I'm wondering if the following strengthening of the Besicovitch covering theorem holds: Suppose $A\subset\mathbb R^n$ is a bounded subset and suppose $x\mapsto r_x$ is a function $A\to(0,\infty)$. Is it possible to choose constants $0<\lambda<1$ and $N\in\mathbb N$ (both depending only on the dimension $n$) so that there exists a countable subset $C\subset A$ such that the following two statements hold?

(1) the collection of balls $\{B(x,\lambda r_x)\}_{x\in C}$ covers $A$ and,

(2) we have the pointwise inequality $\sum_{x\in C} \chi_{B(x,r_x)}\le N$ (where $\chi_E$ denotes the indicator function of a set $E$). In words, no point of $\mathbb R^n$ is contained in more than $N$ of the balls $B(x,r_x)$ (with $x\in C$).

If we replaced $\lambda$ by $1$, this statement would be true as a consequence of the usual Besicovitch covering theorem. I'm somewhat stuck trying to modify that proof to get this stronger statement (I would just like $\lambda$ to be strictly smaller than $1$, it can be as close to $1$ as we like).

Is this stronger version with some $0<\lambda<1$ true? If not, is there an easy counterexample? If it is true, a proof (or sketch/hint/reference) would be appreciated!

If it helps to answer the question, I don't mind adding the additional assumption that $A$ is actually a compact set (i.e., its closed in addition to being bounded) and $x\mapsto r_x$ is a continuous (or even Lipschitz) function.


I think the statement is false. Consider $n = 1$. Let $A = \{x_i\}_{i \in \mathbb{N}}$ where $x_i = 2^{-i}$.

Define $r_i$ as follows: if $k^2 \leq i < (k+1)^2$, let $r_i = 2^{-i} - 2^{-(k+1)^2}$.

Notice that $x_{k^2}$ only belongs to $B(x_{k^2}, r_{k^2})$ and no other ball. So any cover of $A$ must include all of $B(x_{k^2}, \lambda r_{k^2})$.

Let $\eta = \lceil-\log_2(1 - \lambda)\rceil$, for $\lambda < 1$. Note that $\eta \in \mathbb{N}$.

Observe that $B(x_i, \lambda r_i) \subset B(x_i, \lambda x_i)$, which means that $x_{i + \eta} \not\in B(x_i, \lambda r_i)$.

Now, suppose we have a cover of $A$. It must include the point $x_{k^2}$. The argument above implies that $x_{k^2+\eta}$ is not included. By construction if $x_{k^2 + \eta}$ belong to $B(x_i, \lambda r_i)$ for some $i$, then $i \leq k^2 + \eta$. This implies that $x_{k^2 + 2\eta}$ cannot belong to whichever ball that covers $x_{k^2+\eta}$.

This argument can be iterated $\sigma$ times, where $\sigma$ is the biggest integer such that $k^2 + \sigma \eta < (k+1)^2$, or that $\sigma = \lfloor (2k+1) / \eta\rfloor$.

Therefore for any fixed $\lambda < 1$, choosing a starting $k$ sufficiently large we see that the cover for $A$ must include at least $\lfloor (2k+1) / \eta \rfloor$ many $x_i$ with $i$ between $k^2$ and $k^2 + 1$. The corresponding balls for all these $x_i$ have non-empty total intersection, on which set the sum of the characteristic function is at least $\sigma$, which can be made arbitrarily large.

Compactness of $A$ will not help. I can add to the set $A$ the origin which we label as $x_\infty$. For any fixed $\lambda, N$ I can choose a sufficiently large $k$ such that the corresponding $\sigma > N$. Then if I take the corresponding $r_\infty \ll 2^{-(k+2)^2}$ we would have a counterexample with a compact $A$.

As $A$ is discrete, obviously $x_i \mapsto r_i$ is continuous; even in the compact case we can modify the definitions of $r_i$ for $i > (k+2)^2$ where $k$ is the critical value above so that the radius function is continuous at $x_\infty$.

  • $\begingroup$ Very nice! It seems to that you're able to make this construction (and get continuity, even Lipschitzness, of the radius function) because $A$ is "almost discrete" in your example. Do you think an example with $A$ being a compact interval (or ball for $n>1$) can be constructed with the radius function being continuous/Lipschitz? $\endgroup$ – Mohan Swaminathan Jun 20 '19 at 14:09
  • $\begingroup$ @MohanSwaminathan: I am not sure. I think one should be able to fill in the "gaps" between the discrete points, All the gaps are pretty straightforward EXCEPT the ones between $(2^{-k^2}, 2^{- k^2 + 1})$. The naive extension will of course cause the $r_x$ to be no longer continuous. I think in that interval we can probably linearly change $r_x$ from (on the right) $2^{-k^2}$ to (on the left) $2^{-k^2} - 2^{-(k+1)^2}$. Then up to some (pretty annoying looking) detail checking, the argument should still work, maybe with some small additional changes. $\endgroup$ – Willie Wong Jun 21 '19 at 16:28

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