# If $(κ_t)_{t≥0}$ is the transition semigroup of a continuous Markov process, is $t↦(κ_tf)(x)$ continuous for all bounded continuous $f$ and fixed $x$?

Let

• $$(\Omega,\mathcal A,\operatorname P)$$ 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 $$x\in E$$.