Pointwise Maximum of Stopping Times is Stopping Time

Theorem

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

Let $T$ and $S$ be stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.

Let $S \vee T$ be the pointwise maximum of $S$ and $T$.

Then $S \vee T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.

Proof

We have, for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$:

$\map {\paren {S \vee T} } \omega \le t$ if and only if $\map S \omega \le t$ and $\map T \omega \le t$

That is:

$\set {\omega \in \Omega : \map {\paren {S \vee T} } \omega \le t} = \set {\omega \in \Omega : \map S \omega \le t} \cap \set {\omega \in \Omega : \map T \omega \le t}$

Since $S$ and $T$ are stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:

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

and:

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

So, since $\FF_t$ is closed under finite union, we have:

$\set {\omega \in \Omega : \map {\paren {S \vee T} } \omega \le t} \in \FF_t$

$\blacksquare$