A Differentiation Basis without the Vitali Covering Property

— 1. Introduction —

A differentiation basis {\mathcal{B}} is a fine collection of bounded measurable sets {R}; i.e. for every {x\in{{\mathbb R}}^{n}} there exists a sequence {\left\{R_{k}\right\}\subset\mathcal{B}(x)} with diameter tending to zero. Classical examples of differentiation bases are the collections of balls and cubes. Since measures are {\sigma}-additive for countable collections of disjoint measurable sets, we are often interested in extracting disjoint subcollections which cover or “almost cover” some set. As is well-known, balls (cubes) satisfy a strong covering property proven by Vitali: given a measurable set {A} with finite (Lebesgue) measure, there exists a countable disjoint collection {\mathcal{G}} of balls (cubes) satisfying

\displaystyle \left|A\triangle\bigcup_{B\in\mathcal{G}}B\right|=0

We can abstract this covering property, hereafterto referred to as the Vitali covering property, and ask which differentiation bases possess it. It turns out that this question is intimately linked to the differentiation of integrals of functions (in the sense of Lebesgue’s theorem), but we save the exploration of that topic for another day.

In today’s post, we are going to construct an example of a differentiation basis {\mathcal{B}} which does not possess the Vitali covering property: there exists a set of finite measure which cannot be almost covered (i.e. modulo a null set) by any countable disjoint subcollection of {\mathcal{B}}. The proof is not difficult, but perhaps notationally cumbersome. The construction rests on a lemma which tells us that we can cover a set {A} of finite measure by disjoint homothetic copies of any compact set {K}, whose diameters can be taken to be arbitrarily uniformly “small”.

Section 2 reviews the classical Vitali covering theorem for nondegenerate closed balls in {{{\mathbb R}}^{n}}. The reader is welcome to skip over it to Section 3, in which we prove the aforementioned lemma and construct the counterexample. The material for this post is mostly adapted from [L. Evans and F. Gariepy Measure Theory and Fine Properties of Functions] and [M.D. Guzman, Differentiation of Integrals in {{{\mathbb R}}^{n}}]. The latter reference is highly recommended to the reader further interested in the covering properties of differentiation bases.

— 2. Vitali Covering Lemma —

For a fixed constant {C>3}, we let {\hat{B}} to denote the concentric closed ball with radius {C} times the radius of {B}. Explanation of the condition on {C} is provided below.

Lemma 1 Let {\mathcal{F}} be a collection of nondegenerate closed balls in {{{\mathbb R}}^{n}} with {\delta:=\sup\left\{\text{ diam} B : B\in\mathcal{F}\right\}<\infty}. Then there exists a countable family {\mathcal{G}} of disjoint balls in {\mathcal{F}} such that for every {B\in\mathcal{F}} there exists a ball {B'\in\mathcal{G}} with {B\cap B'\neq\emptyset} and {B\subset\widehat{B'}}. In particular,

\displaystyle \bigcup_{B\in\mathcal{F}}B\subset\bigcup_{B\in\mathcal{G}}\hat{B}

For the proof of the lemma, we remind the reader that a collection of nondegenerate disjoint balls in {{{\mathbb R}}^{n}} is necessarily countable by the countability of the rationals.

Proof: We partition {\mathcal{F}} by setting {\mathcal{F}_{j}:=\left\{B\in\mathcal{F} : \delta/c^{j}<\text{ diam} B\leq \delta/c^{j-1}\right\}} for all {j\in\mathbb{N}}, where {C=2c+1>3}. Let {\mathcal{G}_{1}} be a maximal (with respect to inclusion) disjoint collection of balls in {\mathcal{F}_{1}}. Assuming we have defined {\mathcal{G}_{1},\ldots,\mathcal{G}_{k-1}}, we define {\mathcal{G}_{k}} to be any maximal disjoint subcollection of

\displaystyle \left\{B\in\mathcal{F}_{k} : B\cap B'=\emptyset \ \forall B'\in\bigcup_{j=1}^{k-1}\mathcal{G}_{j}\right\}

I.e. the largest collection of disjoint balls in {\mathcal{F}_{k}} which do not intersect any of the balls in the previous {G_{j}}. Set {\mathcal{G}:=\bigcup_{j=1}^{\infty}\mathcal{G}_{j}}, which is by construction a countable collection of disjoint balls in {\mathcal{F}}.

