Lebesgue's density theorem
# 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 $R^n$ and A be a Lebesgue measurable subset of $R^n.$ Define the approximate density of A in a ε-neighborhood of a point x in $R^n$ 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 $R^n.$ However, if μ(A) > 0 and $μ(R^n$ \ A) > 0, then there are always points of $R^n$ 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 $R^n$ instead of Lebesgue measure, see Discussion.

