Given a set , we say has a Besicovitch covering if , there exists a cube(or a ball) , such that is the center of . In other word, we also say covers in the sense of Besicovitch.

We will assume the element in is cube , which is the cube center at of edge-length . When the covering consists of balls, the theorem still holds up to minor changes.

**Thm 1(Besicovitch covering lemma)** Suppose is a bounded set, has a Besicovitch covering . Then one can extract a sequence of cubes(balls) (possibly finite or countable) such that

1). is covered by .

2). Each point of is contained in at most (a number only depends on the dimension ) cubes, i.e. , where is the characteristic function of .

3). The sequence of can be classified in families of cubes. That is there exists subcollections of , namely , , , , such that consists of disjoint balls and .

In order to prove this theorem, let’s see a simpler version first. The idea of proving this theorem is embraced in the following simpler version.

**Thm 2** Let be a decreasinf sequence of cubes centered at the origin of . Assume is a bounded set of , and we take a positive integer and denote . Then there exists a sequence (possibly finte) such that:

1). is covered by .

2). , where is the characteristic function of .

3). The sequence of can be classified in families of cubes. Each family consists of disjoint cubes.

Choose such that has the largest edge-length among all , .

If , we are done. Otherwise choose such that has the largest edge-length.

Suppose we have choose .

If , stop this process and we get a finite sequence of cubes. Otherwise choose such that has the largest edge-length. Now we can get a sequence of cubes (possibly finite), which satisfies the following properties:

Satisfies the following properties:

, .

, . Here . Since is bounded, we can assume all these cubes are contained in a sufficently large cube. So if the sequence is infinite, we must have .

Let’s consider the conclusion 1). If the sequence is finite, then such that , which means is covered by these cubes. If the sequence is infinite and , then there exists such that . This contradicts our principle of choosing points.

Consider conclusion 2). Take , WLOG assume is the origin. There are hyper-quadrants in . At each hyper-quadrant, we can only have one cube centered at this hyper-quadrant and cover the origin at the same time. Otherwise the two cubes may contain one’s center which violates the property .

Consider conclusion 3).

Claim: For every , at most cubes in intersect .

pf: If a cube intersect , then since , must contain a vertex of . Note that has at most vertices and each vertex is covered by at most cubes. We know that at most cubes in intersect .

Construct subfamilies of cubes, namely , by the following procedure: Firstly, allocate into , for . Consider when . If every contains a cube intersect , then this will contradict our previous claim. So at least there is one contains balls all disjoint from . We allocate into the one with the smallest index. Proceed this process inductively.

For the proof of **theorem 1,** we don’t have a decreasing family of cubes. So we have to adjust the principle of chosing points. At the same time and will be different from theorem 2.

Define be the edge-length of a cube . Let .

If , since is bounded, we can find a sufficiently large ball to cover . Otherwise choose such that .

Suppose we have chosen .

Let

choose one cube such that and . Proceed this process inductively, then we can get a sequence of cubes satisfy the following properties:

, ,

, . Here . For the same reason, .

Conclusion 1) holds for the same reason.

Consider conclusion 2). For simplicity, we only consider , other dimension is the same. Choose , WLOG, assume is the origin. Let’s focus on the first quadrant.

Claim: There are at most 4 cubes in this sequence which can center at the first quadrant and cover the origin at the same time.

Pf: Let , we can find the one with smallest index(namely the first one to choose) in . WLOG, suppose it is . Consider four cubes in the first quadrant, let

.

Then no cube in can center outside . Otherwise that cube will have an edge-length bigger than .

Then no cube in can center at , because . And there are at most one can center at respectively. In total contains no more than four cubes.

Since there are four quadrants in the plane, each one can have at most 4 cubes cover the origin, so origin can be covered by at most 16 cubes. In other dimensional case, .

Consider conclusion 3). Since , we know that if and , must contain a vertex of or a middle point of an edge of . There are of them. So as proved before, we can find .

**Special case: **

When the dimension is **,** we can get the best constants and . The proof is following:

Firstly we can find a sequence of intervals (possibly finite) such that covers and their overlapping number is dominated by some constant .

Consider , if , then let , otherwise let .

Then consider , if , set , otherwise let .

For , if , set , otherwise let .

Proceed this procedure inductively. Finally we can get a new sequence .

Then , because every point in is covered by finitely many intervals.

Suppoese there are is covered by there intervals, for instance , , , which are all not emptysets. Then there must have an interval is covered by the other two one, say is covered by and .

Case 1: , then must be during the process.

Case 2: or , then should also be .

Case 3: . Since and are nonempty and cover , then should also be at step .

Contradiction. So is covered by at most two intervals. So .

For , construct and by filling with intervals .

WLOG, assume are all nonempty. Allocate in to .

If disjoint from then allocate it in , otherwise in .

Suppose we have put , consider . If there are two intervals from and respectivly, say and , such that intersects them both. Then and must disjoint from each other, otherwise the overlapping number is at least 3. This contradicts our selection of . So and should be in the same . But this contradicts to the fact that these two intervals are from different . This means we can find an such that all the intervals in it are disjoint from . Then allocate to this . Proceed this procedure inductively.

1. Miguel De Guzman, differentiation of integrals in R^n

2. Emmanuele DiBenedetto. Real analysis.

## Comments

Your presentation of the case where n=1 is very nice and just like what I was thinking of last night. For my dissertation I wanted a published citation for the fact that \xi(1) = 2, and wonder if you know of one, or if you also had to come up with this particular case on your own like I did.

Sorry. I have no impression for the citation. I just thought by myself. Maybe this is trivial that nobody borther to write it down. Good luck to you.

Hi, I have a question about the proof of the n=1 case. Say we have E=[0,10] and the intervals of the sequence in the start are I_1=[0,3], I_2=[7,10], I_3=[2,6],I_4=[5,8]. Then they cover E with overlapping number at most 2, and they are all necessary so I_i’=I_i, but then I_1 in G_1, I_2 is disjoint from I_1 so is in G_1 too, then I_3 must be in G_2 and thus I_4 INTERSECTS BOTH I_3 AND I_1, SO INTERSECTS BOTH G_1 AND G_2! so we cannot put it in either group and we lose.

Did I miss something or is it a gap?

Thanks.

Hi, I have a question about the proof of the n=1 case. Say we have E=[0,10] and the intervals of the sequence in the start are I_1=[0,3], I_2=[7,10], I_3=[2,6],I_4=[5,8]. Then they cover E with overlapping number at most 2, and they are all necessary so I_i’=I_i, but then I_1 in G_1, I_2 is disjoint from I_1 so is in G_1 too, then I_3 must be in G_2 and thus I_4 INTERSECTS BOTH I_3 AND I_1, SO INTERSECTS BOTH G_1 AND G_2! so we cannot put it in either group and we lose.

Did I miss something or is it a gap?

Thanks.