How to show that consecutive hitting times are a sequence of stopping times?

Question

Let $\{X_n:n=0,1,\ldots\}$ be a martingale with respect to a filtration $\{\mathcal F_n\}$. Let $A,B$ be nonempty, disjoint Borel sets and define $T_0=0$, \begin{align} S_n &= \inf\{m\geqslant T_{n-1} : X_m\in A\}\\ T_n &= \inf\{m\geqslant S_n : X_m\in B\}. \end{align} I want to show that $S_n$ and $T_n$ are stopping times, i.e. for each nonnegative integer $m$, $$\{S_n\leqslant m\}, \{T_n\leqslant m\}\in\mathcal F_m. $$ $T_0$ is constant and $S_1$ is a hitting time, so they are stopping times. I tried an inductive argument, but I don't see how to show that $T_{n-1}$ and $S_n$ being stopping times implies that $T_n$ and $S_{n+1}$ are stopping times. Perhaps I am missing something obvious here?

Answer 1

Induction is a good idea.

Suppose that $T_{n-1}$ is a stopping time. Then

$$\{S_n = l, T_{n-1}=k\} = \underbrace{\{T_{n-1}=k\}}_{\in \mathcal{F}_k} \cap \underbrace{\bigcap_{j=k+1}^{l-1} \{X_j \notin A\} \cap \{X_l \in A\}}_{\in \mathcal{F}_l} \in \mathcal{F}_l$$

for any $l \geq k$. Moreover, by the very definition of $S_n$,

$$\{S_n = l, T_{n-1}=k\} = \emptyset \in \mathcal{F}_l$$

for any $l<k$. Consequently,

$$\{S_n = l\} = \bigcup_{k \in \mathbb{N}_0} \underbrace{\{S_n = l, T_{n-1}=k\}}_{\in \mathcal{F}_l} \in \mathcal{F}_l$$

for any $l \in \mathbb{N}$. This shows that $S_n$ is a stopping time.

The argumentation that $S_n$ being a stopping time implies that $T_{n+1}$ is a stopping time is very similar; I leave it to you.