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!

name"convergence", not which topology to use on the set of probability measures on the Skorokhod space... To me it'scompactness, not convergence, be it narrow (the first one) or pointwise. $\endgroup$ – Jean Duchon Nov 12 '15 at 14:50