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Lebesgue's density theorem

In mathematics, Lebesgue's density theorem states that for any Lebesgue measurable set , the "density" of is 0 or 1 at almost every point in . Additionally, the "density" of is 1 at almost every point of . Intuitively, this means that the boundary of , the set of points in for which all neighborhoods are partially in and partially outside , is of measure zero.

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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.

Lebesgue's density theorem, applied to the inside of a square, its corners, edges, inside, and outside
Lebesgue's density theorem, applied to the inside of a square, its corners, edges, inside, and outside source ↗

The definition

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

d ε ( x ) = μ ( A B ε ( x ) ) μ ( B ε ( x ) ) {\displaystyle \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 {\displaystyle x} of A R n {\displaystyle A\subseteq \mathbb {R} ^{n}} the density

d ( x ) = lim ε 0 d ε ( x ) {\displaystyle \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 {\displaystyle A} , the density of A {\displaystyle A} is 0 or 1 almost everywhere1. If 0 < μ ( A ) < {\displaystyle 0<\mu (A)<\infty } , then there are always points of A R n {\displaystyle A\subseteq \mathbb {R} ^{n}} where the density either does not exist or exists but is neither 0 nor 1.2.

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

See also

References

References

  1. Mattila, Pertti (1999). Geometry of Sets and Measures in Euclidean Spaces: Fractals and Rectifiability. ISBN 978-0-521-65595-8.
  2. Croft, Hallard (1982). "Three lattice-point problems of Steinhaus". Quarterly J. Math. Oxford (2). 33: 71–83.

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