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The question is to prove $$P\{\sup_{t\geq 0}M_{t}>x\mid \mathcal{F}_{0}\}=\min\left\{1,\frac{M_{0}}{x}\right\},$$ where $M$ is a positive continuous martingale which converges to 0 almost surely as t tends to infinity. I'd like to use the optional sampling theorem in order to prove this. I therefore introduce the stopping time $$\tau=\inf\{t\geq 0\mid M_{t}\geq x\}$$

As $M$ is a martingale we also have that the stopped process is a martingale $$M_{0}=\mathbb{E}[M_{0}\mid\mathcal{F}_{0}]=\mathbb{E}[M_{\tau}\mid\mathcal{F}_{0}]$$ $$=\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau<\infty\}}\mid\mathcal{F}_{0}]+\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau=\infty\}}\mid\mathcal{F}_{0}]$$ $$=\mathbb{E}[M_{\tau}\mathbb{1}_{\{\tau<\infty\}}\mid\mathcal{F}_{0}]$$ $$=x\mathbb{P}(\sup_{t}M_{t}\geq x \mid\mathcal{F}_{0})=x\mathbb{P}(\sup_{t}M_{t}> x \mid\mathcal{F}_{0})+x\mathbb{P}(\sup_{t}M_{t}= x \mid\mathcal{F}_{0})$$

Here is where I get stuck. I'd like to prove that $\mathbb{P}(\sup_{t}M_{t}= x \mid\mathcal{F}_{0})$ is equal to $0$. Is there anyone that could help me prove this or correct me if i'm approaching this in the wrong manner?

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If $P(S_t\geqslant x\mid\mathcal F_0)=M_0/x$ for every $x\gt M_0$, then $P(S_t\gt x\mid\mathcal F_0)=M_0/x$ for every $x\gt M_0$.

To wit, let $x\gt M_0$. Then, $[S_t\geqslant y]\subseteq[S_t\gt x]\subseteq[S_t\geqslant x]$ for every $y\gt x$ hence, applying the hypothesis, one sees that $M_0/y\leqslant P(S_t\gt x\mid\mathcal F_0)\leqslant M_0/x$. Considering the limit $y\to x$, $y\gt x$, one gets the result.

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