Why are the homeomorphisms from the unit circle to the unit circle preserving measure affine? [closed]

Question

Why are the homeomorphisms from the unit circle to the unit circle preserving measure affine? The affine is composition of rotation and continue automorphism.

Answer 1

Take any homeomorphism $h$ of the unit circle (say $C$). Composing $h$ with an appropriate rotation, we see that, without loss of generality, $1=1+0i$ is a fixed point of $h$. Considering parametrizations $[0,2\pi)\ni t\mapsto e^{it}$ and $(0,2\pi]\ni t\mapsto e^{it}$ of $C$, we see that $h$ is given by a map of one of the following two types:

1. $[0,2\pi)\ni t\mapsto f(t)\in[0,2\pi)$ such that $f$ is a homeomorphism, $f(0)=0$, and $f(2\pi-)=2\pi$;

2. $(0,2\pi]\ni t\mapsto f(t)\in[0,2\pi)$ such that $f$ is a homeomorphism, $f(0+)=2\pi$, and $f(2\pi)=0$.

Consider now type 1. Then, by the intermediate value theorem, the function $f$ is strictly increasing. So, for each $x\in[0,2\pi)$, $f$ maps the interval $[0,x]$ onto the interval $[0,f(x)]$. Since $h$ is measure preserving, we have $f(x)=x$, for each $x\in[0,2\pi)$. So, the homeomorphism $h$ is the identity map and hence a rotation.

Type 2 is considered similarly. Alternatively, it can be reduced to type 1 by composing $h$ with the reflection $e^{it}\mapsto e^{-it}=e^{i(2\pi-t)}$ for $t\in[0,2\pi)$.

Thus, $h$ is necessarily a rotation or the composition of a rotation with the reflection. $\quad\Box$