Convergence in distribution (weak convergence) Let $X_n$ and $X$ be random variables taking values in the metric space $(S,d)$.
The sequence $(X_n)_n$ is convergent to $X$ in distribution (or weakly) if
$E[f(X_n)] \to E[f(X)]$ for all $f:S\to R$ continuous and bounded.
I read somewhere that it's equivalent to consider only uniformly continuous and bounded $f$.
Could you give me a proof of this fact?
 A: We denote $\Bbb P_n$ and $\Bbb P$ the measures associated with $X_n$ and $X$ respectively.  
Assume that $E[f(X_n)]\to E[f(X)]$ for all $f$ uniformly continuous and bounded. Fix $F$ a closed set and let $O_n:=\{x\in S,d(x,F)<n^{-1}\}$. Then the map $f_n\colon x\mapsto \frac{d(x,O_n^c)}{d(x,O_n^c)+d(x,F)}$ is uniformly continuous and bounded. 
Claim: $\limsup_{n\to +\infty}\Bbb P_n(F)\leqslant \Bbb P(F)$. 
Indeed, $f_n(x)=1-\frac{d(x,F)}{d(x,O_n^c)+d(x,F)}$ is monotone, bounded by $1$ and converges pointwise to the characteristic function of $F$. We have for each $n$ and $N$,
$$\Bbb P_n(F)\leqslant \int f_N(x)d\Bbb P_n,$$
so for all $N$,
$$\limsup_{n\to +\infty}\Bbb P_n(F)\leqslant \int f_N(x)dP,$$
and we conclude by monotone convergence. 
Now, fix $f$ a continuous function such that $0\leqslant f\leqslant 1$. Let $F_{n,j}:=\{x\in S,f(x)\geqslant \frac jn\}$.
\begin{align*}
\int_S fd\Bbb P_N-\int_S fd\Bbb P &\leqslant \sum_{k=0}^n\frac  kn\left(\Bbb P_N\left(\frac kn\leqslant 
f(x)<\frac{k+1}n\right)-\Bbb P\left(\frac kn\leqslant f(x)<
\frac{k+1}n\right)\right)+\frac 1n\\\
&=\sum_{j=0}^n\frac jn\Bbb P_N(F_{n,j})-\sum_{j=1}^{n+1}\frac{j-1}n\Bbb P_N(F_{n,j}) 
-\sum_{j=0}^n\frac jn\Bbb P(F_{n,j})\\&+\sum_{j=1}^{n-1}\frac{j-1}n\Bbb P(F_{n,j})+\frac 1n\\
&=\frac 1n\sum_{j=1}^n\left(\Bbb P_N(F_{n,j})-\Bbb P(F_{n,j})\right)+\frac 1n.
\end{align*}
Taking $\limsup_{N\to +\infty}$ and doing the same for $1-f$ instead of $f$, we get the wanted result. 
It's a part of portmanteau theorem. 
A good reference for questions about weak convergence is Billingsley's book Convergence or probability measures. 
