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.



Stronger version of Besicovitch covering theorem
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. 




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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 $x\_\infty$. 


