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

There's a basic idea in measure theory that if $E$ is any Lebesgue measurable set in $\mathbb{R}$ such that $\lambda(E)>0$, then for any $\epsilon>0$, then there is a nontrivial, finite $J=[a,b]$ such that $\lambda(E\cap J)>(1-\epsilon)\lambda(J)$.

I want to extend this to $\mathbb{R}^n$. So I'm curious if

for any measurable $E\subseteq\mathbb{R}^n$ and any $\epsilon>0$, then there is some with equal side lengths $J=(a_1,a_1+\delta]\times\cdots\times(a_p,a_p+\delta]$ such that $\lambda(E\cap J)>(1-\epsilon)\lambda(J)$.

I tried adapting the argument like so: Assume $\lambda(E)<\infty$ and $\epsilon<1$. Then take finite intervals $J_m=(a_{1,m},a_{1,m}+\delta]\times\cdots\times(a_{p,m},a_{p,m}+\delta]$ such that $E\subset\bigcup_m J_m$ and $\sum_{m\geq 1}\lambda(J_m)\leq\lambda(E)/(1-\epsilon)$. Then $$\lambda(E)\leq\sum_{n\geq 1}\lambda(J_m\cap E)$$ and for some $m$ $$(1-\epsilon)\lambda(J_m)\leq\lambda(J_m\cap E).$$

Is this the correct idea? I'm essentially following the approach in the $\mathbb{R}^1$ case that I've read, but I'm unsure of some steps. For instance, why is it possible to find such $J_m$ such that $\sum_{m\geq 1}\lambda(J_m)\leq\lambda(E)/(1-\epsilon)$? It seems to be asking a lot from such sets, and I don't see why they should exist.

share|cite|improve this question
Perhaps you can get $\lambda(E\cap J)\gt (1-\epsilon)^n\lambda(J)$. That's useful in many cases. – leo Dec 8 '11 at 20:58

Your Answer


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

Browse other questions tagged or ask your own question.