 Solution

Use the nonnegative martingale $(Y_n)$ and filtration $({\mathcal{H}}_n)$ established above. Fix a finite $n$ and letFor $j<n$, $\{T\le j\}$ is the union of the threshold-crossing events up to $j$, so $T$ is a bounded stopping time. The optional stopping theorem for a martingale at a bounded stopping time gives $\mathbb{E}{Y}_T=\mathbb{E}{Y}_0=1$. On $H$ one has $Y_T\ge \lambda$; on its complement one still has $Y_T\ge 0$. ThereforeOnly a bounded stopping time has been used, so no extra uniform integrability assumption at an unbounded hitting time is implicit. This is also the Doob maximal inequality for a nonnegative submartingale specialized to a mean-one martingale.