Weak convergence of measures with compact support Suppose $\mu_n$ is a sequence of distributions that converges weakly to some $\mu$. In addition suppose that each $\mu_n$ has compact support $[0,\bar x_n]$, but it is not guaranteed that $\bar x_n$ converges (I couldn't prove it). Is it true that $\mathbb{E}_n(x):=\int xd\mu_n\rightarrow\mathbb{E}(x):=\int xd\mu$? 
I know that, since $f(x)=x$ is continuous but not bounded, this statement would be false in general, however I thought compactness of the support for each $\mu_n$ might impose more structure to the problem. I'd really appreciate any input.
 A: For a simpler example, let $\mu_n$ be a discrete measure that puts mass $1/n$ on the point $n$ and $1-1/n$ on $0$.  Let $\mu$ put mass 1 on 0.  Then $\mu_n$ is supported in the compact interval $[0,n]$ and $\mu_n \to \mu$ weakly, but $\int x\,d\mu_n = 1$ for all $n$ while $\int x\,d\mu = 0$.
Alternatively, let $\mu_n$ put mass $1/n$ on the point $n^2$ and $1-1/n$ on $0$.  Then you get $\int x\,d\mu_n = n \to +\infty$.
A: In general, no. Consider this counterexample. Let $m$ be the lebesgue measure and define $\mu$ by $\mu(A)=m(A\cap[0,1])$ and $\mu_n$ by 
$$
\mu_n(A)=m\left(A\cap\left[0,1-\frac{1}{n}\right]\right)+\frac{1}{n(n-1+\frac{1}{n})}m\left(A\cap\left(1-\frac{1}{n},n\right]\right)
$$
Both are probability measures. The support of $\mu$ is $[0,1]$ while the support of $\mu_n$ is $[0,n]$. 

Claim 1: $\mu_n$ converges weakly to $\mu$.

Proof: Let $f$ be a continuous bounded function, then we have
\begin{eqnarray}
\left\vert \int f d\mu - \int f d\mu_n \right\vert & = & \left\vert \int_{1-\frac{1}{n}}^1 fdm - \int_{1-\frac{1}{n}}^n \left(\frac{1}{n(n+\frac{1}{n}-1)}\right)fdm\right\vert\\
& \leq & \int_{1-\frac{1}{n}}^1\vert f \vert dm + \left(\frac{1}{n(n+\frac{1}{n}-1)}\right)\int_{1-\frac{1}{n}}^n \vert f \vert dm\\
& \leq & \frac{M}{n}+ \frac{M(n+\frac{1}{n}-1)}{n(n+\frac{1}{n}-1)}\\
& = & \frac{2M}{n}
\end{eqnarray}
which goes to 0 as $n$ becomes very large. Here $M=\max_{x\in\mathbb{R}}\vert f(x)\vert$

Claim 2: $\mathbb{E}_n[X]$ doesn't converge to $\mathbb{E}[X]$

Proof: 
$$
\mathbb{E}[X]=\int_0^1 xdx=\frac{1}{2}
$$
while 
\begin{eqnarray}
\mathbb{E}_n[X] & = & \int_0^{1-\frac{1}{n}} xdx+\frac{1}{n(n+\frac{1}{n}-1)}\int_{1-\frac{1}{n}}^n xdx\\
& = & \frac{(1-\frac{1}{n})^2}{2}+\frac{n^2-(1+\frac{1}{n})^2}{2n(n+\frac{1}{n}-1)}
\end{eqnarray}
which converges to 1 as $n$ goes to $\infty$. 

However, we can prove this:

Claim 3: $\liminf_{n\to\infty} \bar{x}_n\geq \bar{x}$

Proof: By contradiction, assume $\bar{y}=\liminf_{n\to\infty} \bar{x}_n<\bar{x}$. Then there is a subsequence $\bar{x}_{k_n}$ such that $\bar{x}_{k_n}\to\bar{y}$. Then, there is some $n_0$ such that for $n\geq n_0$, $\bar{x}_{k_n}<(\bar{x}+\bar{y})/2$. 
Consider a continuous positive function whose support is included in $((\bar{x}+\bar{y})/2,\bar{x})$. Then for all $n\geq n_0$, 
$$
\int fd\mu_{k_n}=0
$$
while 
$$
\int fd\mu >0
$$
This contradicts that $\mu_n$ converges to $\mu$ weakly. 
This also implies that if for all $n$, $\bar{x}_n\leq \bar{x}$, then $\bar{x}_n\to\bar{x}$. In that case we can also conclude $\mathbb{E}_n[X]$ converges to $\mathbb{E}[X]$. To prove this you can consider the continous bounded function $f$ defined by 
$$
f(x)=
\begin{cases}
0 & x<0\\
x & x\in[0,\bar{x}]\\
\bar{x} & x>\bar{x}
\end{cases}
$$
and use $\mu_n$ converges weakly to $\mu$. 
