I have a sequence of continuous time random variables $X\_n(t)$ where $t \in [0,1]$. Suppose that there is a filtration $F\_t$ such that for each $n$, $X\_n$ is a martingale with respect to this filtration. Note that the filtration does not depend on $n$. Also assume that $$\sup\_n E[\sup\_{0 \leq t \leq 1 } |X\_n(t)|] < \infty.$$ 


Finally suppose for every $t \in [0,1]$ $X\_n(t)$ converge uniformly to a limit $X(t)$ and $X\_n(t)$ converges in $L^1$. 


I would like to know what are the weakest conditions I can assume about the $X\_n(t)$ so that $X(t)$ will also be a martingale. If I assume the $X\_n(t)$ are continuous or even left continuous then I believe that the limit will also be a martingale. If I just want to assume right continuity, I think I need that the fixed time distributions of $X(t)$ are non-singular. What I'd really like to know is if some kind of continuity of the $X\_n$'s is a requirement to push the martingale through to the limit. 



When is the limit of Martingales a Martingale?
Posting the answer already given in comments:


All you really need is that $X\_n(t) \to X(t)$ in $L^1$ for each $t$. Conditional expectation with respect to any $\sigma$-field is continuous with respect to $L^1$ convergence; this is an elementary property of conditional expectation which follows from the inequality $\left|E[X \mid \mathcal{F}]\right| \le E [|X| \mid \mathcal{F}]$ and taking unconditional expectation of both sides.


Given this, since for each $s \le t$ and each $n$ we have $X\_n(s) = E[X\_n(t) \mid \mathcal{F}\_s]$, you may pass to the $L^1$ limit on both sides to get $X(s) = E[X(t) \mid \mathcal{F}\_s]$. So $X$ is a martingale.


