Let $S =[0, 1]^2$ denote the unit square in $\mathbb R^{2}$. For any subset $A$ of $S$ let $A^{c}$ denote its complement in $S$, and $\overline{A}$ its closure in $S$. Given a measurable map $g: W \to V$ between measure spaces $(W, \mu)$, $(V, \phi)$, we say it is equally scaling if there exists some $r$ in $\mathbb R$ such that for any measurable set $K$ in $V$, we have $\phi(K) = r\mu(g^{-1}(K))$.

Given any measurable bounded function $f: S \to [0, \infty]$ such that $\int_{S} fd\mu= 1$, does there exist

  • a sequence of measurable sets $A_n$ in $S$ such that for every $n$ in $\mathbb N$, the closures $\overline{A_n}$, $\overline{A_n ^{c}}$ have Lebesgue measure $\frac{1}{2}$ $\overline{A_n}$, $\overline{A_n ^{c}}$ and $S$ are homeomorphic;

  • and two sequences of equally scaling homeomorphisms $h_{n}:$ $\overline{A_n} \to $ $\overline{A_n ^c}$, $s_n:A^c \to S$ such that $$ \lim_{n \to \infty} F_{n}...F_{0}(f) \to 1$$ uniformly a.e.?

Here $F_{n}: M(S) \to M(S)$ is defined by $$F_{n}(f) (x) = \frac{[fs_n^{-1}(x) + fh_n^{-1}s_n^{-1}(x)]}{2} $$ and $M(S)$ is the set of measurable functions on $S$.

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    $\begingroup$ This is a pretty notation-heavy question. You might want to consider adding a simple example, e.g. with $f$ constant equal to $1$, to illustrate this. $\endgroup$ – Wojowu Jan 31 at 13:55

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