Convergence in distribution absolute value of random variables.

Question

How to proof from definition of convergence in distribution that fact:

Let $X_{1}, X_{2}, . . .$ and $X$ be real random variables with respective distribution functions $F_{X_1}, F_{X_2}, . . .$ and $F_{X}$. If $X_{n}$ converges in distribution to random variable $X$ then $|X_{n}|$ converge in distribution to $|X|$.

We say that sequence of random variable $X_{1}, X_{2}, . . .$ converges in distribution to $X$ if and only if $ \displaystyle \lim_{n \rightarrow \infty} F_{X_n}(x)=F_{X}(x)$ for every $x \in \mathbb{R}$ at which $F_X$ is continuous.

I will be very grateful for the tips.

Answer 1

Going by the comment, \begin{align*} \displaystyle \lim_{n \rightarrow \infty} F_{\vert X_n \vert } (x) &= \displaystyle \lim_{n \rightarrow \infty} P(\vert X_n \vert \leq x) \\ &= \displaystyle \lim_{n \rightarrow \infty} P(-x \leq X_n \leq x) \\ &= \displaystyle \lim_{n \rightarrow \infty} \{ P(X_n \leq x) - P(X_n < -x) \} \\ &= \displaystyle \lim_{n \rightarrow \infty} P(X_n \leq x) - \displaystyle \lim_{n \rightarrow \infty} P(X_n < -x) \\ &= P(X \leq x) - P(X < -x) = P(\vert X \vert \leq x), \end{align*} where the fourth inequality follows by the algebraic properties of limits.