= Solution
For a <stopping time> $\tau$, $U_{t\wedge\tau}$ is adapted: split according to $\{\tau=j\}$ for $j<t$ and $\{\tau\geq t\}$, all events belonging to $\mathcal F_t$. It is integrable because $|U_{t\wedge\tau}|\leq\sum_{j=0}^t|U_j|$. Moreover,
$$
U_{(t+1)\wedge\tau}-U_{t\wedge\tau}=\mathbf1_{\{\tau>t\}}(U_{t+1}-U_t).
$$
The indicator is $\mathcal F_t$-measurable, so conditioning and the <supermartingale> property give
$$
\mathbb E[U_{(t+1)\wedge\tau}-U_{t\wedge\tau}\mid\mathcal F_t]=\mathbf1_{\{\tau>t\}}(\mathbb E[U_{t+1}\mid\mathcal F_t]-U_t)\leq0.
$$
Thus $\boxed{(U_{t\wedge\tau})_t\text{ is a supermartingale}}$. This proves <stopping preserves supermartingales in discrete time> without assuming $\tau$ itself bounded or integrable.
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