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.