= Solution
Set $\inf\varnothing=\infty$. Since each $Y_j$ is $\mathcal F_j$-measurable, the event
$$
\{\tau_a\leq n\}=\left\{\sum_{j=1}^nY_j>a\right\}
$$
is 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 $n\geq1$.
If $\tau_a=k<\infty$, the sum before the overshooting increment is $U_k=\sum_{j<k}Y_j\leq a$. If $\tau_a>n$, the sum through time $n$ is at most $a$, and in particular $U_n\leq a$. Thus
$$
\boxed{0\leq A_n\leq a.}
$$
The process is adapted, and boundedness gives integrability even when the increments have no finite <expectations>. Directly from the one-step-lagged sum,
$$
U_{(n+1)\wedge\tau_a}-U_{n\wedge\tau_a}
=Y_n\mathbf1_{\{\tau_a>n\}}.
$$
This decrement is $\mathcal F_n$-measurable, and $0\leq Y_n\mathbf1_{\{\tau_a>n\}}\leq a$. Therefore
$$
\mathbb E[A_{n+1}\mid\mathcal F_n]
=A_n-Y_n\mathbf1_{\{\tau_a>n\}}\leq A_n.
$$
Hence $A$ 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.
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