Does absolutely continuity of probability measures imply absolute continuity of conditional probability measures almost everywhere? I have a Polish space $X$ equipped with a Borel probability measure $\mu$. I have another Polish space $Y$ and a continuous function $f : X \rightarrow Y$. I give $Y$ the pushforward probability measure $\nu(\cdot) = \mu\left(f^{-1}(\cdot)\right)$.
Now suppose I have another Borel probability measure $\mu'$ on $X$ such that $\mu \ll \mu'$ and $\mu\left(f^{-1}(\cdot)\right) = \mu'\left(f^{-1}(\cdot)\right)$.
Is it true that $\mu_y \ll \mu_y'$ $\nu$-almost everywhere?
Where $\mu_y$ and $\mu_y'$ are the conditional probability measures of $\mu$ and $\mu'$.
So far I have tried the following : We know $\mu(A) > 0  \Rightarrow \mu'(A) > 0$ for all measurable set $A \subset X$. Let $Y_A := \{ y \in Y \; | \; \mu_y'(A) > 0\} \cup \{ y \in Y \; | \; \mu_y(A) = 0 \}$. Pick $A \subset X$ such that $\mu(A) > 0$, then $\mu'(A) = \int_{Y} \mu'_y(A) d\nu > 0$ (by the disintegration theorem) so we know $\nu(Y_A) = 1$. I was then able to show
\begin{align*} \nu\left(\bigcap_{A \subset X\text{ with non-emptry interior }} Y_A \right) = 1.
\end{align*} but I am still short of showing \begin{align*} \nu\left(\bigcap_{A\subset X \text{ measurable}} Y_A \right) = 1.
\end{align*} 
I have solved it for $Y$ countable, but it would be great to have the general case.
 A: Not only this is true, but also one can write the "conditional density" explicitly. Let us try first to identify $d\mu_y/d\mu_y'$ informally, and then prove the heuristics. Write 
$$
\mu_y(\cdot) = \mu(\cdot)/\mu(f^{-1}(\{y\}))
$$
and similarly for $\mu'$, so that 
$$
\frac{d\mu_y}{d\mu_y'}(x)= \frac{d\mu}{d\mu'}(x) \frac{\mu'(f^{-1}(\{y\}))}{\mu(f^{-1}(\{y\}))}.
$$
Now 
$$
\frac{\mu(f^{-1}(\{y\}))}{\mu'(f^{-1}(\{y\}))} = \frac{1}{\mu'(f^{-1}(\{y\}))} \int_{f^{-1}(\{y\})} \frac{d\mu}{d\mu'}(z)\mu'(dz),
$$
which looks like the conditional expectation $\mathsf E_{\mu'}[\frac{d\mu}{d\mu'}\mid f = y]$.
Turning to the formal part, let  $\mathcal G$ be the sub-sigma-algebra of $\mathcal B(X)$ generated by $f$. Obviously, $\mu\ll \mu'$ on $\mathcal G$ as well, so there is a Radon-Nikodym density $Z = d\mu|_\mathcal G/d\mu'|_\mathcal G$ (it coincides with the above conditional expectation). But since $\mathcal G$ is generated by $f$, $Z(x) = h(f(x))$ for some measurable $h$. I claim that 
$$
\frac{d\mu_y}{d\mu_y'}(x)= \frac{d\mu}{d\mu'}(x) \frac{\mathbf{1}_{h(y)>0}}{h(y)}.
$$
a.e. This is equivalent to the equality
$$
\mu(A\mid \mathcal G)(w) = \frac{\mathbf{1}_{Z(w)>0}}{Z(w)}\int_A \frac{d\mu}{d\mu'}(x)\mu'(dx\mid \mathcal G)(w)\tag{1}
$$
for a.a. $w$. To show the last, write for any $A\in \mathcal B(X)$, $B\in \mathcal G$ 
$$
\mathsf E_{\mu}\left[\frac{\mathbf{1}_{Z>0}}{Z} \mathsf E_{\mu'}\left[\frac{d\mu}{d\mu'}\mathbf{1}_{A}\,\Big|\,\mathcal G\right]\mathbf{1}_{B} \right] = \mathsf E_{\mu'}\left[Z\frac{\mathbf{1}_{Z>0}}{Z} \mathsf E_{\mu'}\left[\frac{d\mu}{d\mu'}\mathbf{1}_{A}\,\Big|\,\mathcal G\right]\mathbf{1}_{B} \right]  \\
= \mathsf E_{\mu'}\left[\mathbf{1}_{Z>0}\frac{d\mu}{d\mu'}\mathbf{1}_{A}\mathbf{1}_{B} \right] = \mathsf E_{\mu}\left[\mathbf{1}_{Z>0}\mathbf{1}_{A}\mathbf{1}_{B}\right].
$$
(The first equality holds since the expression inside the expectation is $\mathcal G$-measurable.) 
This shows that 
$$
\mathsf E_{\mu}\left[\mathbf{1}_{A}\mid \mathcal G\right] = \frac{\mathbf{1}_{Z>0}}{Z} \mathsf E_{\mu'}\left[\frac{d\mu}{d\mu'}\mathbf{1}_{A}\,\Big|\,\mathcal G\right]
$$
almost surely on $\{Z>0\}$, which is equivalent to (1) (obviously, $d\mu_y/d\mu'_y$ vanishes on $\{Z=0\}$).
