I want to prove this lemma which is left as exercise for the reader in the book Measure Theory and Integration of Micheal E.Taylor. The lemma is used to prove the Besicovitch Lemma. This is the statement:

$\mathbf{Claim}:$ There exists $K_{1}(n)$ such that if $r \in (0,+\infty),$ $B$ is a closed ball of radius $r,$ and $B_{1},\dots,B_{K_{1}(n)}$ are closed balls of radius $ \geq \frac{2}{3}r$ such that $B \cap B_{j} \neq \emptyset,$ then some $B_{j}$ contains the center of $B_{k}$ (for some $k \neq j$).

Until now, I have been able to prove only the statement with the following additional hypothesis: each of the radii is limited by a common constant $R_{0}$. I could stop here because a hypothesis of Besicovitch's lemma is the uniform boundness of the radii of the covering. Despite this, I would be interested in proving the claim above without this further hypothesis or I would like a counterexample.

I include below my proof of the $\mathbf{Claim}$ with the hypothesis of radii limited by a common constant $R_{0}.$

Suppose that $B_{1}, \dots, B_{K}$ are balls such that $B \cap B_{j} \neq \emptyset $ for every $j = 1, \dots, \, K$, and that no $B_{j}$ contains the center of another $B_{i}.$ The proof aims to find a bound on $K$ depending only on $n$ and $R_{0}.$ Firstly, indicate with $x_j$ and $r_{j}$ the center and the radius of $B_{j},$ and indicate $x_{B}$ the center of B. Since $B_{j}$ does not include the center of any other $B_{i}$ we obtain that $B(x_{j},\frac{1}{2}r_{j})$ and $B(x_{i},\frac{r_{i}}{2})$ are disjoint for every $i \neq j.$ In addition, there exists $C$ depending from $R_{0}$ such that $$ B(x_{j},\,\frac{1}{2}) \subseteq B(x_{B},\, C\cdot r) \quad \forall j. $$ But the balls $B(x_{j},\,\frac{r_{j}}{2})$ are all disjointed from each other, thus we have $$ \sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{2}r_{j})\right) = m\left( \bigcup_{j=1}^{k} B(x_{j},\,\frac{1}{2}r_{j})\right) \leq m \left(B(x_{B}, C \cdot r)\right)= C^{2} \cdot \omega_{n} \cdot r^{n}, $$ where $m$ is the Lebesgue measure and $\omega_{n}$ the volume of the unitary ball. But all the radii $r_{j}$ are at least $\frac{2}{3}r$ and thus we obtain $$ K\cdot \frac{r^{n}}{3^{n}} \cdot \omega_{n} =\sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{3}r)\right) \leq \sum_{j=1}^{n} m\left(B(x_{j},\,\frac{1}{2}r_{j}\right)) \leq C^{n} \cdot \omega_{n} \cdot r^{n}. $$ In conclusion, $K \leq 3^{n} \cdot C^{n}.$ Thus each set of closed balls intersecting $B$ having a radius of at least $2/3$ that of $B$, and having more than $3^{n} \cdot C^{n}$ must include a ball that contains the center of another ball in the set.

  • $\begingroup$ Could you clarify what "rays" you are referring to? Did you mean the radii of the balls? $\endgroup$
    – M W
    Commented Sep 18, 2023 at 22:59
  • $\begingroup$ Also, assuming you are talking about a bound on the radius of the balls, then you certainly could NOT stop right there, since in the Besicovitch covering lemma, the constant does not depend on the bound of the radii. The constant is to only depend on the dimension of the space. $\endgroup$
    – M W
    Commented Sep 19, 2023 at 1:56
  • $\begingroup$ I would add that if you allow the constant to depend on the bound on the radii, then the result becomes true in any doubling metric space, and there are many doubling metric spaces where Besicovitch fails. $\endgroup$
    – M W
    Commented Sep 19, 2023 at 2:02
  • $\begingroup$ Yes sorry everyone! By rays I actually meant radii; in my native language, the two terms are synonyms! $\endgroup$ Commented Sep 19, 2023 at 18:48
  • $\begingroup$ @MW could you please give me the definition of $n-$ separated set because I cannot infer it from your post? $\endgroup$ Commented Sep 19, 2023 at 19:04

1 Answer 1


As mentioned in the comments, the constant $K(n)$ needs to be independent of any bound on the radii for the family of balls, in order for the claim to be of any use in the Besicovitch covering lemma.

