So the question is as follows: Let $M$ be a positive continous martingale, converging a.s. to zero as $t \rightarrow \infty$. Prove that for every $x>0$: $\mathbb{P}\{\sup_{\{t \geq 0 \}} M_t > x |\mathcal{F}_0\}=1 \wedge \frac{M_0}{x}.$
We've tried so far to define $\tau=t \wedge \inf\{k \leq t | M_k>x\}$. Then apply optional sampling to find: $EM_0=EM_t=EM_{\tau}.$ Then split into two indicators to find: $$EM_{\tau}=EM_{\tau}1_{\{\max_{k\leq t} M_k \geq x \}}+EM_{\tau}1_{\{\max_{k\leq t} M_k < x \}}= x \mathbb{P}(\max_{k\leq t} M_k \geq x)+0.$$
This would essentially prove the statement when letting $t \rightarrow \infty$ and conditioning on $\mathcal{F}_0$, but we're not sure of our reasoning and steps. It seems somewhat uncarefull. Could someone please help us with this proof.
Hi the Optimal Sampling Theorem works fine here because you don't need bounded stopping times if you know that the process $X$ is uniformly integrable. Indeed in that case Optimal Sampling Theorem works for every stopping times (even unbounded ones) and your argument is simplified.
As you may know a (right-)continuous martingale $(X_t)$ is uniformly integrable if and only if it is converging in $L^1$ as $t\to +\infty$ which is the case as $X_t$ converges almost surely to 0. (reference "Karatzas and Shreve Brownian motion and stochastic calculus" problem 3.20 page 18 condition a and b)).
Let's apply fully this idea, and let's denote $T_a$ the first time $X$ attains $a$ (with $a>X_0>0$ so we discard the trivial case) which is a stopping time but not a bounded one. Nevertheless we don't mind because $X$ is UI, so that by we have by Optimal Sampling Theroem :
$X_0=E[X_0]=E[X_{T_a}]=a.P[T_a < \infty] + X_\infty.P[T_a = \infty]$
$ =a.P[T_a < \infty] + 0.P[T_a = \infty] =a.P[sup_{s\geq 0}(X_s)\geq a]$
Which leads to your result ( taking the sup with 1 if $x\leq X_0$).
Best regards.