Given a Polish space $X$ and a set valued map $\Psi:X\to 2^X$ we way $\Psi$ is weakly Borel if for every open $U\subset X$, $$ \Psi^{-1}(U):=\{x\,|\,\Psi(x)\cap U\ne\varnothing\}\in\mathcal{B}(X). $$ With the additional assumption that $\Psi(x)$ is a closed set for every $x$, one can establish measurable selection theorems, and better even, a collection of selections whom are dense in $\Psi$ (see Castaing representations).
In short, the denseness can be used to get a sequence of measurable maps, $\zeta_n$, who get 'closer' to $\Psi$ (they are not necessarily selections yet) and upon taking a limit the closed-valuedness implies $\lim_n \zeta_n(x)=:\zeta(x)\in \Psi(x)$ and usual measurability arguments give $\zeta$ measurable. Now for the question:
Proposition: Given a closed valued, weakly Borel set map $\Psi$ with the property $y\in\Psi(x)\iff x\in\Psi(y)$, does there exist a Borel measurable selection $\zeta$ such that $\zeta=\zeta^{-1}?$
In general, the symmetry assumption alone is not sufficient to conclude the statement, in fact it is easy to find counterexamples (Take the set map from $\mathbb{R}$ to $\mathbb{R}$ with graph $\{(x,y)\,|\,x+y>(x-y)^2\}$). But instead suppose one relaxes the conclusion of the proposition to only require a Borel measurable involution on a set of positive Lebesgue measure, then counterexamples are less clear.
Furthermore, one realizes that the above (relaxed) question (for $X=\mathbb{R}$) is equivalent to finding a measurable monotone selection of a set valued map and reflecting appropriately. To this end I have read some of the Economics papers on comparative statics which use order theory to get monotone selections. This said, they were not concerned about measurability, and I am having trouble branching the two theories.
In summary, if anyone has insight on the question on finding a monotone measurable selection in $\mathbb{R}\to\mathbb{R}$ for some positive measure set this would be much appreciated. Maybe I am missing something clear
