# question about the sequential continuity of the set of probability measures

I have a question about the sequential continuity of the set of probability measures. Let $\Omega$ be the space of continuous functions defined in $[0,1]$ taking values in $\mathbb{R}$. Let $\mathcal{P}$ be the set of probability measures under which the canonical process $X_t(\omega):=\omega_t$ is a local martingale. \

Let $\mu=(\mu_t)_{0\leq t\leq 1}$ be a familly of marginal distribution such that

$$\int_{\mathbb{R}}x\mu_s(dx)=\int_{\mathbb{R}}x\mu_t(dx),\ \ \forall s, t\in [0,1]$$

Define

$$\mathcal{P}(\mu)=\{\mathbb{P}\in\mathcal{P}: \mathbb{P}\circ X_t^{-1}=\mu_t,\ \ \forall t\in [0,1]\}$$

$$\mathcal{P}^n(\mu)=\{\mathbb{P}\in\mathcal{P}: \mathbb{P}\circ X_{t_k}^{-1}=\mu_{t_k},\ \ \forall 0\leq k\leq 2^n\}$$

where

$$t_k=\frac{k}{n},\ \ 0\leq k\leq 2^n$$

In other words, under any $\mathbb{P}\in\mathcal{P}(\mu)$, $X_t$ has marginal distribution $\mu_t$ and under any $\mathbb{P}\in\mathcal{P}^n(\mu)$, $X_{t_k}$ has marginal distribution $\mu_{t_k}$. My question is whether we have

$$\bigcap_{n\geq 1}\mathcal{P}^n(\mu)=\mathcal{P}(\mu)$$

In particular, for any bounded continuous function $f$ defined on $\Omega$, do we have

$$\lim_{n\rightarrow\infty}\sup_{\mathbb{P}\in\mathcal{P}^n(\mu)}E^{\mathbb{P}}[f(X)]=\sup_{\mathbb{P}\in\mathcal{P}(\mu)}E^{\mathbb{P}}[f(X)] (\ast)$$

I am now reading Convergence of probability measures, and I think of this question for some time. Intuitively I believe the answer is yes, but I do not find a rigorous proof of $(\ast)$. If someone knows the result or can prove it, please help me. Many thanks!

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It is not hard to show $\cap_{n\geq 1}\mathcal{P}^n(\mu)=\mathcal{P}(\mu)$. $\forall\mathbb{P}\in\cap_{n\geq 1}\mathcal{P}^n(\mu)$, we show $X_t\sim\mu_t$, for the tranformation of Fourier of $X_t$ characterize uniquely its distribution. Since there exists a sequence $t_n$ such that $t_k=k_n/2^n$ and $t_n\rightarrow t$, then –  user84572 Sep 6 '13 at 11:58
we have by continuity of $X$ that $E^{\mathbb{P}}[X_{t_n}]\rightarrow E^{\mathbb{P}}[X_{t}]$, which implies that we have necessarily $X_t\sim\mu_t$. –  user84572 Sep 6 '13 at 12:01
To prove $(\ast)$, if and only if for any $\varepsilon>0$, there exists $n$ such that $\forall\mathbb{P}\in\mathcal{P}^n(\mu)$, we can find $\mathbb{P}'\in\mathcal{P}^n(\mu)$ with –  user84572 Sep 6 '13 at 12:04
To prove $(\ast)$, if and only if for any $\varepsilon>0$, there exists $n$ such that $\forall\mathbb{P}\in\mathcal{P}^n(\mu)$, we can find $\mathbb{P}'\in\mathcal{P}^n(\mu)$ with $E^{\mathbb{P}}[f(X)]\leq E^{\mathbb{P}'}[f(X)]$, but I have not found an estimation of their difference, could someone can help me? –  user84572 Sep 6 '13 at 12:22