In this question, a suggested approach is given for improving the constant in a Hardy-Littlewood maximal inequality from 3 to 2, and the following lemma is stated without proof:

Suppose $K$ is a compact set, and for every $x \in K$, we are given an open ball $B(x,r_x)$ that is centered at $x$ and of radius $r_x$. Assume that $$R:= \sup_{x \in K} r_x < \infty.$$ Let $\mathcal{B}$ be this collection of balls, i.e. $$\mathcal{B} = \{B(x,r_x) \colon x \in K\}.$$ Then given any $\varepsilon > 0$, there exists a finite subcollection $\mathcal{C}$ of balls from $\mathcal{B}$, so that the balls in $\mathcal{C}$ are pairwise disjoint, and so that the (concentric) dilates of balls in $\mathcal{C}$ by $(2+\varepsilon)$ times would cover $K$.

The hint in a comment is to include "sufficiently many" balls in the cover so that each epsilon neighborhood includes a center, but I am not sure how to get a finite cover with such a property. A further hint (or complete proof) would be appreciated.

Note: This is not homework.


2 Answers 2


Well, I am almost 2 years late, but I figured I could show my answer. I didn't quite follow the selection procedure in @ydx 's answer.

We need to consider $2+\epsilon$ dilation because we will need $2$ radii to get to a centre and an $\epsilon$ more to cover the ball around said centre. Fix $0<\epsilon<1$.

Around each $x\in K$, there's a ball of radius $r_x$. Cover $K$ by $B(x, \epsilon r_x)$ and obtain a finite subcover $B(x_1, \epsilon r_{x_1}), \dots, B(x_n, \epsilon r_{x_n})$.

Now, from the collection $B(x_i, r_{x_i}), 1\leq i\leq n$ obtain a disjoint subcollection by choosing the largest radius at each step of the selection (similar to the construction in the finite Vitali covering). Let this subcollection be $B(x_1, r_{x_1}), \dots, B(x_m, r_{x_m})$ with $r_{x_1}\geq r_{x_2}\geq\dots\geq r_{x_m}$ (by construction).

Now, for $j>m, B(x_j, r_{x_j})$ intersects some ball in the subcollection, say $B(x_1, r_{x_1})$. By construction $r_{x_1}\geq r_{x_j}$. Therefore, $B(x_1, 2r_{x_1})$ contains the centre $x_j$ and going an $\epsilon$ further will contain $B(x_j, \epsilon r_{x_j})$, i.e., $$B(x_1, (2+\epsilon)r_{x_1})\supseteq B(x_j, \epsilon r_{x_j}).$$

Since $\epsilon<1$, the collection $B(x_i, (2+\epsilon)r_i)$ covers the union of $B(x_i, \epsilon r_i)$, hence covers $K$.


Consider the open cover $\{B(x,\varepsilon r_x):\ x\in K\}$ and get a finite subcover, say $\{B(x_j,\varepsilon r_j):\ j=1,\dots,N\}$. WLOG, assume $r_1\geq r_2\geq\dots\geq r_N$. Then, for each $j=1,\dots,N$, select the open ball $B(x_j,r_j)$ if it does not intersect with any selected ball, otherwise discard it. Say the selected balls form a set $\{B(x_{j_k},r_{j_k}):\ k=1,\dots,M\}$ where $M\leq N$ and $j_{1}\leq j_2\leq\dots\leq j_M$.

For each $j$, if $B(x_j,\varepsilon r_j)\not\subset\bigcup_{k=1}^MB(x_{j_k},(2+\varepsilon)r_{j_k})$, then $|x_j-x_{j_k}|+\varepsilon r_j>(2+\varepsilon)r_{j_k}$ for each $k$. Since $B(x_j,r_j)$ is not selected, we can also find some $k$ such that $B(x_j,r_j)\cap B(x_{j_k},r_{j_k})\neq \varnothing$, so $|x_j-x_{j_k}|< r_j+r_{j_k}$. Note that we may choose the smallest $k$ with such a property, so we may assume $B(x_j,r_j)\cap B(x_{j_l},r_{j_l})= \varnothing$ if $l<k$. But this implies that $(2+\varepsilon)r_{j_k}-\varepsilon r_j<r_j+r_{j_k}$, i.e. $r_{j_k}<r_j$. Here is a contradiction, since we should have chosen $B(x_j,r_j)$ instead of $B(r_{j_k},r_{j_k})$.

  • $\begingroup$ Thank you! This seems to be a really sensible approach. I couldn't figure out how to force enough overlap. Interestingly, this seems to not satisfy the hint, in that not every epsilon neighborhood contains a center. $\endgroup$
    – Zach Boyd
    Dec 16, 2018 at 2:20
  • $\begingroup$ Yes I think there should be another way to prove this lemma using the hint. Here I didn't even use the assumption that sup r_x<\infty, so I guess there should be a proof relying on this assumption. $\endgroup$
    – mathssww
    Dec 16, 2018 at 2:49

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