Let $p:Z\to X$ be a continuous surjective map between compact Hausdorff spaces.
Does there exist a family $m=(m\_x)\_{x\in X}$ of Radon probability measures on $Z$, such that


* the support of $m\_x$ is contained in the fibre $p^{-1}(\{ x\})$,
* for every $f\in C(Z)$ we have $f^m\in C(X)$, where $f^m(x)=\int\_Zf(z)\ dm\_x(z)$.


If not, under which additional restrictions does it exist? I am particularly interested in the case, when $Z$ is not first countable.



Integration along fibres of continuous map on compact Hausdorff spaces
I think that such family might not exist even for rather well-behaved spaces. Consider $Z = [0,3]$ and $X = [0,2]$. Then, let $p \colon Z \to X$ be defined by:
$$
p(z) = 
\begin{cases}
z, &\text{ if } z \in [0,1], \\ 
1, &\text{ if } z \in [1,2], \\ 
z-1, &\text{ if } z \in [2,3].
\end{cases}
$$
For $x \in X \setminus \{1\}$ measure $m\_x$ has to be Dirac's delta $\delta\_{p^{-1}(x)}$ as fibres $p^{-1}(x)$ are singletons for such $x$. Now, let $f \in C(Z)$ be any such that $f \equiv 0$ on $[0,1]$ and $f\equiv 1$ on $[2,3]$. Then
$$
f^m(x) = \int\_Z f \ \mathrm{d}m\_x(z) = \begin{cases}
0, &\text{ for } x \in [0,1), \\
\text{something}, &\text{ for } x =1, \\
1, &\text{ for } x \in (1,2].
\end{cases}
$$
Such $f^m$ cannot be continuous regardless of the value of measure $m\_1$.


