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 $f(x) \equiv 1$.