Let $(E,\mathcal E)$ be a measurable space, $$\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$$ be equipped with the supremum norm and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$. It is easy to verify that $$\kappa_tf:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for }f\in\mathcal E_b\text{ and }t\ge0$$ is a contraction semigroup on $\mathcal E_b$. So, it does make sense to talk about the (integral representation of the) resolvent of $(\kappa_t)_{t\ge0}$. However, we can consider something weaker here. Assume that $$[0,\infty)\to\mathbb R\;,\;\;\;t\mapsto(\kappa_tf)(x)\tag1$$ is Borel measurable for all $(x,f)\in E\times\mathcal E_b$. Now, let $$\rho_{\text w}:=\left\{\lambda\in\mathbb R:\int_0^\infty e^{-\lambda t}\left|(\kappa_tf)(x)\right|\:{\rm d}t<\infty\text{ for all }(x,f)\in E\times\mathcal E_b\right\}$$ and $$R_\lambda(x,B):=\int_0^\infty e^{-\lambda t}\kappa_t(x,B)\:{\rm d}t\;\;\;\text{for }(x,B)\in E\times\mathcal E\text{ and }\lambda\in\rho_{\text w}.$$
Question 1: Let $\lambda\in\rho_{\text w}$. What do we need to ensure that $R_\lambda$ is actually a Markov kernel on $(E,\mathcal E)$? Is it sufficient (by Fubini's theorem) to assume that $$[0,\infty)\times E\to\mathbb R\;,\;\;\;(t,x)\mapsto(\kappa_tf)(x)\tag2$$ is Borel measurable for all $f\in\mathcal E_b$?
Question 2: What is the weakest assumption we need to impose in order to show that $$[0,\infty)\to\mathcal E_b\;,\;\;\;t\mapsto\kappa_tf\tag3$$ is Borel measurable for all $f\in\mathcal E_b$? Is it sufficient to assume Borel measurablility of $(2)$ for all $f\in\mathcal E_b$?
Now let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$ and $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued $(\mathcal F_t)_{t\ge0}$-adapted process on $(\Omega,\mathcal A,\operatorname P)$.
Question 3: Let $s\ge0$ and $f\in\mathcal E_b$. How can we infer from$^1$ $$\operatorname E\left[\int_0^\infty e^{-\lambda t}f(X_{s+t})\:{\rm d}t\mid\mathcal F_s\right]=(R_\lambda f)(X_s)\tag4\;\;\;\text{for all }\lambda\ge0$$ that $$\operatorname E\left[f(X_{s+t})\mid\mathcal F_s\right]=(\kappa_tf)(X_s)\;\;\;\text{for all }t\ge0\tag5$$ (i.e. that $(X_t)_{t\ge0}$ is a time-homogeneous $(\mathcal F_t)_{t\ge0}$-Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup $(\kappa_t)_{t\ge0}$)?
First of all, I guess we need to assume that $$\Omega\times[0,\infty)\to E\;,\;\;\;(\omega,t)\mapsto X_t(\omega)\tag6$$ is $\mathcal A\otimes\mathcal B([0,\infty))$-measurable in order to ensure that the left-hand side of $(4)$ is well-defined.
I would like to find an elegant way to prove question 3. Note that I'm familiar with operator theory and functional analysis. I guess it's easier to argue with this knowledge.
Let $A\in\mathcal F_s$ and \begin{align}\mu((a,b])&:=\operatorname E\left[\int_a^bf(X_{s+t})\:{\rm d}t;A\right];\\\nu((a,b])&:=\operatorname E\left[\int_a^b(\kappa_tf)(X_s)\:{\rm d}t;A\right]\end{align} for $0\le a<b$. We know that $\mu$ and $\nu$ are uniquely determined by their Laplace transform $\mathcal L_\mu$ and $\mathcal L_\nu$; \begin{align}\mathcal L_\mu(\lambda)&=\int\mu({\rm d}t)e^{-\lambda t}=\operatorname E\left[\int_0^\infty e^{-t\lambda}f(X_{s+t})\:{\rm d}t;A\right]\\\mathcal L_\nu(\lambda)&=\int\nu({\rm d}t)e^{-\lambda t}=\operatorname E\left[\int_0^\infty e^{-t\lambda}(\kappa_tf)(X_s)\:{\rm d}t;A\right]=(R_\lambda f)(X_s)\end{align} for all $\lambda\ge0$.By $(4)$, $$\mathcal L_\mu=\mathcal L_\nu\tag7.$$ From this we can infer that $$\int_a^b\operatorname E\left[f(X_{s+t});A\right]\:{\rm d}t=\int_a^b\operatorname E\left[(\kappa_tf)(X_s);A\right]\:{\rm d}t\;\;\;\text{for all }0\le a<b\tag8$$ and hence $$\operatorname E\left[f(X_{s+t});A\right]=\operatorname E\left[(\kappa_tf)(X_s);A\right]\;\;\;\text{for Lebesgue-almost all }t\ge0\tag9.$$ However, I have no idea how to conclude from here (especially since the Lebesgue-null set in $(9)$ depends on $A$).