Equivalence of Definitions of Stopping Time in Discrete Time

Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.

Let $T : \Omega \to \Z_{\ge 0} \cup \set {\infty}$ be a function.

Definition 1 implies Definition 2
Suppose that:


 * $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

for all $t \in \Z_{\ge 0}$.

Setting $t = 0$ this certainly implies:


 * $\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$

Now take $t \ge 1$ a positive integer.

We have:


 * $\set {\omega \in \Omega : \map T \omega = t} = \set {\omega \in \Omega : \map T \omega \le t} \setminus \set {\omega \in \Omega : \map T \omega \le t - 1}$

with:


 * $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

and:


 * $\set {\omega \in \Omega : \map T \omega \le t - 1} \in \FF_{t - 1}$

Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have:


 * $\FF_{t - 1} \subseteq \FF_t$

and so:


 * $\set {\omega \in \Omega : \map T \omega \le t - 1} \in \FF_t$

So from Sigma-Algebra Closed under Set Difference, we have:


 * $\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$

Definition 2 implies Definition 1
Suppose that:


 * $\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$

for all $t \in \Z_{\ge 0}$.

We can write:


 * $\ds \set {\omega \in \Omega : \map T \omega \le t} = \bigcup_{0 \le s \le t, \, t \in \Z} \set {\omega \in \Omega : \map T \omega = s}$

We have from hypothesis:


 * $\set {\omega \in \Omega : \map T \omega = s} \in \FF_s$

for $0 \le s \le t$.

Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have:


 * $\set {\omega \in \Omega : \map T \omega = s} \in \FF_t$

for $0 \le s \le t$.

Since $\sigma$-algebras are closed under finite union, we have:


 * $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$