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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!

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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.

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  • $\begingroup$ Wow thank you so much for putting me out of my agony! $\endgroup$ – Derpsilon Dec 27 '19 at 5:42
  • $\begingroup$ @Derpsilon Most welcome! $\endgroup$ – Kavi Rama Murthy Dec 27 '19 at 5:43
  • $\begingroup$ To obtain $\frac1{m(B(x,r))} \int_{B(x,r)}f\ \mathsf dm$ we have that $$\int_{B(x,r)} f\ \mathsf dm = \mu(B(x,r)) = m(B(x,r)\cap E)\leqslant \frac13 m(B(x,r)),$$ right? Which Lebesgue's theorem are you referring to (there are many theorems attributed to Lebesgue)? And the conclusion follows from $m(E\cap E)= m(E)\leqslant \frac 13 m(E)$, which implies that $m(E)=0$? $\endgroup$ – Math1000 Dec 27 '19 at 6:33
  • $\begingroup$ For any integrable function $f$ $\lim_{r \to 0} \frac 1 {m(B(x,r))} \int_{B(x,r)} fdm \to f(x)$ for almost all $x$ . This is the measure theoretic version of the elementary result you have when $f$ is a continuous function . Proof can be found in Rudin's RCA. @Math1000 $\endgroup$ – Kavi Rama Murthy Dec 27 '19 at 6:37
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    $\begingroup$ You don't need RNT; clearly $\mu(A)= \int_A\chi_E\,dm.$ $\endgroup$ – zhw. Dec 27 '19 at 6:47
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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.

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  • $\begingroup$ Thanks for the answer. I didn't know of the Lebesgue's density theorem, but it seems to be a handy corollary to Lebesgue diff thm to have in mind when dealing with these type of problems! $\endgroup$ – Derpsilon Dec 27 '19 at 20:38

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