Suppose I have a stochastic process $(X_r)$ with $r\in\mathbb{Q}\cap [0,T]$. Furthermore I know that there is a sequence of stochastic processes $Y^n$ (each a supermartingale for every $n$) such that for all rationals $r\in [0,T]$ we have $Y^n_r\to X_r$ $P$-a.s. simultaneously. For another stochastic process (RCLL) $Z$ I know that $Y^nZ$ is a supermartingale for every $n$. Hence I can show, using Fatou, that $XZ$ is a supermartingale over $\mathbb{Q}\cap [0,T]$.
Then a theorem from Dellacherie and Meyer should guarantee (as my script says) an RCLL supermartingale $H$ over $[0,T]$ with $H_r=X_r$ for rational $r$. One can take $H_t:=\lim_{r\downarrow t}X_r$. Now, why is $HZ$ again a supermartingale (over $[0,T]$). I'm able to prove this for $s\in \mathbb{Q}$, since $E[H_tZ_t|\mathcal{F}_s]\le \lim\inf E[X_rZ_r|\mathcal{F}_s]=H_sZ_s$. But what about a not rational $s$?
Without loss of generality, we may assume that the filtration $(\mathcal{F}_t)_{t \in [0,T]}$ is right-continuous, i.e. $\mathcal{F}_{t+} = \mathcal{F}_t$ (see the lemma below). By the backward martingale convergence theorem, this implies in particular
$$\mathbb{E}(U \mid \mathcal{F}_s) = \mathbb{E}(U \mid \mathcal{F}_{s+}) = \lim_{r \downarrow s} \mathbb{E}(U \mid \mathcal{F}_r)$$
for any $U \in L^1$. Applying this to $U=H_t \cdot Z_t$ yields
$$\mathbb{E}(H_t \cdot Z_t \mid \mathcal{F}_s) = \lim_{\mathbb{Q} \ni r \downarrow s} \mathbb{E}(H_t \cdot Z_t \mid \mathcal{F}_r) \leq \lim_{\mathbb{Q} \ni r \downarrow s} H_r \cdot Z_r = H_s \cdot Z_s.$$
Proof: Let $s,t \in \mathbb{Q}$, $s<t$, and choose a sequence $\mathbb{Q} \ni s_n \downarrow s$ such that $s_n<t$. Then,
$$\mathbb{E}(X_t \mid \mathcal{F}_{s_n}) \to \mathbb{E}(X_t \mid \mathcal{F}_{s+}). \tag{1}$$
On the other hand,
$$\mathbb{E}(X_t \mid \mathcal{F}_{s_n}) \geq X_{s_n} \to X_{s+} =X_s \tag{2}$$
by the right-continuity of $X$. Combining $(1)$ and $(2)$, we get
$$\mathbb{E}(X_t \mid \mathcal{F}_{s+}) \geq X_s.$$