We claim that for every {B\in\mathcal{F}}, there exists a ball {B'\in\mathcal{G}} such that {B\cap B'\neq\emptyset} and {B\subset\widehat{B'}}. Indeed, {B\in\mathcal{F}_{k}}, for some {k}, whence by the maximality of {\mathcal{G}_{k}}, {B\cap B'\neq\emptyset} for some {B'\in\mathcal{G}_{k}}. It follows from the triangle inequality that {B} is contained in the ball concentric to {B'} of radius

\displaystyle r=\text{ diam}(B')/2+\text{ diam}(B)\leq \text{ diam}(B')/2+c(\delta/c^{j})<\text{ diam}(B')(1/2+c)


Lemma 2 Let {\mathcal{F}} be a fine cover of a measurable set {A} by closed balls with uniformly bounded diameter. Then there exists a countable subcollection {\mathcal{G}} of disjoint balls in {\mathcal{F}} such that for every finite subset {\left\{B_{1},\ldots,B_{m}\right\}\subset\mathcal{F}},

\displaystyle A\setminus\bigcup_{j=1}^{m}B_{j}\subset\bigcup_{B\in\mathcal{G}\setminus\left\{B_{1},\ldots,B_{m}\right\}}\hat{B}

Proof: Let {\mathcal{G}} be as in Lemma 1, and let {\left\{B_{1},\ldots,B_{m}\right\}\subset\mathcal{F}}. If {A\subset\bigcup_{i=1}^{m}B_{i}}, then there is nothing to prove. Otherwise, let {x\in A\setminus\bigcup_{i=1}^{m}B_{i}}. Since {\bigcup_{i=1}^{m}B_{i}} is closed and {\mathcal{F}} is fine, there exists a ball {B\in\mathcal{F}} containing {x} and {B\cap B_{i}=\emptyset} for {1\leq i\leq m}. But there exists {B'\in\mathcal{G}} with {B\cap B'\neq\emptyset}, whence {B\subset\widehat{B'}}. \Box

 Theorem 3 Let {U\subset{{\mathbb R}}^{n}} be open and {\delta>0}. With {\mathcal{F}} as above, there exists a countable subcollection {\mathcal{G}} of disjoint closed balls in {U} such that {\text{ diam} B\leq\delta} for all {B\in\mathcal{G}} and

\displaystyle \left|U\setminus\bigcup_{B\in\mathcal{G}}B\right|=0

Proof: We first assume that {\left|U\right|<\infty}. Fix {1-C^{-n}<\theta<1}. We claim that there is a finite collection {\left\{B_{i}\right\}_{i=1}^{M_{1}}} of disjoint closed balls in {U} such that {\text{ diam}(B_{i})<\delta} for all {1\leq i\leq M_{1}} and

\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{1}}B_{i}\right|\leq\theta\left|U\right|

Indeed, let {\mathcal{F}_{1}:=\left\{B : B \text{ is an open ball}, B\subset U, \text{ diam} B<\delta\right\}}. By Lemma 1, there exists a countable disjoint subcollection {\mathcal{G}_{1}\subset\mathcal{F}_{1}} satisfying

\displaystyle U\subset\bigcup_{B\in\mathcal{G}_{1}}\hat{B}

Whence by the dilation property of Lebesgue measure and {\sigma}-additivity,

\displaystyle \left|U\right|\leq\sum_{B\in\mathcal{G}_{1}}\left|\hat{B}\right|=C^{n}\sum_{B\in\mathcal{G}_{1}}\left|B\right|=C^{n}\left|\bigcup_{B\in\mathcal{G}_{1}}B\right|


\displaystyle \left|U\setminus\bigcup_{B\in\mathcal{G}_{1}}B\right|\leq\left(1-\dfrac{1}{C^{n}}\right)\left|U\right|<\theta\left|U\right|

By the continuity of measure, there exists an {M_{1}} which satisfies the claim above.

Now set {U_{2}:=U\setminus\bigcup_{i=1}^{M_{1}}B_{i}} and

\displaystyle \mathcal{F}_{2}:=\left\{B : B\text{ is open}, B\subset U_{2}, \text{ diam} B<\delta\right\}

Repeating the argument, we can find finitely many disjoint balls {B_{M_{1}+1},\ldots,B_{M_{2}}\in\mathcal{F}_{2}} satisfying

\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{2}}\right|=\left|U_{2}\setminus\bigcup_{i=M_{1}+1}^{M_{2}}B_{i}\right|\leq\theta\left|U_{2}\right|\leq\theta^{2}\left|U\right|

We continue in this fashion to obtain a countable collection of disjoint balls and an increasing sequence {(M_{k})} such that

\displaystyle \left|U\setminus\bigcup_{i=1}^{M_{k}}B_{i}\right|\leq\theta^{k}\left|U\right|,\indent\forall k\in\mathbb{N}

Since {\theta^{k}\rightarrow 0} as {k\rightarrow\infty}, the proof is complete.

