Cross-posting from math stack-exchange since it's not getting any visibility there.


I am given a function $F: \{[0, y]: y \in I\} \to \Sigma(I)$, such that $\lambda(F([0, y])) = y$, and $F([0, y]) \subseteq F([0, z])$ for $y \le z$. Here $\lambda$ is the Lebesgue measure on the unit interval $I$ and $\Sigma(I)$ denotes the completion of the Borel $\sigma$-algebra on $I$. How do I show that there exists a measurable bijection $f: I \to I$ with $f([0, y]) = F([0, y])$ for all $y \in I$? What about just a measurable function?


One naive idea is to consider dyadic expansion of $I$. Each $x \in I$ lives in one nested infinite sequence of such dyadic subsets, say $I\_j(x)$, $j=1, 2, \ldots$. We simply define $f(x) = \lim\_j \sup \{y: y \in F(I\_j(x))\}$. Note that the outer limit exists because the sequence is monotone non-increasing. This however may lead to pathological example with $f(x) \equiv 1$.



Automorphism on the unit interval compatible with a measure preserving set function
Are you sure that you are not missing an hypothesis? If I take
$$F([0,y]) = [0,y] \setminus {\bf Q}$$
there can be no such bijective $f$ because then
$$f(\{y\}) = f([0,y]) \setminus \bigcup\_{z<y} f([0,z]) = 
([0,y]\setminus {\bf Q}) \setminus \bigcup ([0,z]\setminus {\bf Q}) = \{y\}$$
and thus $f$ would be the identity.


You may want to look at your objects modulo sets of zero measure.
See e.g. Glasner, "*ergodic theory via joinings*" th. 2.13 for the conditions under which a function between two $\sigma$-algebras comes from a function between the underlying spaces. In short, you need 
$F(A\cup B) = F(A) \cup F(B) $ and $F(A^c) = F(A)^c$, together with measure preservation.


