It's kind of a classical problem with simple formulation but it turned out to be quite difficult for me.
$(\Omega,\mathscr{F},P)$ is a probability space. If $(\mathscr{D}_n,\;n\geqslant 0)$ is a decreasing sequence of sub-$\sigma$-fields of $\mathscr{F}$: $\mathscr{D}_0\supset\mathscr{D}_1\supset\ldots$, $\mathscr{C}$ is another sub-$\sigma$-field of $\mathscr{F}$ independent of $\mathscr{D}_0$, then
$\bigcap\limits_n(\mathscr{C}\vee\mathscr{D}_n)=\mathscr{C} \vee \Bigl(\bigcap\limits_n\mathscr{D}_n\Bigr).$
Note 1: $\mathscr{C}\vee\mathscr{D}_n$ is a $\sigma$-algebra generated by $\mathscr{C}\cup\mathscr{D}_n$.
Inclusion $\bigcap\limits_n(\mathscr{C}\vee\mathscr{D}_n)\supseteq\mathscr{C} \vee \Bigl(\bigcap\limits_n\mathscr{D}_n\Bigr)$ is obvious. But the other one is not.
Note 2: I found this problen in Revuz, Yor and there is a hint: show, that if $C\in\mathscr{C}$ and $D\in\mathscr{D}_0$ then $\lim\limits_{n\rightarrow\infty}P(CD|\mathscr{C}\vee\mathscr{D}_n)$ is $[\mathscr{C}\vee(\cap_n\mathscr{D}_n)]$-measurable.
First, I don't really understand how to prove that the limit will be [some $\sigma$-algebra]-measurable. Second, with martingale convergence $E(I_C I_D|\bigcap\limits_n(\mathscr{C}\vee\mathscr{D}_n))$ is $[\mathscr{C} \vee (\cap_n\mathscr{D}_n)]$-measurable, what's next?