On the existence of exhausting sequences for thin sets

Question

I'm having trouble understanding the proof of the existence of exhausting sequences for thin sets in Cohen and Elliott (2015), and would appreciate it if someone could help shed some light on the situation, or if anyone could furnish (a reference to) an alternative proof.

Let $(\Omega,\mathcal F, \{\mathcal F_t\},P)$ be a filtered probability space satisfying the usual conditions. Denote the optional $\sigma$-algebra on $[0,\infty[ \times \Omega$ by $\Sigma$. The graph of a stopping time $T$ is $$ [T] = \{ (t,\omega) \in [0,\infty[\times\Omega : T(\omega)=t < \infty \}. $$

A set $A \in \Sigma$ is called thin if there exists a sequence of stopping times $\{ S_n \}_{n\in\mathbb N}$ such that $A \subseteq \cup_{n\in\mathbb N}[S_n]$. Moreover, if $[S_n]\cap[S_m]=\emptyset$ whenever $n\neq m$, then $\{ S_n \}_{n\in\mathbb N}$ is called an exhausting sequence for $A$.

My question is regarding the existence of exhausting sequences for thin sets. Cohen and Elliott state the following theorem.

Theorem. Any thin set admits an exhausting sequence.

I don't see how their proof establishes the existence of such a sequence, unfortunately. Here is their proof.

Proof. Suppose $A \subseteq \cup_{n\in\mathbb N}[S_n]$. Write $T_n$ for the stopping time whose graph is given by $$ \left(A \setminus \bigcup_{k<n}[S_k]\right) \bigcap [S_n]. \label{1}\tag{1}$$ Then the $T_n$ have disjoint graphs and $A = \cup_n [T_n]$.

It is easy to see that the $T_n$ have disjoint graphs and that $A = \cup_n [T_n]$. However, I do have the following question.

Question. How does one know that there exists a stopping time with a graph given by \eqref{1}?

Since \eqref{1} lies inside $[S_n]$, it seems natural to try to modify $S_n$ so that its graph is indeed given by \eqref{1}, yet I don't see a way of doing this.

One thought I had was to call the set in \eqref{1} $B_n$, and to consider the restriction of the stopping time $S_n$ to the set $\pi_\Omega(B_n)$. ($\pi_\Omega$ is the projection operator.) For this to work (that is, for the restriction to be a stopping time), I need $\pi_\Omega(B_n)$ to be $\mathcal F_{S_n}$-measurable, which I believe is not necessarily the case.

Answer 1

Use your notation and change a little of your description. Let $L=\{\omega: 1_{B_n}(S_n(\omega),\omega)=1 \} $. Since $B_n\in\Sigma$ and $S_n$ is a stopping time, then $ L\in \mathscr{F}_{S_n}$ and the restriction $T_n=(S_n)_L $ is a stopping time with a graph given by (1).

Answer 2

It appears I was overthinking how to construct the appropriate stopping time. My problem is resolved by the following construction.

Let $T$ be a stopping time, and consider any $\Sigma$-measurable $A \subseteq [T]$. We construct a stopping time $S$ such that $[S]=A$ by setting $S(\omega) = t$ whenever $(t,\omega) \in A$. On the other hand, when $(t,\omega) \notin A$ for all $t\in[0,\infty[$, set $S(\omega) = \infty$.

It is straightforward to verify that $S$ is a well-defined map from $\Omega$ to $[0,\infty]$, and that $[S]=A$. The fact that $S$ is a stopping time then follows from the measurable projection theorem and this lemma in Cohen and Elliott.

Lemma 7.3.6. Suppose $T$ is a random time with values in $[0,\infty]$. Then, $T$ is a stopping time if and only if $[T]\in \Sigma$.

Thus, we only need to check that $S$ is $\mathcal F$-measurable. Note that for any $t \in [0,\infty[$, \begin{align} \{S \le t \} &= \{ \omega:S(\omega) \le t \} \\ &= \pi_\Omega (A \cap [0,t]\times\Omega). \end{align}

Since $A$ is $\mathcal B \otimes \mathcal F$-measurable, the measurable projection theorem guarantees that $S$ is a random time.