In our homework assignment, we were supposed to find an example showing that the assumption of right-continuity in the statement of the stopping theorem cannot be omitted in general (cf. http://www.math.ethz.ch/education/bachelor/lectures/fs2012/math/bmsc/bmsc_fs12_04.pdf exercise 4-2 c)).
In the hint it said: For a standard exponentially distributed random variable T, consider the process $ M = (M_t)_{t \geq 0} $ given by $ M_t = (T \wedge t ) + 1_{ \{ t \leq T \} } $ together with the $P$-augmentation of the filtration generated by the process $ (T \wedge t) $.
Moreover, we were told that we should try to prove that $ M_t = E[T | \widetilde{\mathcal{F}}_t] $, were $ \widetilde{\mathcal{F}}_t $ denotes the $P$-augmentation of the sigma algebra $ \mathcal{F}_t = \sigma (T \wedge s ; s \leq t) $.
Well, I know that once if proven $ M_t = E[T | \widetilde{\mathcal{F}}_t] $, it follows that $ M $ is a uniformly integrable $ \widetilde{\mathcal{F}}_t $-martingale. Also, I can show that $ T $ is a $ \widetilde{\mathcal{F}}_t $-stopping time. Therefore, if the assumption of right-continuity were not necessary, the stopped process $ M^T $ with $ M_t^T = (T \wedge t ) + 1_{ \{ T \wedge t \leq T \} } = (T \wedge t ) + 1 $ would also be a uniformly integrable martingale (by the stopping theorem). Then, the difference $ N_t := M_t^T - M_t = 1_{ \{ t > T \}} $ would also be a uniformly integrable martingale. But $ E[N_0] = 0 $ whereas $ E[N_{\infty}] = E[1] = 1 $, a contradiction.
Can anybody help me prove $ M_t = E[T | \widetilde{\mathcal{F}}_t] $? Or would anybody happen to know a different counterexample?
Thanks a lot!
Regards, Si