198 CHAPTER 8. POSITIVE LINEAR FUNCTIONALS

So let K be the set of open rectangles along with /0 and Rp and let G consist of all mea-surable sets such that mp (E ∩Rn) = mp (E ∩Rn + z) with E ∩Rn + z measurable. HereRn ≡∏

pi=1 (−n,n). Then, similar to the proof of Lemma 8.4.2, you show that G is closed

with respect to countable disjoint unions and complements. Then you use Dynkin’s lemmato conclude that G contains the Borel sets. Next let n→ ∞ to obtain the conclusion for anyE Borel. Now suppose E is just an arbitrary measurable set in Fp. Apply Proposition 8.3.2to get F,G as described there, an Fσ and Gδ set with F ⊆ E ⊆ G and all three having thesame Lebesgue measure. Thus, from the first part,

mp (E) = mp (G) = mp (G+ z)≥ mp (E + z)≥ mp (F + z) = mp (F) = mp (E)

and so all inequalities are equal signs. ■

8.6 The Vitali Covering TheoremsThese theorems are remarkable and fantastically useful. They are covering theorems be-cause they have to do with covering sets with balls. These balls may be open, closed, orneither open nor closed.

Lemma 8.6.1 In a normed linear space, B(x,r) = {y : ∥y−x∥ ≤ r}

Proof: y→∥y−x∥is continuous and so {y : ∥y−x∥ ≤ r} is a closed set which containsB(x,r). Therefore,

B(x,r)⊆ {y : ∥y−x∥ ≤ r} (8.11)

Now let y be in the right side. It suffices to consider y such that ∥y−x∥ = 1. Considerx+ n−1

n (y−x)≡ xn. Then

∥xn−y∥=∥∥∥∥x+

n−1n

(y−x)−y∥∥∥∥= 1

n∥x−y∥

and so y is a limit point of B(x, t) and is therefore in B(x,r) so the two sets in 8.11 areequal. ■

Thus the usual way we think about the closure of a ball is completely correct in anormed linear space. This lemma is not always true in the context of a metric space. Recallthe discrete metric for example in which the distance between different points is 1 anddistance between a point and itself is 0. In what follows we will use the result of thislemma without comment. Balls will be either open, closed or neither. I am going to use theHausdorff maximality theorem because it yields a very simple argument.

Recall the following definition of a partially ordered set. A nonempty set is partiallyordered if there exists a partial order, ≺, satisfying x≺ x and if x≺ y and y≺ z then x≺ z.An example of a partially ordered set is the set of all subsets of a given set and≺ is definedas ⊆. Note that two elements in a partially ordered set may not be related. In other words,just because x, y are in the partially ordered set, it does not follow that either x≺ y or y≺ x.A subset of a partially ordered set C , is called a chain if x, y ∈ C implies that either x≺ yor y≺ x. If either x≺ y or y≺ x then x and y are described as being comparable. A chain isalso called a totally ordered set. C is a maximal chain if whenever C̃ is a chain containingC , it follows the two chains are equal. In other words C is a maximal chain if there is nostrictly larger chain. It turns out that every nonempty partially ordered set has a maximalchain. This is the Hausdorff maximal theorem discussed in Section 1.4. I will need to usethis major result a few other times, so this might be a good place to introduce it.