# Probabilistic statement

I'm confused about a conclusion in a proof:

Suppose we have a sequence $(X_n)$ of r.v. with values in $\mathbb{N}$, which onverges to a r.v. $X\in[0,\infty)$ a.s. Therefore for large $n$ we know $X_n=X$.

Further suppose for any $l\ge1$ and $N\ge0$ we have:

$$P\left(\bigcap_{n\ge N}\{X_n=l\}\right) =0$$

Now the question is, why can I deduce from this facts, that $P(X\ge1)=0$?

Is the property true for any $l\geq 1$ and $N\leq 0$? In this case, the set $\{X=l\}$ is contained in the set $\{\omega\in\Omega, \exists N(\omega)\mid X_n(\omega)=l \forall n\geq N(\omega)\}$. Since $P(\bigcup_N\bigcap_{n\geq N}\{X_n=l\})=0$, you can conclude. – Davide Giraudo Jan 14 '12 at 17:28
it's true that for any $l\ge 1$ and $N\ge0$! I edited my question. But I do not quite understand your comment. Do you mean $\{X_n=l\}$? $P(\cup_N\cap_{n\ge N}\{X_n=l\})=0$ follows from sigma additivity, right? – user20869 Jan 14 '12 at 17:40
Let $l\geq 1$. Since $X_n$ gets its values on $\mathbb N$, the set $\{\omega\in\Omega, X(\omega)=l\}$ is equal the set $\{\omega\in\Omega, \exists N(\omega)\mid X_n(\omega)=l \,\forall n\geq N(\omega)\}$. Since $$P(X=l)=P\left(\bigcup_N\bigcap_{n\geq N}\{X_n=l\}\right)=0,$$ we got $P(X=l)=0$ for all $l\geq 1$, and since $X(\omega)$ is almost everywhere a limit of integer, $X(\omega)$ is an integer. So $P(X\geq 1)=\sum_{l=1}^{+\infty}P(X=l)=0$.