It is wellknown that in the classical Besicovitch covering theorem, the point being covered should be the center of some ball. However in Javier Duoandikoetxea's book, Fourier Analysis, it says that the point $x$ need not to be the center of the ball or cube but must be uniformly close to the center. For more details, you can see the following picture

Besicovitch-Morse lemma

Where I don't understand is the meaning of "uniformly close to the center". Does it mean that $$\sup_{x}|x-c_{B_x}|<\infty$$ or the ratio of $|x-c_B|$ and $r_B$ is not very large, e.g. $$\sup_x\frac{|x-c_{B_x}|}{r_{B_x}}\leq 1/2 ?$$ I have seen the proof of the Besicovitch covering lemma, but have no progress. I wonder if there are some related references or papers? Thanks for your warm help.


1 Answer 1


I haven't been able to find a reference that addresses this specific point, but I'm farily sure that "uniformly close to the center" has the second meaning from the question, i.e. $$ \sup_{x \in A} \frac{|x - c_{B_x}|}{r_x} < 1. $$

Here's an adaptation of a proof of the centered Besicovitch covering lemma from some notes that handles the uncentered case for balls.

Step 0: Parameters
We begin by defining a few parameters that will be important as we go along. First, let $$ \gamma = \sup_{x \in A} \frac{|x - c_{B_x}|}{r_x}, $$ so that $\gamma < 1$. Select $\alpha$ satisfying $(1 + \gamma)/2 < \alpha < 1$, and further define $\delta = (\alpha - \gamma) / (\alpha + 1)$, so that $0 < \delta < 1$. Finally, let $R = \sup_{x \in A} |x|$.

Step 1: Selecting the cover
Now we iteratively select a sequence of balls from our collection. Define $R_1 = \sup_{x \in A} r_x$. We can assume that $R_1$ is finite, as otherwise we can find a single ball in the collection which covers $A$. (Since we're allowing uncentered balls, the uniformity condition comes into play here.) Pick the first ball $B_1 = B(c_1, r_1)$, associated with the point $x_1 \in A$, so that $r_1 > \alpha R_1$. Having selected $B_1, B_2, \ldots, B_m$ (and assuming they do not cover $A$), let $$ R_{m+1} = \sup_{x \in A \setminus \bigcup_{k=1}^m B_k} r_x $$ and select $B_{m+1} = B(c_{m+1}, r_{m+1})$ associated to $x_{m+1} \in A \setminus \bigcup_{k=1}^m B_k$ so that $r_{m+1} > \alpha R_{m+1}$. Note that for $i < j$, we have $r_i > \alpha R_i \geq \alpha R_j \geq \alpha r_j.$

The balls $\{ B'_k = B(c_k, \delta r_k) \}$ are then disjoint: assuming $i < j$, we have $x_j \notin B(c_i, r_i)$ and as a result \begin{align*} |c_i - c_j| &\geq |c_i - x_j| - |x_j - c_j| \\ &\geq r_i - \gamma r_j \\ &> \delta r_i + (1 - \delta) \alpha r_j - \gamma r_j \\ &= \delta r_i + \delta r_j. \end{align*} We can also say that every $B'_k$ is contained in the ball $B(0, 2R_1 + R)$ since $|c_k| \leq |c_k - x_k| + |x_k| \leq R_1 + R$. It then follows that the radii $r_k \to 0$, and so $R_k \to 0$ as well. We can in turn conclude that the main balls $\{ B_k = B(c_k, r_k) \}$ cover $A$, as otherwise we obtain a contradiction to $R_k \to 0$.

Step 2: Bounding intersections, small radii
Going forward, we let $M$ denote a (potentially large) parameter whose value we defer selecting until later. (Its value will ultimately depend on the values of $\gamma$ and $\alpha$, but otherwise there won't be more specific dependence on the starting collection of balls.) We now show that for any $m$, there are at most $(2M + 1)^n \alpha^{-n} \delta^{-n}$ of the balls $B_k$, $k = 1, 2, \ldots, m-1$ which both intersect $B_m$ and satisfy $r_k \leq M r_m$.

To see this, fix an $m$ and suppose that $B_{k_1}, \ldots, B_{k_{N}}$ satisfy the given criteria. Then the associated dilated balls $B'_{k_1}, \ldots, B'_{k_{N}}$ are disjoint (as showed in Step 1) and are further contained in $B(c_m, (2M+1) r_m)$: for $x \in B(c_{k_i}, \delta r_{k_i})$ we have \begin{align*} |x - c_m| &\leq |x - c_{k_i}| + |c_{k_i} - c_m| \\ &\leq \delta r_{k_i} + r_{k_i} + r_m \\ &\leq 2 M r_m + r_m. \end{align*} Measure considerations now give \begin{align*} (2M + 1)^n r_m^n \geq \delta^n \sum_{i = 1}^{N} r_{k_i}^n \geq \delta^n \sum_{i = 1}^{N} \alpha^{n} r_m^n = N \alpha^{n} \delta^n r_m^n, \end{align*} since $r_{k_i} > \alpha r_m$ as showed in Step 1. This establishes the claimed bound.

