High density interval for a positive Lebesgue measure set in higher dimensions

Asked Sep 17 '15 at 22:24

Active Sep 17 '15 at 22:24

Viewed 61 times

I was reading 1d case saying for $E \subset R$ with $\mu (E) >0$ and $0< \rho <1$ $\quad \exists$ open interval $I$ s.t. $\mu (E \cap I) > \rho \cdot \mu (I)$.

How can we prove higher dimensional analog of this saying: let $E \subset R^n$ with $\mu (E) >0$ and $0< \rho <1$ then $\quad \exists$ product of intervals $I=I_1 \times ... \times I_n$ s.t. $$\mu (E \cap I) > \rho \cdot \mu (I)$$.

edited Apr 13 '17 at 12:19

Community

asked Sep 17 '15 at 22:24

user16015

1,020

This does not hold with $I$ being a product of intervals. It is true, though, if $I$ is a union of products of intervals, for example. – detnvvp Sep 17 '15 at 22:36
@detnvvp There is a cube (product of equal intervals) with this property, by Lebesgue's density theorem – Sep 19 '15 at 04:16

High density interval for a positive Lebesgue measure set in higher dimensions

0 Answers0