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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!

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1 Answer 1

up vote 2 down vote accepted

The functions $F_n$ have the special property that $F_n(x)$ can only be zero or one, for every $x$. For every $x$ in the dense set where $F_n(x)\to F(x)$, the same is true for $F(x)$, i.e., either $F(x)=0$ or $F(x)=1$. Since $F$ is right continuous, we have $F(x)=0$ or $F(x)=1$ for every $x\in \mathbb{R}$.

The function $F$ is non-decreasing and non-constant, and has $F(x)\in\{0,1\}$ for every $x$. The only such functions are $\chi_{[\alpha,\infty)}$ for $\alpha\in\mathbb{R}$.


Added: (1) You do not need to show a specific relationship. The proof does not assume that $\lim_n \alpha_n$ exists. As in my solution, you first show that $\alpha$ exists, and then prove that $\alpha_n\to\alpha$. You yourself said that this is easy.

(2) Any dense set will do.

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Thank you, Byron. I edited the question to make it more specific. Could you please further clarify it? Thanks. –  user16859 Apr 9 '12 at 1:38
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