Besicovitch Covering Theorem

In mathematical analysis, a Besicovitch cover, named after Abram Samoilovitch Besicovitch, is an open cover of a subset ''E'' of the Euclidean space R^''N'' by balls such that each point of ''E'' is the center of some ball in the cover.

The Besicovitch covering theorem asserts that there exists a constant ''c''_N depending only on the dimension ''N'' with the following property:

* Given any Besicovitch cover F of a bounded set ''E'', there are ''c''_''N'' subcollections of balls ''A''₁ = , …, ''A''_''c''''N'' =  contained in F such that each collection ''A''_i consists of disjoint balls, and
:$E \subseteq \bigcup_^ \bigcup_ B.$

Let G denote the subcollection of F consisting of all balls from the ''c''_''N'' disjoint families ''A''₁,...,''A''_''c''''N''.
The less precise following statement is clearly true: every point ''x'' ∈ R^''N'' belongs to at most ''c''_''N'' different balls from the subcollection G, and G remains a cover for ''E'' (every point ''y'' ∈ ''E'' belongs to at least one ball from the subcollection G). This property gives actually an equivalent form for the theorem (except for the value of the constant).

* There exists a constant ''b''_N depending only on the dimension ''N'' with the following property: Given any Besicovitch cover F of a bounded set ''E'', there is a subcollection G of F such that G is a cover of the set ''E'' and every point ''x'' ∈ ''E'' belongs to at most ''b''_''N'' different balls from the subcover G.

In other words, the function ''S''_G equal to the sum of the indicator functions of the balls in G is larger than 1_''E'' and bounded on R^''N'' by the constant ''b''_''N'',
:$\mathbf_E \le S_ := \sum_ \mathbf_B \le b_N.$

Application to maximal functions and maximal inequalities

Let μ be a Borel non-negative measure on R^''N'', finite on compact subsets and let $f$ be a $\mu$-integrable function. Define the maximal function $f^*$ by setting for every $x$ (using the convention $\infty \times 0 = 0$)
:$f^*(x) = \sup_ \Bigl( \mu(B(x, r))^ \int_ , f(y), \, d\mu(y) \Bigr).$
This maximal function is lower semicontinuous, hence measurable. The following maximal inequality is satisfied for every λ > 0 :
:$\lambda \, \mu \bigl( \ \bigr) \le b_N \, \int , f, \, d\mu.$

;Proof.
The set ''E''_λ of the points ''x'' such that $f^*(x) > \lambda$ clearly admits a Besicovitch cover F_λ by balls ''B'' such that
:$\int \mathbf_B \, , f, \ d\mu = \int_ , f(y), \, d\mu(y) > \lambda \, \mu(B).$
For every bounded Borel subset ''E''´ of ''E''_λ, one can find a subcollection G extracted from F_λ that covers ''E''´ and such that ''S''_G ≤ ''b''_''N'', hence
:$\begin \lambda \, \mu(E') &\le \lambda \, \sum_ \mu(B)\\ &\le \sum_ \int \mathbf_B \, , f, \, d\mu = \int S_ \, , f, \, d\mu \le b_N \, \int , f, \, d\mu, \end$
which implies the inequality above.

When dealing with the Lebesgue measure on R^''N'', it is more customary to use the easier (and older) Vitali covering lemma in order to derive the previous maximal inequality (with a different constant).

See also

*Vitali covering lemma

References

* .
** .
* .
* {{citation, title=On the best constant for the Besicovitch covering theorem, jstor=2161215, first1=Z, last1=Füredi, author1-link=Zoltán Füredi, first2=P.A., last2=Loeb, journal=Proceedings of the American Mathematical Society, volume=121, year=1994, issue=4, pages=1063–1073, doi=10.2307/2161215.

Covering lemmas
Theorems in analysis