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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\sequence {\FF_t}_{t \ge 0}$ be a continuous-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 $\sequence {\FF_t}_{t \ge 0}$.


Then:

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


Proof

Note that we have $\map T \omega < \infty$ if and only if $\map T \omega \le N$ for some $N \in \N$.

So, we have:

$\ds \set {\omega \in \Omega : \map T \omega < \infty} = \bigcup_{N \in \N} \set {\omega \in \Omega : \map T \omega \le N}$

Since $T$ is a stopping time, we have:

$\set {\omega \in \Omega : \map T \omega \le N} \in \FF_N$

For each $N \in \N$, we have:

$\ds \FF_N \subseteq \bigcup_{t \in \hointl 0 \infty} \FF_t$

from Set is Subset of Union, so that:

$\FF_N \subseteq \FF_\infty$

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

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

$\ds \bigcup_{N \in \N} \set {\omega \in \Omega : \map T \omega \le N} \in \FF_\infty$

So:

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

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

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

$\blacksquare$


Sources