Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Is it obvious that a point cannot be a density point for $E$ and $E^c$ simultaneously ? – Epsilon Aug 19 '15 at 23:11
up vote 1 down vote accepted

Hint: the difference between $(t+t_n - r, t + t_n+r)$ and $(\tau - r, \tau + r)$ has measure at most $2|\tau - t - t_n|$. In particular this can be made less than $\epsilon r$.

share|cite|improve this answer
Oh, by difference you mean symmetric difference? Then, the difference between $E\cap(t+t_n-r,t+t_n+r)$ and $E\cap(\tau-r,\tau+r)$ also has measure less than $\epsilon r$. Thus the desired limit holds. – dls Jun 29 '12 at 17:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.