I have the following problem, but I'm unsure if my proof is correct:
Suppose that $\{U_t\}_{t\in \mathbb{N}}$ is a uniformly integrable supermartingale and denote $U_{t \wedge\tau} \equiv U_t^\tau$.
Show that $E(U_\tau) = E(U_0) \iff \{U_t^\tau\}_{t \in \mathbb{N}}$ is a martingale.
$(\impliedby)$ Since $\{U_t\}_{t\in \mathbb{N}}$ is uniformly integrable, so is $\{U_{t \wedge \tau}\}_{t\in \mathbb{N}}$. Thus, if it's a martingale, the martingale convergence theorem for uniformly integrable martingales implies that $E(U_\tau | \mathcal{F}_0) = U_0^\tau = U_0 \implies E(U_\tau) = E(U_0)$
$(\implies)$ This part is where I'm really unsure. We would have that $U_{t \wedge\tau} \rightarrow U_\tau$ almost surely and in $L^1$ by the (sub/super)martingale convergence theorem, and that $\{U_{t \wedge\tau}\}$ is a supermartingale:
$E(U_{t \wedge\tau} | \mathcal{F}_{m \wedge \tau}) \leq U_{m \wedge\tau} \quad \forall m \leq t$
We have $E(U_{t \wedge\tau} | \mathcal{F}_{m \wedge \tau}) \xrightarrow{L^1} E(U_{\tau} | \mathcal{F}_{m \wedge \tau})$, so that some subsequence $E(U_{t_j \wedge\tau} | \mathcal{F}_{m \wedge \tau})$ converges almost surely.
The subsequence converging almost surely implies that $E(U_{\tau} | \mathcal{F}_{m \wedge \tau}) \leq U_{m \wedge \tau}$
Define $Y_m \equiv E(U_\tau|\mathcal{F}_{m \wedge \tau})$ for ease in notation.
Our inequality, assumption and supermartingale property imply that $$E(U_\tau) = E(Y_m) \leq E(U_m^\tau) \leq E(U_0^\tau) = E(U_0) = E(U_\tau)$$
And therefore $Y_m = E(U_\tau|\mathcal{F}_{m \wedge \tau}) = U_m^\tau \quad a.s.$
It is obvious that $E(U_\tau|\mathcal{F}_{m \wedge \tau})$ is a martingale with respect to the filtration $\{\mathcal{F_{m \wedge \tau}}\}_{m \in \mathbb{N}}$
Any help would be greatly appreciated.