I want to prove that: if $E[M_t\mid\mathcal{F}_s]=0$ where $\mathcal{F}_s$ is the filtration generated by a stochastic process X knowing that $E[M_t\prod_0^n h_i(X_{t_i})]=0$ for all $n\in N,\quad 0\leq t_0<t_1<\dots<t_n,\quad h_i$ bounded functions.
It seems obvious if $\mathcal{F}_t$ is generated by the finite dimensional rectangles $(X_{t_1}\in A_1,\dots,X_{t_n}\in A_n$) but is this true and how can it be proved? I would appreciate it if you could help. Thank you in advance.
On
Hint It is not difficult to see that
$$\mathcal{F}_s := \sigma(X_u; u \leq s) = \sigma \left( \bigcup_{n \geq 1} \bigcup_{u_1<\ldots<u_n \leq s} \sigma(X_{u_1},\ldots,X_{u_n}) \right),$$
i.e. that
$$\mathcal{G} := \bigcup_{n \geq 1} \bigcup_{u_1<\ldots<u_n \leq s} \sigma(X_{u_1},\ldots,X_{u_n})$$
is $\cap$-stable generator of $\mathcal{F}_s$. Now prove that
$$\mathcal{D} := \{A \in \mathcal{F}; \mathbb{E}(M_t \cdot 1_F)=0\}$$
is a Dynkin system. Then, since $\mathcal{G} \subset \mathcal{D}$, it follows that $$\mathcal{F}_s = \delta(\mathcal{G}) = \sigma(\mathcal{G}) \subset \mathcal{D}.$$
(Here $\delta(\mathcal{G})$ denotes the Dynkin system generated by $\mathcal{G}$.)
You noticed that $\mathbb E[M_t\cdot \chi_S]=0$ for any $S$ of the form $\{X_{t_1}\in A_1\}\cap\cdots\cap \{X_{t_n}\in A_n\}$ (choosing $h_i:=\chi_{A_i}$). It also holds for finite unions of such sets. These finite unions form an algebra, hence by an approximation argument we can show that the equality $\mathbb E[M_t\cdot \chi_S]=0$ holds for each $S\in\sigma(X_{t_1},\dots,X_{t_n})$. By an other approximation argument, we extend this equality to $S$ in the $\sigma$-algebra generated by countably many $X_{t_i}$, $t_i\lt s$.
To conclude, observe that elements of $\mathcal F_s$ may be written as $\{\omega,(X_{t_i}(\omega))\in B\}$, where $(t_i)_{i\in\mathbf N}$ is a sequence of real numbers smaller than $s$ and $B$ is a Borel subset of $\mathbf R^{\infty}$.