$P[\lim_{n \rightarrow \infty} X_n = \infty]=0$ for tight sequence of real valued random variables? Let $X_1,X_2,...$ be a tight sequence of real valued random variables.
Is it true that $P\left[\lim\limits_{n \rightarrow \infty} X_n = \infty\right]=0$ ?
I think this fact is used in the paper I am reading, but I can not see why this holds. I would be grateful for hints or a solution.
 A: Assuming all random variables are defined on a common probability space $(\Omega,\mathscr{F},\mathbb{P})$,
$$\{\omega\in\Omega:\lim_nX_n(\omega)=\infty\}=\bigcap^\infty_{n=1}\bigcup^\infty_{m=1}\bigcap^\infty_{k=m}\{\omega\in\Omega:X_k(\omega)>n\}$$
Then, by the monotonicity of the sequences $E(n)=\bigcup^\infty_{m=1}\bigcap^\infty_{k=m}\{\omega\in\Omega:X_k(\omega)>n\}$, and the monotonicity of the sequence $E(n,m)=\bigcap^\infty_{k=m}\{\omega\in\Omega:X_k(\omega)>n\}$ in $m$,
$$\begin{align}
\mathbb{P}\big(\lim_n X_n=\infty\big)=\lim_n\mathbb{P}\Big(\bigcup^\infty_{m=1}\bigcap^\infty_{k=m}\{X_k>n\}\Big)&=\lim_n\lim_m\mathbb{P}\Big(\bigcap^\infty_{k=m}\{X_k>n\}\Big)\\
&\leq\lim_n\liminf_m\mathbb{P}\big(X_m>n)
\end{align}
$$
Can you finish from this?
A: By the Fatou's Lemma, for any $a > 0$,
\begin{align*}
\liminf_{n\to\infty} \mathbf{P}(X_n > a)
&\geq \mathbf{P}\Bigl(\liminf_{n\to\infty} \, \{X_n > a\}\Bigr) \\
&= \mathbf{P}(X_n > a \text{ eventually in }n) \\
&\geq \mathbf{P}\Bigl( \lim_{n\to\infty} X_n = +\infty \Bigr).
\end{align*}
Since $(X_n)$ is (uniformly) tight, for each $\varepsilon > 0$ we can find $a$ such that $\mathbf{P}(|X_n| > a) < \varepsilon$ for any $n$, which in turn implies
$$\mathbf{P}\Bigl( \lim_{n\to\infty} X_n = +\infty \Bigr) \leq \varepsilon$$
for any $\varepsilon > 0$. Therefore the claim follows by letting $\varepsilon \downarrow 0$.
(Of course, this is more or less the same as what @Oliver Diaz explained, formulated in a slightly more human friendly way.)
A: Lazy proof:
Let $A = \{X_n \to \infty\}$ and let $Y_n = 1_A X_n$, so that $Y_n \to \infty 1_A$ a.s.  Now $|Y_n| \le |X_n|$, so the sequence $Y_n$ is also tight.  By Prokhorov's theorem, passing to a subsequence, $Y_n$ converges weakly to some random variable $Y$ which is a.s. finite.  But $Y_n \to \infty 1_A$ a.s., so $Y = \infty 1_A$.  We conclude $P(A)=0$.
