Confusion on the proof of continuous time optional stopping time theorem.

Question

In the proof of Doob's (continuous version) optional stopping time theorem, most of the literates claims the follows:

Given an $\mathcal{F}_{t+}-$stopping time $\tau$, define $$\tau_{\ell}=\dfrac{[2^{\ell}\tau]+1}{2^{\ell}}=\sum_{k\in\mathbb{N}}\dfrac{k}{2^{\ell}}\mathbb{1}_{\{\tau\in[\frac{k-1}{2^{\ell}},\frac{k}{2^{\ell}}]\}}, \ \ \ \ell\in\mathbb{N}.$$

Then it is clear that $\tau_{\ell}$ is a stopping time with respect to $\mathcal{F}_{t}$, $\mathcal{F}_{\tau_{n}}\supset\mathcal{F}_{\tau+}$ and $\tau_{n}\searrow\tau $.

However, I don't quite see why this claim is true.

One of the note online provides the proof of $\tau_{\ell}$ being a $\mathcal{F}_{t}-$stopping time:

Fixing $\ell\in\mathbb{N}$, let $\mathbb{Q}^{(2,\ell)}=\{k2^{-\ell}:k\in\mathbb{Z}_{+}\}$ By construction of $\tau_{\ell}$, it takes values in $\mathbb{Q}^{(2,\ell)}\cup\{\infty\}$. With $\{\tau<t\}\in\mathcal{F}_{t}$ for any $t\geq 0$ it follows that for any $q\in\mathbb{Q}^{(2,\ell)}$, we have $$\{\omega:\tau_{\ell}(\omega)=q\}=\{\omega:\tau(\omega)\in [q-2^{\ell},q)\}\in\mathcal{F}_{q}.$$ Thus, $\tau_{\ell}$ is an $\mathcal{F}_{t+}-$stopping time.

I understand most of it, but it only proves $\{\tau_{\ell}(\omega)=q\}\in\mathcal{F}_{q}$, why does this implies $\tau_{\ell}$ is a $\mathcal{F}_{q}-$stopping time? Wouldn't we need to show $\{\tau_{\ell}\leq q\}\in\mathcal{F}_{q}$?

Also, I have no idea about how to show $\mathcal{F}_{\tau_{n}}\supset\mathcal{F}_{t+}$ and $\tau_{n}\searrow\tau$, any idea?

Thank you!

Answer 1

Note that $\tau_\ell$ is a way to discretize $\tau$. You create $\tau_\ell$ as follows:

• Split $[0,\infty)$ evenly into intervals of length $1/2^{\ell}$, i.e. intervals of the form $[(k-1)/2^\ell,k/2^\ell]$ for $k=1,2,\ldots$
• If $\tau$ takes value in $[(k-1)/2^\ell,k/2^\ell]$, define $\tau_\ell$ as $k/2^\ell$.

For example, if $\ell=1$, you first split $[0,\infty)$ into the intervals $[0,1/2]$, $[1/2,1]$, $[1,3/2]$, and so on. You then check where $\tau$ lies. If it lies in $[3,7/2]$, for example, you define $\tau_\ell$ as $\tau_\ell = 7/2$.

Note that by this construction, $\tau_\ell$ will be larger than $\tau$. And as you increase $\ell$, the split of $[0,\infty)$ into the intervals $[(k-1)/2^\ell,k/2^\ell)$ will become finer and finer (i.e. the intervals will become smaller and smaller), and the value you select for $\tau_\ell$ will become closer and closer to the actual value of $\tau$. This is why $\tau_\ell \downarrow \tau$.

I add that this is a well-known discretization/approximation technique, and it is usually used to prove the monotone convergence theorem in measure-theoretic probability. You might want to look that up to understand this construction better.

The reason it is sufficient to show that $\{\tau_\ell=q\} \in {\cal F}_q$ is that $\tau_\ell$ takes only countably many values (it can only be $k/2^\ell$ for some $k \in \mathbb{N}$), so $\{\tau_\ell \leq q\} = \cup_{x \leq q} \{\tau_\ell=x\}$ is a countable union of sets from ${\cal F}_q$, which is in ${\cal F}_q$.

For the claim ${\cal F}_{\tau+} \subset {\cal F}_{\tau_\ell}$, do you know the defition of ${\cal F}_{\tau_\ell}$?