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Let $(X,\mathscr{B},\mu,T)$ be a measure-preserving system, where $X$ is a compact metric space, $\mathscr{B}$ its Borel $\sigma$-algebra, $\mu$ a Borel probability measure and $T$ continuous. Let $f:X \to Y$ be a measurable map where $Y$ is another compact metric space. Let $\mathscr{A}$ be the $\sigma$-algebra generated by the level sets of $f$. Denote by $\mu_x^\mathscr{A}$ the conditional measure of $\mu$ at $x$ with respect to $\mathscr{A}$.
Is there any reason why for $\mu$-almost every $x$, $\mu_x^\mathscr{A}(f^{-1}(x))=1$? If the $\sigma$-algebra $\mathscr{A}$ is countably generated, then this would be true, as the atom of $x$ (the smallest element of $\mathscr{A}$ containing $x$) is contained in $f^{-1}(x)$ and $\mu_x^\mathscr{A}([x]_\mathscr{A})=1$, where $[x]_\mathscr{A}$ is the atom of $x$. But is this still true if $\mathscr{A}$ is not countably generated? Or is there a argument why the $\sigma$-algebra $\mathscr{A}$ is always countably generated?
In my case the function $f$ has the following additional property: If the ergodic component of $\mu$ in $x$ is the same as the ergodic component in $y$ then $f(x)=f(y)$, i.e. the function $f$ depends only on the ergodic components. By this we get that the conditional measure are invariant under $T$, but I don't think (or I just don't see) how this property can be helpful to solve the above question.

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