Suppose $X$ is a $\mathbb{R}$-valued stochastic process defined on $(\Omega,\mathscr{G},P)$ such that
- $X$ has cadlag sample paths
- $X$ is Markov with respect to its natural filtration $\mathscr{G}_t^X$
- $M_t^f := f(X_t)-\int_0^t(Af)(X_s)ds$ is a $\mathscr{G}_t^X$ martingale.
Since $X$ has cadlag paths, I want to view it in the canonical set-up $(\mathbb{D}[0,\infty),\mathscr{F},Q)$ where $\mathbb{D}$ is the Skorohod space, $\mathscr{F}$ is the Borel $\sigma$-algebra of $\mathbb{D}$ and $Q$ is the probability measure induced by $P$. The $\sigma$-algebra $\mathscr{F}$ is same as the one generated by the coordinate random variables of $\mathbb{D}$ (Prop III.7.1 in Ethier-Kurtz).
To be clear, the coordinate random variables are defined as follows: for any $t\geq 0$, the map $\pi_t:\mathbb{D}\to \mathbb{R}$ given by $\pi_t(\omega)=\omega(t)$ is a coordinate random variable.
Let the filtration generated by $\{\pi_t\}_{t\geq 0}$ be denoted by $\mathscr{F}_t$. Note that $\mathscr{F}_t$ is intrinsic to $\mathbb{D}$ and does not depend on what process we are studying on $\mathbb{D}$.
My question is this: When we want to view $X$ in the canonical set-up, what is the filtration used on the canonical set-up? Is it obtained by some operation on $\mathscr{G}_t^X$? OR, is it the filtration $\mathscr{F}_t$? These filtrations might not be same because the process $X$ might not be capable of exploring all sample paths of $\mathbb{D}$.
Would $Q$ be Markov if the filtration $\mathscr{F}_t$ is used? Would $M_t^f$ remain martingales if $\mathscr{F}_t$ is used?
In section IV.3 (page 174) of Ethier-Kurtz, the concept of martingale problem on $\mathbb{D}$ is defined. But it is not clear what is the filtration that is being used.