Lebesgue's density theorem
In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set , the "density" of A is 0 or 1 at almost every point in . 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
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
exists and is equal to 1.
In other words, for every measurable set A, the density of A is 0 or 1 almost everywhere in Rn. However, it is a curious fact that if μ(A) > 0 and μ(Rn \ 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.
- Hallard T. Croft. Three lattice-point problems of Steinhaus. Quart. J. Math. Oxford (2), 33:71-83, 1982.