Let $\mathbb{T} = \{z \in \mathbb{C}: |z| = 1\}$. Define $\phi:[0,1] \to \mathbb{T}$ by $\phi(t) = e^{2\pi i t}$, let $m$ be the Lebesgue measure on $[0,1]$, and define $\mu$ by $\mu(E)= m(\phi^{-1}(E))$ for each Borel set $E$ of $\mathbb{T}$.
Suppose $\theta$ is irrational and $E$ is a Borel set of $\mathbb{T}$. For any $n \in \mathbb{N}$, define $E(n) = \{ze^{2n\theta\pi i}: z \in E\}$. Show that $\mu(E) = 0)$ iff $\mu(E \cap E(n)) =0$ for all $n\in \mathbb{N}$.
I've already shown that $\mu$ is a measure, and the forward direction of this is trivial. My intuition is that I might be able to show that if $E \neq \emptyset$, $$\mu\left(\bigcup_{n \in \mathbb{N}}E(n)\right) = 1$$ I think this has something to do with $\theta$ being irrational, and that if I fix a $z \in E$, the set $\{ze^{2k\theta \pi i}:k \in \mathbb{N}\}$ should be dense in $\mathbb{T}$. Then from there I think that I could get $$\mu(E) = \mu\left(E \cap \bigcup_{n \in \mathbb{N}}E(n)\right) \leq \mu(E \cap E(n)) = 0$$ but I don't know how to fill in the details or even if that is a good idea.