 Solution

Fix $a>1$. Continuity ensures $M_{T_a}=a$ when $T_a<\infty$, even though the defining inequality is strict. Before that infimum the process is at most $a$. Thus $M^{T_a}$ is a bounded nonnegative local martingale and hence a true martingale, with expectation one. For deterministic $t$,The stopped process converges to $a$ on $\{T_a<\infty \}$ and to zero on its complement. It is bounded by $a$, so dominated convergence gives $1=a\mathbb{P}(T_a<\infty )$. The crossing event is exactly $\{{\sup }_{t\ge 0}{M}_t>a\}$. ThereforeThis is the maximal identity for a continuous nonnegative local martingale tending to zero. It also shows there is no atom at a level greater than one. The tail tends to one as $a\downarrow 1$, so the overall maximum has no atom at its lower endpoint either.