Lebesgue's density theorem

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 Bε(x) denote the open ball of radius \varepsilon centered at x. Define :\qquad\qquad d_\varepsilon(x)=\frac{\mu(A\cap B_\varepsilon(x))}{\mu(B_\varepsilon(x))} '''Lebesgue's density theorem asserts that for almost every point x of A\subseteq \R^n the density''' :\qquad\qquad\qquad d(x)=\lim_{\varepsilon\to 0} d_{\varepsilon}(x) exists and is equal to 0 or 1. ==What the Lebesgues density theorem states==

For every measurable set A, the density of A is 0 or 1 almost everywhere. If 0, then there are always points of A\subseteq \R^n where the density either does not exist or exists but 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 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==