I'm working through Katznelson's An Introduction to Harmonic Analysis. Currently, I'm looking at an exercise in the first chapter: Prove that if $E$ is a measurable set on $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$ and $E$ is invariant under translation by infinitely many $\tau\in\mathbb{T}$, then either $E$ or its complement has measure zero. A similar question, but doesn't really clear my confusion. My attempt:
Denote by $G$ the subgroup of elements of $\mathbb{T}$ under which $E$ is invariant, that is, such that $E+\tau=E$. By a previous exercise, $G$ must be dense in $\mathbb{T}$ as it is infinite. Now if $\mu(E)=0$, then there is nothing to show, so assume $\mu(E)>0$. The Lebesgue density theorem implies that $E$ has a point of density, say $t$. I believe it suffices to show that every point of $\mathbb{T}$ is a point of density of $E$, as then $E^c$ would have no points of density and hence have measure zero.
Fixing $\tau\in\mathbb{T}$, we want to show $$\lim_{r\rightarrow0}\frac{\mu(E\cap(\tau-r,\tau+r))}{2r}=1.$$ At this point, I want to use the density of $G$ to translate $t$ to $\tau$. As $G$ is dense, there exists a sequence $\{t_n\}$ in $G$ approaching $\tau-t$. For each $t_n$ and $r>0$, we know that $$\mu(E\cap(t+r,t-r))=\mu(E\cap(t+t_n-r,t+t_n+r))$$ since $E \pm t_n=E$. This shifts the point of density of $E$ arbitrarily close to $\tau$, but I am not sure how to conclude that $\tau$ itself has metric density 1. Thanks!
