Suppose $X$ and $Y$ are two real-valued random variables with a specified joint probability distribution $P\_{X,Y}.$ I wish to determine if there is a $\sigma$-finite measure $\mu$ on the real line such that $P\_{Y|X=x} << \mu$ for $P\_X$-almost all $x\in\mathbb{R}$. Call this property $Q.$ Property $Q$ does not always hold e.g. if $X$ is a standard normal variable and $Y=X.$


Is there an equivalent description of property $Q$ that is easy to check for a given probability distribution $P\_{X,Y}$?



Conditions for existence of dominating $\sigma$-finite measure for all conditional distributions
In one direction it's clear: If $P\_{X,Y}\ll P\_X\otimes P\_Y$, then there is a jointly measurable density $f$, and (a version of) the conditional distribution $P\_{Y|X=x}$ is given by $f(x,y)P\_Y(dy)$, so the choice $\mu=P\_Y$ suffices.


Conversely, suppose that $Q$ holds. Then there is a jointly measurable function $g$ such that $P\_{Y|X=x}(dy) = g(x,y)\mu(dy)$, for $P\_X$-a.e. $x$. Define $h(y)=\int\_R g(x,y)\,P\_X(dx)$, and observe that if $h(y)=0$ then $g(x,y)=0$ for $P\_X$-a.e. $x$. Define $B=\{y: h(y)=0\}$, and $f(x,y) = 1\_{B^c}(y)g(x,y)/h(y)$. I claim that $f\cdot P\_X\otimes P\_Y=P\_{X,Y}$. Indeed,
$$
\eqalign{
P\_{X,Y}(C)
&=\int\int 1\_C(x,y)P\_X(dx)P\_{Y|X=x}(dy)\cr
&=\int\int 1\_C(x,y)P\_X(dx)g(x,y)\mu(dy)\cr
&=\int\int 1\_C(x,y)1\_{B^c}(y)P\_X(dx)g(x,y)\mu(dy).\cr
}
$$
In particular, taking $C=R\times A$, for $A$ a Borel subset of $R$,
$$
\eqalign{
P\_Y(A)=P\_{X,Y}(R\times A)
&= \int\int 1\_A(y)1\_{B^c}(y)P\_X(dx)g(x,y)\mu(dy)\cr
&= \int 1\_A(y)1\_{B^c}(y)h(y)\mu(dy)\cr
&= \int 1\_A(y)h(y)\mu(dy).\cr
}
$$
That is, $P\_Y=h\cdot\mu$. Using this to continue the computation in the last-but-one display (multiply and divide by $h(y)$) we have
$$
\eqalign{
P\_{X,Y}(C)
&=\int\int 1\_C(x,y)1\_{B^c}(y)P\_X(dx)g(x,y)\mu(dy).\cr
&=\int\int 1\_C(x,y)P\_X(dx)f(x,y)P\_Y(dy),\cr
}
$$
and the asserted absolute continuity.


