Let $X_n \sim \mathcal{N}(\mu_n,\sigma_n^2)$. Prove that if $X_n \rightarrow X$ in distribution, then either $X$ is normally distributed or there exists a constant $c$ such that $X = c$ almost surely.
Here is my attempt. We have that
$$\lim_{n\rightarrow\infty} \int_{-\infty}^x\frac{1}{\sqrt{2\pi}\sigma_n}e^{-\frac{(t-\mu_n)^2}{2\sigma_n^2}}dt = P(X \leq x)$$ for all $x$ at which $F(x) = P(X \leq x)$ is continuous.
Now let $v = \frac{t-\mu_n}{\sigma_n}$ and define $G(x)$ as the cumulative distribution function of a standard normal variable, i.e. $G(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{v^2}{2}}dv$. Then the above limit expression becomes
$$\lim_{n\rightarrow\infty} G\left(\frac{x-\mu_n}{\sigma_n}\right) = P(X \leq x)$$
At this point a "miracle" happens and I get $$\lim_{n\rightarrow\infty} \frac{x-\mu_n}{\sigma_n} = G^{-1}(P(X \leq x))$$
Since $$\lim_{n\rightarrow\infty} \frac{x-\mu_n}{\sigma_n}$$ exists everywhere according to what I wrote above, I claim that the limits $\lim_{n\rightarrow\infty}\sigma_n$ and $\lim_{n\rightarrow\infty}\mu_n$ must also exist and from that I "show" that $X$ is normal. I have no justification whatsoever why these two limits must exist by the way.
Reflecting back on what I wrote I don't see any reason why the statement in the problem should be restricted to normal r.v.'s except the transformation I used, which preserves normality. I don't know if normal is the only distribution that is invariant under this transformation. Anyway, I would appreciate if someone could tell me if I am working in the right direction or just give a hint that will put me on the right path. Thanks!
