# Proving $( X_t)$ is a Martingale by proving equivalent statements.

Let $$(X_t)$$ and $$(Y_t)$$ be two stochastic processes taking values in $$E$$ and $$F$$ respectively. Consider the sigma algebra $$(\mathcal{F}_t)$$ generated by $$Z_t=( X_t,B_t,Y_t)$$ where $$B_t$$ is the Brownian motion. Prove that showing $$(X_t)$$ is an $$(\mathcal{F}_t)$$ martingale is equivalent to showing for fixed $$0\le s $$E( h_s X_t)=E(h_s X_s)$$ for all $$h_s$$ that is $$\mathcal {F}_s$$ measurable. And as $$(\mathcal{F}_s)$$ is generated by $$(X_r,B_r,Y_r)_{0\le r\le s}$$, it is sufficient to show for all $$0 \le t_1<\cdots t_n\le s$$, and for all continuous functions $$f$$ on $$\Gamma^n$$ where $$\Gamma = E \times \mathbb{R}^d \times F$$, one has $$E[ f(Z_{t_1}, \cdots, Z_{t_n})X_t] =E[ f(Z_{t_1}, \cdots, Z_{t_n})X_s]$$

Proving $$(X_t)$$ is a martingale is showing $$E(1_A X_t)= E(1_A X_s)$$ for all $$A \in \mathcal{F}_s$$. And as any $$h_s$$ which is $$\mathcal{F}_s$$ measurable can be approximated by indicator functions we have $$E( h_s X_t)=E(h_s X_s)$$ by passing to the limit. How can I prove the other part consisting of continuous function?

Since $$f$$ is continuous, we know that $$f(Z_{t_1}, ..., Z_{t_n})$$ is $$\cal{F}_s$$-measurable. Thus the only thing we need to show is that for $$X$$ to be a martingale it is sufficient to have $$$$\label{eqn:answer} \mathbb{E}[f(Z_{t_1}, ..., Z_{t_n}) X_t] = \mathbb{E}[f(Z_{t_1}, ..., Z_{t_n}) X_s].$$$$ By approximating indicator functions and using dominated convergence you can show that the above implies $$\mathbb{E}[\chi_{E} X_t] = \mathbb{E}[\chi_{E} X_s]$$ for all $$E \in \bigcup\limits_{n \geq 1} \bigcup\limits_{0 \leq t_1 \leq ... \leq t_n \leq s} \sigma(Z_{t_1}, ..., Z_{t_n}) =: \mathcal{A}.$$ Note that $$\mathcal{A}$$ is closed under intersections, so it is a $$\pi$$-system. Now let $$\mathcal{M}$$ denote the collection of sets $$E$$ for which $$\mathbb{E}[\chi_{E} X_t] = \mathbb{E}[\chi_{E} X_s]$$ so that in particular $$\mathcal{A} \subseteq \mathcal{M}$$. Using the MCT you can show that $$\mathcal{M}$$ is closed under countable monotone unions, and therefore is a monotone class. By the monotone class theorem we have $$\sigma(\mathcal{A}) \subseteq \mathcal{M}$$. Since $$\sigma(\mathcal{A}) = \mathcal{F}_s$$ this proves the claim.