# The Weak Besicovitch Covering Property and the Lebesgue Differentation Property

## Some preliminary terminology.

Before making the question let me introduce some terminology.

Notation. Let $$X$$ be a set and $$A$$ a subset of $$X$$. I denote by $$\chi_A\colon X\to\{0,1\}$$ the characteristic function of $$A$$.

Definition $$1$$. We say that a metric space $$(X,d)$$ satisfies the Weak Besicovitch Covering Property $$1$$ (WBCP1) if there exists a constant $$K\in\mathbb{N}^+$$ such that every finite Besicovitch family of balls of $$(X,d)$$ has cardinality $$\le K$$.

Recall that a Besicovitch family of balls of $$(X,d)$$ is a family $$\mathcal{F}$$ of closed balls of $$(X,d)$$ such that the centre of each ball does not belong to any other ball of the family and such that $$\bigcap\mathcal{F}\ne\emptyset$$.

Proposition $$a$$. The WBCP$$1$$ is equivalent to require that there exists a constant $$K\in\mathbb{N}^+$$ such that for every finite family of closed balls of $$(X,d)$$ there exists a subfamily $$\mathcal{G}\subseteq\mathcal{F}$$ such that $$$$\chi_C\le\sum_{G\in\mathcal{G}}\chi_G\le K$$$$ where $$C$$ is the set of all the centres of the balls of $$\mathcal{F}$$.

Definition $$2$$. We say that a metric space $$(X,d)$$ satisfies the Weak Besicovitch Covering Property $$2$$ (WBCP2) if there exists a constant $$N\in\mathbb{N}^+$$ such that for every bounded set $$A$$ of $$X$$ and for every family of closed balls $$\mathcal{F}$$ of $$(X,d)$$ such that each point of $$A$$ is the centre of some ball of $$\mathcal{F}$$ and such that

• either $$\sup\{r_B\mid B\in\mathcal{F}\}=+\infty$$
• or $$\{r_B\mid B\in\mathcal{F}\}$$ is a discrete subset of $$(0,+\infty)$$,

there exists a countable subfamily $$\mathcal{G}\subseteq\mathcal{F}$$ such that it holds $$$$\chi_A\le\sum_{G\in\mathcal{G}}\chi_G\le N.$$$$

## Now, we are ready for the question.

How can I prove the following proposition? It is taken from page 109 of the " New Trends on Analysis and Geometry in Metric Spaces ", Levico Terme, Italy 2017.

Proposition $$b$$. The WBCP$$1$$ implies the WBCP$$2$$ in every metric space $$(X,d)$$.

A possible way to prove it could be to follow the idea of the proof of the fact that in every doubling metric space $$(X,d)$$ the WBCP$$1$$ is equivalent to the BCP (see Proposition 3.7 of the Article " BESICOVITCH COVERING PROPERTY ON GRADED GROUPS AND APPLICATIONS TO MEASURE DIFFERENTIATION " of Le Donne and Rigot.). I recall the BCP in the following definition.

Definition 3. We say that a metric space $$(X,d)$$ satisfies the Besicovitch Covering Property (BCP) if there exists a constant $$N\in\mathbb{N}^+$$ such that for every bounded set $$A$$ of $$X$$ and for every family of closed balls $$\mathcal{F}$$ of $$(X,d)$$ such that each point of $$A$$ is the centre of some ball of $$\mathcal{F}$$, there exists a countable subfamily $$\mathcal{G}\subseteq\mathcal{F}$$ such that it holds $$$$\chi_A\le\sum_{G\in\mathcal{G}}\chi_G\le N.$$$$

## Why I need the proof of Proposition b?

Because I'm trying to prove the implication 1 $$\implies$$ 2 of the Preiss Theorem that is the following one.

Theorem. Let $$(X,d)$$ be a separable complete metric space. Then the following are equivalent.

1. $$(X,d)$$ is $$\sigma$$-finite dimensional, that is to say, there exists a sequence $$\{X_n\}_{n\in\N}$$ of subsets of $$X$$ such that $$X=\bigcup_{n=0}^{+\infty}X_n$$ and a sequence $$\{s_n\}_{n\in\N}\subseteq(0,+\infty]$$ such that every set $$X_n$$ has finite Nagata dimension inside $$X$$ on scale $$s_n$$.
2. $$(X,d)$$ has the Lebesegue Differentation Property, that is to say, for every locally finite Borel measure $$\mu$$ on $$X$$ it exists $$$$\lim_{r\to0^+}\frac{1}{\mu(\mathbb{B}(x,r))}\int_{\mathbb{B}(x,r)}f\,d\mu=f(x)\quad\text{ for }\mu-a.e.\, x\in X$$$$ for every $$\mu$$-measurable function $$f\colon X\to\overline{\mathbb{R}}$$ such that $$$$\int_{A}|f|\,d\mu<+\infty\text{ for all \mu-measurable bounded set A\subseteq X}.$$$$

If anyone knows where I can find a detalied Proof of the latter theorem, please let me know becasuse it would be very usefull for me. I know that David Preiss proved it the "Dimension of metrics and differentiation of measures", General topology and its relations to modern analysis and algebra, V (Prague, 1981), Sigma Ser. Pure Math., vol. 3, Heldermann, Berlin, 1983, pp. 565–568. But I cannot find this article anywhere.

