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Let $X \in L^1,(\mathcal{F_k})_k$ a sequence of sub-$\sigma$-algebra and converging to $\mathcal{W},$ i.e. $$\mathcal{W}=\bigcap_{k \in \mathbb{N}}\sigma\left(\bigcup_{q \geq k} \mathcal{F}_q\right)=\sigma\left(\bigcup_{k \in \mathbb{N}}\bigcap_{q \geq k}\mathcal{F}_q\right).$$

Do we have the almost sure convergence of $E[X\mid\mathcal{F}_k]$ to $E[X\mid\mathcal{W}]$ ? Justify.

The following was asked before: If $\lim_k \mathcal{F}_k=\mathcal{W}$ then $E[X\mid\mathcal{F}_k]$ converges to $E[X\mid\mathcal{W}]$, where convergence in $L^1$ was proved, letting $\mathcal{Q}^1_k=\sigma(\bigcup_{q \geq k}\mathcal{F}_q),\mathcal{Q}_k^2=\bigcap_{q \geq k}\mathcal{F}_q,$ then $E[X\mid\mathcal{Q}_k^1]$ and $E[X\mid\mathcal{Q}_k^2]$ converges a.s and in $L^1$ to $E[X\mid\mathcal{W}].$ The question is solved using the inequality: $$E[|E[X|\mathcal{F}_k]-E[X|\mathcal{W}]|] \leq E[|E[X|\mathcal{F}_k]-E[X|\mathcal{Q}_k^2]|]+E[|E[X|\mathcal{Q}_k^2]-E[X|\mathcal{W}]|] \leq E[|E[X|\mathcal{Q}_k^1]-E[X|\mathcal{Q}_k^2]|]+E[|E[X|\mathcal{Q}_k^2]-E[X|\mathcal{W}]|].$$

What about the almost sure convergence ? It seems it doesn't hold in general, in this case what counter-example do you suggest?

Kurt.W.X
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    Could you clarify whether $(\mathcal F_k)k$ is an arbitrary sequence of sub-$\sigma$-algebras or a filtration, i.e. do we have $\mathcal F{k} \subseteq \mathcal F_{k+1}$ for all $k$? – user6247850 Sep 28 '21 at 14:36
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    It's arbitrary such that $\bigcap_{k \in \mathbb{N}}\sigma\left(\bigcup_{q \geq k} \mathcal{F}q\right)=\sigma\left(\bigcup{k \in \mathbb{N}}\bigcap_{q \geq k}\mathcal{F}_q\right),$ not necessarily a filtration or a reversed filtration and this should be considered if we want to provide a counter-example. – Kurt.W.X Sep 28 '21 at 21:33

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