Step 3: Bounding intersections, large radii
Next suppose that $i < j < m$ and that the balls $B_i$ and $B_j$ both intersect $B_m$, while also $r_i, r_j > M r_m$. We establish a lower bound for the angle made by the rays from $x_m$ to each of $c_i$ and $c_j$.

Since $x_m \notin B_i$ by construction, we first have the lower bound $| x_m - c_i | \geq r_i$, and likewise $| x_m - c_j | \geq r_j$. We also have the upper bound $$ | x_m - c_i | \leq | x_m - c_m | + | c_m - c_i | \leq (1 + \gamma) r_m + r_i, $$ and likewise replacing $i$ by $j$. Third, we have a lower bound \begin{align*} | c_i - c_j | &\geq | c_i - x_j | - | x_j - c_j | \\ &\geq r_i - \gamma r_j \\ &\geq (\alpha - \gamma) r_j \\ &> 0, \end{align*} relying on the fact that $x_j \notin B_i$.

Letting $\theta$ denote the angle between the rays, the law of cosines now gives \begin{align*} \cos \theta &= \frac{|x_m - c_i|^2 + |x_m - c_j|^2 - |c_i - c_j|^2}{2 |x_m - c_i| |x_m - c_j|} \\ &\leq \frac{[(1 + \gamma) r_m + r_i]^2 + [(1 + \gamma) r_m + r_j]^2 - [r_i - \gamma r_j]^2} {2 r_i r_j} \\ &= \gamma + \frac{1 - \gamma^2}{2} \frac{r_j}{r_i} + (1 + \gamma) \left( \frac{r_m}{r_i} + \frac{r_m}{r_j} \right) + (1 + \gamma)^2 \frac{r_m}{r_i} \frac{r_m}{r_j}. \end{align*} We further have $r_j / r_i < 1 / \alpha$ since $i < j$, and by hypothesis both $r_m / r_i$ and $r_m / r_j$ are less than $1/M$. Therefore $$ \cos \theta \leq \gamma + \frac{1 - \gamma^2}{2 \alpha} + \frac{2 (1 + \gamma)}{M} + \frac{(1 + \gamma)^2}{M^2}. $$ Our initial choice of $\alpha$ ensures that $$ \gamma + \frac{1 - \gamma^2}{2 \alpha} < 1, $$ and so by now selecting $M = M(\gamma, \alpha)$ sufficiently large we can also bound $\cos \theta$ away from $1$, thereby bounding $\theta$ away from $0$. That is, we can say that there is an $\epsilon = \epsilon(\gamma, \alpha)$ such that $\theta \geq \epsilon$.

To conclude this step, suppose that $m$ is fixed and that among the balls $B_k$, $k = 1, 2, \ldots, m-1$, there are $N$ which both intersect $B_m$ and satisfy $r_k > M r_m$. The angular separation condition just established then implies that there is a constant $C = C(n, \gamma, \alpha)$ such that $N \leq C$.

Step 4: Wrapping up
Putting together Steps 2 and 3, we can say that there is a constant $C' = C'(n, \gamma, \alpha)$ so that for any $m$, at most $C'$ of the balls $B_1, B_2, \ldots, B_{m-1}$ intersect $B_m$.

Finally, it follows that $$ \sum_{k} \chi_{B_k}(x) \leq C' + 1 $$ for all $x$. For otherwise, we can select $x$ and $m$ so that $x \in B_m$ and $$ \sum_{k = 1}^{m} \chi_{B_k}(x) > C' + 1. $$ So then $x$ lies in more than $C'$ of the balls $B_1, \ldots, B_{m-1}$, meaning that more than $C'$ of these balls intersect $B_m$.

Here's a 1-dimensional counterexample to show that a Besicovitch-type result isn't ensured by the condition $\sup_x |x - c_{B_x}| < \infty$.

Let $A = (0, 1)$, and for $x \in A$ let $B_x = (0, (1+x)/2)$. Note that $|x - c_{B_x}| \leq 1/2$ for every $x$. It's straightforward to see that no finite collection of these balls (intervals) covers $A$. But the interval $(0, 1/2)$ is contained in each of the balls, so if $A' \subseteq A$ is such that $\{ B_x : x \in A' \}$ covers $A$ then $\sum_{x \in A'} \chi_{B_x}$ is infinite on $(0, 1/2)$.

  • $\begingroup$ This answer is very useful, thank you very much! By the way, actually I want a covering theorem that has the first meaning rather than the second. Is there a covering theorem that has a uncentered version and don't have to enlarge the balls, like the Vitali's covering lemma? $\endgroup$
    – 123
    Dec 21, 2020 at 6:53
  • $\begingroup$ @Burning I can think of a counterexample to the first meaning, so I'll make an edit to add that to the answer. Beyond that, I'd suggest a deeper dive into the literature if you can. A starting point that I'm a little familiar with is Mattila's Geometry of Sets and Measures in Euclidean Spaces. The beginning of Chapter 2 on covering and differentiation has a few references to more in-depth discussions of covering theorems. $\endgroup$
    – Jason
    Dec 21, 2020 at 19:37
  • $\begingroup$ Thank you very much for your answer and the counterexample! I think I should read the book you suggest to me carefully and then think about this again. $\endgroup$
    – 123
    Dec 22, 2020 at 10:53

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