To prove this, we will need to use geometry specific to Euclidean space, and particularly the Euclidean plane.

First, with no loss of generality we may take $r=1$, as once the result is proved for $r=1$, scaling the entire picture gives the result for arbitrary $r$, and likewise, we may translate the picture so that $B$ is centered at the origin.

Thus all the balls in our family have radius at least $\frac{2}{3}$ and intersect the unit ball. As in your proof, we suppose none of the balls contain another's center, and must show a bound depending only on $n$ for the number of balls.

Now, since no ball contains another's center, the centers of the balls form a $\frac{2}{3}$-separated set, so by the doubling property of $\mathbb R^n$ (see here for example), only $C(n)$ of the centers can lie inside $3B$, the ball around the origin of radius $3$. It therefore suffices to suppose that each $B_i$ has a center outside of $3B$, and hence has radius at least $2$, since if we can bound the number of these balls, then adding $C(n)$ gives a bound for the original collection.

Moreover, we may shrink the balls so that they each intersect $B$ at a single point without changing the condition that no ball contains another's center, and still the radii must be at least $2$, since the centers lie outside of $3B$ and the balls intersect $B$.

We claim that suffices to show that given any two balls $B_i$,$B_j$, the angle $\angle POQ$, where $O$ is the origin and $P$ and $Q$ are the centers, is bounded away from $0$ by an absolute constant $\theta_0$. Indeed, if this is the case, then the projections onto the unit sphere $\mathbb S^{n-1}$ of all of the centers of the balls $B_i$ form a $\theta_0$-separated subset of $\mathbb S^{n-1}$ (in the arc-length metric), and by compactness of $\mathbb S^{n-1}$ the size of such a set has a bound depending only on $\mathbb S^{n-1}$, hence only on $n$.

Finally, to complete the proof, we observe that if $r_1\leq r_2$ are the radii of the balls, then we have $|OP|=1+r_1$, $|OQ|=1+r_2$, and $|PQ|>r_2$. Then by the law of cosines, we have \begin{align*} \cos(\angle POQ) &= \frac{|OP|^2+|OQ|^2-|PQ|^2}{2|OP|\cdot|OQ|} \leq \frac{(1+r_1)^2+(1+r_2)^2-r_2^2}{2(1+r_1)(1+r_2)}\\ & = \frac{(1+r_1)^2+(1+2r_2)}{2(1+r_1)(1+r_2)} \leq \frac{1}{2} + \frac{1}{(1+r_1)}\\ & \leq \frac{1}{2} +\frac{1}{3} = \frac{5}{6}\text{.} \end{align*} Thus the angle $\angle POQ$ is at least $\theta_0:=\cos^{-1}(\frac{5}{6})>0$, and the proof is concluded.

  • $\begingroup$ I'm wondering about the assumption that the balls are closed. In my attempted proof, it appears I didn't utilize this assumption at any point. I was trying to prove the facts you mentioned about doubling spaces, but I couldn't succeed. However, I've come up with a proof of my own that seems to work. My only uncertainty is whether the closedness of the balls is necessary. Could you provide further guidance, or perhaps point me to a source for the proof of the doubling spaces statements? $\endgroup$ Commented Oct 13, 2023 at 13:55
  • $\begingroup$ @MatteoAldovardi This fact doesn’t require the balls to be closed, you could make the same statement about open balls and it would work just fine, I think people like to just specify open vs. closed in many of these theorems just to have a precise statement of the theorem. As for the facts about doubling spaces see here, for example math.stackexchange.com/questions/4755471/doubling-metric-spaces/… Though the original link in the main post should also be good enough for this case I think. $\endgroup$
    – M W
    Commented Oct 13, 2023 at 19:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .