martingale problem: $M_t = f(X_t) - \int_0^t A f(X_s) ds$, $t \geq 0$

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I am considering the Martingale Problem. Therefore, we consider $(X_t,\mathcal{F}_t, P_x)$ as a progessively measurable Markov process with values in $(E, \mathcal{E})$ with associated semigroup $(P_t)_{t \geq 0}$ and generator $A$ on $B(E)$ (the Banach space of all bounded and measurable functions from $E$ to $\mathbb{R}$) with domain $D(A)$.

(We defined $D(A) := \{ f \in L : lim_{h \to \infty} \frac{1}{h} [P_h f - f] \text{ exists in } L \}$)

It's about the theorem that for every $f \in D(A)$ :

$M_t = f(X_t) - \int_0^t A f(X_s) ds$, $t \geq 0$ is again an $\{\mathcal{F}_t\}$-martingale

My question is why should $M_t$ be integrable and adapted?

I noticed that the Markov process is progessively measurable, but I still can't see why the whole term should be adapted. I am struggling to understand the exactly relation to this claim. Tithus.