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Let $\mu_n$ be sequence of probability measures on a polish space $S$ such that for any bounded and continuous $f:S \to \Bbb R$ we have $$\int fd\mu_n \to \int fd\mu$$

Then I have seen in some place claiming the following: $$\int h(x_n,y)\mu_n(dy) \to \int h(x,y)\mu(dy)$$

for $h:\Bbb R \times S \to \Bbb R$ and bounded, continuous with $x_n \to x$.

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Edit: Previous answer was bogus.

Let me change your notation a little to let $x_0$ be the limit of the $x_n$.

Yes, this is true. It follows from tightness and the general fact that $h(x_n, \cdot) \to h(x_0, \cdot)$ uniformly on compact sets.

Without loss of generality, let's suppose that $h(x_0, \cdot) = 0$ (replace $h(x,y)$ by $h(x,y) - h(x_0,y)$).

Let $\epsilon > 0$ and let $M$ be the sup norm of $h$. By the easier direction of Prohorov's theorem, the sequence $\{\mu_n\}$ is tight, so there is a compact $K \subset S$ such that for every $n$ we have $\mu_n(K^C) < \epsilon$. Now we have $$\begin{align*}\left|\int h(x_n, y) \mu_n(dy)\right| &\le \int_K |h(x_n,y)|\,\mu_n(dy) + \int_{K^C} |h(x_n, y)| \mu_n(dy) \\ &\le \sup_{y \in K} |h(x_n, y)| + M \epsilon.\end{align*}$$ So it suffices to show that $h(x_n, \cdot) \to 0$ uniformly on $K$. Suppose not; then passing to a subsequence if necessary, we can find $\delta > 0$ so that $\sup_{y \in K} |h(x_n, y)| > \delta$ for all $n$. Thus for each $n$ we can find $y_n \in K$ such that $|h(x_n, y_n)| > \delta$. By compactness, we can pass to a further subsequence so that $y_n$ converges to some $y_0$. Now by continuity of $h$ we have $|h(x_n, y_n)| \to |h(x_0, y_0)| = 0$, a contradiction.

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  • $\begingroup$ @podu: Sorry about that. For probability measures it is true. See my edit. $\endgroup$ – Nate Eldredge Jan 19 '15 at 17:53

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