This post is adapted from a question I posted on Math.StackExchange approximately two weeks ago, which has not received a satisfactory answer. If you have a solution or a counterexample, please add it to Math.SE instead of here.
Let be a differentiation basis. Given a measurable set , we say that is a Vitali covering of if for every there exists a sequence containing and with diameter tending to zero.
Question 1 Suppose that for every measurable set of finite measure, every Vitali covering of , and every , there exists a countable subcollection with the following properties:
- (Underflow condition)
- (Overflow condition)
- (Overlap condition)
where is a positive constant independent of , , and . Does it follow that satisfies the density property
where for all and the diameter of tends to zero.
My motivation for posing this question is that when and , a variant of this theorem holds when replace the overlap condition above with and (*) by
for every sequence as above and . (There is no issue in considering only simple functions above, as opposed to the entire space , since the differentiaton of the integrals of each space are equivalent.) In fact, the above three conditions are a special case of what is called the () covering property:
Definition 1 A differentiation basis has the covering property if for every measurable set with , every Vitali covering of , and every , there exists a countable subcollection satisfying
where is a constant independent of , , and .
Briefly we introduce the notation
where the supremum and infimum are taken over all sequences in containing and with diameter tending to zero. When both and are equal, we simply write .
Proof: It suffices to show that the set
has measue zero for all . Without loss of generality we may assume that has compact support. Let be a continuous function of compact support, such that . Observe that
So if , then , where
Intersecting with a compact set if necessary, we may assume that . By definition, for each there exists a sequence containing and with diameter tending to zero. The collection of these sets forms a Vitali cover of , whence there exists a subcollection satisfying the conditions. By Hölder’s inequality,
Dividing both sides of the inequality by gives the desired estimate. By Chebyshev’s inequality,
Putting these two estimates together and noting that was arbitrary completes the proof. The proof that a.e. is completely analogous.
It is easy to see that the above proof breaks down in the endpoint case . Continuous functions are not dense in , so an estimate for and Hölder’s inequality is not good enough. If we had some “better” control over the overlap of the , then we would be able to show that the density property holds.
Suppose that instead of the overlap condition for the covering property stated above, we have condition 3′
Given a measurable set with and , we will show that
has measure zero. It is clear that the union of all forms a Vitali cover of , whence there exists a countable subcollection satisfying conditions 1,2, and 3′. Observe that
since . Since was arbitrary, we conclude that , whence has measure zero. We conclude that for a.e. . We apply the same argument to the set and conclude that for a.e. , which implies for a.e. .
If we could somehow pass from condition 3 to a countable subcollection satisfying conditions 1,2 and 3′, then the question would be resolved as the argument above shows. But it is not clear to this author how to do that, nor do any potential counterexamples come to mind.