• Probably best to send Preiss an email or use interlibrary loan (I've seen this series of books in many U.S. libraries). For what it's worth, I wasn't able to find it either, but the earlier volumes of this book/conference series are available here. I thought it might be in this book, but it doesn't appear to be here (only cited), at least I couldn't find it in my (hard)copy of the book. Aug 22 at 15:58
• Ok, we might need additional assumptions and/or clarifications. If I'm allowed to take the closed ball of radius $1/2$ around all the points of an uncountable discrete space of points separated by distance $1$, then that seems to contradict the proposition, since said space has WBPC1 with $K=1$. I might be missing something though.
– M W
Aug 22 at 20:04
• Also, once we establish whatever additional assumptions are necessary, I'd say that the condition WBCP2 that the radii must be discrete strikes me as maybe too weak. In the case the radii are bounded (unboundedness is trivial), you probably want $\{r_B\}$ to form a decreasing sequence so you can use a transfinite induction argument that respects $r_B$. So if the radii form a set like $\{1+1/2^n\}$ then that would be tough to pull off. But who knows, maybe they do some magic.
– M W
Aug 22 at 20:16
• Sorry, that last comment should have had that if the radii form a set like $\{1-1/2^n\}$, then its potentially a problem for transfinite induction.
– M W
Aug 22 at 20:27

[I have added an argument via Zorn's lemma as an alternative to the transfinite recursion argument. I have also formalized the latter somewhat.]

First, as I mentioned in the comments, proposition b pretty clearly needs a separability assumption in order for the sub cover $$\mathcal G$$ to be countable. I can prove the proposition, provided we interpret the discreteness criterion of the set of radii in a strong sense. That is, in the case where the radii are bounded, I need the second condition to imply that $$\{r_B\mid B\in\mathcal{F}\}$$ is well ordered by $$\geq$$, i.e., every nonempty subset of radii has a maximal element.

This is fine if by “discrete” the author meant "has no limit points in $$(0,\infty)$$", but I mention it because some people (including myself) prefer to use “discrete” to simply mean "has the discrete topology".

We also should note that the word "finite" is not necessary for WBCP1, since if $$\mathcal F$$ is an arbitrary Besicovitch family as you have defined it, then so is every finite subfamily.

Finally, we formally adopt the convention that a (closed) ball $$B$$ is just a pair $$(x_0,r)$$ consisting of a center $$x_0 \in X$$ and a radius $$r>0$$, even though we will often abuse notation and identify $$B$$ with the point set $$\{x\in X\mid d(x,x_0)\leq r\}$$. Thus we will write things like “$$x\in B$$“, but behind the scenes, it is important that we have a designated center and radius for every ball, since in a metric space a ball’s center and/or radius need not be uniquely determined by its point set. We denote by $$c(B)$$ and $$r_B$$ the respective center and radius of the ball $$B$$.

With that out of the way, here's a proof, which invokes separability only to establish countability of $$\mathcal G$$:

### Proof

Firstly the proposition is trivially satisfied when the radii are unbounded - in this case just take a single ball that covers the whole set $$A$$, since $$A$$ is bounded.

Therefore we can assume $$\mathcal R=\{r_B\mid B\in\mathcal{F}\}$$ is bounded, so by our strong interpretation of "discrete", every subset of $$\mathcal R$$ has a largest element.

Now, we will be done if we can construct a subfamily $$\mathcal G\subseteq \mathcal F$$ such that

1.) $$\mathcal G$$ covers every point in $$A$$.

2.) No ball in $$\mathcal G$$ contains the center of another ball in $$\mathcal G$$.

To see this, observe that the first condition implies $$\chi_A\leq \sum_{B\in\mathcal G}\chi_B\text{,}$$ while on the other hand, the second condition implies that for any $$x\in\bigcup \mathcal G$$, $$\{B\in\mathcal G \mid x\in B\}$$ is a Besicovitch family, and so has cardinality at most $$K$$, proving that $$\sum_{B\in\mathcal G}\chi_B\leq K\text{.}$$

Moreover, if $$X$$ is separable, then $$\mathcal G$$ includes at most countably many balls of radius $$r$$ for any fixed $$r>0$$, since the centers of said balls are $$r$$-separated by the second condition, and a separable space cannot have an uncountable $$r$$-separated set for any $$r>0$$. Since $$\mathcal R$$ is discrete, we conclude $$\mathcal G$$ is a countable union of countable families, and is hence countable itself. (Note that we really only need the weaker requirement that $$A$$ is separable, and moreover, we can relax the strong discreteness of $$\mathcal R$$ to the condition that $$\mathcal R$$ is well-ordered by $$\geq$$, as that is already enough for countability.)

It remains to construct $$\mathcal G$$, and we offer two equivalent methods, both dependent on variants of the axiom of choice.

