Under which conditions on a Markov process $X$ does $\frac 1t\lim_{t\downarrow 0}\operatorname P_x\left[X_t\in B\right]$ exist?

Question

Let

• $I\subseteq[0,\infty)$ be closed under addition and $0\in I$
• $(\Omega,\mathcal A)$ be a measurable space
• $E$ be a Polish space and $\mathcal E$ be the Borel $\sigma$-algebra on $E$
• $X=(X_t)_{t\in I}$ be a stochastic process on $(\Omega,\mathcal A)$ with values in $(E,\mathcal E)$
• $(\mathcal F_t)_{t\in I}$ be the filtration generated by $X$

$X$ is called a Markov process with distributions $(\operatorname P_x)_{x\in E}$ $:\Leftrightarrow$

1. $\operatorname P_x$ is a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x\left[X_0=x\right]=1\;,$$ For all $x\in E$
2. $\kappa : E\times\mathcal E^{\otimes I}\to[0,1]$ with $$\kappa(x,A):=\operatorname P_x\left[X\in A\right]$$ is a stochastic kernel
3. $X$ has the weak Markov property, i.e. $$\operatorname P_x\left[X_{s+t}\in B\mid\mathcal F_s\right]=\kappa_t(X_s,B)\;\;\;\text{for all }x\in E\;s,t\in I\;\text{and }B\in\mathcal E\;\tag 1$$ where $$\kappa_t:E\times\mathcal E\to[0,1]\;,\;\;\;(x,B)\mapsto\kappa\left(x,\left\{y\in E^I:y(t)\in B\right\}\right)=\operatorname P_x\left[X_t\in B\right]$$

Question:$\;\;\;$Let $x\in E$ and $B\in\mathcal E$. Under which conditions does $$q(x,B):=\frac 1t\lim_{t\downarrow 0}\kappa_t(x,B)$$ exist?

Probably, we need at least right-continuity of $X$, but I'm unsure which further assumptions we need in detail.