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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?

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hint: continuous functions are uniformly continuous on compact sets. –  Alex R. Nov 13 '12 at 19:11
2  
Thank you Alex, but how can I use this fact? In general the space $S$ can't be seen as an increasing union of compact sets, am I right? –  qwertyuio Nov 13 '12 at 19:58
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1 Answer

up vote 3 down vote accepted

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.

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Thank you for your answer. I guess $F_{n,j}=\{x|f(x)\geq \frac{j}{n}\}$. I understood this proof which is part of Portmanteau theorem. But isn't it possible to give a direct proof, maybe approximating a continuous function by a uniformly continuous one? –  qwertyuio Nov 14 '12 at 12:07
    
Could you make the argument "By monotone convergence, $\liminf_{n\to +\infty}\mu_n(F)\leqslant \mu(F)$." a little bit more explicit? –  Per Nov 14 '12 at 13:34
    
@per I've added the details –  Davide Giraudo Nov 14 '12 at 13:53
    
@DavideGiraudo I'd like to ask you a new question. Suppose $X_n=(A_n,B_n)$ and $X=(A,B)$. On the left-hand side take the expectation only w.r.t. $B_n$, while on the right-hand side take the expectation w.r.t. both (A,B): $$E_{B_n}[f(A_n,B_n)]\to E_{A,B}[f(A,B)]$$ so that the left-hand side is a random variable. Do you think the result remains true? It seems to me that your proof reamins valid, but I'm not sure.. –  qwertyuio Nov 14 '12 at 16:50
    
@DavideGiraudo. Thanks Davide. –  Per Nov 14 '12 at 17:08
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