# Lebesgue density theorem

Let $\mu $ be the Lebesgue measure^{} on ${\mathbb{R}}^{n}$, and for a
measurable set^{} $A\subset {\mathbb{R}}^{n}$ define the density of $A$ in
$\u03f5$-neighborhood^{} of $x\in {\mathbb{R}}^{n}$ by

$${d}_{\u03f5}(x)=\frac{\mu (A\cap {B}_{\u03f5}(x))}{\mu ({B}_{\u03f5}(x))}$$ |

where ${B}_{\u03f5}(x)$ denotes the ball of radius $\u03f5$ centered at $x$.

The Lebesgue density theorem asserts that for almost every point of $A$ the density

$$d(x)=\underset{\u03f5\to 0}{lim}{d}_{\u03f5}(x)$$ |

exists and is equal to $1$.

In other words, for every measurable set $A$ the density of $A$ is $0$ or $1$ almost everywhere. However, it is a curious fact that if $\mu (A)>0$ and $\mu ({\mathbb{R}}^{n}\setminus A)>0$, then there are always points of ${\mathbb{R}}^{n}$ where the density is neither $0$ nor $1$ [1, Lemma 4].

## References

- 1 Hallard T. Croft. Three lattice-point problems of Steinhaus. Quart. J. Math. Oxford (2), 33:71–83, 1982. http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0499.10035Zbl 0499.10035.

Title | Lebesgue density theorem |
---|---|

Canonical name | LebesgueDensityTheorem |

Date of creation | 2013-03-22 13:21:02 |

Last modified on | 2013-03-22 13:21:02 |

Owner | bbukh (348) |

Last modified by | bbukh (348) |

Numerical id | 7 |

Author | bbukh (348) |

Entry type | Theorem |

Classification | msc 28A75 |