A brief break occured while I moved 2700 miles away. The important thing is I’m back, and we’re going to prove a big one today. First let’s define a Vitali covering. A set is Vitali covered by the collection of sets if for any and any x in the set, there exists a set such that and . So note that this is sort of stringent in that we always have to find one of the covering sets to be of arbitrarily small diameter at any point.
Vitali’s Covering Theorem (easy version): If is a sequence of intervals that Vitali covers an interval , then there is a countable disjoint subcollection of that covers except for a set of Lebesgue measure 0. (Note I call this the easy version because it can be extended to finite dimensional space using balls and such, but this is easier to prove).
Proof: Suppose the hypothesis with same notation. The claim is that we can find disjoint such that . All interval types have same measure, so WLOG assume the intervals are closed in . Define to be the collection of finite unions of disjoint intervals from .
Claim: If and then there is a such that and .
Proof of claim: Since is a Vitali covering we can choose intervals of small enough diameter so that each (since it has positive measure there will be at least one of these). Since we don’t care about overlap right now, we can do this at enough points so that . Now by the Vitali Covering Lemma of last time we can find a disjoint subset so that .
Now simply apply this inductively and use countable additivity of measure to get that , i.e. . We are done.
The generalization is exactly the same, except where you use Vitali Covering Lemma, you replace 3 with (notice that this relies on n being finite). It is not true in infinite dimensional spaces. Also, you can reformulate the the statement to use Hausdorff measure.