Let $(X,\Sigma_X,\mu)$ and $(Y,\Sigma_Y,\nu)$ be finite measure spaces (the measures are finite, not the sets). Let $k:X\times\Sigma_Y\to[0,1]$ be a Markov kernel which disintegrates $\nu$, i.e. with the property that for every $B\in\Sigma_Y$, $$ \int_X k(B|x)\,\mu(dx) = \nu(B) . $$ This way, we can form the measure $\mu k$ on $X\times Y$, given for all measurable $A\subseteq X\times Y$ by $$ \mu k (A) = \int_X k(A_x|x) \,\mu(dx) , $$ where $$ A_x = \{y\in Y : (x,y)\in A\} . $$ The measure $\mu k$ has $\mu$ and $\nu$ as marginals.
Suppose now that $f:X\times Y\to\Bbb{R}$ is $\mu k$-integrable. That is, $$ \int_{X\times Y} |f(x,y)|\,\mu k(dx\,dy) < +\infty . $$ Can we conclude that the integrals $$ \int_Y |f(x,y)| \,k(dy|x) $$ are finite for $\mu$-almost all $x$, and so that $$ \int_{X\times Y} f(x,y)\,\mu k(dx\,dy) = \int_X \left( \int_Y f(x,y)\,k(dy|x) \right) \mu(dx) , $$ as in the usual Fubini theorem?
If so, what is the name of this theorem?
(Note that the usual Fubini theorem, at least the one I know about, is the special case where $k$ does not depend on $x$.)