Lebesgue's density theorem

In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set A⊆ℝn, the "density" of A is 0 or 1 at almost every point in ℝn. Additionally, the "density" of A is 1 at almost every point of A. Intuitively, this means that the boundary of A, the set of points in A for which all neighborhoods are partially in A and partially outside A, is of measure zero.

Lebesgue's density theorem, applied to the inside of a square, its corners, edges, inside, and outside
Lebesgue's density theorem, applied to the inside of a square, its corners, edges, inside, and outside

Statement

Let μ be the Lebesgue measure on the Euclidean space and A⊆ℝn be a Lebesgue measurable set. Let x∈ℝn and let Bε(x) denote the open ball of radius ε centered at x. Define the density at a point x

dA(x)=limε→0μ(A∩Bε(x))μ(Bε(x))

Lebesgue's density theorem—For a measurable set A⊆ℝn, the density of A is 0 or 1 almost everywhere. That is, the set of points which do not have density 0 or 1, {x∈ℝn∣da(x)∉{0,1}}, has measure 0. Furthermore if 0<μ(A)<∞, then there are always points of A where the density either does not exist or exists but 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 of measure zero.

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 A⊆ℝn instead of Lebesgue measure, as proven in sections 2.8–2.9 of Federer's Geometric Measure Theory, 1969.

See also

References

  1. ↑ Mattila, Pertti (1999). Geometry of Sets and Measures in Euclidean Spaces: Fractals and Rectifiability. ISBN 978-0-521-65595-8.
  2. ↑ Croft, Hallard (1982). "Three lattice-point problems of Steinhaus". Quarterly J. Math. Oxford (2). 33: 71–83.

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