Limit of a sequence of $L^2$ converging, a.s. continuous martingales has a continuous modification. Given a sequence $(X^n)_{n \in \mathbb{N}}$ of almost surely continuous martingales, such that for for every $t \in \mathbb{R}$ the $X^n _t$ converge to a $X_t$ in $L^2$, show that the limit $X$ is a martingale, and has an almost surely continuous modification $\bar{X}$.
I have shown that $X$ is a martingale using the $L^2$ convergence without any bigger problems. But the only result I know that implies the existence of a continuous modification is Kolmogorov's continuity criterion, and I don't see how this would be applied here. My only other idea would be to try and define $\bar{X}$ as the limit of a sequence of linear interpolation of $X$, but this seems like a fairly difficult approach . Is there something I am missing or is that the way to go?
 A: There is some ambiguity in the question.
This is actually false if you interpret the question as only assuming that the $X^n$ are martingales in their own filtration. Indeed let us define $\gamma_N:[0,1]\to[0,1]$ by $\gamma_N(t):= 0$ on $[0,1/2-N^{-1}]$ and $\gamma_N(t) = 1$ on $[1/2,1]$ and linearly interpolated in between. Then for a standard Brownian motion $W$ the processes $W(\gamma_N(t))$ are continuous martingales in their own filtration but the limit (which exists a.s. and in $L^2)$ is discontinuous.
However it is true if you make the stronger assumption that the $X^n$ are all adapted to some joint filtration and they are all martingales in that joint filtration. Indeed in this case the differences $X^m-X^n$ are martingales and we have by Doob that $$\Bbb E[\sup_{t\in [0,1]} |X^m(t)-X^n(t)|] \leq\Bbb E[\sup_{t\in [0,1]} (X^m(t)-X^n(t))^2]^{1/2} \leq 2 \Bbb E[(X^m(1)-X^n(1))^2]^{1/2}.$$ Choosing a subsequence $n_k$ so that $\Bbb E[(X^{n_{k+1}}(1)-X^{n_k}(1))^2]<2^{-k}$ we then find that $$\Bbb E[ \sum_k \|X^{n_{k+1}}-X^{n_k}\|_{C[0,1]} ] <\infty.$$The expectation being finite implies that the random variable itself is a.s. finite. This implies that we may realize $\bar X$ as a.s. a uniform limit of continuous functions and it is therefore continuous.
