Lebesgue's density theorem

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

Statement

Let $\mu$ be the Lebesgue measure on the Euclidean space and $A\subseteq {ℝ}^n$ be a Lebesgue measurable set. Let $x\in {ℝ}^n$ and let $B$_ε$(x)$ denote the open ball of radius $\epsilon$ centered at $x$. Define the density at a point $x$

$$d_A(x)=\underset{\epsilon \to 0}{lim}\frac{\mu (A\cap B_{\epsilon }(x))}{\mu (B_{\epsilon }(x))}$$

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 $A\subseteq {ℝ}^n$ instead of Lebesgue measure, as proven in sections 2.8–2.9 of Federer's Geometric Measure Theory, 1969.

See also

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