weak convergence $\mathbb{R}^2$ When we have two independent sequences or random variables $\{X_{n}\}$ and $\{Y_{n}\}$ for which $X_{n}$ converges weakly to $X$ ( $X_n \overset{w}{\rightarrow}X$) and $Y_n \overset{w}{\rightarrow}Y$ I want to show that holds
$$(X_n,Y_n) \overset{w}{\rightarrow}(X,Y)$$
I know that holds for $f\in C_{b}$:  $\mathbb{E}_{X}f(X_n)\rightarrow \mathbb{E}f(X)$ and the same for $Y$. I want to show that $\mathbb{E}_{X\times Y}f(X_n,Y_n)\rightarrow \mathbb{E}_{X\times Y}f(X,Y)$
As $X_n$ and $Y_n$ are independent I can write $\mathbb{E}_{X\times Y} = \mathbb{E}_X \times \mathbb{E}_Y$, the product measure.
So $$\mathbb{E}_{X\times Y}f(X_n,Y_n)=\mathbb{E}_X (X_n,Y_n)\times \mathbb{E}_Y (X_n,Y_n)=\mathbb{E}_{X}(X,Y_{n})\times \mathbb{E}_Y (X_n,Y)$$
I now need some help making the last step to get $\mathbb{E}_{X\times Y}(X,Y)$. Could anyone help me with this?
I'd really prefer to do it without using characteristic functions if possible.
 A: In general your assertion isn't true (see this post). But you can prove that there always exist independent random variables $U,V$ such that $X \sim U$, $Y \sim V$ and $(X_n,Y_n) \stackrel{w}{\to} (U,V)$ (see also this post).
Follows like this: There exist independent random variables $U$, $V$ such that $X \sim U$, $Y \sim V$, $X_n \stackrel{w}{\to} U$, $Y_n \stackrel{w}{\to} V$ (proof). We have $$\mathbb{E}e^{\imath \, (\xi,\eta) \cdot (X_n,Y_n)} = \mathbb{E}e^{\imath \, \xi \cdot X_n} \cdot \mathbb{E}e^{\imath \, \eta \cdot Y_n} \to \mathbb{E}e^{\imath \, \xi \cdot U} \cdot \mathbb{E}e^{\imath \, \eta \cdot V} = \mathbb{E}e^{\imath \, (\xi,\eta) \cdot (U,V)}$$ By Lévy's theorem: $(X_n,Y_n) \stackrel{w}{\to} (U,V)$.
Remark: Your last equation above isn't correct: Since $f: \mathbb{R}^2 \to \mathbb{R}$ you can't simply write the expectation value as a product:
$$\mathbb{E}(f(X_n,Y_n)) = \int f(x,y) \, d\mathbb{P}_{X_n,Y_n}(x,y) = \int f(x,y) \, d\mathbb{P}_{X_n}(x) \, d\mathbb{P}_{Y_n}(y) \\ \not= \int f(x,y) \, d\mathbb{P}_{X_n}(x) \cdot \int f(x,y) \, d\mathbb{P}_{Y_n}(y)$$
This works only if $f(x,y)=g_1(x) \cdot g_2(y)$ where $g_1,g_2 \in C_b$.
