[sam-n]

a measurable is said to have density at if the .define the set at which has density .prove that .

[Kent Merryfield]

This is an immediate corollary of the Lebesgue theorem on the differentiation of the integral: if is locally integrable, then

almost everywhere.

Just apply this theorem to the characteristic function of

Do you want a proof of the differentiation of the integral theorem? Or are you trying to prove the density without quoting that theorem?

[candy]

Kent Merryfield wrote:

This is an immediate corollary of the Lebesgue theorem on the differention of the integral...

where might I find the statement of that theorem? do you know if it is in bartle or royden?

[candy]

(by the way, a proof that does not rely on the theorem that kent mentioned is in "Measure and Category" by Oxtoby page 17)

[Kent Merryfield]

I have a 2nd edition of Royden: the key chapter is Chapter 5, Differentiation. The result is never quite stated in exactly that way, but the gist of it is in that chapter.

I don't have a copy of that Bartle, so I can't help you there. I know it's a skinny book, so I don't know what's included and what isn't.

In Wheeden and Zygmund, it's given in generality as Thoerem 7.2 (p.98) and the proof and several corollaries (including the statement on points of density that started this topic) last for the next 10 pages. This proof proceeds by way of the Hardy-Littlewood maximal function, which is my preferred proof.

Somewhere, somehow, there has to be a "covering lemma" (Vitali, or some other) in the proof of this theorem.

[grobber]

I'll use Vitali's Theorem (discussed on the forum), as Kent suggested. I think we should restate the problem first, since it seems a bit awkward :

Let be a measurable subset of the real line with positive measure, and let be the set of points of where the density is . Then .

For and , define . Let be the set of points of which satisfy . It's easy to see that is measurable. If we fix an arbitrary and prove that , we're done. We can also assume WLOG that all the measures involved are finite, i.e. . In fact, we can assume that is bounded.

For every , we can find a sequence which decreases towards such that . Let . If we assume that has positive measure, we'll be able to find an interval such that . Now , the subcollection of consisting of those which are contained in , Vitali-covers . This means that we can find a disjoint sequence such that , according to Vitali's Theorem. However, we have and the 's are disjoint, so , contradiction.