If $X_n \overset{a.s.}\to X$,$Y_n\overset{a.s.}\to Y$, when are $X$ and $Y$ independent? Consider two sequences of r.v's $\{X_n\}$ and $\{Y_n\}$. Suppose $X_n$ and $Y_m$ are independent for any $m,n\in\mathbb{N}$. Suppose $X_n\overset{a.s.}\to X$ and $Y_n\overset{a.s.}\to Y$. Is it true that $X$ and $Y$ are independent? What if we weaken the condition from a.s. convergence to convergence in probability?
I can show that for any continuous bounded function $f$ and $g$, $\mathbb{E}\left[ f(X) g(Y)\right] = \mathbb{E}\left[f(X)\right] \mathbb{E}\left[g(Y)\right]$. But I don't know how to proceed from here (since identity function is not bounded).
 A: Equality $$ \tag{*}\mathbb{E}\left[ f(X) g(Y)\right] = \mathbb{E}\left[f(X)\right] \mathbb{E}\left[g(Y)\right]$$
for all continuous and bounded functions $f$ and $g$
is sufficient to guarantee independence between $X$ and $Y$. Indeed, fix real numbers $s$ and $t$ and consider for a fixed integer $n$ a continuous and bounded function $f_n$ defined by $f_n(x)=1$ if $x\leqslant t$, $0$ if $x\geqslant t+n^{-1}$ and affine interpolation between $t$ and $t+n^{-1}$. Let $g_n$ be defined similarly with $t$ replaced by $s$. Apply (*) with $f_n$ and $g_n$ instead of $f$ and $g$ respectively and use the monotone convergence theorem to get that
$$\mathbb P\left(\left\{X\leqslant s\right\}\cap \left\{Y\leqslant t\right\}\right)=\mathbb P\left(\left\{X\leqslant s\right\}\right)\cdot \mathbb P\left(\left\{Y\leqslant t\right\}\right).$$

What if we weaken the condition from a.s. convergence to convergence in probability?

Find an increasing sequence $\left(n_j\right)_{j\geqslant 1}$ of integers 
such that the sequences $\left(X_{n_j}\right)_{j\geqslant 1}$ and $\left(Y_{n_j}\right)_{j\geqslant 1}$ converge to $X$ and $Y$ respectively.
