For a cadlag submartingale $X$ the function $t \mapsto E[X_{t}]$ is right continuous. I want to show that for a cadlag submartingale $(X_{t})_{t \geq 0}$ the function $t \mapsto E[X_{t}]$ is right continuous. So far I've only noticed, that this function is monotonously increasing. I would be really grateful for hints.
 A: Let $\{t_n\}_{n=1}^\infty$ be a sequence of rational numbers in $(t, \infty)$, monotonically decreasing to $t\ge 0$ as $n\to\infty.$ We want to show that $\lim_{n\to\infty} \mathbb{E} [X_{t_n}]=\mathbb{E} [X_t].$ To this end, we use the theory of backward martingales:

Definition: Let $\{\mathcal{F}_n\}_{n=1}^\infty$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$ (i.e., $\mathcal{F}_{n+1}\subset\mathcal{F}_n\subset \mathcal{F}$, $\forall n\ge 1$). Then $\{Y_n,\mathcal{F}_n\}$ is a backward submartingale if $\mathbb{E} [Y_n]<\infty$, $Y_n$ is $\mathcal{F}_n$-measurable, and $\mathbb{E} [Y_n\mid \mathcal{F_{n+1}}]\ge Y_{n+1}$ a.s., for all $n$.

Notice that if $\{X_t,\mathcal{F}_t\}$ is a submartingale, then $\{X_{t_n},\mathcal{F}_{t_n}\}$ is a backward submartingale (for $\{t_n\}$ defined above). Moreover, backward submartingales are uniformly integrable. Hence,  we deduce $$\lim_{n\to\infty} \mathbb{E} [X_{t_n}]=\mathbb{E} [X_t],$$
using the fact that $\lim_{n\to\infty} X_{t_n}=X_t$ a.s. (from right-continuity of $X_t)$ and the uniform integrability of $\{X_{t_n}\}_n$ (see, for example, Theorem 0.2 here). The right-continuity of the function $t\mapsto \mathbb{E} [X_t]$ follows.
