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.
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 $m_1$.
