Disintegration of measure: a convergence problem. Let $\Omega$ and $X$ be compact metric spaces endowed with the Borel $\sigma$-algebra. Consider the product $\Omega\times X$ endowed with the product $\sigma$-algebra.
Let $\mathbb{P}$ be a probability measure on $\Omega$ and let $\mu$ be a probability measure 
on $\Omega\times X$ whose marginal on $\Omega$ is $\mathbb{P}$, i.e, $\pi_{*}\mu=\mathbb{P}$, where $\pi$ is the projection on the first factor, $\pi(\omega,x)=\omega$. 
It is known that there is  a family of probability measures 
$\{\mu_{\omega}\}_{\omega\in \Omega}$  on $X$, called the disintegration of $\mu$ with respect to $\mathbb{P}$ such that
$$
\mu(A)=\int_{\Omega}\mu_{\omega}(A^{\omega})\,d\mathbb{P}(\omega),
$$
 where $A^{\omega}=\{x\colon (\omega,x)\in A\}$.
My question is the following: 
Let $F_{n}\colon \Omega\times X\to \mathbb{R}$ be a sequence of measurable
functions such that for every $x\in X$ 
$$
F_{n}(\omega,x)\to 0 \quad \mbox{for}\quad \mathbb{P-}\mbox{almost every} \,\,\omega.
$$
I would like to know  if it is true that 
$$
F_{n}(\omega,x)\to 0 \quad \mbox{for}\quad \mathbb{\mu-}\mbox{almost every} \,\,(\omega,x),
$$
for every probability measure $\mu$ on $\Omega\times X$ whose marginal is $\mathbb{P}$.
I know this is true for product measures $\mathbb{P}\times\nu$. It is just a direct consequence of Fubini's theorem.
 A: No, this won't hold for every measure on the product. This is because there might be sets of positive measure in $\Omega \times X$ such that every "slice" obtained by fixing $x$ and projecting to $\Omega$ will have measure zero in $\Omega$. A simple example is the following:
Take both $\Omega$ and $X$ to be the space $[0,1]$ with the Borel $\sigma$-algebra and equip $\Omega$ with Lebesgue measure $m$ (I'll still write them as $\Omega$ and $X$ to distinguish the two factors). Let $\pi: \Omega \times X \to \Omega$ be the projection on the first factor. Let $i: \Omega \to \Omega \times X$ be the embedding of $[0,1]$ as the diagonal $\Delta$, that is $i(\omega) = (\omega, \omega)$. Define the measure $\mu$ on the product as $\mu = i_* m$. Then $\mu( \Delta) =1$ and $\mu$ projects to Lebesgue measure on the first factor:
$$
\pi_* \mu = \pi_*( i_* m ) = (\pi \circ i)_* m = (id_{\Omega})_* m =m.
$$
Now consider for example the indicator function of the diagonal, that is 
$$
F(\omega, x) = \begin{cases} 1 & \text{if} \; x= \omega \\ 0 & \text{otherwise} \end{cases}
$$
Then for any fixed $x_0 \in X$ we have 
$F(\omega, x_0) = 1_{\{x_0\}}( \omega)$ so the restricted map on $\Omega$ is zero $m$-almost everywhere. However, $F$ is equal to $1$ $\mu$-almost everywhere since $\mu( \Delta) =1$ by construction.
