I'm stuck with the next problem
Let $\mu_n,\mu$ be probability measures; show that if
$$\int f\,d\mu_n\to\int f \,d\mu $$
for all continuous $f$ with bounded support, then $\mu_n\to \mu.$
$\mu_n\to\mu$ means $\mu_n(-\infty,t]\to\mu(-\infty,t]$ for all $t$ such $\mu(\{t\})=0$, I already know that this is equivalent to,
$\int f\,d\mu_n\to\int f \,d\mu$ for all continuous and bounded $f$
$ \mu_n(A)\to\mu(A)$ for $A$ continuity for $\mu$
I had the idea to use the first assertion to prove that $\int f\,d\mu_n\to\int f\, d\mu$ for all $f$ continuous and bounded, with the help of the dominated convergence theorem, but I couldn't do it.
Let $g_n$ be a continuous function with the following properties: $0\leq g(x) \leq 1$ for all $x$, $g_n(x)=0$ of $|x| >n+1$ and $g_n(x)=1$ if $|x| \leq n$. [ Draw a picture. Use a straight line graph between $n$ and $n+1$ as well as $-n-1$ and $-n$ to get a continuous graph]. Now let $\epsilon >0$ There exists $n$ such that $\mu (-n,n) >1-\epsilon $. Now $g_n$ is continous with compact support, so $\int g_n d\mu_k \to \int g_n d\mu > 1-\epsilon $ becasue $g_n=1$ on $(-n,n)$. But
$\int g_n d\mu_k \leq \mu_n (-n-1,n+1)$ so we get $\mu_n (-n-1,n+1) > 1-\epsilon $ for $k$ sufficiently large. We have found a compact set $K$ (namely $[-n-1,n+1]$) such that $\mu_k(K) >1-\epsilon$ and $\mu (K) >1-\epsilon$ for $k$ sufficiently large. Let $f$ be any bounded continuous function. I leave it to you to use that fact that $\int fg_n \, d\mu_k \to \int fg_n \, d\mu $ as $k \to \infty$ to prove that we actually have $\int f \, d\mu_k \to \int f \, d\mu $. [ You have to estimate $|\int fg_n \, d\mu_k -\int f \, d\mu_k|$ as well as $|\int fg_n \, d\mu -\int f \, d\mu|$].