Suppose that $\{\mathcal{F}_{t},$ $t \in [0,\infty)\}$ be a filtration satisfying the usual conditions on a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\{(X_{t},\mathcal{F}_{t}),$ $t \in [0,\infty)\}$ be a submartingale. It is a well known result that if the map $t \mapsto \textbf{E}(X_{t}):[0,\infty) \to \mathbb{R}$ is right-continuous, then the submartingale has a $\textbf{r.c.l.l (cadlag)}$ modification. My question is, does there exist a similar result for a $\textbf{l.c.r.l}$ modification if the $t \mapsto \textbf{E}(X_{t})$ is left-continuous.
I know for the given submartingale, $\textbf{E}(|X_{t-}|)<\infty$ and $\textbf{E}[X_{t}|\mathcal{F}_{t-}]\geq X_{t-}$ a.s . Using this result, I tried to prove $X_{t-}$ is $\mathcal{F}_{t-}-$measurable and that $\{(X_{t-},\mathcal{F}_{t-}),$ $t \in [0,\infty)\}$ is a submartingale. This, paired with the assumption that the filtration is left-continous would give me a $\textbf{l.c.r.l}$ modification. However, I was unable to prove these results. Any help would be appreciated.