Suppose $(M_t)_{t \in [0,\infty)}$ is a submartingale (on a complete probability space $(\Omega, \mathcal{F},P)$) with respect to a filtration $(\mathcal{F}_t)_{t \in [0,\infty)}$ that satisfies the usual conditions.
$(M_t)_{t \in [0,\infty)}$ is a submartingale means: $\forall s \le t \in [a,b]$, $$E[M_t \mid \mathcal{F}_s] \ge M_s \qquad a.s.,$$ where $E [\cdot \mid \mathcal{F}_s]$ denotes conditional expectation given $\mathcal{F}_s$.
The filtration $(\mathcal{F}_t)_{t \in [0,\infty)}$ satisfies the usual conditions means: [i] $P$-null sets belong to $\mathcal{F}_0$, and [ii] $\forall t \in [0,\infty)$, $$\mathcal{F}_t = \bigcap_{s > t} \mathcal{F}_s.$$
Also, let the process $(S_t)_{t \in [0,\infty)}$ be given by $$S_t := \lim_{\mathbb{Q} \ni r \to t^{+}} M_r,$$ where the term $\lim_{\mathbb{Q} \ni r \to t^{+}}$ means that $r$ are rational numbers that approaches to $t$ from the right.
Question: given that the map $t \mapsto E[M_t]$ is right continuous, how can I show that $(S_t)_{t \in [0,\infty)}$ is a modification of $(M_t)_{t \in [0,\infty)}$?
($(S_t)_{t \in [0,\infty)}$ is a modification of $(M_t)_{t \in [0,\infty)}$ means that $P (S_t = M_t) = 1$.)
This is part of Lemma 1.4.2 of the book "Multiparameter Resources: An Introduction to Random Fields" by Davar Khoshnevisan. I couldn't figure out the proof. Any help is much appreciated!