# Infima of conditional densities after disintegration

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.

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Indeed, let us define the auxiliary measure $m$ by $$m(B) = \int_{(0,1)} \mathrm{Leb}_y(B\cap L_y) dP(y),$$ where $\mathrm{Leb}_y$ stands for the Lebesgue measure on the horizontal interval $L_y$. (In fact, $m$ is the product measure of the Lebesgue measure on $(0,1)$ and of $P$.) We show that $\mu$ is absolutely continuous with respect to $m$. So, let $B$ be such that $m(B)=0$. Then $\mathrm{Leb}_y(B\cap L_y)=0$ for $P$-almost-every $y\in(0,1)$. But we assumed that the conditional measure $\mu_y$ of $\mu$ is absolutely continuous with respect to $\mathrm{Leb}_y$. It follows that $\mu_y(B\cap L_y)=0$ for $P$-almost-every $y\in(0,1)$, or $\mu(B)=0$. This proves absolute continuity. Accordingly, there exists an integrable Borel measurable function $\rho:(0,1)\times(0,1)\to[0,\infty)$, for which $$\int_B d\mu = \int_B\rho dm = \int_{(0,1)}\int_{L_y} (1_B\rho)(x,y)d\mathrm{Leb}_y(x)dP(y)$$ is true for any Borel set $B$. This implies that, for $P$-a.e. $y\in(0,1)$, the function $\rho(\cdot,y)$ is the density $\rho_y$ of $\mu_y$. $\square$
(2) Coming to Question 2, the answer is affirmative in the mentioned special case that each $\rho_y$ is continuous, because in that case $\inf_{x\in L_y}\rho_y(x) = \inf_{x\in (0,1)\cap\mathbb{Q}}\rho(x,y)$, where in the last expression we are taking the infimum of countably many Borel measurable functions of $y$.