Is the following statement and its proof correct? Do you know this or related results and where I could read more about such things?
Lemma:
If $X$ is a right-continuous process, and the collection $\{X_t\}_{t\in[a,b]}$ is uniformly integrable, then $t\mapsto \mathbb E X_t$ is right-continuous on $[a,b)$.
Proof:
It suffices to show show $\lim_{n\rightarrow \infty} \mathbb E X_{t_n} = \mathbb E X_{a}$ for any sequence $(t_n)>a$ with limit $a$.
Right-continuity of $X$ implies almost sure convergence of $X_{t_n}\rightarrow X_a$. Due to uniform integrability we have $L^1$ convergence, thus:
$$\left|\mathbb E X_{t_n} - \mathbb E X_{a}\right| \le \mathbb E|X_{t_n} - X_{a}| \rightarrow 0$$
PS: If this is preventing anyone from answering, I will accept of course a simple 'yes' answer.