TITLE: Independence is preserved under taking almost sure limits
QUESTION [10 upvotes]: I´m not sure why this theorem is right, how can i prove it?
Let $X_1,X_2.... $ and $Y_1,Y_2...$ be random variables such that $X_n,Y_n$ are independent for every $n∈\mathbb N$ and such that X, resp Y, are the almost sure limits of $(X_n)$ and $(Y_n)$. Then also X and Y are independant.
Thanks for advice

REPLY [14 votes]: For every continuous bounded functions $u$ and $v$, $u(X_n)\to u(X)$, $v(Y_n)\to v(Y)$ and $u(X_n)v(Y_n)\to u(X)v(Y)$ almost surely. All these random variables are uniformly bounded hence 
$$
\mathbb E(u(X_n))\to\mathbb E(u(X)),\quad  
\mathbb E(v(Y_n))\to\mathbb E(v(Y)),
$$
and
$$
\mathbb E(u(X_n)v(Y_n))\to\mathbb E(u(X)v(Y)).
$$
By independence of $X_n$ and $Y_n$, for each $n$, $\mathbb E(u(X_n)v(Y_n))=\mathbb E(u(X_n))\mathbb E(v(Y_n))$. By identification of the limits, 
$$
\mathbb E(u(X)v(Y))=\mathbb E(u(X))\mathbb E(v(Y)).
$$
This holds for every continuous bounded functions $u$ and $v$, hence the distribution of $(X,Y)$ is the product of their distributions, QED.