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).


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.

  • $\begingroup$ 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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