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 $x\_n \to x$.



Problem on convergence in probability measres
**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.


