then $P(N_S=\infty)=1$

*What happens for $S=V$ ? I think that $P(N_V=\infty)=0$ and $P(N_V=0)<1$. Am I right and can we obtain some more precise results ?


Thank you for your answers ! 
 A: It is a bit surprising to see that this post is still alive... but here we go. The question is:

What happens for $S=V$?

In a nutshell, and as was already explained in the comments, what happens is that nothing guarantees that the sets $[N_V=\infty]$ or $[N_V=0]$ are measurable, hence neither $P(N_V=\infty)$ nor $P(N_V=0)$ is defined. Thus, asking if these probabilities are $0$ or $\lt1$ or whatever has no sense.
Let us recall why (the function) $N_S$ is measurable when (the subset) $S$ is measurable. One writes
$$
N_S=\sum_{i=1}^{+\infty}\mathbf 1_{A^S_i},\qquad A^S_i=[X_i\in S],
$$
and, by the definition of the measurability of $X_i$ and $S$, each set $A^S_i$ is measurable hence each function $\mathbf 1_{A^S_i}$ is measurable and, by measurability of pointwise limits, the function $N_S$ is measurable.
When $V$ is not measurable, the reasoning above breaks down at the moment when one needs each $A^V_i$ to be measurable. For example,
$$
[N_V=0]=\bigcap_{i=0}^{+\infty}(\Omega\setminus A_i^V)=\bigcap_{i=0}^{+\infty}[X_i\notin V],
$$
and none of the subsets $[X_i\notin V]$ is measurable, a priori. If you find a way to prove that these subsets are in fact measurable, or only that their whole intersection is measurable (something which could happen without every $[X_i\notin V]$ being measurable), please go ahead. Otherwise...
