Countable-state martingale-problem characterization

Suppose $X$ is a $p$-chain. By [F1], [F1, F2, F3] $\mathrm{\mid }{M}_n^f\mathrm{\mid }\le (1+2n)\parallel f{\parallel }_{\mathrm{\infty }},$ so $M^f$ is integrable; it is adapted because $X$ is. Its increment is $M_{n+1}^f-M_n^f=f(X_{n+1})-f(X_n)-Lf(X_n)=f(X_{n+1})-Pf(X_n)\mathrm{.}$ By [F2] this increment has conditional mean zero given ${\mathcal{F}}_n$. Therefore [F3] makes $M^f$ a martingale. This includes $n=0$ and constant $f=0,1$, for which the compensator vanishes.

Conversely, suppose every $M^f$ is a martingale. The finite preceding sum [F1, F2, F3] in its definition is ${\mathcal{F}}_n$-measurable and integrable. Expanding the identity in [F3] and cancelling that sum gives $\mathbb{E}[f(X_{n+1})\mid {\mathcal{F}}_n]=f(X_n)+Lf(X_n)=Pf(X_n)\mathrm{.}$ By [F2], $X$ is a $p$-chain. Equivalently, choosing $f=1_A$ for every $A\subseteq S$ gives the conditional transition probability $p(X_n,A)={\sum }_{y\in A}p(X_n,y)$; $A=\mathrm{\varnothing }$ and $A=S$ give zero and one. This proves both implications. Choice is used precisely for the conditional expectations in [F2]--[F3].