Consider the measurable partition of the open unit square $(0,1)\times(0,1)$ into horizontal intervals $L_y=(0,1)\times\{y\}$. Let $\mu$ be a Borel probability measure with the disintegration

$$ \mu(B) = \int_{(0,1)} \mu_y(B\cap L_y) dP(y) $$

holding for all Borel sets $B\subset (0,1)\times(0,1)$. Here $\mu_y$ is the conditional probability measure on $L_y$ and $P$ is the factor probability measure on $(0,1)$. Let us assume that $\mu_y$ has a density $\rho_y$ on $L_y$ for each $y\in(0,1)$. (However, $P$ may well be singular. We can set $\mu_y=\text{Leb}$ as necessary to define a conditional probability for all $y$ without affecting the measure.)

Question 1: Is the function $\rho:(0,1)\times(0,1)\to\mathbb{R}:\rho(x,y)=\rho_y(x)$ Borel measurable?

Question 2: Is the function $f:(0,1)\to\mathbb{R}:f(y)=\inf_{x\in L_y}\rho_y(x)$ Borel measurable?

I would appreciate an answer even in the special case where each $\rho_y$ is assumed continuous.