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 $\epsilon$-neighborhood  of $x\in\mathbb{R}^{n}$ by

 $d_{\epsilon}(x)=\frac{\mu(A\cap B_{\epsilon}(x))}{\mu(B_{\epsilon}(x))}$

where $B_{\epsilon}(x)$ denotes the ball of radius $\epsilon$ centered at $x$.

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

 $d(x)=\lim_{\epsilon\to 0}d_{\epsilon}(x)$

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 LebesgueDensityTheorem 2013-03-22 13:21:02 2013-03-22 13:21:02 bbukh (348) bbukh (348) 7 bbukh (348) Theorem msc 28A75