Proof that $(X_n \Rightarrow\mathcal{N}(0,1)) \implies (-X_n \Rightarrow \mathcal{N}(0,1))$, where $"\Rightarrow"$ convergence in distribution Assume that a sequence of random variables $(X_n)$ converges in distribution to $\mathcal{N}(0,1)$, i.e. that for every $x \in \mathbb {R}$ we have
$$\lim_{n \to \infty}\mathbb{P}(X_n \leq x) = \Phi(x).$$
The question is does that imply that the sequence $(-X_n)$ also converges in distribution to $\mathcal{N}(0,1)$. I'm not sure how to prove that it does (?). This is my attempt so far.
For any $x \in \mathbb{R}$,
\begin{split}
& \lim_{n \to \infty}\mathbb{P}(X_n \leq- x)=\Phi(-x) \\
\implies & \lim_{n \to \infty}\mathbb{P}(X_n \leq -x)=1-\Phi(x) \\
\implies & \lim_{n \to \infty}\left[\mathbb{P}(X_n \leq -x)-1\right]=-\Phi(x) \\
\implies & \lim_{n \to \infty}\left[1-\mathbb{P}(X_n \leq -x)\right]=\Phi(x) \\
\implies & \lim_{n \to \infty}\mathbb{P}(X_n > -x)=\Phi(x) \\
\implies & \lim_{n \to \infty}\mathbb{P}(-X_n < x)=\Phi(x), \\
\end{split}
so we yet have to prove that for every $x \in \mathbb{R}$
$$ \lim_{n \to \infty}\mathbb{P}(-X_n < x)= \lim_{n \to \infty}\mathbb{P}(-X_n \leq x).$$
Let's assume opposite, i.e. that there's an $x \in \mathbb{R}$ such that
$$\lim_{n \to \infty}\left[\mathbb{P}(-X_n \leq x)-\mathbb{P}(-X_n < x)\right] \neq 0.$$
Now, we have
$$\lim_{n \to \infty} \mathbb{P}(-X_n=x)\neq 0,$$
i.e.
$$\lim_{n \to \infty} \mathbb{P}(X_n=-x)\neq 0,$$
which means that the limiting cumulative distribution has a discontinuity at the point $-x$. But I'm not sure what to do with that fact - how to get a contradiction with my starting assumption. Thanks for any help, I appreciate it.
 A: It is sufficient to show that: $Z \sim N(0,1) \Rightarrow -Z \sim N(0,1)$. 
\begin{align}
\Phi(z) &= P(Z < z) \\
&= P(-Z > -z) \\
&= 1 - P(-Z < -z) \\
&= 1- F_{-Z}(-z)\\
&= 1 - \Phi(-z)
\end{align}
This implies that $Z$ and $-Z$ have the same CDF. So if $F_n \Rightarrow F_Z$, then clearly $F_n \Rightarrow F_{-Z}$.
A: You have $\lim_{n \to \infty}\mathbb{P}(-X_n < x)=\Phi(x)$ and want to prove that 
$$\lim_{n \to \infty}\mathbb{P}(-X_n < x)= \lim_{n \to \infty}\mathbb{P}(-X_n \leq x).$$
You cannon get the contradiction without using continuity of limiting function.
Choose small $\delta>0$ and bound the r.h.s.:
$$ 
\mathbb{P}(-X_n < x)\leq\mathbb{P}(-X_n \leq x) \leq \mathbb{P}(-X_n < x+\delta).
$$
Take limits of all sides and get
$$
\Phi(x)=\lim_{n\to\infty}\mathbb{P}(-X_n < x)\leq\liminf\mathbb{P}(-X_n \leq x)\leq\limsup\mathbb{P}(-X_n \leq x) \leq 
$$
$$
\leq\lim_{n\to\infty}\mathbb{P}(-X_n < x+\delta)=\Phi(x+\delta).$$
Since the above inequalities hold for any $\delta>0$, one can consider $\delta\to0$ and use the continuity $\Phi(x+\delta)\to\Phi(x)$ as $\delta\to0$.
