Why $\mu_n(A)=\frac{1}{\sqrt{2\pi n}}\int_A e^{-\frac{x^2}{2n}}dx$ doesn't converges weakly? Let $\mu_n(A)=\frac{1}{\sqrt{2\pi n}}\int_A e^{-\frac{x^2}{2n}}dx$. In my lecture, we take $\varphi _n(x)=e^{-\frac{nx^2}{2}}\to \boldsymbol 1_{\{0\}}(x)$. $\varphi _n$ are the characteristic function of $\mathcal N(0,n)$ law. In my leture, they say that since $\mu_n(-\infty ,x]\to 1/2$ for all $x$, $\mu_n$ doesn't converges weakly, and thus $\boldsymbol 1_{\{0\}}$ is not the caracteristic function of any r.v.
Question 
1) Why $\mu_n$ doesn't converges weakly ? I don't understand the argument.
2) Why the fact that $\varphi _n\to \boldsymbol 1_{\{0\}}$ and $\mu_n$ doesn't converges weakly yield that $\boldsymbol 1_{\{0\}}$ is not the caracteristic function of a random variable ? 

$\mu_n$ is the distribution of $X_n=\sqrt {n} X$ where $X$ has standard normal distribution. Since $|X_n| \to \infty$ a.s. it follows that $\mu_n$ does not converge weakly.
Details: $P(X_n >M)=P(X >\frac  M {\sqrt n}) \to \frac  12 $  for every $M>0$. If $X_n \to Y$ weakly then we get $P(Y>M) \geq \frac  1 2$ whenever $M>0$ and $M$ is a continuity point for $Y$. This is clearly impossible since $P(Y>M)$ must tend to $0$ as $ M \to \infty$. 
