(GTM 259 Exercise 2.1.6.)Prove that the circle rotation $R_\alpha$ from Example 2.2 is not measurably isomorphic to the circle-doubling map $T_2$ from Example 2.4.?
Suppose there is a measurably isomorphic $\varphi$ to the circle-doubling map, then $\forall t\in\mathbb T$, $\varphi(t+\alpha)=2\varphi(t)\pmod 1$. How to deduce a contradiction?
Note that $t\mapsto\varphi(t)$, $t+\alpha\mapsto 2\varphi(t)$ and $t+2\alpha\mapsto 4\varphi(t)$. Since $\varphi$ is measure-preserving, then $\alpha=\varphi(t)$ and $2\alpha=3\varphi(t)$ if $\varphi(t)<1/3$ and $\alpha\le1/2$ then this is a contradiction. Since $\varphi$ is surjective, we can always find such $t$ with $\varphi(t)<1/3$. As for $\alpha>1/2$, we can take $-\alpha$ to get the same thing.
Is that correct?
Edit:
Since $\varphi$ is an isomorphism, and we know that $\varphi^{-1}$ preserves the measure, we have $\mu(t+\alpha-t)=\mu(2\varphi(t)-\varphi(t))$ where $\mu$ is the Lebesgue measure. Hence $\alpha=\varphi(t)$ under the condition $\varphi(t)<1/3$ and $\alpha\le 1/2$.