# Lebesgue measure is ergodic on a circle with irrational rotation.

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 $$\mu(A) = 1$$.

"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.