Let a continuous adapted process $Z= (Z_t)_{t \geq 0}$ be of class DL if \begin{equation} \{ Z_{\tau \wedge t} : \, \tau \text{ is a stopping time } \} \end{equation} is uniformly integrable, for each $t \geq 0$.
I am not sure about this fact (my textbook seems to be using it without proof):
If $(M_t)_{t \geq 0}$ is a local martingale such that $ M_t \geq X_t, $ for some process $(X_t)_{t \geq 0}$ in DL, then $(M_t)_{t \geq 0}$ is a supermartingale.
(I only know how to prove it if $(X_t)_{t \geq 0}$ is a constant process.)
Fix $s \le t$. Let $\{\tau_k\}$ be a localizing sequence for $M_t$; i.e. $\tau_k$ are stopping times with $\tau_k \uparrow \infty$ almost surely and $M_{t \wedge \tau_k}$ is a martingale for each $k$.
By uniform integrability, we have $X_{t \wedge \tau_k} \to X_t$ in $L^1$, and hence $E[X_{t \wedge \tau_k} \mid \mathcal{F}_s] \to E[X_t \mid \mathcal{F}_s]$ in $L^1$. Passing to a subsequence of the $\tau_k$ (which will again localize $M_t$), we can assume further that $E[X_{t \wedge \tau_k} \mid \mathcal{F}_s] \to E[X_t \mid \mathcal{F}_s]$ almost surely.
Notice that since $M_{t \wedge \tau_k} \ge X_{t \wedge \tau_k}$ which is integrable, the conditional expectation $E[M_{t \wedge \tau_k} \mid \mathcal{F}_s]$ is well defined, though it may be $+\infty$ with positive probability. Moreover, the subtraction $$E[M_{t \wedge \tau_k} - X_{t \wedge \tau_k} \mid \mathcal{F}_s] = E[M_{t \wedge \tau_k} \mid \mathcal{F}_s] - E[X_{t \wedge \tau_k} \mid \mathcal{F}_s]$$ is valid. The same holds for $E[M_t \mid \mathcal{F}_s]$.
Now $(M_{t \wedge \tau_k} - X_{t \wedge \tau_k})_{k=1}^\infty$ is a sequence of nonnegative random variables which converges almost surely to $M_t - X_t$, so by the conditional Fatou lemma, $$\begin{align*}E\left[M_t - X_t \mid \mathcal{F}_s\right] &\le \liminf_{k \to \infty} E\left[ M_{t \wedge \tau_k} - X_{t \wedge \tau_k} \mid \mathcal{F}_s \right] \\ &= \liminf_{k \to \infty} \left(E[M_{t \wedge \tau_k} \mid \mathcal{F}_s] - E[X_{t \wedge \tau_k} \mid \mathcal{F}_s]\right) \\ &= \liminf_{k \to \infty} \left( M_{s \wedge \tau_k} - E[X_{t \wedge \tau_k} \mid \mathcal{F}_s]\right) \end{align*}$$ since $M_{t \wedge \tau_k}$ is a martingale for each $k$. But as $k \to \infty$, we have $M_{s \wedge \tau_k} \to M_s$ almost surely, and as argued above, $E[X_{t \wedge \tau_k} \mid \mathcal{F}_s] \to E[X_t \mid \mathcal{F}_s]$ almost surely. So the right side is simply $M_s - E[X_t \mid \mathcal{F}_s]$. Hence we have shown
$$E[M_t \mid \mathcal{F}_s] - E[X_t \mid \mathcal{F}_s] = E[M_t - X_t \mid \mathcal{F}_s] \le M_s - E[X_t \mid \mathcal{F}_s]$$ and the result follows by adding $E[X_t \mid \mathcal{F}_s]$ (which is a.s. finite) to both sides.