Consider a probability measure on the product of Polish spaces $X \times Y$. The disintegration of measure theorem allows one to write probability measures on $X \times Y$ as an average of probability measures on $\{x\} \times Y$ for $x \in X$. In particular, if $\mu$ is the marginal of $\pi$ on $X$ then there exists a measurable map $\pi_x$ from $X$ to probability measure on Y such that $$ \pi = \int_{X}{(\delta_x \otimes \pi_x)}{\; \text{d}\,\mu(x)}. $$
The exercise (7.8, page 210) asks for the statement of disintegration of measure when all the measures in consideration are absolutely continuous with respect to a fixed reference measure, for instance the Lebesgue measure. I will further add the requirement that they have a continuous density. In particular what is the form of $\pi_x$?
My attempt
Write $\lambda_U$ the Lebesgue measure on $U$ and suppose that $\pi$ (resp. $\mu$) is absolutely continuous w.r.t. the Lebesgue measure with continuous density $w_\pi: X\times Y \rightarrow \mathbb R_+$ (resp. $w_\mu:X \rightarrow \mathbb R_+$). Then we have that the measurable map from $X$ to $P(Y)$ has the following form: $$ \begin{aligned} & X \rightarrow P(Y)\\ & x \mapsto \pi_x = w^\text{des}(x,-)\; \text{d}\, \lambda_Y \end{aligned} $$ where $$ w^\text{des}(x,-):\begin{aligned} & Y \rightarrow \mathbb R_+ \\ & y \mapsto \begin{cases} \frac{w_\pi(x,y)}{w_\mu(x)}& & \text{ if } w_\mu(x) \neq 0 \\ 0 & &\text{ otherwise} \end{cases} \end{aligned} $$
I am failing to see whether $w^\text{des}$ is indeed still continuous (or at least $L^1$) so that $\pi_x$ is also absolutely continuous w.r.t. the Lebesgue measure (which I assume is true). Having continuity preserved is also very desirable for me.
EDIT : there was a typo in the expression of $w^\text{des}$ (I had written $w_\mu(y)$ instead of $w_\mu(x)$). This makes de facto the map continuous so I think it closes the question.