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I would like to prove or give a counterexample for the following statement:

Let $(S,d)$ be a complete and separable 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 there is some $a \in S$ for which: $$\lim_{M \rightarrow \infty} \sup_{n\in \mathbb{N}} \int_{\{(d(a,\cdot) > M\}} d(a,x) \, dP_n =0.$$

Note that it follows easily by taking $f/k$ if $f$ is Lipschitz with Lipschitz constant $k$ that we have convergence $\int f \, dP_n \rightarrow \int f \, dP$ for all Lipschitz functions $f$ and thus by the Portmanteau theorem it follows that we have weak convergence (but this convergence is stronger than weak convergence since we also have convergence for unbounded functions).

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Because of the weak convergence, there is a probability space $(\Omega,\mathcal F,\Bbb P)$ and random variables $X_1,X_2,\ldots,X$ defined thereon, with values in $S$, such that (i) $P_n$ is the distribution of $X_n$ (i.e., $\Bbb P(X_n\in B)=P_n(B)$ for all Borel $B\subset S$), (ii) $P$ is the distribution of $X$, and (iii) $\lim_{n\to\infty} X_n(\omega)=X(\omega)$ for all $\omega\in\Omega$. (This is Skorokhod's Representation Theorem.) Now fix $a\in S$ and define non-negative random variables $Y_n=d(a,X_n)$, $Y=d(a,X)$. To this sequence we can apply Scheffé's Lemma to deduce that $\|Y_n-Y\|_1\to 0$ as $n\to\infty$. This $L^1$ convergence implies the requested uniform integrability.

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  • $\begingroup$ Thanks man! This is very very appreciated and that's a very nice proof! $\endgroup$ – HolyMonk Nov 25 '15 at 7:57

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