= Solution
Backward induction makes every $U_t$ <adapted> and <integrable>: the <conditional expectation> of an <integrable random variable> is <integrable>, and $|\max(a,b)|\leq|a|+|b|$. The <Snell envelope> recursion directly gives $U_t\geq\mathbb E[U_{t+1}\mid\mathcal F_t]$, so $U$ is a <supermartingale>.
If $Z$ is a <submartingale>, iterating its defining inequality gives $\mathbb E[Z_T\mid\mathcal F_t]\geq Z_t$. Starting at $U_T=Z_T$ and applying the <tower property of conditional expectation> backwards shows
$$
U_t=\max\{Z_t,\mathbb E[Z_T\mid\mathcal F_t]\}
=\mathbb E[Z_T\mid\mathcal F_t].
$$
This last <conditional expectation> is a true <martingale>. \b[The conclusions are]
$$
\boxed{U\text{ is a supermartingale};\qquad
Z\text{ submartingale}\ \Longrightarrow\ U_t=\mathbb E[Z_T\mid\mathcal F_t]\text{ is a martingale}.}
$$
In the second case waiting until $T$ attains the <optimal stopping> value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a <martingale>.
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