In the book *The Geometry of Domains in Spaces* by Krantz and Parks, the authors proved the weak $(1,1)$-type estimate of the maximal function $M\_\mu f$, where $\mu$ is a Radon measure, using their version of the [Besicovitch covering theorem](https://en.wikipedia.org/wiki/Besicovitch_covering_theorem).



> 
> Let $d$ be a positive integer. Then there exists a constant $C=C(d)$ such that for any finite collection of balls $\mathcal B = \{B\_i\}\_{i=1}^m$ in $\Bbb R^d$ with the property that no ball contains the center of any other ball, we can partition the family $\mathcal B$ into 
>  $$
> \mathcal B = \mathcal B\_1 \cup\mathcal B\_2 \cup \dots \cup \mathcal B\_C,
> $$
>  where each subfamily $\mathcal B\_j$ consists of disjoint balls.
> 
> 
> 


This version of the covering theorem seems pretty restrictive, especially the part about no ball contains the center of any other balls. Indeed, in theor proof of the weak $(1,1)$-type estimate, they relied on a certain claim that they did not prove.


**Edit** As Skeeve mentioned in the comment, this claim is not explicitly stated in the book but more of a paraphrasing of the part the authors left out in a proof.



> 
> **Claim**: Let $K\subset \Bbb R^d$ be a compact set such that each $x\in K$ is associated with a real number $r\_x>0$. Then $K$ can be covered by a family of balls 
>  $$
> \mathcal B = \{ B(x\_i,r\_i) : i=1,\dots,k\ \},
> $$
>  where $r\_i := r\_{x\_i}$, such that for any distinct $i,j \le k$, we have 
>  $$
> x\_i\notin B(x\_j,r\_j) \quad\text{and}\quad x\_j\notin B(x\_i,r\_i).
> $$
> 
> 
> 


I don't find this claim to be trivial at all. In fact, I tried many different methods but failed to prove it. Note that the mapping $x\mapsto r\_x$ doesn't enjoy any nice property like continuity of any kind.


While the usual version of Besicovitch covering theorem circumvents this problem, I still would like to know how to prove the above claim (or a counter example if it is actually false).



A Besicovitch-type Covering Theorem
This trivial counterexample in $\mathbb R^2$ should have taken me five minutes. Instead, I spent almost two days. The moral is the usual one: after 50 you'd better give up on mathematics.


Let $y,z$ be 2 points at distance $1$ from each other. We shall construct by induction a sequence of points $x\_j$ and radii $r\_j>\max(d(x\_j,y),d(x\_j,z))$ such that $x\_j\to y$ when $j$ is odd, $x\_j\to z$ when $j$ is even, $x\_j$ do not lie on the line $yz$, $\max(d(x\_j,y),d(x\_j,z))<1$, the disk $D(x\_j,\max(d(x\_j,y),d(x\_j,z)))$ contains $x\_1,\dots, x\_j$ but $x\_{j+1}\notin D(x\_j,r\_j)$ . If you choose the radius $\rho$ for $y$ and $z$ small enough so that the corresponding disks do not contain $x\_1$, you'll get a bad configuration.


Indeed, an attempt to choose $y$ or $z$ as one of the centers results in the exclusion of all the centers $x\_j$, after which covering $x\_1$ gets impossible.


Out of $x\_i$, we can choose only one (if $i<j$, then $x\_i\in D(x\_j,r\_j)$). But then, if we choose $x\_i$, the point $x\_{i+1}$ is not covered. 


Now the sequences. Start with any $x\_1$ very close to $y$ and not on the line $yz$ so that $d(x\_1,z)<1$. Assume that $x\_1,x\_2,\dots x\_j$ and $r\_1,\dots,r\_{j-1}$ are already constructed and, say $j$ is odd, so $x\_j$ is close to $y$. Then the circles centered at $x\_j$ and $y$ containing $z$ cross at an angle, so $D(y,1)\setminus \bar D(x\_j,d(x\_j,z))$ is an open set containing points arbitrarily close to $z$. Choose $x\_{j+1}$ to be any point in that difference that doesn't lie on the line $yz$ and satisfies $d(y,x\_{j+1})\ge 1-d(z,x\_{j+1})>\max\_{i\le j}d(z,x\_i)+d(z,x\_{j+1})\ge \max\_{i\le j}d(x\_i,x\_{j+1})$ and choose $r\_j$ anywhere between $d(x\_j,z)$ and $d(x\_j,x\_{j+1})$.


Clearly, we can keep $x\_j$ with odd indices at the distance $<1/3$ to $y$ and converging to $y$ and similarly for even indices and $z$.


