Automorphism on the unit interval compatible with a measure preserving set function

Question

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 $f(x) \equiv 1$.

Answer 1

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.