 Solution

For a finite horizon $N$, let $U_N[a,b]$ be the number of completed upcrossings by time $N$. Use the predictable strategy that holds one unit of the process after a visit below $a$ until the next visit above $b$. For a supermartingale, the expected gain of this nonnegative predictable martingale transform is nonpositive. Pathwise, the completed trades earn at least $(b-a)U_N[a,b]$, while an unfinished final trade can lose at most $(X_N-a{)}^{-}$. HenceTaking expectations and using $X_N\ge 0$ gives the Doob upcrossing inequalityAs $N\to \infty$, monotone convergence yields