Density of Kernel Operating on a measure Suppose $(S, \mathcal{S})$ is a measurable space, $K:S\times \mathcal{S}\to [0, 1]$ is a Markov Kernel and $\mu:\mathcal{S}\to [0, 1]$ is a probability measure. The kernel operates on measures so the following is a measure on $(S, \mathcal{S})$
$$
\mu K(A) = \int \mu(dx) K(x, A) \qquad \qquad A\in\mathcal{S}
$$

Assuming all the regularity conditions needed, what is the density of this new measure?

For instance, suppose that $\mu$ and $K$ have densities
$$
\frac{d \mu}{d \lambda} = p_\mu \qquad \text{and} \qquad \frac{d K(x, \cdot)}{d \lambda} 
 = p_K 
$$
with respect to some dominating (Lebesgue) measure $\mu \ll \lambda$ and $K\ll \lambda$. What is the density of $\mu K$?
$$
\frac{d \mu K}{d \lambda} = \frac{d}{d\lambda} \int K(x, A) d\mu(x) = ?
$$
 A: First of all, keep in mind that existence of a common ($\sigma$-)finite measure $\lambda$ for $K(x,\cdot)$ is a relatively strong condition, and it does not have to happen at all. For example, consider $K(x,\cdot) = \delta_{f(x)}(\cdot)$ where $f$ is some measurable map. If you are dealing with uncountable spaces, there won't exist such $\lambda$.
So, let's make an assumption that there exists such $\lambda$ that $K(x,\cdot) \ll \lambda$ for all $x\in S$. We may need to check whether $\varphi(x,\cdot) = \frac{\mathrm dK(x, \cdot)}{\mathrm d\lambda(\cdot)}$ is a jointly measurable map, here I'm not sure whether that's guaranteed. Ok, suppose that it is. Then you can easily check that $\int_S \varphi(x,\cdot)\mu(\mathrm dx)$ satisfies the definition of density for $\mu K$.
A couple of comments:

*

*What do you mean by saying that $\lambda$ is not only dominating, but also perhaps Lebesgue measure? The latter is only defined on $\Bbb R^n$, whereas here you seem to deal with general measurable spaces.


*$\mu(\cdot)$ and $K(x, \cdot)$ are measures on different spaces, so the condition $\mu\ll\lambda$ and $K(x,\cdot)\ll\lambda(\cdot)$ does not make sense


*As you can see, nowhere in the derivation a dominating measure for $\mu$ was used. One intuitive explanation may be - we only need dominating measures, when we talk about at least two measures on the same space. On $S$ you have only one measure to cake about, and that's $\mu$ itself
A: If $(S,\mathscr{S},\mu)$ is a probability space, $K$ is a kernel from $(S,\mathscr{S})$ into itself, and both $\mu$ and $K$ are dominated by a $\sigma$-finite measure $\lambda$ on $(S,\mathscr{S})$.
If $\mathscr{S}$ is countably generated (there is a countable collection $\mathcal{C}$ such that $\mathscr{S}=\sigma(\mathcal{C})$, then de Possel-Doob's theorem (for example: Kallenberg, O., Random measures theory and applications, Springer 2010, section 1.5) shows that there is a measurable function $X:(S,\mathscr{S})\times(S,\mathscr{S})\rightarrow[0,\infty]$ such that
$$K(s,B)=\int_B X(s, t)\mathbb{1}_{\{X<\infty\}}(s, t)\,\lambda(dt) + K(s,B\mathbb{1}_{\{X=\infty\}})$$
and that $X(s, t)\mathbb{1}_{\{X<\infty\}}(s, t)\,\lambda(dt)\ll\lambda$ and $\mathbb{1}_{\{X=\infty\}}\cdot K\perp\lambda$. Since by assumption in your OP $K\ll\lambda$, then $X<\infty$ $\lambda$-a.s. and so
$\frac{dK}{d\lambda}(s, t)=X(x,t)$ is measurable $\mathscr{S}\otimes\mathscr{S}$-measurable.
It  follows that $\mu K$ is also dominated by $\lambda$. Moreover, if $\frac{d\mu}{d\lambda}=p_\mu$, then by Funini's theorem
$$
\mu K(B)=\int_S\Big(\int_B X(s, t)\,\lambda(dt)\Big)p_\mu(s)\,\lambda(ds)=\int_B\Big(\int_S p_\mu(s)\,X(s, t)\,\lambda(ds)\Big)\,\lambda(dt)
$$
whence one concludes that $\frac{d\mu K}{d\lambda}(s)= \mu(p_\mu K)(t)=\int_S p_\mu(s)\,X(s, t)\,\lambda(ds)$
