If $M$ is a martingale and $σ_n↓$ and $τ_n↓τ$, then $\text E[M_{τ_n}\mid\mathcal F_{σ_n}]\xrightarrow{n→∞}\text E[M_τ\mid\mathcal F_σ]$ Let


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*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$(\mathcal F_t)_{t\ge0}$ be a filtration of $\mathcal A$

*$M$ be a $\operatorname P$-almost surely right-continuous $(\operatorname P,\mathcal F)$-martingale

*$\sigma,\tau$ be $\mathcal F$-stopping times with $$\tau\le T\;\;\;\operatorname P\text{-almost surely}\tag1$$ for some $T\ge0$

*$\mathcal F_\varsigma:=\left\{A\in\mathcal A:A\cap\left\{\varsigma\le t\right\}\in\mathcal F_t\text{ for all }t\in I\right\}$ for any $\mathcal F$-stopping time $\varsigma$


Now, let $$\sigma_n:=\frac1{2^n}\lceil 2^n\sigma\rceil\text{ and }\tau_n:=\frac1{2^n}\lceil 2^n\tau\rceil\;\;\;\text{for }n\in\mathbb N\;.$$ Note that $\sigma_n,\tau_n$ are $(\mathcal F_{k2^{-n}})_{k\in\mathbb N_0}$-stopping times and $(\sigma_n)_{n\in\mathbb N},(\tau_n)_{n\in\mathbb N}$ are nonincreasing with $$\sigma_n\xrightarrow{n\to\infty}\sigma\text{ and }\tau_n\xrightarrow{n\to\infty}\tau\tag2\;.$$

I want to conclude that $$\operatorname E\left[M_{\tau_n}\mid\mathcal F_{\sigma_n}\right]\xrightarrow{n\to\infty}\operatorname E\left[M_\tau\mid\mathcal F_\sigma\right]\tag3\;\;\;\operatorname P\text{-almost surely}\;.$$

Since $M$ is $\operatorname P$-almost surely right-continuous, we obtain $$M_{\tau_n}\xrightarrow{n\to\infty}M_\tau\;\;\;\operatorname P\text{-almost surely}\tag4\;.$$ However, I don't know how I need to proceed (especially with the indexed $\sigma$-algebra in the conditional expectation).
 A: In my answer I assume that
$$\sup_{t \leq T+1} |M_t| \in L^1 \tag{5}$$
(for $T>0$ specified in $(1)$). By Doob's maximal inequality, this is, for instance, satisfied if $M_{T+1} \in L^p$ for some $p>1$.

Clearly,
$$|\mathbb{E}(M_{\tau_n} \mid \mathcal{F}_{\sigma_n})-\mathbb{E}(M_{\tau} \mid \mathcal{F}_{\sigma})| \leq I_1+I_2$$
where
$$\begin{align*} I_1 &:= |\mathbb{E}(M_{\tau} \mid \mathcal{F}_{\sigma_n})-\mathbb{E}(M_{\tau} \mid \mathcal{F}_{\sigma})| \\ I_2 &:= |\mathbb{E}(M_{\tau_n}-M_{\tau} \mid \mathcal{F}_{\sigma_n})|. \end{align*}$$
Since $\sigma_n \downarrow \sigma$, we have $\mathcal{F}_{\sigma_n} \downarrow \mathcal{F}_{\sigma}$, and therefore it follows from Lévy's convergence theorem that $I_1 \to 0$ almost surely as $n \to \infty$. Now fix some number $m \geq 1$. For $n \gg 1$ sufficiently large we have
$$I_2 \leq \mathbb{E} \left( \sup_{k \geq m} |M_{\tau}-M_{\tau_k}| \mid \mathcal{F}_{\sigma_n} \right).$$
Letting $n \to \infty$ it follows once again from Lévy's backward convergence theorem that
$$\limsup_{n \to \infty} I_2 \leq \mathbb{E} \left( \sup_{k \geq m} |M_{\tau}-M_{\tau_k}| \mid \mathcal{F}_{\sigma} \right). \tag{6}$$
As
$$\sup_{k \geq m} |M_{\tau}-M_{\tau_k}| \xrightarrow[]{m \to \infty} 0$$
and, by $(5)$,
$$\sup_{k \geq m} |M_{\tau}-M_{\tau_k}| \leq 2 \sup_{t \leq T+1} |M_t| \in L^1$$
an application of the dominated convergence theorem (for conditional expectations) shows that the right-hand side of $(6)$ converges to $0$ as $m \to \infty$. This finishes the proof.
