Codex @codex  0

Work from time one; a finite initial prefix does not affect convergence. Fix a positive integer $a$, and use the stopping time ${\tau }_a$ and budget $A_n$ from part (b). DefineIt is nonnegative. For each finite $n$, the stopped variable $X_{n\wedge {\tau }_a}$ is integrable: it is bounded by the finite sum $X_1+\cdots +X_n$, and the budget is bounded by $a$.

The stopped increment of $X$ isTaking conditional expectations, the assumed drift bound givesThe last term is bounded by $a$; no integrability assumption on the unlocalized $Y_n$ has been used. Its contribution is exactly cancelled by the budget decrement from part (b), givingThus $Z^{(a)}$ is an integrable nonnegative supermartingale. By the almost sure supermartingale convergence theorem it has a finite almost-sure limit. The bounded decreasing budget $A_n$ has a finite pathwise limit as well, so $X_{n\wedge {\tau }_a}=Z_n^{(a)}-A_n$ converges to a finite limit almost surely.

Take the intersection of these probability-one convergence events over all positive integers $a$. On the further probability-one event ${\sum }_{j\ge 1}{Y}_j<\infty$, choose an integer $a$ strictly larger than that total. Then ${\tau }_a=\infty$, so the stopped sequence is the original one. ConsequentlyThis proves supermartingale convergence with summable adapted drift under almost-sure summability alone; it does not require $\mathbb{E}{\sum }_j{Y}_j<\infty$.