Supremal distribution of positive continuous martingale, which converges to zero a.s. 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.
 A: 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.
