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I'm looking for an example of a seperable and complete metric space $(S,d)$ such that there exist some $\varepsilon > 0$ and $P$ a probability measure on the Borel open sets $\mathcal{B}_S$ for which $\int d(x,a) P(dx) < \infty$ for some $a \in S$ such that for any finite collection $F$ of non-expansive maps from $S$ to $\mathbb{R}$, there exists some probability measure $Q$ on $B_S$ for which there exists some $a \in S: \int d(x,a) P(dx) < \infty$ and: $$ \sup_{f \in \kappa(S,\mathbb{R})} \left| \int f dP - \int f dQ \right| > \sup_{f\in F} \left| \int f dP - \int f dQ \right|+ \varepsilon. $$ Here we call a map $f$ from $S$ to $\mathbb{R}$ non-expansive if for any $x,y \in S$ we have $|f(x) - f(y)| \leq d(x,y)$. I have tried a bunch of things, but nothing worked, so I'm starting to wonder that this might be impossible but that seems unlikely in my eyes, but a proof that such an example can't exist would of course also be appreciated!

A very closely related question:

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..

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  • $\begingroup$ (1) What is $\kappa(S,\mathbb{R})$? (2) Strictly speaking a Polish space does not come with a metric, so perhaps you just want to say "complete separable metric space $(S,d)$? Or is, say, $(0,1)$ with its Euclidean metric allowed (it's homeomorphic to a complete separable metric space, namely $\mathbb{R}$)? (3) "Contraction" / "non-expansive" here means $|f(x)-f(y)| \le d(x,y)$? $\endgroup$ Nov 22, 2015 at 15:46
  • $\begingroup$ I editted my question, also I added a very closely related question which may be more intuitive. I'll anwer your 3 questions here aswel (I'm sorry for not defining everything in my question) (1) the non-expansive maps from $S$ to $\mathbb{R}$ where non-expansive means $|f(x) - f(y)| \leq d(x,y)$ (2) I meant seperable metric space, I do appolagize for this. (3) Exactly! $\endgroup$
    – HolyMonk
    Nov 22, 2015 at 19:32
  • $\begingroup$ Any hints which I could try to work out would also be appreciated even if they don't lead to anything! $\endgroup$
    – HolyMonk
    Nov 23, 2015 at 18:29
  • $\begingroup$ I have been able to show that for the second claim it suffices to show that if we have convergence for each $f \in \kappa(S,\mathbb{R})$ of $\int f dP_n$ to $\int f dP$ that then we also have $\lim_{k\rightarrow \infty} \sup_{n\in \mathbb{N}} \int_{\{d(a,\cdot) > k\}} d(a,x) dP_n(x) = 0$ for some $a \in S$ and also that if this doesn't hold the claim must be false. $\endgroup$
    – HolyMonk
    Nov 23, 2015 at 19:27
  • $\begingroup$ The second part of the question has been solved! $\endgroup$
    – HolyMonk
    Nov 25, 2015 at 8:55

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