Convergence of probability measures problem 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 $x_n \to x$.
It looks like convergence in product topology is used, I am not much aware of such things
 A: By Prokhorov's Theorem (http://en.wikipedia.org/wiki/Prokhorov%27s_theorem), the sequence $(\mu_n)_n$ is (uniformly) tight, i.e. for every $\varepsilon > 0$, there is a compact set $K_\varepsilon \subset S$ such that $\mu_n ( S\setminus K_\varepsilon) < \varepsilon$ for all $n \in \Bbb{N}$.
Now use that the set $L := \{x_n \mid n \in \Bbb{N}\} \cup \{x\}$ is also compact, so that $L \times K_\varepsilon$ is compact.
We can now use that continuous functions on compact sets are uniformly continuous.
Together with the boundedness of $h$, which ensures that
$$
\bigg| \int_{S \setminus K_\varepsilon} h(x_n, y) \, d\mu_n(y)\bigg| \leq C \cdot \mu_n (S \setminus K_\varepsilon) \leq C \cdot \varepsilon,
$$
this should allow you to conclude the proof.
EDIT: Putting all pieces together, we arrive at
\begin{eqnarray*}
 &  & \left|\int h\left(x_{n},y\right)\,{\rm d}\mu_{n}\left(y\right)-\int h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\
 & \leq & \left|\int_{S\setminus K_{\varepsilon}}h\left(x_{n},y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{K_{\varepsilon}}h\left(x_{n},y\right)-h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{K_{\varepsilon}}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\
 & \leq & C\cdot\varepsilon+\sup_{y\in K_{\varepsilon}}\left|h\left(x_{n},y\right)-h\left(x,y\right)\right|+\left|\int_{S\setminus K_{\varepsilon}}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\
 & \leq & 2C\cdot\varepsilon+\sup_{y\in K_{\varepsilon}}\left|h\left(x_{n},y\right)-h\left(x,y\right)\right|+\left|\int_{S}h\left(x,y\right)\,{\rm d}\mu_{n}\left(y\right)-\int_{S}h\left(x,y\right)\,{\rm d}\mu\left(y\right)\right|\\
 & \xrightarrow[n\to\infty]{} & 2C\cdot\varepsilon,
\end{eqnarray*}
where the last step used the uniform continuity of $h$ on $L\times K_\varepsilon$ and the assumption on the convergence $\mu_n \to \mu$ (in the sense of your assumption). As $\varepsilon > 0$ was arbitrary, this establishes the claim.
