Convergence in distribution and integration Let $X_n\in\mathbb{R}$ with $n\in\mathbb{N}$ be a sequence of random variables converging in distribution to $X$. Then, for any bounded Lipschitz function $f$ we have that $\lim_{n\rightarrow\infty}E[f(X_n)]=E[f(X)]$.
Can we show that
$$
\lim_{n\rightarrow\infty} \int_0^\infty E[g(y,X_n)]dy = \int_0^\infty E[g(y,X)]dy
$$
where $g$ is a bounded Lipschitz function with $\int_0^\infty g(y,x)dy<\infty$ for any finite $x$.
If not, what additional conditions are needed.
 A: Here is a counter-example (modified from my comments): Take the deterministic case $X=0, X_n=1/n$ for $n \in \{1, 2, 3, ...\}$. 
It will be easier to define $g(x,y)$ over the domain $D = \{(x,y): x \in [0,1], y\geq 0\}$: 
$$ g(x,y)=xe^{-xy} \quad \forall (x,y) \in D$$
Then:
$$ \int_{0}^{\infty} g(x,y)dy = \left\{ \begin{array}{ll}
0 &\mbox{ if $x =0$} \\
1  & \mbox{ if $x>0$} 
\end{array}
\right.$$
Thus
$$ \int_0^{\infty} g(X,y)dy =0$$
But 
$$ \int_0^{\infty} g(X_n, y)dy = 1 \quad \forall n \in \{1, 2, 3, ...\}$$
Note $g$ is bounded because $0\leq g(x,y)\leq 1$ for all $(x,y)\in D$. Finally, note that $g$ is  Lipschitz continuous (see Note below). $\Box$

Note: To show $g$ is Lipschitz we note it is continuously differentiable and for all $(x,y) \in D$: 
$$[\partial g/\partial x; \partial g/\partial y] = [(1-xy)e^{-xy}; -x^2e^{-xy}] $$
and so 
\begin{align}
||[\partial g/\partial x; \partial g/\partial y]||^2&= (1-xy)^2e^{-2xy} + x^4e^{-2xy}\\
&\overset{(a)}{\leq} (1-xy)^2e^{-2xy} + e^{-2xy}\\
&\overset{(b)}{\leq} \sup_{t\geq 0} \left\{(1-t)^2e^{-2t} + e^{-2t}\right\}\\
&\overset{(c)}= 2
\end{align}
where (a) holds because $x \in [0,1]$ so $x^4\leq 1$; (b) holds because $xy \geq 0$; (c) holds because the supremum is achieved at $t=0$.
