Let $X$ be a continuous non-negative local martingale with $X_0=1$ and $X_t\to0$ almost surely as $t\to\infty$. For $a>1$, let $\tau_a=\inf\{t\geq0:X_t>a\}$.
I am tasked with showing that $\mathbb{P}(\tau_a<\infty)=\mathbb{P}(\sup_{t\geq0}X_t>a)=1/a$. The hint given is to compute the expected value of $X_{t\wedge\tau_a}=a1_{\tau_a\leq t}+X_t1_{\tau_a>t}$, but I still don't really have any idea how to proceed, so any advice would be greatly appreciated!
Using the hint: $$\mathbb{E}(X_{t\wedge\tau_a})=a\mathbb{P}(\tau_a\leq t)+\mathbb{E}(X_t1_{\tau_a>t}).$$
Now $0\leq X_t1_{\tau_a>t}\leq X_t\to0$ almost surely as $t\to\infty$. Additionally $X_t1_{\tau_a>t}\leq a$, so by dominated convergence we have $\mathbb{E}(X_t1_{\tau_a>t})\to0$.
We also have that $X_{t\wedge\tau_a}\leq a$, so $X^{\tau_a}$ is a bounded local martingale, thus a true martingale, implying that $\mathbb{E}(X_{t\wedge\tau_a})=\mathbb{E}(X_0)=1$ (by OST because $t\wedge\tau_a$ is bounded for a fixed $t$). The above then becomes $$a\mathbb{P}(\tau_a\leq t)+\mathbb{E}(X_t1_{\tau_a>t}) = 1,$$ and taking the limit as $t\to\infty$ we obtain the result.
Please let me know if I've made a mistake in the solution above or could have done this more easily.