Lebesgue's density theorem

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In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set A, 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 of A the density

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

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.

[edit] References

  • Hallard T. Croft. Three lattice-point problems of Steinhaus. Quart. J. Math. Oxford (2), 33:71-83, 1982.

This article incorporates material from Lebesgue density theorem on PlanetMath, which is licensed under the GFDL.

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