Lebesgue's density theorem

In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set $A\subset \mathbb {R} ^{n}$, the "density" of A is 0 or 1 at almost every point in $\mathbb {R} ^{n}$. 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 B_ε 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 \ 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

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

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