In the context of research in the field of $f$-divergences, I am using measure theory to prove some nice properties. The following result is really useful and so I want to be sure I understand it fully, if it is a known result with an associated name as well as check if the notations are correct or could be improved. Any comment would be appreciated.
Suppose that we have two probability measures $\mu$ and $\nu$ on measure space $(X,\mathcal A)$ and we have a Markov kernel $\kappa$ from measure space $(X,\mathcal A)$ to $(Y,\mathcal B)$. Suppose further more that $\nu\ll\mu$. Define the measure $\mu\otimes\kappa$ on measure space $(X\times Y,\mathcal A\otimes \mathcal B)$ to be for any $A\times B\in \mathcal A\otimes \mathcal B$ \begin{align*} \mu\otimes\kappa(A\times B) = \int_A \kappa(B|x) d\mu(x) \end{align*}
and likewise for $\nu\otimes\kappa$. Since for any $A\times B\in \mathcal A\otimes \mathcal B$ \begin{align} \nu\otimes\kappa(A\times B) &= \int_A \kappa(B|x) d\nu(x)\\ &=\int_A \kappa(B|x) \frac{d\nu}{d\mu}(x) d\mu(x) \end{align} we have that $\nu\otimes\kappa\ll\mu\otimes\kappa$ and for any $(x,y)\in X\times Y$, $\frac{d\nu\otimes\kappa}{d\mu\otimes\kappa}(x,y) = \frac{d\nu}{d\mu}(x)$ is the Radon-Nykodym derivative.
Is this reasoning correct ? does this result have a name ? are there some fun facts or technicalities I should hear about ?
Last question, if I call $\tilde\mu$ the measure on $(Y,\mathcal B)$ defined as $\tilde\mu(\cdot)=\mu\times\kappa(X\times\cdot)$ and likewise for $\tilde\nu$, does $\nu\otimes\kappa\ll\mu\otimes\kappa$ implies $\tilde\nu\ll\tilde\mu$ ?
Thank you very much.
I think you mean $\nu\times\kappa(A\times B)$ instead of $\kappa\times\nu(A\times B)$. And I think you mean $\overline\mu(\cdot)=\mu\times\kappa(X\times\cdot)$ instead of $\mu\times\kappa(\cdot,X)$.
I agree with your reasoning. I don't think this result has a name since it is quite straightforward. As far as I am concerned, I write $\mu\otimes\kappa$ instead of $\mu\times\kappa$ but there does not seem to be a consensus on this notation: $\mu\times\kappa$ is fine.
On your last question: yes. It derives from the fact that for all $B\in\mathcal B$, $\overline\mu(B)=\mu\times\kappa(X\times B)$ and $\overline\nu(B)=\nu\times\kappa(X\times B)$. Therefore, $$ \overline\mu(B)=0\implies\mu\times\kappa(X\times B)=0\implies\nu\times\kappa(X\times B)=0\implies\nu(B)=0. $$