Convergence in distribution (weak convergence) of sum of real-valued random variables Suppose that $\{(X_n,Y_n)\}^\infty_{n=1}$ is a sequence of pairs of real-valued random variables that converge in distribution to $(X,Y)$. Show that $X_n + Y_n$ converges in distribution to X+Y.
Attempt at a solution: 
If $h(x,y) = x+y$, then $h(x,y)$ is continuous, and so is any $f(h(x,y))$ where $f$ is a continuous function. I know that the class of bounded, continuous functions is measure-determining, which I think would be useful, but I just can't wrap my head around how to apply it.
 A: You are very close.
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ an arbitrary continuous and
bounded function. 
Note that the function $h:\mathbb{R}^{2}\rightarrow\mathbb{R}$
prescribed by $\langle x,y\rangle\mapsto x+y$ is continuous so that
the composition $f\circ h:\mathbb{R}^{2}\rightarrow\mathbb{R}$ is
continuous. 
Also $f\circ h$ is bounded, since $f$ is bounded. 
Then $\lim_{n\rightarrow\infty}\mathbb{E}f\left(h\left(X_{n},Y_{n}\right)\right)=\mathbb{E}f\left(h\left(X,Y\right)\right)$
as a consequence of $\left(X_{n},Y_{n}\right)\stackrel{d}{\rightarrow}\left(X,Y\right)$.
This comes to the same as $\lim_{n\rightarrow\infty}\mathbb{E}f\left(X_{n}+Y_{n}\right)=\mathbb{E}f\left(X+Y\right)$.
This is true for any continuous and bounded function $f:\mathbb{R}\rightarrow\mathbb{R}$,
so we are allowed to conclude that $X_{n}+Y_{n}\stackrel{d}{\rightarrow}X+Y$.
A: http://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_distribution
According to this wikipedia article, this statement is false. I don't know what the counterexample would be. It might take me a little bit to think of a counterexample.
