Lebesgue's density theorem
In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set [math]\displaystyle{ A\subset \R^n }[/math], the "density" of A is 0 or 1 at almost every point in [math]\displaystyle{ \R^n }[/math]. 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
- [math]\displaystyle{ d_\varepsilon(x)=\frac{\mu(A\cap B_\varepsilon(x))}{\mu(B_\varepsilon(x))} }[/math]
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
- [math]\displaystyle{ d(x)=\lim_{\varepsilon\to 0} d_{\varepsilon}(x) }[/math]
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.[1] However, 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.
See also
- Lebesgue differentiation theorem – Mathematical theorem in real analysis
References
- ↑ Mattila, Pertti (1999). Geometry of Sets and Measures in Euclidean Spaces: Fractals and Rectifiability. ISBN 978-0-521-65595-8.
- Hallard T. Croft. Three lattice-point problems of Steinhaus. Quart. J. Math. Oxford (2), 33:71-83, 1982.
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