In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set A ⊆ R n {\displaystyle A\subseteq \mathbb {R} ^{n}} , the "density" of A {\displaystyle A} is 0 or 1 at almost every point in R n {\displaystyle \mathbb {R} ^{n}} . Additionally, the "density" of A {\displaystyle A} is 1 at almost every point of A {\displaystyle A} . Intuitively, this means that the boundary of A {\displaystyle A} , the set of points in A {\displaystyle A} for which all neighborhoods are partially in A {\displaystyle A} and partially outside A {\displaystyle A} , is of measure zero.
Statement Let μ {\displaystyle \mu } be the Lebesgue measure on the Euclidean space and A ⊆ R n {\displaystyle A\subseteq \mathbb {R} ^{n}} be a Lebesgue measurable set. Let x ∈ R n {\displaystyle x\in \mathbb {R} ^{n}} and let B {\displaystyle B} ε ( x ) {\displaystyle (x)} denote the open ball of radius ε {\displaystyle \varepsilon } centered at x {\displaystyle x} . Define the density at a point x {\displaystyle x}
d A ( x ) = lim ε → 0 μ ( A ∩ B ε ( x ) ) μ ( B ε ( x ) ) {\displaystyle \qquad \qquad d_{A}(x)=\lim _{\varepsilon \to 0}{\frac {\mu (A\cap B_{\varepsilon }(x))}{\mu (B_{\varepsilon }(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 ⊆ R n {\displaystyle A\subseteq \mathbb {R} ^{n}} instead of Lebesgue measure, as proven in sections 2.8–2.9 of Federer's Geometric Measure Theory, 1969.
See also Lebesgue differentiation theorem – Mathematical theorem in real analysis
References
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