 Solution

Backward induction first proves that the Snell envelope is adapted and integrable. At each step, $|U_t|\le |Y_t|+\mathbb{E}[|U_{t+1}|\mid {\mathcal{F}}_t]$, so integrability follows from that of $Y_t$ and $U_{t+1}$. Its defining maximum immediately giveswhich proves that $U$ is a supermartingale.

If $Y$ is a submartingale, the sharper backward identity isIt holds at $T$. If it holds at $t+1$, the tower property of conditional expectation gives $\mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]=\mathbb{E}[Y_T\mid {\mathcal{F}}_t]$. Iterating the submartingale inequalities shows that this is at least $Y_t$, so the maximum chooses this continuation value. The displayed identity then makes $U$ a martingale.