If $\lim_k \mathcal{F}_k=\mathcal{W}$ then $E[X\mid\mathcal{F}_k]$ converges to $E[X\mid\mathcal{W}]$

Question

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

Prove that $E[X\mid\mathcal{F}_k]$ converges in $L^1$ to $E[X\mid\mathcal{W}].$ Does the almost sure convergence hold? Justify.

If we let $\mathcal{Q}^1_k=\sigma(\bigcup_{q \geq k}\mathcal{F}_q),\mathcal{Q}_k^2=\bigcap_{q \geq k}\mathcal{F}_q,$ then $E[X\mid\mathcal{Q}_k^1]$ and $E[X\mid\mathcal{Q}_k^2]$ converges a.s and in $L^1$ to $E[X\mid\mathcal{W}].$

How to prove that $E[X\mid\mathcal{F}_k]$ converges in $L^1$? What about the a.s convergence ?

Remark: the problem can be solved using the inequality: $$E[|E[X|\mathcal{F}_k]-E[X|\mathcal{W}]|] \leq E[|E[X|\mathcal{F}_k]-E[X|\mathcal{Q}_k^2]|]+E[|E[X|\mathcal{Q}_k^2]-E[X|\mathcal{W}]|] \leq E[|E[X|\mathcal{Q}_k^1]-E[X|\mathcal{Q}_k^2]|]+E[|E[X|\mathcal{Q}_k^2]-E[X|\mathcal{W}]|].$$

Answer 1

First observe that since $\mathcal F_k\subset \mathcal Q_k^1$, $$ \mathbb E\left[X\mid\mathcal F_k\right]-\mathbb E\left[\mathbb E[X\mid\mathcal W]\mid\mathcal F_k\right]=\mathbb E\left[\mathbb E[X\mid\mathcal Q_k^1]- \mathbb E[X\mid\mathcal W]\mid\mathcal F_k\right] $$ hence $$ \left\lVert \mathbb E\left[X\mid\mathcal F_k\right]-\mathbb E\left[\mathbb E[X\mid\mathcal W]\mid\mathcal F_k\right]\right\rVert_1\leqslant \left\lVert \mathbb E\left[X\mid\mathcal Q_k^1\right] -\mathbb E\left[X\mid\mathcal W\right] \right\rVert_1 $$ hence letting $X':= \mathbb E[X\mid\mathcal W]$, it suffices to prove the convergence in $\mathbb L^1$ when $X$ is $\mathcal W$-measurable.

Approximating $X'$ in $\mathbb L^1$ by linear combinations of indicator functions, we are reduced to prove that for each $A\in\mathcal W$, $\mathbb E\left[\mathbf{1}_A\mid\mathcal F_k\right]\to \mathbf{1}_A$ in $\mathbb L^1$.

Let $\mathcal A:= \bigcup_{k \in \mathbb{N}}\bigcap_{q \geq k}\mathcal{F}_q$; then $\mathcal A$ is an algebra generating $\mathcal W$ by assumption. Moreover, the convergence $\mathbb E\left[\mathbf{1}_A\mid\mathcal F_k\right]\to \mathbf{1}_A$ is obvious if $A\in\mathcal A$, as $\mathbb E\left[\mathbf{1}_A\mid\mathcal F_k\right]=\mathbf{1}_A$ for $k$ large enough.

To extend the convergence to the elements of $\mathcal W$, we need approximation : for each positive $\varepsilon$ and each $A\in\mathcal W$, there exists $A'\in\mathcal A$ such that $\mathbb P(A\Delta A')<\varepsilon$. Then $$ \lVert \mathbb E\left[\mathbf{1}_A\mid\mathcal F_k\right]=\mathbf{1}_A\rVert_1\leqslant \lVert \mathbb E\left[\mathbf{1}_{A'}\mid\mathcal F_k\right]-\mathbf{1}_{A'}\rVert_1+2\lVert \mathbf 1_A-\mathbf{1}_{A'}\rVert_1\leqslant \lVert \mathbb E\left[\mathbf{1}_{A'}\mid\mathcal F_k\right]-\mathbf{1}_{A'}\rVert_1+2\varepsilon. $$ For the almost sure convergence, I need to think more; all the tricks here really use properties of $\mathbb L^1$-norm. Of course, if a counter-example exists, the sequence $(\mathcal F_k)$ should be neither increasing nor decreasing.