I am given a fixed Borel measurable set $E\subset\mathbb [0,1]^n$, $|E|>0$, and I want to show that there exists some $\epsilon_0>0$ such that for all $\delta>0$ there exists a cube $C_\delta\subset [0,1]^n$ with $|C_\delta|\le\delta$ such that $$ \epsilon_0\le\frac{|C_\delta\cap E|}{|C_\delta|}\le 1-\epsilon_0. $$ This would hold true if there existed a point at which the metric density of $E$ was neither zero nor one. My question therefore is: Does there always exist such a point? And if not, how else can I prove the claim above?
A simple example is $E = [0,\tfrac 1 2]$ in $[0,1]$ (i.e., $n=1$). Then we can choose $C_\delta = (\tfrac 1 2-\tfrac\delta 2,\tfrac 1 2+\tfrac\delta 2)$.
For any Borel-measurable set, it is known that the almost-all points are Lebsgue Points, see also Lebesgue differentiation theorem.
It is easy to find simple counterexamples for the fact that not all points are Lebesgue Points: Take just $E=[1/3,2/3]$, then for any $\delta >0$ we have with $C_\delta := [1/3-\delta,1/3+\delta]$ $$\frac{\lambda(C_\delta \cap E)}{\lambda(C_\delta)} = \frac{1}{2}.$$