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Let $\mu_n, \mu$ be a sequence of probability measures on a Polish space $S$ and $\mu_n', \mu'$ be some kind of extension of $\mu_n, \mu$ on $\bar{S}$ such that all the boundary points of $S$ gets a zero measure. Now if

$$\int fd\mu_n' \to \int fd\mu'$$ for every $f \in C(\bar{S})$

then I think this claims that

for any bounded $g \in C(S)$

$$\int gd\mu_n \to \int gd\mu$$

is true. I think to claim that we need continuous extension of $g$ over $\bar{S}$. Am I correct ?

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  • $\begingroup$ How are exactly $S$ and $\overline S$ related? (I'm not in my office, and I cannot access the paper: maybe you can write the main relevant elements of the paper) $\endgroup$
    – Davide Giraudo
    Commented Jan 10, 2015 at 13:23
  • $\begingroup$ @DavideGiraudo: $S$ is a proper subset of some metric space $X$ and $\bar{S}$ its closure. $\endgroup$
    – user56932
    Commented Jan 11, 2015 at 3:53

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It seems that in the paper, $S$ is a Polish space, which is homeomorphic to a dense subset of a compact metric space denoted by $\overline S$. Without loss of generality, we shall work with this subspace instead of the original $S$.

By portmanteau theorem, it suffices to show the wanted convergence when $g$ is bounded and uniformly continuous on $S$. Such a function can be extended to a continuous (and bounded) function on $\overline S$, denoted $\overline g$. Then we deduce convergence $$\lim_{n\to \infty } \int_{\overline S} \overline g\mathrm d\mu_n=\int_{\overline S} \overline g\mathrm d\mu,$$ and since for each $n$, $\mu_n(\overline S\setminus S)=\mu(\overline S\setminus S)=0$ (by definition of measures on $\overline S$), we get the wanted convergence.

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  • $\begingroup$ Why "Such a function can be extended to a continuous (and bounded) function on $\overline S$, denoted $\overline g$." ? That is my main doubt. $\endgroup$
    – user56932
    Commented Jan 13, 2015 at 11:49
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    $\begingroup$ If $g$ is uniformly continuous and $x\in \overline S\setminus S$, take $x_n\in S$ such that $d(x,x_n)\to 0$. Since $g$ is uniformly continuous, the sequence $(g(x_n))$ is Cauchy and converges to a value $l$. It remains to show that this $l$ does not depend on the choice of the approximating sequence $(x_n)$. $\endgroup$
    – Davide Giraudo
    Commented Jan 13, 2015 at 12:09
  • $\begingroup$ Thanks, understood that. So, if one has a sequence of convergent probability measure on $\overline{\phi(S)}$, then the corresponding probability measures (with zero measure on the boundary) converge in $\phi(S)$. $\phi$ is the homeomorphism. With this I am not able to verify the claim made in line 8 of the same page, same column. I think we need a uniform bound for $h$. $\endgroup$
    – user56932
    Commented Jan 13, 2015 at 13:46

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