Lebesgue's Density Theorem - intuition and weaker forms 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:


*

*Other, perhaps more correct interpretations

*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).

*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?)

*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?

*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.

 A: 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. 
A: 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.
