(1) The answer to Question 1 is affirmative.

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 $L_y$. 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 $y$.

In other words, Question 2 is still open in the general case. Can a Lusin-type argument ("every measurable function is nearly continous") work there?

(1) The answer to Question 1 is affirmative.

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 $y$.

In other words, Question 2 is still open in the general case. Can a Lusin-type argument ("every measurable function is nearly continous") work there?

(1) The answer to Question 1 is affirmative.

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 $L_y$. 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 $$ \int_B d\mu = \int_B\rho d\mu = \int_{(0,1)}\int_{L_y} (1_B\rho)(x,y)d\mathrm{Leb}_y(x)dP(y) $$ 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 $y$.

In other words, Question 2 is still open in the general case. Can a Lusin-type argument ("every measurable function is nearly continous") work there?

(1) The answer to Question 1 is affirmative.

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 $L_y$. 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 $y$.

In other words, Question 2 is still open in the general case. Can a Lusin-type argument ("every measurable function is nearly continous") work there?