Lebesgue's density theorem

In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set $$A\subset \R^n$$, the "density" of A is 0 or 1 at almost every point in $$\R^n$$. Additionally, the "density" of A is 1 at almost every point in A. Intuitively, this means that the "edge" of A, the set of points in A whose "neighborhood" is partially in A and partially outside of A, is negligible.

Let μ be the Lebesgue measure on the Euclidean space Rn and A be a Lebesgue measurable subset of Rn. Define the approximate density of A in a ε-neighborhood of a point x  in Rn as


 * $$ d_\varepsilon(x)=\frac{\mu(A\cap B_\varepsilon(x))}{\mu(B_\varepsilon(x))}$$

where Bε denotes the closed ball of radius ε centered at x.

Lebesgue's density theorem asserts that for almost every point x of A the density


 * $$ d(x)=\lim_{\varepsilon\to 0} d_{\varepsilon}(x)$$

exists and is equal to 0 or 1.

In other words, for every measurable set A, the density of A is 0 or 1 almost everywhere in Rn. However, if μ(A) > 0 and μ(Rn&thinsp;\&thinsp;A) > 0, then there are always points of  Rn where the density is neither 0 nor 1.

For example, given a square in the plane, the density at every point inside the square is 1, on the edges is 1/2, and at the corners is 1/4. The set of points in the plane at which the density is neither 0 nor 1 is non-empty (the square boundary), but it is negligible.

The Lebesgue density theorem is a particular case of the Lebesgue differentiation theorem.

Thus, this theorem is also true for every finite Borel measure on Rn instead of Lebesgue measure, see Discussion.