Let

- $(E,\mathcal E)$ be a measurable space with $\{x\}\in\mathcal E$ for all $x\in E$
- $\mathcal E_b:=\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\}$
- $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$
- $(\Omega,\mathcal A)$ be a measurable space
- $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$
- $\left(\mathcal F^X_t\right)_{t\ge0}$ denote the filtration generated by $(X_t)_{t\ge0}$
- $\operatorname P_x$ be a probability measure on $(\Omega,\mathcal A)$ with $$\operatorname P_x[X_0=x]=1\tag1$$ and $$\operatorname E\left[f(X_{s+t})\mid\mathcal F^X_s\right]=(\kappa_tf)(X_s)\tag2\;\;\;\text{for all }f\in\mathcal E_b\text{ and }s,t\ge0$$
- $c:E\to[0,\infty)$ be $\mathcal E$-measurable

Assume that $X:\Omega\times[0,\infty)\to E$ is $(\mathcal A\otimes\mathcal B([0,\infty)),\mathcal E)$-measurable and hence $$Y_t:=\int_0^tc(X_s)\:{\rm d}s$$ is a well-defined $[0,\infty]$-valued random variable on $(\Omega,\mathcal A)$ for all $t\ge0$.

I'm searching for a mild additional assumption, ensuring that $$\frac{\operatorname E_x[Y_t]}t\xrightarrow{t\to0+}c(x)\tag3.$$ For example, I could imagine that we need to assume that

- $E$ is a topological space and $\mathcal E=\mathcal B(E)$
- $c$ is locally bounded
- $(X_t)_{t\ge0}$ is càdlàg
Would this be enough to conclude?

We may note that $(\kappa_t)_{t\ge0}$ is a contraction semigroup on $\mathcal E_b$. If $(\kappa_t)_{t\ge0}$ would be strongly continuous, we could conclude that $$\frac1t\int_0^t\kappa_sf\:{\rm d}s\xrightarrow{t\to0+}f\tag4$$ for all $f\in\mathcal E_b$. However, $(\kappa_t)_{t\ge0}$ doesn't need to be strongly continuous and hence we cannot apply this. On the other hand, $(4)$ is clearly stronger than what we need for $(3)$ to hold.

To begin with, we might want to note that $$c_n:=\min(c,n)\in\mathcal E_b\;\;\;\text{for all }n\in\mathbb N.$$

localboundedness of $c$. $\endgroup$1more comment