Problem on convergence in probability measres 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 $x_n \to x$.
 A: 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.
