— 1. Introduction —
A differentiation basis is a fine collection of bounded measurable sets ; i.e. for every there exists a sequence with diameter tending to zero. Classical examples of differentiation bases are the collections of balls and cubes. Since measures are -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 with finite (Lebesgue) measure, there exists a countable disjoint collection of balls (cubes) satisfying
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 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 . The proof is not difficult, but perhaps notationally cumbersome. The construction rests on a lemma which tells us that we can cover a set of finite measure by disjoint homothetic copies of any compact set , whose diameters can be taken to be arbitrarily uniformly “small”.
Section 2 reviews the classical Vitali covering theorem for nondegenerate closed balls in . 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 ]. The latter reference is highly recommended to the reader further interested in the covering properties of differentiation bases.
— 2. Vitali Covering Lemma —
For the proof of the lemma, we remind the reader that a collection of nondegenerate disjoint balls in is necessarily countable by the countability of the rationals.
Proof: We partition by setting for all , where . Let be a maximal (with respect to inclusion) disjoint collection of balls in . Assuming we have defined , we define to be any maximal disjoint subcollection of
I.e. the largest collection of disjoint balls in which do not intersect any of the balls in the previous . Set , which is by construction a countable collection of disjoint balls in .
We claim that for every , there exists a ball such that and . Indeed, , for some , whence by the maximality of , for some . It follows from the triangle inequality that is contained in the ball concentric to of radius
Proof: Let be as in Lemma 1, and let . If , then there is nothing to prove. Otherwise, let . Since is closed and is fine, there exists a ball containing and for . But there exists with , whence .
Proof: We first assume that . Fix . We claim that there is a finite collection of disjoint closed balls in such that for all and
Indeed, let . By Lemma 1, there exists a countable disjoint subcollection satisfying
Whence by the dilation property of Lebesgue measure and -additivity,
By the continuity of measure, there exists an which satisfies the claim above.
Now set and
Repeating the argument, we can find finitely many disjoint balls satisfying
We continue in this fashion to obtain a countable collection of disjoint balls and an increasing sequence such that
Since as , the proof is complete.
If , then we can apply the preceding argument to the disjoint sets , for .
As an aside, we remark that 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 . Observe that every ball contains the origin, hence any disjoint subcollection of consists of one ball. It is clear that , but is a proper subset of .
— 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 dimensions, when .
Proof: Let be a half-open cube such that is contained in the interior of , and let with . Using the hypothesis that is open together with a stopping-time argument, we can partition into a sequence of disjoint half-open cubes of diameter less than . For each cube , let be the homothecy mapping onto , and let . Note that by construction . For an integer , define an open set and observe that
Now if we take sufficiently large, then the last quantity is . Set for . Suppose we constructed sets and such that
for . Repeating the argument above, we see that we can find sets so that
The conclusion of the lemma follows by induction.
Theorem 5 Let be the unit square in . There exists a subset with with the following property: there exists a sequence of open rectangles containing and with diameter tending to zero, such that for each disjoint sequence from ,
Before diving into the proof of Theorem 5, we give an auxillary geometric construction which we will need. Let be an integer greater than , and consider in the rectangles where the vertices of are , , , and . It is evident that for all and . If we set , then it is easy to verify that
Proof: Let be an increasing sequence of natural numbers. For fixed, we can use Lemma 4 to almost cover the open unit cube with a disjoint sequence of sets homothetic to (as defined above), with each contained in and with diameter .
Let be the open rectangles constituting , homothetic to the intervals of . Define a family of rectangles . If we denote the union of the elements of by , then it is clear that . Observe that if is any disjoint subcollection of , then contains at most one rectangle of those constituting , since for . Also observe that since
we have that
Now set . Since and , it follows that . Consider the family of rectangles . It is evident that for each there is a sequence of rectangles of containing and having diameter tending to zero. For any disjoint sequence , we have
Since , we can chose our original sequence to satisfy
To generalize the counterexample to higher dimensions, we simply modify the rectangles in the auxilary construction by , so has volume and has volume . We can then repeat the argument in the proof above with now the unit cube in .
Suppose we replace the Lebesgue measure on by a Radon measure . Does there exist a differentiation basis for which does not possess the Vitali covering property? The above construction relied on the fact that if two measurable sets and have measures have ratio , then for any homothety . 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.