Let $\mu$ be a (finite) Radon measure on $S^1$ such that:
- $\mu$ is continuous, that is, $\mu(\{z\})=0$ for every $z \in S^1$.
- $\mu$ is positive in the sense that if $f$ is a positive measurable function, then $\int_{S^1} f d\mu \geq 0$.
- $\mu$ is square-root invariant, meaning that for all measurable $E \subseteq S^1$, $\mu(E) = \mu(\sqrt E)$, where $\sqrt E = \{z \in S^1:z^2 \in E\}$.
I want to prove that $\mu$ is absolutely continuous with respect to the Lebesgue measure $\lambda$ on $S^1$.
I don't know if this result is true, I have tried many attempts envolving defining a larger set $E'$ by applying successively the square root on $E$, and an equivalence relation on $S^1$ as follows: $z \sim w$ iff $\exists n,m \in \mathbb{N}: z^{2^n}=w^{2^n}$, these led nowhere. Furthermore, I believe that irrational numbers play a big role in here since each one defines a dense "orbit" by applying the square root.
If this result is false then the counter example would be a square root invariant version of the Cantor meansure obtained from the Cantor distribution, but I doubt such an object exists.