If $(\kappa_t)$ is a Markov semigroup, is $t\mapsto\left\|\kappa_tf\right\|_\infty$ Borel measurable?

Question

Let $(E,\mathcal E)$ be a measurable sapce, $$\mathcal E_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$$ be equipped with the surepmum norm and $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$. Moreover, let $$(\kappa_tf)(x):=\int\kappa_t(x,{\rm d}y)f(y)\;\;\;\text{for }f\in\mathcal E_b.$$ Note that $$\kappa_tf\in\mathcal E_b\;\;\;\text{for all }f\in\mathcal E_b\tag1.$$

If $f\in\mathcal E_b$, are we able to show that $$[0,\infty)\to[0,\infty)\;,\;\;\;t\mapsto\left\|\kappa_tf\right\|_\infty\tag2$$ is Borel measurable? If not, are we able to impose a suitable assumption ensuring that?

Answer 1

Yes, in fact the function $t \mapsto \|\kappa_t f\|_\infty$ is monotone decreasing, and every monotone function is Borel.

The operator $\kappa_t$ is Markovian, and so $\|\kappa_t f\|_\infty \le \|f\|_\infty$. To see this directly, note that for each $x$, the measure $\kappa_t(x, \cdot)$ is a probability measure, and therefore $$|\kappa_t f(x)| =\left| \int f(y) \kappa_t(x,dy) \right| \le \int |f(y) |\kappa_t(x,dy) \le \|f\|_\infty$$ then take the supremum over $x$.

Now for any $t > s$ we can use the semigroup property and the Markovian property of $\kappa_{t-s}$ to write $$\|\kappa_t f\|_\infty = \|\kappa_{t-s} \kappa_{s} f\|_\infty \le \|\kappa_s f\|_\infty$$ and so the function $t \mapsto \|\kappa_t f\|_\infty$ is indeed monotone decreasing.