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Let $X$ be a Borel space with a Borel measure $\mu$. Suppose $\xi: X\times X\to\mathbb R_{\geq 0}$ is a continuous function and put $s(x) = \{y\in X:\xi(x,y) = 0\}$. For any set $b\in\mathcal B(X)$ we put $ \mathcal Sb = \bigcup\limits_{x\in b}s(x). $ I am interested in the solutions of an equation $\mathcal Sb\subseteq b$, or even the generalized one: $ \mu(\mathcal Sb\setminus b) = 0. $ Could we say that these equations are fixpoint problems? If there is a literature for such problems?

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    @Willie: thank you very much2012-01-18

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The following is not so much an answer as it is a pointer to some connections with other areas that one might exploit.

Assume for simplicity that $X$ is a compact metric space. All the results I use can be found in the book Infinite Dimensional Analysis by Aliprantis and Border.

Then every closed subset of $X\times X$ is the zero-set of some function $\xi$ (just let $\xi$ be the distance to the subset), so we can dispense with the function $\xi$. This closed subset is the graph of the correspondence (or multifunction) $s$. That the graph is closed is equivalent to $s$ being an upper hemicontinuous correspondence (uhc) with closed values. Then $\mathcal{S}$ is simply the usual forward image of a set under a correspondence. The coordinatewise union of two uhc correspondences is uhc, so $s^*$ given by $s^*(x)=s(x)\cup\{x\}$ is uhc and compact-valued.

You are looking for the Borel sets $B$ such that $s^*(B)=B$. A compact valued uhc correspondence maps compact sets to compact sets, so $s^*(B)$ is an increasing function on the complete lattice of compact sets in $X$. So the set of compact fixed points forms a complete lattice, by the Tarski fixed point theorem. It might, however, be the case that $\emptyset$ and $X$ are the only such fixed points.

For general Borel sets, the problem is that uncountable unions of Borel sets are usually not Borel, so $s^*$ doesn't map $\mathcal{B}(X)$ to itself.

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    +1 Michael, thanks a lot for the answer. I hope you can excuse me if I write a comment in a couple of days - I have to figure out something w.r.t. this problem.2012-01-18