Suppose $\{\mu_{n}\}$ is a sequence of probability measure on $\mathbb{R}$ such that $\mu_{n}(A)=\delta_{\alpha_{n}}(A)=\begin{cases} 1, & \alpha_{n}\in A\\ 0, & \alpha_{n}\notin A \end{cases}.$ Suppose $\delta_{\alpha_{n}}\overset{w}{\rightarrow}\mu$. Show that $\exists\alpha$ such that $\mu=\delta_{\alpha}$ and $\alpha_{n}\rightarrow\alpha$ .
Proof: $F_{n}(x)=\mu_{n}(-\infty,x]=\chi_{[\alpha_{n},\infty)}$
$F(x)=\mu(-\infty,x]$
I don't understand the following step:
$\mu_{n}\overset{w}{\rightarrow}\mu\Rightarrow F_{n}(x)\rightarrow F(x)$ on a dense subset$\Rightarrow F(x)=\mu(-\infty,x]=\chi_{[\alpha,\infty)}$ where $\alpha_n\rightarrow\alpha$.
Question: (1) Do I need to show the specific relationship between $\alpha_n$ and $\alpha$? For example, take the sup or inf. The above proof seems to naturally assumes that $\lim \alpha_n$ exists. (2) Does the dense set have to be any specific dense set?
Assume that we can change the question. Assume $\delta_{\alpha_{n}}\overset{w}{\rightarrow}\delta_{\alpha}$. Then, I can easily prove $\alpha_{n}\rightarrow\alpha$. So I guess the real problem for me is $\mu=\delta_{\alpha}$.
Thank you!