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 $\sequence {\FF_n}_{n \ge 0}$.