everyone.

Given a probabilistic space $(\Omega, \mathcal{F}_t, \mathbb{P})$ and a martingale $(M_t)_{t\leq 1}$ on it. Suppose

$$M_1\stackrel{\mathbb{P}}{\sim}\mu$$

where $\mu$ is a probability distribution on $\mathbb{R}$. Now given another probability distribution $\nu$ such that $d(\mu,\nu)<\epsilon$ where $d$ is Levy metric. We want to construct another martingale $N$ such that $M_1$ is "close" to $N_1$.

Consider $\tilde{\Omega}=\Omega\times \mathcal{C}[0,1]$ and $\tilde{\omega}=(\omega,w)\in\tilde{\Omega}$. Let $\mathcal{G}_t$ be the natural filtration generated by $w\in\mathcal{C}[0,1]$ and set $\mathcal{H}_t=\sigma(\mathcal{F}_t,\mathcal{G}_t)$. Let $\mathbb{P}_0$ be Wiener measure and set $\tilde{\mathbb{P}}=\mathbb{P}\times\mathbb{P}_0$. Since $\mathcal{F}$ and $\mathcal{G}$ are independent, thus $\tilde{M}_t(\tilde{\omega}):=M_t(\omega)$ is a $\mathcal{H}$ martingale and $B_t(\tilde{\omega}):=w_t$ is a Brownian motion.

By a general result cited in A.V. Skorokhod, On a representation of random variables, there exists a measurable function $f$ such that

$$N:=f(M_1, B_1)$$

and

$$N\stackrel{\tilde{\mathbb{P}}}{\sim}\nu,\ \ \tilde{\mathbb{P}}(|N-M_1|\geq\epsilon)\leq\epsilon$$

then set $\tilde{N}_t=E^{\tilde{\mathbb{P}}}[N|\mathcal{H}_t]$, then we get the need martingale. Now my question is if we have two marginals $\mu_1$ and $\mu_2$

$$M_1\stackrel{\mathbb{P}}{\sim}\mu_1,\ \ M_2\stackrel{\mathbb{P}}{\sim}\mu_2$$

$$d(\mu_1,\nu_1)<\epsilon,\ \ d(\mu_2,\nu_2)<\epsilon$$

and $\nu_1$, $\nu_2$ satisfy the condition called "increasing convex order": for any convex function $f$, we have $\nu_1(f)\leq\nu_2(f)$, which ensures the existence of martingales having respectively marginal distributions $\nu_1$ and $\nu_2$.

Can we get another martingale $N$ such that(defined maybe on another space)

$$N_i\stackrel{\mathbb{P}}{\sim}\nu_i,\ \ i=1,2$$

and

$$\mathbb{P}(|M_2-N_2|\geq\epsilon)\leq\epsilon$$

Thanks a lot for your help!