Show that the set in $\Bbb{R}^n$ whose intersection with each ball has measure one third of the ball has Lebesgue measure 0

Question

I had problems approaching the following problem:

Let $E$ be a Lebesgue measurable set such that $m(E\cap B)\leq \frac{1}{3}m(B)$

for every ball $B\subset \mathbb{R}^n$, where $m$ is the Lesbesgue measure in $\mathbb{R}^n$. I am asked to prove that $E$ has Lebesgue measure zero.

I am aware of the version of the problem with n=1. However, the proofs (I've seen so far) rely on the fact that in $\mathbb{R}$ any open set is a countable disjoint union of intervals, which is not true for $n\geq 2$.

I tried using Vitali's covering lemma, which might be useful (in deriving some contradiction with disjoint union of open balls) if upper bound has the ratio $\frac{1}{3^n}$ instead of $\frac{1}{3}$, but couldn't make any meaningful progress with the problem as is.

I've also tried using the inner regularity by assuming $E$ has a positive measure and approximating it from below with some compact set (positive measure). Again, that didn't take me that far.

Could anyone help me with this problem? Even confirming that the statement is true would be helpful, as the source (some past paper) can contain type-os. Many thanks in advance!

Answer 1

We may asssume that $E$ is Borel set of finite measure. Let $\mu (A)=m(A\cap E)$. Then $\mu << m$. We can write $\mu (A)=\int_A fdm$ where $f=I_E$. Now $\frac 1 {m(B(x,r))} \int_{B(x,r)} fdm \leq \frac 1 3$ for every open ball $B(x,r)$. By Lebesgue's Theorem this implies that $f \leq \frac 13$ a.e.. Hence $m(A\cap E)\leq \frac 1 3 m(A)$ for every Borel set $A$. Now put $A=E$ to finish the proof.

Answer 2

Lebesgue's density theorem states that for almost all $x \in E$ we have $$\lim_{r\to 0^+} \frac{m(E \cap B(x,r))}{m(B(x,r))} = 1$$

However, our assumption states that the above limit (if it even exists) is $\le \frac13$ for all $x \in E$ so it follows that $E$ has measure zero.