Lebesgue measure is ergodic on a circle with irrational rotation.

Question

Let $X$ be the circle $\mathbb{R}/\mathbb{Z}$ and let $R_{\theta}(x) = x + \theta$ mod $1$ be the rotation by an irrational angle $\theta$. Prove that the Lebesgue measure $\mu$ is ergodic.

This is taken from Foundations of ergodic theory by Viana and Oliveira. I have found a few different proofs of this, but here authors give the following hint, which I don't really know how to use:

Let $A$ be an invariant set with positive measure. Recalling that the orbit $\{R_{\theta}^n(a): n \in \mathbb{Z}\}$ is dense in $X$ for every $a$, show that no point in $X$ is a density point of $A^c$. Conclude that $\mu(A) = 1$.

Answer 1

"Which I don't really know how to use". I think it is quite obvious how to use the hint; indeed, ergodicity is equivalent to any invariant set of positive measure having measure $1$, so the hint immediately implies ergodicity.

The question is how to prove the hint. Before I say the proof, I'd like to say the idea of it. By "density point", the authors are (presumably) referring to a density point in the sense of Lebesgue. Since $A$ has positive measure, there is some $a \in A$ (in fact almost every point in $A$) such that $A$ basically contains an interval around $a$ in a measure-theoretic sense. Specifically, $\lim_{r \downarrow 0} \frac{|A \cap B_r(a)|}{|B_r(a)|} = 1$. The usefulness is that no $x \in A^c$ can be a point of density for $A^c$; indeed, any interval around $x$ contains part of a translate of the interval around $a$ by a multiple of $\theta$ (this is the fact that the orbits of $a$ are dense). Therefore, by Lebesgue's density theorem, $A^c$ must have measure $0$, which is what we want.

In fact, I'll leave it there. Hopefully the ideas came across. If things are still unclear or you need some details worked out, just let me know.