Let be $X_n\sim\mathcal{N}(\mu_n,\sigma^2_n)$ and suppose that $X_n\to X$ in distribution. I have to show that $X$ is normal itself. It's a classical exercise and I found many posts related to this question (in particular this one), but I wouldn't want to use the notion of median, so I followed a different approach.
From the convergence in law I obtain the tightness of the sequence of the probability distributions $\mathbb{P}_{X_n}$, and the convergence of characteristic functions provides me the convergence (and so the boundedness) of variances $\sigma^2_n$.
Which I have to proove to conclude is that the sequence of means $\mu_n$ is bounded.
Let we suppose that $(\mu_n)_{n\ge1}$ is not bounded; $\forall M>0$ it will exist a subsequence $(\mu_{n_k})_{k\ge1}$ such that: $$|\mu_{n_k}|\ge M \ \ \forall k.$$
Since $$|\mu_{n_k}|=|\mu_{n_k}-X_{n_k}+X_{n_k}|\le|\mu_{n_k}-X_{n_k}|+|X_{n_k}|$$ we have that $\forall M>0$ it will exist a subsequence $(\mu_{n_k})_{k\ge1}$ such that $$1=\mathbb{P}\left\{|\mu_{n_k}|>M\right\}\le\mathbb{P}\left\{|\mu_{n_k}-X_{n_k}|>\frac M2\right\}+\mathbb{P}\left\{|X_{n_k}|>\frac M2\right\}.$$
From Chebyshev's inequality we obtain that $$\mathbb{P}\left\{|\mu_{n_k}-X_{n_k}|>\frac M2\right\}\le\frac{4\sigma^2_{n_k}}{M^2}\le\frac{C}{M^2}$$ so
$$\mathbb{P}\left\{|X_{n_k}|>\frac M2\right\}\ge1-\frac{C}{M^2}$$ which contradicts the hypotesis of tightness.
Is my proof correct?
