Does converge in distribution imply limit being finite almost surely? If I have a sequence of random variable $x_n$ converge in distribution to a standard normal random variable, i.e.,
$$x_n\overset{d}{\to}N(0,1)$$
Does the following hold? Why or why not?
$$p(\lim_{n\to\infty}x_n<\infty)=1$$

Edited on 02/22/2018
More specifically, think about $x_n=\dfrac{1}{\sqrt{n}}\sum_{i=1}^{n}y_{I}$ where $y_i$'s are i.i.d. rv with zero mean and variance 1. By CLT you would have
$$x_n\overset{d}{\to}N(0,1)$$
The reason that I care about $p(\lim_{n\to\infty}x_n<\infty)=1$ is because that I am trying to find the limit of the following integral
$$\int f_n(t,x_n) dt$$
and I can show that $\lim_{n\to\infty}f_n(t,x_n)=f_{0}(t, x_n)$ and $f_{0}(t,x_n)$ is integrable no matter what $x_n$ is. I want to apply DCT and I have been able to show that for sufficiently large $n$,
$$f_n(t,x_n)<g(t)\exp\{x_n\},$$
where $g(t)$ is integrable. Hence it seems to me that as long as $\int g(t)\exp\{x_n\} dt<\infty$ holds almost surely I can apply DCT.

The Law of Iterated Logarithm:
https://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm
show that the opposite is true: $P\{\limsup x_n =\infty\} =1$
