 Solution

Fix $x>0$ and let $A=\{M_0<x\}\in {\mathcal{F}}_0$. On $A$, continuity of the paths implies $0\le M_{t\wedge T_x}\le x$, and $M_{T_x}=x$ on $\{T_x<\infty \}$. Bounded-time optional stopping gives, for every $t$,On $A\cap \{T_x=\infty \}$, the stopped process is $M_t\to 0$; on $A\cap \{T_x<\infty \}$ it eventually equals $x$. The bound by $x$ permits conditional dominated convergence, yieldingOn $A^c$, $T_x=0$, so the conditional hitting probability is one. Finally, a continuous nonnegative path tending to zero attains any positive supremum: after a sufficiently late time it is below half that supremum, and its maximum on the earlier compact interval is attained. Thus $\{M^{\ast }\ge x\}=\{T_x<\infty \}$, including equality at the level. We have proved the conditional form of the maximal identity for a continuous nonnegative local martingale tending to zero:The restriction to $A$ before taking the limit is important: $M_0$ need not be bounded, and the unstopped martingale need not have uniform integrability.