Lebesgue's density theorem

In mathematics, Lebesgue's density theorem states that given a Lebesgue measurable set $A$, the "density" at almost every point in $A$ set is 1. 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 $\mu$ 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 in a $\epsilon$-neighborhood of a point $x$ in Rⁿ as

$d_{\epsilon }(x)={\frac {\mu (A\cap B_{\epsilon }(x))}{\mu (B_{\epsilon }(x))}}$

where $B_{\epsilon }(x)$ denotes the closed ball of radius $\epsilon$ centered at $x.$

Lebesgue's density theorem asserts that for almost every point of $A$ the density

$d(x)=\lim _{\epsilon \to 0}d_{\epsilon }(x)$

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 $\mu (A)>0$ 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.

References

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

Lebesgue density theorem at PlanetMath.