# automorphisms of a measurable space can be approximated by continuous measure preserving maps?

Suppose first that $D=[0,1]$ is equipped with the usual Lebesgue measure, and that $\varphi$ is a measure-preserving transformation $\varphi:D\to D$ that is bijective and whose inverse is also measure preserving (called automorphism).

Is it true that there exists a sequence of continuous measure-preserving transformations $\varphi_n:D\to D$ converging in measure to $\varphi$?

• If you want the continuous measure-preserving transformations to be automorphisms also, then certainly not... – Anthony Quas Jun 20 '18 at 15:34
• @AnthonyQuas I know, I removed that condition – user49870 Jun 20 '18 at 15:35
• You probably should edit the title also. One interesting example to keep in mind is where $\phi$ is the map that takes a point $x$ and interchanges the $2n-1$st and $2n$th binary digit of $x$ for all $n$. This is an automorphism, and I believe it can be approximated by continuous mpts. – Anthony Quas Jun 20 '18 at 17:52

In fact, sequences of continuous piecewise linear transformations of $[0,1]$ suffice to approximate a.e. any measure preserving transformation of $[0,1]$ as proved in thm 2.1 in this paper (therefore in measure, by Severini-Egorov theorem; but in fact in that proof convergence in measure is first proved).