Constant Function is Stopping Time
Theorem
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T_\ast$ be an extended natural number.
Define $T : \Omega \to \Z_{\ge 0} \cup \set \infty$ by:
- $\map T \omega = T_\ast$
for each $\omega \in \Omega$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Proof
First, if $T_\ast = \infty$, we have:
- $\set {\omega \in \Omega : \map T \omega \le t} = \O$
for all $t \in \Z_{\ge 0}$.
Since each $\FF_t$ is a $\sigma$-algebra we therefore have:
- $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$ in the case $T_\ast = \infty$.
Now let $T_\ast < \infty$.
For $t < T_\ast$, we have:
- $\set {\omega \in \Omega : \map T \omega \le t} = \O$
and so:
- $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$ again.
For $t \ge T_\ast$, we have:
- $\set {\omega \in \Omega : \map T \omega \le t} = \Omega$
Since each $\FF_t$ is a $\sigma$-algebra we therefore have:
- $\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$.
So $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
$\blacksquare$