• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(E,\mathcal E)$ be a measurable space
  • $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$ and $$\kappa_tf:=\int\kappa_t(\;\cdot\;,{\rm d}y)f(y)$$ for bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $t\ge0$
  • $(X_t)_{t\ge0}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A,\operatorname P)$ with $$\operatorname P\left[X_t\in B\mid X_0\right]=\kappa_t(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal E\tag1$$ for all $t\ge0$

Let $f:E\to\mathbb R$ be bounded and $\mathcal E$-measurable and assume $\left(f(X_t)\right)_{t\ge0}$ is right-continuous. Are we able to show that $$[0,\infty)\ni t\mapsto(\kappa_tf)(x)\tag2$$ is continuous at $t=0$ for all $x\in E$?

I've seen this claim in many books (e.g. here). Let's take a look: Let $(t_n)_{n\in\mathbb N}\subseteq[0,\infty)$ with $t_n\xrightarrow{n\to\infty}0$. By the dominated convergence theorem, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0\right]\xrightarrow{n\to\infty}\operatorname E\left[f\left(X_0\right)\mid X_0\right]=f(X_0)\;\;\;\text{almost surely}\tag3.$$ By $(2)$, $$\operatorname E\left[f\left(X_{t_n}\right)\mid X_0=\;\cdot\;\right]=\kappa_{t_n}f\;\;\;\text{for all }n\in\mathbb N\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}\tag4$$ and hence $$\kappa_{t_n}f\xrightarrow{n\to\infty}f\;\;\;\operatorname P\circ\:X_0^{-1}\text{-almost surely}\tag5.$$

I don't see how we obtain continuity of $(1)$ at $t=0$ for a fixed $x\in E$ by $(5)$. However, by $(5)$, $\operatorname E\left[\left(\kappa_{t_n}f\right)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_0)\right]$. So, we obtain the desired result at least if $\operatorname P\circ\:X_0^{-1}=\delta_x$ for a fixed $x\in E$.


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