Conditional independence of random variables Let $(X_t),$ $(Y_t)$ be independent bounded martingales (for filtration $ \{ \mathcal{F}_t \}$ )which converge to $X_\infty$ and $Y_\infty$ respectively, by the martingale convergence theorem. Let $\{ \mathcal{F}^{X,Y}_t \}$ denote the filtration generated by processes $X$ and $Y$. Then the book claims that

$$ \mathbb{E} [X_\infty Y_\infty | \mathcal{F}^{X,Y}_t ] = \mathbb{E} [X_\infty  | \mathcal{F}^{X,Y}_t ] \mathbb{E} [Y_\infty  | \mathcal{F}^{X,Y}_t ] = X_t Y_t \quad  \text{ a.s..}$$

Can anyone explain to me why $ \mathbb{E} [X_\infty Y_\infty | \mathcal{F}^{X,Y}_t ] = \mathbb{E} [X_\infty  | \mathcal{F}^{X,Y}_t ] \mathbb{E} [Y_\infty  | \mathcal{F}^{X,Y}_t ] $ ? Does this conditional independence property follow from the independence of the stochastic processes $X$ and $Y$?
 A: The answer is quite nontrivial and requires $X,Y$ to be continuous independent martingales. See 
http://mech.math.msu.su/~cherny/pmt.pdf
where it is laid out in full detail. I will present an an easy version in which $X,Y$ are bounded.
Theorem: If $X,Y$ are continuous bounded $\mathscr{F}_t$-martingales, then $XY$ is a continuous $\mathscr{F}_t$- martingale.
Proof: It suffices to show $\langle X,Y\rangle = 0$. Since $X,Y$ are continuous, $\langle X,Y \rangle_t$ is the limit in probability of sums like
$
\sum (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i})
$
as the partition size $\Delta$ tends to $0$. We'll show these sums converge in $L^2$ to $0$.
$$
E\left(\sum (X_{t_{i+1}}-X_{t_i})(Y_{t_{i+1}}-Y_{t_i})\right)^2 =\sum E(X_{t_{i+1}}-X_{t_i})^2E(Y_{t_{i+1}}-Y_{t_i})^2
$$
$$
\leq \max_i E(X_{t_i+1}^2-X_{t_i}^2) \cdot \sum E(Y_{t_{i+1}}^2-Y_{t_i}^2) =
\max_i E(X_{t_i+1}^2-X_{t_i}^2) E(Y_t^2 - Y_0^2).
$$
The max term tends to $0$ as $\Delta\to 0$ since $EX_t^2$ is continuous in $t$ (monotone/dominated convergence). Thus $\langle X,Y\rangle_t =0$.
If you wanted to extend this to continuous local martingales (showing $XY$ is a local marginale), you would then take a reducing sequence of stopping times where the local martingales are bounded (hence true martingales) and approximate.
To then extend to continuous martingales, stop the martingale at times which it is bounded and apply the result for local martingales. Then use standard estimates to get $L^1$ convergence and show $XY$ is a martingale.
