I read somewhere that if we have $X_n \Rightarrow X$ and $Y_n \Rightarrow Y$ it does not implies $X_n + Y_n \Rightarrow X + Y$ (where $\Rightarrow$ means weak convergence/convergence in distribution).
Unfortunately it seems that I "proved" this fact and even worse it seems that I proved that for every continuous function f from $\mathbb{R}^2$ to $\mathbb{R}$ we will have $f(X_n,Y_n) \Rightarrow f(X,Y)$ which cannot be true.
I used a result from Durrett "Probability: Theory and examples" (Theorem 3.2.8 of version 5a).
"If $F_n \Rightarrow F$ then there is a sequence of random variables $Y_n$ with distribution $F_n$ so that $Y_n \rightarrow Y$ a.s where $Y$ has distribution $F$."
Hence I can find $\tilde{X}_n \rightarrow \tilde{X} $ a.s and $\tilde{Y}_n \rightarrow \tilde{Y}$ a.s where $\tilde{X}_n,\tilde{Y}_n,\tilde{X},\tilde{Y}$ have the same distribution as $X_n,Y_n,X,Y$. Now, for every bounded continuous function g, since g and f are continuous, we have $g(f(\tilde{X}_n,\tilde{Y}_n)) \rightarrow g(f(\tilde{X},\tilde{Y}))$ a.s and since g is bounded by the dominated convergence theorem :
$\mathbb{E}[g(f(X_n,Y_n))] = \mathbb{E}[g(f(\tilde{X}_n,\tilde{Y}_n))] \rightarrow \mathbb{E}[g(f(\tilde{X},\tilde{Y}))] = \mathbb{E}[g(f(X,Y))]$
And hence, $f(X_n,Y_n) \Rightarrow f(X,Y)$.
I can't find where my reasoning fails.