A question from the weak convergence of probability measure Show that the probability measure satisfy $\mu_n \Rightarrow \mu$ if $\mu_n(a,b] \rightarrow \mu(a,b]$ whenever $\mu\{a\}=\mu\{b\}=0$.
My attempt: Let $F_n$ and $F$ be the corresponding distribution functions, and $x_0$ and $x_1$ are continuity points of $F$ such that $F(x_0) < \epsilon$ and $F(x_1)>1-\epsilon$.
Thus,
\begin{eqnarray*}
\limsup_{n \rightarrow \infty}|F_n(x)-F(x)|&\leq \limsup_{n \rightarrow \infty}|F_n(x)-F_n(x_0)-(F(x)-F(x_0)|+|F_n(x_0)+F(x_0)| \\
 &\leq \limsup_{n \rightarrow \infty}|\mu_n(x_0,x]-\mu(x_0,x]|+\limsup_{n \rightarrow \infty} F_n(x_0)+F(x_0)
\end{eqnarray*}
The first term in the R.H.S of the above inequality goes to zero. How do I show that $\limsup_{n \rightarrow \infty} F_n(x_0)+F(x_0) <\epsilon$.
Can anyone give some hints for this part?
 A: Denote the c.d.f. associated with $\mu_{n}$ and $\mu$ by $F_{n}$
and $F$ respectively. Firstly, we go to show that $\{\mu_{n}\mid n\in\mathbb{N}\}$
is tight, in the following sense:
For any $\varepsilon>0$, there exists a compact set $K\subseteq\mathbb{R}$
such that $\mu_{n}(K)>1-\varepsilon$ for all $n\in\mathbb{N}$.
Let $\varepsilon>0$ be given. Choose $a<b$ such that $F$ is continuous
at $a$ and $b$ and that $F(b)-F(a)>1-\varepsilon$. Hence, $\mu(a,b]=F(b)-F(a)>1-\varepsilon$.
By assumption, $\mu_{n}(a,b]\rightarrow\mu(a,b]$, so there exists
$N$ such that $\mu_{n}(a,b]>1-\varepsilon$ whenever $n>N$. Choose
$K'=[a-1,b+1]$, then $\mu_{n}(K')>1-\varepsilon$ for all $n>N$.
For each $n=1,\ldots,N$, choose compact set $K_{n}$ such that $\mu_{n}(K_{n})>1-\varepsilon$.
Finally, define $K=K_{1}\cup\cdots\cup K_{N}\cup K'$, which is compact.
Clearly, we have that $\mu_{n}(K)>1-\varepsilon$ for all $n$.
Now, we go back to your question. We go to show that $F_{n}(b)\rightarrow F(b)$
whenever $F$ is continuous at $b$. Let $b\in\mathbb{R}$ be arbitrary
such that $F$ is continuous at $b$. Let $\varepsilon>0$ be aribtrary.
Choose a compact set $K$ such that $\mu_{n}(K)>1-\varepsilon$ and
$\mu(K)>1-\varepsilon$. By choosing $M>0$ sufficiently large, we
may assume that $K\cup\{b\}\subseteq[-M,M]$ and hence $\mu_{n}([-M,M])>1-\varepsilon$
and $\mu([-M,M])>1-\varepsilon$. Choose $a\in\mathbb{R}$ such that
$F$ is continuous at $a$, $a<-M$, and $F(a)<\varepsilon$ (This
is possible because $F(x)\rightarrow0$ as $x\rightarrow-\infty$
and $F$ has at most conuntably many discontinuous points). By assumption,
$F_{n}(b)-F_{n}(a)=\mu_{n}(a,b]\rightarrow\mu(a,b]=F(b)-F(a)$, so
there exists $N$ such that
$$
\left|\mu_{n}(a,b]-\mu(a,b]\right|<\varepsilon
$$
whenever $n\geq N$. Note that $F(a)=\mu(-\infty,a]\leq\mu\left([-M,M]^{c}\right)<\varepsilon$
and similarly $F_{n}(a)=\mu_{n}(-\infty,a]\leq\mu_{n}\left([-M,M]^{c}\right)<\varepsilon$
for all $n$. Hence, for any $n\geq N$
\begin{eqnarray*}
 &  & \left|F_{n}(b)-F(b)\right|\\
 & = & \left|\left[F_{n}(b)-F_{n}(a)\right]-\left[F(b)-F(a)\right]+F_{n}(a)-F(a)\right|\\
 & \le & \left|\mu_{n}(a,b]-\mu(a,b]\right|+F_{n}(a)+F(a)\\
 & < & 3\varepsilon.
\end{eqnarray*}
This shows that $\mu_{n}\Rightarrow\mu$.
