Let $X$ be a right continuous Markov process with left limits and generator $L$. Why is $f(X_t)-f(X_0) - \int Lf(X_s) ds$ a martingale for every $f \in D(L)$?
Let s<t. $E^x[M_t^f|F_s]= E^x[f(X_t)-f(X_0)- \int_0^t Lf(X_r) dr |F_s]$ $= E^x[f(X_t)-\int_s^t Lf(X_r) dr |F_s]-f(X_0)- \int_0^s Lf(X_r) dr\\$.
It remains to show that $E^x[f(X_t)-\int_s^t Lf(X_r) dr |F_s]=f(X_s)$. I'm not sure how to go on from here. Perhaps I can start with $ E^x[f(X_t)|F_s]-E^x[\int_s^t Lf(X_r) dr |F_s]= E^x[f(X_t)|F_s]-E^x[\int_0^{t-s} Lf(X_{r+s}) dr |F_s] =E^{X_s}[f(X_{t-s})]-\int_0^{t-s}E^{X_s} (Lf(X_r)) dr $.
Let $\{T_t\}$ be the semigroup of $X$, i.e. $T_tf(x) = E_x f(X_t)$. Then by Markov property $$E_x [f(X_t)\mid F_s] = T_t f(X_s), \qquad s<t. $$
As $Lf = \frac{d}{du}T_u f\mid_{u=0}$, by the semigroup property of $\{T_t\}_t$ we have $T_r L f = \frac{d}{du}T_u f |_{u=r}$.
So $$ E [ \int_s^t Lf(X_r) d r \mid F_s] = \int_s^t T_r Lf(X_s) d r = T_tf(X_s) - T_sf(X_s).$$