A question about stochastic processes and stopping times (Galmarino's test) Working through the book "Brownian Motion & Stochastic Processes" by Karatzas and Shreve, 
I found the following problem (page 6, Problem 2.2):
Let $ X $ be a stochastic process and $ T $ a stopping time of $ \{ \mathcal{F}_t^X \}$, where  $ \mathcal{F}_t^X := \sigma (X_s, 0 \leq s \leq t) $.
Suppose that for any $ \omega, \omega ' \in \Omega $, we have $ X_t (\omega) = X_t ( \omega '),
 $ for all $ t \in [0, T ( \omega )] \cap [ 0, \infty ) $ .
Show that $ T ( \omega ) = T ( \omega ') $.
Does anybody know how to prove this? Thanks a lot for your efforts!
Regards, Si
 A: I'm studying the same problem. I found the accepted solution hard to follow and I believe I have a much simpler solution:
Seeking a contradiction, suppose that $T(\omega) \neq T(\omega')$. Then one must be greater than the other, it doesn't matter which. We will suppose $T(\omega) < T(\omega')$ although the logic for the other case is the same. Then label $T(\omega) = t_0$ such that $T(\omega) = t_0 < T(\omega')$. Then $\omega \in \{T \le t_0\}$ and $\omega' \not\in \{T \le t_0\}$.
$\{T \le t_0\} \in \mathscr{F}_{t_0}^X$. Every set in a $\sigma$-algebra $\mathscr{F}_{t_0}^X$ is of the form $\cap_i \{X_{t_i} \in B_i\}$ for countable indices $0 \le t_1 < t_2 < \cdots \le t_0$ and Borel sets $B_i$. So $\{T \le t_0\}$ is one such $\cap_i \{X_{t_i} \in B_i\}$. Since all $t_i \le t_0$, then $X_{t_i}(\omega) = X_{t_i}(\omega')$. Therefore, it is a contradiction that $\omega \in \cap_i \{X_{t_i} \in B_i\}$ and $\omega' \not\in \cap_i \{X_{t_i} \in B_i\}$.
A: Here is a proof coming from a french Math forum :
http://www.les-mathematiques.net/phorum/read.php?12,376956,377123#msg-377123
I translate the solution for the non french readers.
So here comes the solution (credit goes to egoroff) :
For $T(\omega)<\infty$, fix $\mathcal{H}$ as the collection of sets that do not separate $\omega$ and $\omega'$, i.e. sets $A$  s.t. either  $\{\omega,\omega'\}\in A$ or $ \in A^c$. Then it is easy to see that  $\mathcal{H}$ is a $\sigma$-field.
This was the first step. Next for every $(n+1)$-tuple $t_0<...<t_n\le T(\omega)$ and every Borel sets $A_{t_i}$, the set $(X_{t_i})_{i=0,...,n}\in \Pi_{i=0}^n A_{t_i}$ is in $\mathcal{H}$, by hypothesis over $\omega$ and $\omega'$, so $\mathcal{F}_t\subset \mathcal{H}$ for every $t\le T(\omega)$ as those set generate $\mathcal{F}_t$ .
Now $T(\omega)$ is known and finite we have :
-$S=T\wedge T(\omega)$ is a stopping time and moreover $S\in \mathcal{F}_{T(\omega)}\subset \mathcal{H}$ we have $S(\omega)=S(\omega')$, and so $T(\omega)\le T(\omega')$.
-On the other hand the event $\{T\le T(\omega)\}$ is in $\mathcal{F}_{T(\omega)}$, as $T$ is a stopping time so it is in $\mathcal{H}$, and $\omega\in \{T\le T(\omega)\}$ and so $\omega'$ too, and $T(\omega')\le T(\omega)$.
Finally we have shown that $T(\omega)=T(\omega')$ over $T(\omega)<\infty$ which was the claim to be proved.
Best regards
PS :
I also have a solution of mine based on a variant of Doob's lemma but as it is longer, more technical and far less elegant than this one, I do not post it here.
