Proof of Lebesgue Density Theorem

Asked Dec 13 '13 at 01:12

Active Dec 13 '13 at 01:12

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If E is a Borel set in $\Bbb R^n$, the density $D_E (x)$ of $E$ at $x$ is defined as

$$D_E(x) = \lim_{r \to 0} \frac{m(E \cap B(r,x))}{m(B(r,x))}$$

Show that $D_E(x) = 1$ for almost everywhere $x \in E$ and $D_E(x) = 0$ for almost everywhere $x \in E^C$.

I am not sure what to do to prove this statement. As far as I can tell, this is the Lebesgue Density Theorem, but I can only find proofs of this in the one-dimensional case (which is by far the easiest case).

asked Dec 13 '13 at 01:12

retupmoc258

This follows from the Lebesgue differentiation theorem by considering the characteristic function of $E$. This is a standard result found in measure theory books. There is also a proof on Wikipedia. – Ayman Hourieh Dec 13 '13 at 01:25
Ah. I see it now. I started to try that, but got stumped. I focused too much on trying to prove it with epsilons and such. Not useful. Thanks. – retupmoc258 Dec 13 '13 at 03:23

Proof of Lebesgue Density Theorem

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