Let $G$ be a group action on the set $X$, and let $f:(X,\Sigma_X)\to (Y,\Sigma_Y)$ be a measurable maximal invariant ($f$ is constant on each orbit and take different values on different orbits). Am asked to show that if $(X,\Sigma_X)$ and $(Y,\Sigma_Y)$ are standard Borel then
$$\sigma(f)=\{B\in \Sigma_X:g^{-1}(B)=B \text{ for all } g \in G\}$$
The inclusion $\subset$ is easy. How can I show the other inclusion?
It suffices to show that for any measurable invariant (constant on each orbit) $g:(X,\Sigma_X)\to [0,1]$ there exists measurable $h:(Y,\Sigma_Y)\to [0,1]$ such that $g=h\circ f$.
This is because we can then take $g=1_B$ so that $B=g^{-1}(1)=f^{-1}(h^{-1}(1))\in \sigma (f)$.
Any invariant function $g$ is function of $f$. But how to choose $h$ in a measurable way?
Can I use a measurable selection theorem?