Let $(E,\mathcal E)$ be a measurable space, $B(E,\mathcal E)$ denote the set of bounded $\mathcal E$-measurable functions from $E$ to $\mathbb R$ equipped with the supremum norm and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$.
We know that $(\kappa_t)_{t\ge0}$ is contractive, but not necessarily strongly continuous, on $B(E,\mathcal E)$. Let $(\mathcal D(B),B)$ denote the generator of $(\kappa_t)_{t\ge0}$ considered as an operator semigroup on $B(E,\mathcal E)$.
I want to characterize $(\mathcal D(B),B)$ at least on a subset of $\mathcal D(B)$. In order to do that, let $C$ be a closed subspace of $B(E,\mathcal E)$ and $(\mathcal D(A),A)$ be a densely-defined linear operator on $C$.
One way to show that $(\kappa_t)_{t\ge0}$ is strongly continuous on $C$ and its generator (considered as an operator semigroup on $C$) is $(\mathcal D(A),A)$ (in particular, $(\mathcal D(A),A)$ is an extension of $(\mathcal D(B),B)$) is given by a martingale problem approach.
However, I'm really struggling to see what we really need to assume for this.
To elaborate on that, I'm not sure if it's really necessary to assume this, but I think we need to assume that $(E,\mathcal E)$ is a Borel space (i.e. isomorphic to $([0,1],\mathcal B([0,1]))$, e.g. Polish), since then we know that for every probability measure $\mu$ on $(E,\mathcal E)$, there is an unique probability measure $\operatorname P_\mu$ on $\left(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)}\right)$ with $$\operatorname P_\mu\circ\:\pi_{\{t_0,\:\ldots\:,\:t_n\}}=\mu\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\tag1$$ for all $n\in\mathbb N_0$ and $0=t_0<\cdots<t_n$, where $\pi_I$ denotes the projection from $E^{[0,\:\infty)}$ onto $E^I$ for $\emptyset\ne I\subseteq[0,\infty)$.
Given that, let $\pi_t:=\pi_{\{t\}}$ for $t\ge0$ and $\delta_x$ denote the Dirac measure on $(E,\mathcal E)$ at $x$, $\operatorname P_x:=\operatorname P_{\delta_x}$ and $E_x$ denote the expectation associated to $\operatorname P_\mu$ for $x\in E$.
We easily see that $(\pi_t)_{t\ge0}$ is a time-homogeneous Markov process on $\left(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)},\operatorname P_x\right)$ with transition semigroup $(\kappa_t)_{t\ge0}$ and initial distribution $\delta_x$ for all $x\in E$.
Now, assuming that $$\left(f(\pi_t)-\int_0^t(Af)(\pi_s)\:{\rm d}s\right)_{t\ge0}\tag2$$ is an $\mathcal F^\pi$-martingale$^1$ on $\left(E^{[0,\:\infty)},\mathcal E^{\otimes[0,\:\infty)},\operatorname P_x\right)$ for all $f\in\mathcal D(A)$ and $x\in E$, we easily see that $$(\kappa_tf)(x)-f(x)=\int_0^t(\kappa_sAf)(x)\:{\rm d}s\;\;\;\text{for all }x\in E\tag3$$ for all $f\in\mathcal D(A)$ and $t\ge0$. Since $(\kappa_t)_{t\ge0}$ is contractive on $C$, this implies $$\left\|\kappa_tf-f\right\|_\infty\le t\left\|Af\right\|_\infty\xrightarrow{t\to0+}0\tag4$$ for all $f\in\mathcal D(A)$. And since $\mathcal D(A)$ is a dense subset of $C$, we conclude that $(\kappa_t)_{t\ge0}$ is strongly continuous on $C$.
In order to obtain the claim about the generator, I guess we need to assume that $$\kappa_t\mathcal R(A)\subseteq C\;\;\;\text{for all }t\ge0\tag5,$$ since then the strong continuity on $C$ yields $$\left\|\frac{\kappa_tf-f}t-Af\right\|_\infty\le\sup_{x\in E}\frac1t\int_0^t\left|(\kappa_sAf)(x)-(Af)(x)\right|{\rm d}s\xrightarrow{t\to0+}0\tag6.$$
Question 1: Am I missing any crucial assumption? I've found similar results and therein it is assumed that the process $(\pi_t)_{t\ge0}$ is right-continuous and $C$ is separating$^2$. However, I can't see any stage of my proof above where these assumptions are needed.
Question 2: Are we able to drop the assumption that $(E,\mathcal E)$ is Borel? For example, if we replace the time scale $[0,\infty)$ by $\mathbb N_0$, we know that the existence of the probability measures $\operatorname P_\mu$ are ensured for arbitrary measurable spaces. Is the assumption of being Borel a necessary assumption when we are going from a countable time scale to a continuous one?
$^1$ $\mathcal F^\pi$ denotes the filtration generated by $\pi$, i.e. $\mathcal F^\pi_t=\sigma(\pi_s,s\le t)=\mathcal E^{\otimes[0,\:t]}$ for all $t\ge0$.
$^2$ i.e. if $\mu,\nu$ are probability measures on $(E,\mathcal E)$ with $$\int f\:{\rm d}(\mu-\nu)=0\;\;\;\text{for all }f\in C\tag7,$$ then $\mu=\nu$.
