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:

\(\ds \set {\omega \in \Omega : \map T \omega < \infty}\)

\(=\)

\(\ds \set {\omega \in \Omega : \map T \omega \in \Z_{\ge 0} }\)

\(\ds \)

\(=\)

\(\ds \bigcup_{t \in \Z_{\ge 0} } \set {\omega \in \Omega : \map T \omega = t}\)

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:

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

$\blacksquare$