Procedure to show martingale property When one's trying to prove the martingale property (assuming it's adapted and integrable), why does it suffice for a stochastic process $(X_t)_{t\geq0}$ satisfy
$$ \mathbb{E}[(X_t-X_s)\prod_{i=1}^n f_i(X_{s_i})]=0,$$
where $f_i$ are bounded continuous functions and $s_1\leq s_2\leq\dots s_n\leq s<t$.
Is it a simple application of the Monotone Class theorem or is the more/less to it?
 A: Martinagle property of $(X_t)$ is equivalent to the condition $E(X_t-X_s)|\mathcal F_s)=0$ for $s <t$. By definition of conditional expectation this is equivalent to the condition $\int_A (X_t-X_s)dP=0$ for all $A \in \mathcal F_s$. By the $\pi-\lambda$ theorem it is enough to verfy this for some class of sets which is closed under finite intersection and which generates $\mathcal F_s$. One such class is sets of the form $X_{s_1}^{-1}(A_1) \cap X_{s_2}^{-1}(A_3)\cap...\cap X_{s_n}^{-1}(A_n)$ where $s_1\leq s_2\leq ...\leq s_n \leq s$ and $A_i$'s are closed sets in $\mathbb R$. I will now leave it to you to show that this is indeed true under the given hypothesis using the fact that for any closed set $C$ there exists a sequence of uniformly bounded continuous functions $f_n$ converging pointwise to $I_C$.
A: From the functional form of the montone class theorem, you can conclude immediately from the identity you have displayed that
$$
\Bbb E[(X_t-X_s)\cdot F]=0
$$
for all bounded $\mathcal F_s:=\sigma(X_u: 0\le u\le s)$-measurable $F$. (See Theorem 1 here: https://almostsuremath.com/2019/10/27/the-functional-monotone-class-theorem/.)
