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 $\mu K$?
$$
\frac{d \mu K}{d \lambda} = \frac{d}{d\lambda} \int K(x, A) d\mu(x) = ?
$$
Best Answer
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 $\frac{d\mu K}{d\lambda}(s)= \mu(p_\mu K)(t)=\int_S p_\mu(s)\,X(s, t)\,\lambda(ds)$