If {\left|U\right|=\infty}, then we can apply the preceding argument to the disjoint sets {U_{m}:=\left\{x\in U: m-1<\left|x\right|\leq m\right\}}, for {m\geq 1}. \Box

As an aside, we remark that {3} cannot be taken as the constant in Vitali’s lemma, as the following counterexample in one dimension due to O. Schramm shows that it cannot. Let {\mathcal{F}:=\left\{B(x,r):\left|x\right|< 1/2,\left|x\right|<r<(\left|x\right|+1)/3\right\}}. Observe that every ball {B\in\mathcal{F}} contains the origin, hence any disjoint subcollection of {\mathcal{F}} consists of one ball. It is clear that {\bigcup_{\mathcal{F}}B=(-1,1)}, but {3B} is a proper subset of {(-1,1)}.

— 3. Counterexample —

We present a version of a counterexample due to H. Bohr which has been simplified by M.D. Guzman. At the end of the section, we will make some remarks on how to extend the counterexample to {n} dimensions, when {n\geq 2}.

Lemma 4  Let {U\subset{{\mathbb R}}^{n}} be any bounded open set, and let {K} be a compact set with positive measure. For {r>0} there exists a disjoint sequence {\left\{K_{k}\right\}} of sets homothetic to {K} and contained in {U} such that {\left|U\setminus\bigcup K_{k}\right|=0} and {\text{ diam}{K_{k}}<r}.

Proof: Let {Q} be a half-open cube such that {K} is contained in the interior of {Q}, and let {\alpha\left|Q\right|=\left|K\right|} with {0<\alpha<1}. Using the hypothesis that {U} is open together with a stopping-time argument, we can partition {U} into a sequence of disjoint half-open cubes {\left\{Q_{j}\right\}} of diameter less than {r}. For each cube {Q_{j}}, let {T_{j}} be the homothecy mapping {Q} onto {Q_{j}}, and let {K_{j}'=T_{j}(K)}. Note that by construction {K_{j}'\subset Q_{j}}. For an integer {N_{1}\geq 1}, define an open set {U_{1}:=U\setminus\bigcup_{i=1}^{N_{1}}K_{j}'} and observe that

\displaystyle \begin{array}{lcl}\displaystyle\left|U_{1}\right|=\left|\bigcup_{j=1}^{\infty}Q_{j}\setminus\bigcup_{j=1}^{N_{1}}K_{j}'\right|&=&\displaystyle\left|\bigcup_{j=1}^{N_{1}}Q_{j}\setminus K_{j}'\right|+\left|\bigcup_{j=N_{1}+1}^{\infty}Q_{j}\right|\\[2 em]&=&\displaystyle(1-\alpha)\left|\bigcup_{j=1}^{N_{1}}Q_{j}\right|+\left|\bigcup_{j=N_{1}+1}^{\infty}Q_{j}\right|\end{array}

Now if we take {N_{1}} sufficiently large, then the last quantity is {<(1-\alpha/2)\left|U\right|}. Set {K_{j}:=K_{j}'} for {j=1,\ldots,N_{1}}. Suppose we constructed sets {U_{1},\ldots,U_{k}} and {\left\{K_{j}\right\}_{j=1}^{N_{k}}} such that

\displaystyle U_{i}:=U_{i-1}\setminus\bigcup_{i=N_{j-1}+1}^{N_{j}}K_{j}, \left|U_{i}\right|\leq\left(1-\dfrac{\alpha}{2}\right)^{i}\left|U\right|

for {2\leq i\leq k}. Repeating the argument above, we see that we can find sets {\left\{K_{j}\right\}_{j=N_{k}+1}^{N_{k+1}}} so that

\displaystyle \left|U_{k+1}\right|=\left|U_{k}\setminus\bigcup_{j=N_{k}+1}^{N_{k+1}}K_{j}\right|\leq\left(1-\dfrac{\alpha}{2}\right)^{k+1}\left|U\right|

The conclusion of the lemma follows by induction. \Box

 Theorem 5 Let {Q} be the unit square in {{{\mathbb R}}^{2}}. There exists a subset {F\subset Q} with {\left|F\right|=1} with the following property: there exists a sequence {\left\{I_{k}(x)\right\}} of open rectangles containing {x} and with diameter tending to zero, such that for each disjoint sequence {\left\{R_{k}\right\}} from {\left\{I_{k}(x)\right\}_{x\in F,k\geq 1}},

\displaystyle \left|F\setminus\bigcup_{k}R_{k}\right|>\dfrac{1}{2}

Before diving into the proof of Theorem 5, we give an auxillary geometric construction which we will need. Let {N} be an integer greater than {1}, and consider in {{{\mathbb R}}^{2}} the rectangles {I_{1},\ldots,I_{N}} where the vertices of {I_{j}} are {(0,0)}, {(j,0)}, {(0,N/j)}, and {(j,N/j)}. It is evident that {\left|I_{j}\right|=N} for all {j} and {\left|E:=\bigcap_{j=1}^{N}I_{j}\right|=1}. If we set {J_{N}:=\bigcup_{j=1}^{N}I_{j}}, then it is easy to verify that

