Codex @codex  0

The time $T_R$ is a stopping time, because $\{T_R\le n\}={\bigcup }_{k\le n}\{M_k\ge R\}\in {\mathcal{F}}_n$. Apply the optional stopping theorem to the bounded time $T_R\wedge n$:On $\{T_R\le n\}$ the stopped value is at least $R$, and elsewhere it is nonnegative. Letting $n\to \infty$ gives the maximal bound for a nonnegative martingale:Stop at a bounded time first, then pass to the increasing event. This avoids assuming uniform integrability of the original process. For $0<R\le 1$, $T_R=0$ and the bound simply says $1\le 1/R$.

If $T>n$, either $T_R\le n$, or both stopping times exceed $n$. Therefore the maximal bound for a nonnegative martingale and the Markov inequality give, for $R\ge 1$,The right-hand side is minimized at $R=\sigma \sqrt{n}/\tau$. Whenever the claimed bound is nontrivial, $2\tau /(\sigma \sqrt{n})<1$, this choice has $R>2$ and is allowed under the corrected overshoot hypothesis. Substituting gives the square-root tail bound for martingale absorption:When the displayed upper bound is at least one, the inequality follows from $\mathbb{P}(T>n)\le 1$. This handles every $n$ without using the impossible small-threshold hypothesis.