Event of Stopping Time Equal to Infinity is Measurable in Limit of Filtration/Discrete Time

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\sequence {\FF_n}_{n \ge 0}$ be a discrete-time filtration of $\Sigma$.

Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.

Let $\FF_\infty$ be the limit of the filtration $\sequence {\FF_n}_{n \ge 0}$.

Then:


 * $\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$

Proof
Note that we have:

From the definition of a stopping time, we have:


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

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

For each $t \in \Z_{\ge 0}$, we have:


 * $\ds \FF_t \subseteq \bigcup_{s \in \Z_{\ge 0} } \FF_s$

from Set is Subset of Union, so that:


 * $\FF_t \subseteq \FF_\infty$

from the definition of the $\sigma$-algebra generated by collection of subsets.

Then we have:


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

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

Now, since $\sigma$-algebras are closed under countable union, we have:


 * $\ds \bigcup_{t \in \Z_{\ge 0} } \set {\omega \in \Omega : \map T \omega = t} \in \FF_\infty$

So:


 * $\set {\omega \in \Omega : \map T \omega < \infty} \in \FF_\infty$

Then since $\FF_\infty$ is closed under relative complement, we have:


 * $\set {\omega \in \Omega : \map T \omega = \infty} \in \FF_\infty$