Let $B\_b(E)$ be the space of bounded measurable functions on some Polish space $E$ endowed with the supremum norm. It seems quite classical that Markov semigroups $P\_t:B\_b(E)\to B\_b(E)$ are in one to one correspondence with Markov processes on $E$.


By Markov semigroup I mean a strongly continuous semigroup $P\_t:B\_b(E)\to B\_b(E)$ satisfying


* $P\_t$ is conservative for all $t$, i.e. $P\_t1=1$
* $\|P\_t\|=1$, for all $t$
* $P\_t$ is a positive operator for all $t$.


From such an object, the natural way to build the transition functions $p\_t(x,dy)$ of an eventually associated Markov process is to set
$p\_t(x,A):=P\_t\mathbb{1}\_A(x)$, for all measurable set $A$ and all $x$ in $E$.


However, it does not seem direct that $p\_t(x,\cdot)$ is $\sigma$-additive and I was not able to find a proper reference showing this (in the particular context of semigroups on $B\_b(E)$).


In fact, the only references I found assume that $P\_t$ is bounded-pointwise (bp) continuous, i.e.
for any sequence $(f\_n)\_n$ of $B\_b(E)$ converging pointwise to a function $f$ and being uniformly bounded, then $$P\_t f\_n(x)\to P\_tf(x)$$ for all $x$ as $n$ goes to infinity. Property which is in fact equivalent to the $\sigma$-additivity of $p\_t$.


So, are Markov semigroup bp-continuous ? If yes, how can it be derived from the assumptions? if no, is there a counterexample ?



Bounded-pointwise continuity of Markov operators / semigroups
Some continuity assumption is needed, as shown by the following "Markov" semigroup on the set of natural numbers $\mathbb{N}$.


Let $\omega$ be an ultrafilter in $\mathbb{N}$. Every element $f \in B\_b(\mathbb{N})$ (that is, every bounded sequence $f(n)$) has a finite "generalised limit" along $\omega$, that we denote by $\omega(f)$. Furthermore, $\omega(c f) = c \omega(f)$, $\omega(f + g) = \omega(f) + \omega(g)$, if $f$ is non-negative, then $\omega(f) \geqslant 0$, and if $f$ is constant $1$, then $\omega(f) = 1$.


We define $$P\_t f(n) = e^{-t} f(n) + (1 - e^{-t}) \omega(f).$$ It is straightforward to see that $P\_t$ is a strongly continuous semigroup of operators on $B\_b(\mathbb{N})$ with all the properties listed in the question. However, $P\_t f(n)$ is not given as an integral of $f$ with respect to a $\sigma$-additive measure (or, in other words, $p\_t(n, A)$ is not $\sigma$-additive with respect to $A$).


