If $(\kappa_t)_{t\ge0}$ is a Markov semigroup with invariant measure $μ$, under which assumption is $t\mapsto\kappa_tf$ measurable for $f\in L^p(μ)$?

Question

Let

• $(E,\mathcal E)$ be a measurable space;
• $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$;
• $\mu$ be a finite measure on $(E,\mathcal E)$ which is subinvariant with respect to $(\kappa_t)_{t\ge0}$;
• $p\ge1$.

Under this assumptions, $(\kappa_t)_{t\ge0}$ has a unique extension to a contraction semigroup on $L^p(\mu)$.

If $(\kappa_t)_{t\ge0}$ is strongly continuous at $f\in L^p(\mu)$, the orbit $$\operatorname{orb}f:[0,\infty)\to L^p(\mu)\,,\;\;\;t\mapsto\kappa_tf$$ is continuous and hence Borel measurable.

Is there a suitable assumption under which every orbit is Borel measurable? For example, is it sufficient to assume that $$[0,\infty)\times E\;,\;\;\;(t,x)\mapsto(\kappa_tf)(x)\tag1$$ is Borel measurable for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$?