Covering Theorem (we use past Lemmas)


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 \mathcal{V} if for any \epsilon>0 and any x in the set, there exists a set V\in\mathcal{V} such that x\in V and diam(V)<\epsilon. 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 \mathcal{I} is a sequence of intervals that Vitali covers an interval E\subset \mathbb{R}, then there is a countable disjoint subcollection of \mathcal{I} that covers E 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 I_n\in\mathcal{I} disjoint such that m\left(E\setminus\cup I_n\right)=0. All interval types have same measure, so WLOG assume the intervals are closed in \mathcal{I}. Define \mathcal{I}^* to be the collection of finite unions of disjoint intervals from \mathcal{I}.

Claim: If A\in\mathcal{I}^* and m(E\setminus A)>0 then there is a B\in\mathcal{I}^* such that A\cap B=\emptyset and m(E\setminus(A\cup B))<\frac{3}{4}m(E\setminus A).

Proof of claim: Since \mathcal{I} is a Vitali covering we can choose intervals of small enough diameter \{J_i\}_1^n\subset\mathcal{I} so that each J_i\subset E\setminus A (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 m(E\setminus(A\cup J_1\cup\cdots\cup J_n))<\frac{1}{12}m(E\setminus A). Now by the Vitali Covering Lemma of last time we can find a disjoint subset \{I_j\}_1^k\subset \{J_i\}_1^n so that m(\bigcup I_j)\geq \frac{1}{3}m(\bigcup J_i).

Then m(E\setminus (A\cup I_1\cup\cdots\cup I_k))

<\frac{2}{3}m(J_1\cup\cdots\cup J_n)+\frac{1}{12}m(E\setminus A)

\leq \frac{2}{3}m(E\setminus A)+\frac{1}{12}m(E\setminus A)

=\frac{3}{4}m(E\setminus A).

Thus B=I_1\cup\cdots\cup I_k\in\mathcal{I}^* works.

Now simply apply this inductively and use countable additivity of measure to get that m(E\setminus A_1\cup\cdots \cup A_k)\leq \left(\frac{3}{4}\right)^km(E), i.e. \displaystyle m\left(E\setminus \bigcup_{k=1}^\infty A_k\right)=0. We are done.

The generalization is exactly the same, except where you use Vitali Covering Lemma, you replace 3 with 3^n (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.

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