1

I recently read over Nathaniel Martin's thesis, which has the following theorem on Page 19

Theorem 3.1. Let $a$ be any point in $\mathbf{R}$ and let $\delta$ be any real number with $0 \leq \delta \leq 1$. Then there exists a set $G$ whose density exists at $a$ and has the value $\delta$.

Here density defined to be the

$$ D_E(x) = \lim_{\epsilon\to 0} \frac{m(E \cap (x-\epsilon, x+\epsilon))}{2\epsilon}. $$

I can mostly follow his proof, but I was wondering if the problem is meaningfully simplified if instead of letting $a$ be any point in $\mathbf{R}$, it is instead set to be $0$? I am looking for an application of Vitali's covering lemma in the proof, in a similar way to the proof for the Lebesgue density theorem in the line, but am having a hard time progressing.

I saw elsewhere a suggestion for a constructive approach, using $$E_n = \pm (1/(n+1), (n+\alpha)/(n(n+1))]$$

I am struggling to find an application for these pairs of intervals, though.

1 Answers1

2

The proof of Theorem 3.1 does in fact do exactly what you are asking for. It first solves the problem in the case $a=0$, and then observes that if $G$ works for $a=0$, then $G+a$ will work for any other value of $a$. This doesn't actually simplify the proof in any meaningful way though; it just slightly simplifies some of the notation.

As for understanding the proof given, I would suggest you first just focus on understanding the first paragraph (up through the definition of $G$), as that is by far the most important part. The idea behind defining $G$ is extremely simple: just take a sequence of intervals approaching $0$ and then let $G$ consist of a $\delta$ fraction of each one of them. The rest of the proof is just some technical calculations to check that this idea actually works.

You should not expect the Vitali covering lemma or anything like the proof of the Lebesgue density theorem to be relevant here. You aren't proving a theorem about arbitrary measurable sets. You're just proving there exists one particular set, which has the property that it contains about a $\delta$ fraction of any small interval around $a$. So, naturally, the proof involves describing how to construct such a set, and then doing a calculation to check that it actually does have the required property.

Eric Wofsey
  • 330,363
  • As I asked Dave above, if you have a moment: does Martin cover the case when $\delta$ is equal to $0$ or $1$ and we look at $D_E(0)$? I cannot see it in his proof. – user546077 Mar 29 '18 at 02:20
  • He addresses those cases at the end. Those cases are completely trivial, though: you should stop and think about how to prove them yourself if you do not see it immediately. – Eric Wofsey Mar 29 '18 at 02:49