Compactness of cadlag martingales w.r.t. to the point-wise topology Given a sequence of cadlag (right-continuous with left limits) martingales $X^n=(X^n_t)_{0\le t\le 1}$, we may use the well known criteria to determine whether it is weakly convergent, i.e. subtract a subsequence $(X^{n_k})_{k\ge 1}$ such that for all Skorokhod-continuous and bounded function $f$ one has
$$\lim_{k\to\infty}E[f(X^{n_k})]~~=~~E[f(X)],$$
where the process $X$ is called the weak limit, that is again a martingale. Now, let us consider a different convergence. The sequence $(X^{n})_{n\ge 1}$ is said to be point-wise weakly convergent iff for any subsequence $(X^{n_k})_{k\ge 1}$ there exists a cadlag process $X$ (which is also a martingale) such that 
$$(X_{t_1}^{n_k},\ldots, X_{t_m}^{n_k})~~\stackrel{Law}{\longrightarrow}~~(X_{t_1},\ldots, X_{t_m}) \mbox{ for all } 0\le t_1\le \cdots t_m\le 1.$$ 
My question is the following: Assume that the sequence of martingales have same marginal distributions, i.e. $Law(X_t^n)=\mu_t$ for all $n\ge 1$, where $(\mu_t)_{0\le t\le 1}$ is a sequence of distributions on $\mathbb R$, then could we show that the sequence $(X^{n})_{n\ge 1}$ is point-wise weakly convergent? I believe strongly that it is not true, but cannot find a counterexample. Does some give an example or prove this claim? Thanks a lot for the reply!
 A: Since that this question does not draw much attention, I would like to present what I'm thinking about. First, thanks for the reply of Jean Duchon, I decide to make my question clearer. 
Denote $D:=D([0,1],\mathbb R)$ the Skorokhod space of all cadlag functions $\omega=(\omega_t)_{0\le t\le 1}$ defined on $[0,1]$. Now let us endow $D$ with the topology induced by the point-wise convergence, i.e. $\omega^n$ converge to $\omega$ iff $\omega^n_t$ converges to $\omega_t$ for all $t\in [0,1]$. 
For the sequence of martingales described above $(X^n)_{n\ge 1}$, I wonder whether it is compact w.r.t. to this topology, i.e. find some subsequence $(X^{n_k})_{k\ge 1}$ and another martingale $X$ such that 
$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } 0\le t_1\le \cdots\le t_m\le 1.$$
How to prove or disapprove this claim? A natural idea is that we take first a arbitrary countable set $Q\subset [0,1]$ and define
$$D_{Q}~~:=~~\left\{\bar\omega=(\bar\omega_t)_{t\in Q}: \exists \omega\in D \mbox{ s.t. } \omega_t=\bar\omega_t \mbox{ for all } t\in Q\right\}.$$
If we denote by $\mathbb R^Q$ the product space of $\mathbb R$ and use again $\bar\omega$ to represent its elements, then we have the following result:
Given $\bar\omega\in\mathbb R^Q$, it belongs to $D_Q$ iff the map $t\mapsto \bar\omega_t$ is cadlag and the number of upcrossings of $\bar\omega$ through the interval $[0,1]$ is finite, see e.g. for the definition of the number of upcrossings in the blog of Almost Sure
https://almostsure.wordpress.com/2009/12/06/upcrossings-downcrossings-and-martingale-convergence/
Then fix such a countable set $1\in Q\subset [0,1]$, it is easy to show that there exists a subsequence $(X^{n_k})_{k\ge 1}$ and a martingale $X$ such that 
$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } t_1\le \cdots\le t_m\in Q.$$
Now, with the supplementary condition that $Law(X^{n_k}_t)=\mu_t$ for all $k\ge 1$, we may deduce that $Law(X_t)=\mu_t$ for all $t\in [0,1]$. My question is whether we may show further that  
$$Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m})~~\longrightarrow~~Law(X^{n_k}_{t_1},\ldots,X^{n_k}_{t_m}) \mbox{ for all } 0\le t_1\le \cdots\le t_m\le 1?$$
Thanks again Jean Duchon for his remarks!
