 Solution

Set $\inf \varnothing =\infty$. Since each $Y_j$ is ${\mathcal{F}}_j$-measurable, the eventis in ${\mathcal{F}}_n$, so ${\tau }_a$ is a stopping time. The stopped predictable budget for adapted increments is $A_n=a-U_{n\wedge {\tau }_a}$, for $n\ge 1$.

If ${\tau }_a=k<\infty$, the sum before the overshooting increment is $U_k={\sum }_{j<k}{Y}_j\le a$. If ${\tau }_a>n$, the sum through time $n$ is at most $a$, and in particular $U_n\le a$. ThusThe process is adapted, and boundedness gives integrability even when the increments have no finite expectations. Directly from the one-step-lagged sum,This decrement is ${\mathcal{F}}_n$-measurable, and $0\le Y_n{1}_{\{{\tau }_a>n\}}\le a$. ThereforeHence $A$ is a nonnegative supermartingale; in fact its paths are nonincreasing. At time zero one may set $U_0=0$ and $A_0=a$; then $A_1=a$ too. Stopping the sum including $Y_{{\tau }_a}$ would lose nonnegativity, which is why the shifted definition of $U$ matters.