Uniform convergence from pointwise convergence for uniform Lipshitz functions I would like to prove or give a counterexample for the following statement:
Let $(S,d)$ be a complete and seperable space. We define:
$$
\mathcal{P}^1(S) := \{P: \mathcal{B}_S \rightarrow [0,1] \mid P \mbox{ probability measure, }\exists a \in S: \int d(x,a) P(dx) < \infty\}
$$
Let $(P_n)_n, P$ all be in $\mathcal{P^1}(S)$ and suppose we have for any $f:S \rightarrow \mathbb{R}$ with $\forall x,y \in S: |f(x) - f(y)| \leq d(x,y): \int f dP_n \rightarrow \int f dP$ then we also have $\sup_{f \in L^1(S,\mathbb{R})} \left| \int f dP_n - \int f dP \right| \rightarrow 0$.
where $L^1(S,\mathbb{R})$ is the collection of Lipschitz functions with Lischitz constant 1.
I have both tried proving and disproving this statement, as a counterexample I tried to take $S := \mathbb{R}$ and $P(\{n\}) := \frac{1}{2^n}, n \in \mathbb{N}$ and taking $P_n(\{n\}) := 1$ and stuff like that but it didn't work out (I also tried to fiddle with uniform distributions but this didn't give me a counterexample either.
Then I tried to prove it, by regularity we can find a compact set $K$ for which $P(S\setminus K)$ is ''small enough'' then we can get the uniform convergence on $K$ but the problem is that I can't bound $\sup_n P_n(S\setminus K)$, for this it would seem that I need that $(P_n)_n$ is tight but I don't think I have this..
 A: I think we want to assume maybe that $S$ is locally compact and the measures are regular; with the problem as stated I don't see why for example there exists a compact $K$ with $P(S\setminus K)$ small. 
The argument might use a little cleaning up, but I think it's right:
We need the existence of  a "Lebesgue number" for an open cover of a compact subset of a metric space. There's been some confusion about that, since the lemma is sometimes stated instead for an open cover of a compact metric space. So:
Lemma: Suppose $K$ is a compact subset of a metric space and $O$ is an open cover of $K$. There exists $r>0$ so that for every $x\in K$ there exists $V\in O$ with $B(x,r)\subset V$.
Proof: Define $\phi:K\to(0,1]$ by saying $\phi(x)$ is the supremum of the $r\in(0,1]$ such that there exists $V\in O$ with $B(x,r)\subset V$. It's clear that $|\phi(x)-\phi(y)|\le d(x,y)$, so $\phi$ is continuous. So there exists $\delta>0$ with $\phi(x)\ge\delta$ for all $x\in K$. Let $r=\delta/2$. QED.
Corollary: If $K$ is a compact subset of a locally compact metric space then there exists $r>0$ such that $K_r=\bigcup_{x\in K}\overline{B(x,r)}$ is compact.
Proof: Cover $K$ by finitely many balls with compact closure. Let $r$ be a Lebesgue number for this cover. QED.
Fix $a\in S$. Let $F$ be the space of all $f$ with $d(f(x),f(y))\le d(x,y)$ and let $F_0$ be the space of all $f\in F$ with $f(a)=0$. 
Suppose the conclusion fails. Replacing $(P_n)$ by a subsequence, there exists $\epsilon>0$ and $f_n\in F_0$ such that $$|\int f_ndP_n-\int f_ndP|\ge100\epsilon$$for all $n$. Again taking a subsequence, we can assume that $f_n(x)$ converges as $n\to\infty$ for all $x$ in our countable dense subset of $S$. The equicontinuity shows that $f_n(x)\to f(x)$ for all $x\in S$, and in fact $f_n\to f$ uniformly on compact sets.
Let $C=\int d(x,a)dP$. Choose $K$ compact so $$\int_{S\setminus K}d(x,a)dP<\epsilon.$$
Restricting to $n$ large enough we can assume that $$\int d(x,a)dP_n<C+\epsilon.$$ Choose $r>0$ so that $K_r$ is compact, as in the corollary.
Now, there exists a Lipschitz function $\phi$ which equals $1$ on $K$, vanishes off $K_r$, and satisfies $0\le \phi\le 1$ everywhere. Considering the product $\phi(x)d(x,a)$ shows that $$\liminf \int_{K_r}d(x,a)dP_n\ge \int\phi(x)d(x,a)dP\ge\int_K d(x,a)dP.$$ Hence if $n$ is large enough we have $$\int_{S\setminus K_r}d(x,a)dP_n<2\epsilon.$$
Now since $f_n\to f$ uniformly on $K_r$ we have $$|\int_{K_r}|f_n-f|dQ<\epsilon$$if $n$ is large enough, for $Q=P$ and also for $Q=P_n$ If $g\in F_0$ then $|g(x)|\le d(x,a)$ and hence $$\int_{S\setminus K_r}|g|dQ<\epsilon$$again for $Q=P$ or $Q=P_n$, by the inequalities above. Hence $$\int_{S\setminus K_r}|f-f_n|dQ<8\epsilon.$$So $$\int_S|f-f_n|dQ<10\epsilon$$for large $n$.
Finally, if $n$ is large enough we have $$|\int fdP_n-\int fdP|<\epsilon.$$And if I didn't leave anything out, the triangle inequality plus the above now contradict $|\int f_ndP_n-\int f_ndP|>100\epsilon$.