### Transfinite recursion

For each $$r\in\mathcal R$$, let $$\mathcal F_r=\{B\in \mathcal F \mid r_B=r\}$$. By the well-ordering theorem, we may select a well-ordering for each collection $$\mathcal F_r$$. Then since $$\mathcal R$$ is well ordered by $$\geq$$, by concatenating the well ordered sets $$\mathcal F_r$$ in descending order we obtain a well ordering of $$\mathcal F$$, which we denote by $$\preceq$$, such that the radii of the balls are non-increasing.

Now via transfinite recursion, we may for each $$B\in \mathcal F$$ define the family $$\mathcal G_B$$ as

$$$$\mathcal G_B = \begin{cases} \displaystyle\bigcup_{B'\prec B} \mathcal G_{B'} & \text{if } c(B)\in \displaystyle\bigcup\left(\displaystyle\bigcup_{B'\prec B} \mathcal G_{B'}\right)\text{,}\\ \left(\displaystyle\bigcup_{B'\prec B} \mathcal G_{B'}\right) \cup \{B\} & \text{otherwise.} \end{cases}$$$$

Then since the radii are nonincreasing, each family $$\mathcal G_B$$ has the property that no ball contains the center of another ball. Moreover, every point $$a\in A$$ is the center of some ball $$B\in \mathcal F$$, whereby $$a$$ is covered by $$\mathcal G_B$$. Letting $$\mathcal G:=\bigcup_{B\in\mathcal F} \mathcal G_B$$, easily gives that $$\mathcal G$$ satisfies 1.), and condition 2.) follows from the fact that it holds on each $$\mathcal G_B$$ and $$\mathcal G$$ is an increasing union of these sets.

### Zorn's lemma

Alternatively, we may construct $$\mathcal G$$ through Zorn's lemma as follows.

Define a subfamily $$\mathcal H\subseteq \mathcal F$$ to be admissible if

a.) No ball in $$\mathcal H$$ contains the center of another ball in $$\mathcal H$$,

and

b.) If there is some ball $$B\in\mathcal H$$ with radius $$r$$, then every ball in $$\mathcal F$$ of radius strictly greater than $$r$$ has its center covered by $$\mathcal H$$.

Let $$\mathbb G$$ denote the set of admissible subfamilies of $$\mathcal F$$, partially ordered by inclusion. If $$\mathbb G'\subseteq \mathbb G$$ is a chain, then it is easy to verify that the union $$\bigcup_{\mathcal G'\in\mathbb G'}\mathcal G'$$ is again an admissible subfamily, and is hence an upper bound for $$\mathbb G'$$. Since every chain has an upper bound, we conclude by Zorn's Lemma that there is a maximal admissible subfamily $$\mathcal G\subseteq F$$.

Now we claim $$\mathcal G$$ satisfies properties 1.) and 2.) above. Indeed, 2.) is the same as a.), and if 1.) fails, let $$B\in \mathcal F$$ be a ball of maximum radius whose center is not in $$\bigcup \mathcal G$$. Then $$\{B\}\cup \mathcal G$$ is also admissible, contradicting maximality of $$\mathcal G$$.

#### Remark

The method of the above proof, and the fact that finiteness in WBCP1 is superfluous, could be used to considerably strengthen proposition a as well, to say that the following are equivalent:

1. $$X$$ has WBCP1.
2. Every finite family $$\mathcal F$$ of balls in X has a subfamily $$\mathcal G$$ satisfying (for the set of centers $$C$$ of the balls of $$\mathcal F$$) $$\chi_C\leq\sum_{B\in \mathcal G}\chi_B\leq K\text{.}$$
3. Every family $$\mathcal F$$ of balls in X wherein the set of radii $$\{r_B\mid B\in \mathcal F\}$$ is well ordered by $$\geq$$ has a subfamily $$\mathcal G$$ satisfying $$\chi_C\leq\sum_{B\in \mathcal G}\chi_B\leq K\text{,}$$ and if $$C$$ is separable we may further select $$\mathcal G$$ to be countable.
• @M W I read your proof but I'm not so much familiar with transfinite induction. Is it possible to give a more simple proof in the case $X$ is separable by using only the classical countable inducton? Aug 31 at 12:40
• @Grace53 I added a Zorn's Lemma argument. Also made a few things more rigorous regarding centers of balls.
– M W
Sep 1 at 1:09
• Also, for what its worth, I removed all the ordinals from the transfinite recursion argument, so it only requires understanding a), the well-ordering theorem (equivalent to axiom of choice) and b) the principle that when a set is well-ordered you can use it for recursive definitions. But it is still optional now that the Zorn's Lemma alternative is there.
– M W
Sep 1 at 1:20
• @M W I read your proof by using the Zorn's Lemma and everything is clear, you had a great and original idea! I don't know how to thank you! Sep 1 at 15:11
• @Grace53 well, I strongly suspect that what I did is probably the same as what Preiss did, so maybe not exactly original:). Nevertheless, you’re quite welcome, it was my pleasure to help with this, was a fun problem!
– M W
Sep 1 at 17:27