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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?

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up vote 1 down vote accepted

The solution can be found on p.46 of the book "Exercises in Probability A Guided Tour from Measure Theory to Random Processes, via Conditioning" by L. Chaumont and M. Yor. I must notice that it's really a very interesting and useful book.

If we want to show that $\sigma$-algebras are equal up to $P$-negligible sets then the equality $$E(X|\mathscr{A})=E(X|\mathscr{B}),\;\forall X - \text{bounded}, \mathscr{A}\vee\mathscr{B}\text{-measurable r.v.}$$ is what we need (it's nearly obvious, try to replace $X$ with indicators).

So we want to show that $$E(X|\cap_n(\mathscr{C}\vee\mathscr{D}_n))=E(X|\mathscr{C}\vee\mathscr{D}),\; \mathscr{D}=\cap_n\mathscr{D}_n$$ holds for all $\mathscr{C}\vee\mathscr{D}_0$-measurable r.v. Using monotone class theorem for functions (Revuz, Yor p. 3) and the equality $\mathscr{C}\vee\mathscr{D}_0 = \sigma\{AB,\;A\in\mathscr{C},\,B\in\mathscr{D}_0\}$ we can show that it's enough to consider $X=fg$, $f$ and $g$ are bounded and respectively $\mathscr{C}$-measurable and $\mathscr{D}_0$-measurable.

\begin{multline} E(X|\mathscr{C}\vee\mathscr{D}_n)=\text{$f$ is $\mathscr{C}\vee\mathscr{D}_n$-measurable}=fE(g|\mathscr{C}\vee\mathscr{D}_n) \end{multline}

Again, using monotone class theorem and conditional independence of $\mathscr{C}$ and $\mathscr{D}_0$ given $\mathscr{D}_n, \forall\,n$, which holds because $\mathscr{D}_n\subset \mathscr{D}_0$, we can show that $E(g|\mathscr{C}\vee\mathscr{D}_n)=E(g|\mathscr{D}_n)$. So we have $$E(X|\mathscr{C}\vee\mathscr{D}_n)=fE(g|\mathscr{D}_n).$$

By martingale convergence theorem $E(g|\mathscr{D}_n)\rightarrow E(g|\mathscr{D})$, therefore $$E(X|\mathscr{C}\vee\mathscr{D}_n)\rightarrow fE(g|\mathscr{D})=E(X|\mathscr{C}\vee\mathscr{D}),$$ because $\mathscr{C}$ and $\mathscr{D}_0$ are independent conditionally on $\mathscr{D}$. Martingale convergence theorem implies that $$E(X|\mathscr{C}\vee\mathscr{D}_n)\rightarrow E(X|\cap_n(\mathscr{C}\vee\mathscr{D}_n))$$ We deduce the equality from the uniqueness of limit in $L^1$.

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