\displaystyle \left|J_{N}\right|=N\underbrace{\sum_{j=1}^{N}\dfrac{1}{j}}_{\alpha(N)}

Proof: Let {\left\{N_{k}\right\}} be an increasing sequence of natural numbers. For {N_{i}} fixed, we can use Lemma 4 to almost cover the open unit cube {Q} with a disjoint sequence {\left\{S_{k}^{i}\right\}_{k\geq 1}} of sets homothetic to {J_{H_{i}}} (as defined above), with each {S_{k}^{i}} contained in {Q} and with diameter {<2^{-i}}.

Let {\left\{I_{k,j}^{i}\right\}_{j=1}^{N_{i}}} be the {N_{i}} open rectangles constituting {S_{k}^{i}}, homothetic to the intervals {I_{j}} of {J_{N_{i}}}. Define a family of rectangles {A_{i}:=\left\{I_{k,j}^{i}\right\}_{k=1,j=1}^{\infty, N_{i}}}. If we denote the union of the elements of {A_{i}} by {F_{i}}, then it is clear that {\left|F^{i}\right|=\left|Q\right|=1}. Observe that if {\left\{R_{l}^{i}\right\}=\left\{I_{k_{l},j_{l}}^{i}\right\}} is any disjoint subcollection of {A _{i}}, then {\left\{R_{l}^{i}\right\}} contains at most one rectangle {I_{k,j}^{i}} of those constituting {S_{k}^{i}}, since {I_{k,j}^{i}\cap I_{k,m}^{i}\neq\emptyset} for {1\leq j\leq m\leq N_{i}}. Also observe that since

\displaystyle \dfrac{\left|I_{k,j}^{i}\right|}{\left|S_{k}^{i}\right|}=\dfrac{1}{\alpha(N_{i})}\forall k\geq 1,1\leq j\leq N_{i} \text{ and }\sum_{k}\left|S_{k}^{i}\right|=1

we have that

\displaystyle \sum_{l=1}^{\infty}\left|R_{l}^{i}\right|=\sum_{l=1}^{\infty}\dfrac{\left|I_{k_{l},j_{l}}^{i}\right|}{\left|S_{k_{l}}^{i}\right|}\cdot\left|S_{k_{l}}^{i}\right|\leq\dfrac{1}{\alpha(N_{i})}

Now set {F:=\bigcap_{i=1}^{\infty}F_{i}}. Since {F_{i}\subset Q} and {\left|Q\setminus F_{i}\right|=0}, it follows that {\left|F\right|=1}. Consider the family of rectangles {A:=\bigcup_{i=1}^{\infty}A_{i}}. It is evident that for each {x\in F} there is a sequence of rectangles {\left\{I_{k}(x)\right\}} of {A} containing {x} and having diameter tending to zero. For any disjoint sequence {\left\{R_{k}\right\}\subset A}, we have

\displaystyle \sum_{k=1}^{\infty}\left|R_{k}\right|\leq\sum_{k=1}^{\infty}\dfrac{1}{\alpha(N_{k})}

Since {\alpha(N_{k})>\log(N_{k})/2}, we can chose our original sequence {\left\{N_{i}\right\}} to satisfy

\displaystyle \sum_{i=1}^{\infty}\dfrac{1}{\alpha(N_{i})}<\dfrac{1}{2},

whence {\left|F\setminus\bigcup R_{k}\right|>1/2}. \Box

To generalize the counterexample to higher dimensions, we simply modify the rectangles {I_{j}} in the auxilary construction by {\tilde{I}_{j}:=I_{j}\times[0,1]^{n-2}}, so {\bigcap_{j=1}^{N}\tilde{I}_{j}} has volume {1} and {\tilde{J}_{N}:=\bigcup_{j=1}^{N}\tilde{I}_{j}} has volume {N\alpha(N)}. We can then repeat the argument in the proof above with {Q} now the unit cube in {{{\mathbb R}}^{n}}.

Suppose we replace the Lebesgue measure \left|\cdot\right| on {\mathbb{R}}^{n} by a Radon measure \mu. Does there exist a differentiation basis for \mu which does not possess the Vitali covering property? The above construction relied on the fact that if two measurable sets A and B have measures have ratio \alpha:=\left|A\right|/\left|B\right|, then \alpha=\left|T(A)\right|/\left|T(B)\right| for any homothety T. This is a consequence of the translation and dilation properties of the Lebesgue measure. It is not clear to us how to modify the above construction to work without these properties. We are curious to hear anyone else’s thoughts on the question.

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