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Lebesgue's Density Theorem states that given a measurable set $E$ on the real line, then the set of points $E'$ for which $\lim_{h \to 0} \frac{m(E \cap (x-h,x+h) )}{2h} = 1$ is $E$ up to a nullset, i.e., $m(E \Delta E') = 0$. I interpret it as follows: a measurable set behaves locally as an interval, measure-wise.

I ask for the following:

  1. Other, perhaps more correct interpretations
  2. Intuition for why this is true? The only intuition I know is that the theorem is easy consequence in the case of an open set, and a measurable set is almost a $G_{\delta}$ set (intersection of countably many open sets).
  3. Interesting proofs (I know a proof that uses the regulairy of Lebesgue measure, i.e., the ability to approximate the measure of set $E$ arbitrarily close by closed\open sets contained\containing $E$. Are there any different proofs?)
  4. A direct consequence of the lemma is that for almost all $x\in E$ we have: $\forall h>0 m(E\cap (x-h,x+h)) > 0$. Is there a simple proof for this, not using Lebesgue's Density Theorem?
  5. With my intuition, a nowhere dense closed set (closed set that doesn't contain an interval) of positive measure (say, 'thick' Cantor set) might contradict the theorem (but it doesn't). I know that topological denseness and positive measure don't imply each other, but still - it's not trivial for me to see how the theorem works in this case.
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I'm not sure it's useful to try to convert the Lebesgue density theorem into a catchy slogan. But regarding question 4, certainly the (countable) union of the basic open sets which have null intersection with $E$ again has null intersection with $E$, and the points leftover satisfy your condition. (Also, your $H$ seems to play no role in the statement.) –  user83827 Aug 17 '11 at 20:36
    
You're right, I simplified the statement now. Also, thanks for the solution, it is indeed trivial (I didn't give it enough thought) –  Ofir Aug 17 '11 at 22:39
    
Well oftentimes these sorts of questions are difficult until you happen to rephrase it in a particular way that makes it almost immediate! Anyway, I apologize if the first sentence of my comment was overly blunt: I meant it as a simple, lighthearted statement of opinion (i.e., "I think the statement of the theorem is straightforward as is without need of a soft version"), but upon rereading it seems more negative than intended. And of course your opinion may differ. –  user83827 Aug 17 '11 at 23:04
    
Catchy slogan: "The edge of a measurable set is negligible." –  Marcel T. Jul 24 at 18:10

2 Answers 2

up vote 4 down vote accepted

I don't know if this is a more correct interpretation or a more interesting proof, but I always pictured the density theorem as a corollary of Lebesgue's Differentiation Theorem, which says that for $f \in L^1(\mathbb{R}^n)$, $Q_x \subset \mathbb{R}^n$ a cube centered at $x$, and $|\cdot|$ denoting Lebesgue measure,

$$\lim_{|Q_x|\searrow 0} \frac{1}{|Q_x|} \int_{Q_x} f(y) dy = f(x) \text{ for a.e. $x \in \mathbb{R}^n$}.$$

Since the indicator function $\chi_E(x)$ of a measurable set $E$ with finite measure is in $L^1$, we can use this theorem to say that

$$\lim_{|Q_x|\searrow 0} \frac{|E \cap Q_x|}{|Q_x|} = \chi_E(x) \text{ for a.e. $x\in \mathbb{R}^n$}.$$

By breaking up any measurable set into measurable pieces with finite measure, I think the result follows for general measurable sets as well.

In this sense, the density theorem basically says that the indicator function of a measurable set obeys the (Lebesgue version of the) fundamental theorem of calculus.

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I like to think of it in terms of a probability: any event that can happen (i.e. any set with non-zero probability) has arbitrarily high probability of happening in some interval.

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