Let $M$ be a martingale with continuous sample paths with $M_0=x\ge 0$. We assume that $M_t\ge 0$ for every $t\ge 0$ and $M_t\to 0$ as $t\to \infty$ a.s. Show that for every $y>x$, $$ P(\sup_{t\ge 0}M_t\ge y)=\frac{x}{y}. $$
My solution is as follows. Let $T_y:=\inf\{t: M_t=y\}$. Denote by $X_t:=M_{t\land T_y}$ the stopped process (this is a martingale). Since $|X_t|\le y$ is bounded, $X_t$ is uniformly integrable. From optional stopping time theorem, $$ x=EM_0=EX_0=EX_{T_y}=E M_{T_y} $$
Since it is enough to show that $$ yP(T_y\le t)=x $$ where $P(\sup_{t\ge 0}M_t\ge y)=P(T_y\le t)$.
[Comment: this seems not correct? Is it $P(\sup_{t\ge 0}M_t\ge y)=P(T_y\le \infty)$?]
I do not know how to go to the next step. I guess $$ E M_{T_y}=yP(T_y\le \infty) ? \, (*) $$
Since for any stopping time $$ M_{T_y}=1_{T_y<\infty}M_{T_y}+1_{T_y=\infty}M_{\infty}=1_{T_y<\infty}M_{T_y} $$ where the second part is zero.
The LHS in (*) becomes $$ E M_{T_y}=E(1_{T_y<\infty}M_{T_y})?=yE(1_{T_y<\infty}) $$
