Why is there $M$ s.t. $\mathbb P\{X_\infty \notin [-M,M]\}<\varepsilon$? Let $X_n$, $n\in [1,\infty ]\cap \mathbb N$ v.a. s.t. $\mathbb P\{X_n\in \mathbb Z\}=1$ for all $n\in [1,\infty ]$. I want to show that if $\mathbb P\{X_n=m\}\longrightarrow \mathbb P\{X_\infty =m\}$ then $X_n\longrightarrow X_\infty $ in distribution. The proof start by : Let $x\in\mathbb R$ and $\varepsilon>0$. Let $M\in\mathbb Z$, $M>x$ s.t. $$\mathbb P\{X_\infty \notin[-M,M]\}<\varepsilon.$$
My question : Why such an $M$ exist ? 

Let $\varepsilon>0$. Suppose that $$\forall M>x, \mathbb P\{X_\infty \notin[-M,M]\}>\varepsilon.$$
In particular, $$\mathbb P\{X_\infty \notin \mathbb R\}>\varepsilon,$$
what contradict $\mathbb P\{X_\infty \in\mathbb Z\}=1$. 
