998 CHAPTER 28. HAUSDORFF MEASURE

Lemma 28.2.3 Every open set U in Rp is a countable disjoint union of half open boxes ofthe form Q≡∏

pi=1[ai,ai +2−k) where ai = l2−k for l some integer.

Proof: It is clear that there exists Qk a countable disjoint collection of these half openboxes each of sides of length 2−k whose union is all of Rp. Let B1 be those sets of Q1which are contained in U, if any. Having chosen Bk−1, let Bk consist of those sets of Qkwhich are contained in U such that none of these are contained in Bk−1. Then ∪∞

k=1Bk isa countable collection of disjoint boxes of the right sort whose union is U . This is becauseif R is a box of Qk and R̂ is a box of Qk−1, then either R⊆ R̂ or R∩ R̂ = /0.

Theorem 28.2.4 By choosing β (p) properly, one can obtain H p = mp on all Lebesguemeasurable sets.

Proof: I will show H p is a positive multiple of mp for any choice of β (p) . Define k =mp(Q0)H p(Q0)

where Q0 = [0,1)p is the half open unit cube inRp. I will show kH p (E) = mp (E)for any Lebesgue measurable set. When this is done, it will follow that by adjusting β (p)the multiple can be taken to be 1.

Let Q = ∏pi=1[ai,ai +2−k) be a half open box where ai = l2−k. Thus Q0 is the union of(

2k)p of these identical half open boxes. By translation invariance, of H p and mp,(

2k)p

H p (Q) = H p (Q0) =1k

mp (Q0) =1k

(2k)p

mp (Q) .

Therefore, kH p (Q) = mp (Q) for any such half open box and by translation invariance,for the translation of any such half open box. It follows that kH p (U) = mp (U) for allopen sets because each open set is a countable disjoint union of such half open boxes. Itfollows immediately, since every compact set is the countable intersection of open setsthat kH p = mp on compact sets. Therefore, kH p = mp on all closed sets because everyclosed set is the countable union of compact sets. Now let F be an arbitrary Lebesguemeasurable set. I will show that F is H p measurable and that kH p (F) = mp (F). LetFl = B(0, l)∩F. By Proposition 11.7.3, there exists H a countable union of compact setsand G a countable intersection of open sets such that H ⊆ Fl ⊆G and mp (G\H) = 0 whichimplies by Lemma 28.2.2 that mp (G\H) = kH p (G\H) = 0. Then by completeness ofH p it follows Fl is H p measurable and kH p (Fl) = kH p (H) = mp (H) = mp (Fl) . Nowtaking l→∞, it follows F is H p measurable and kH p (F) = mp (F). Therefore, adjustingβ (p) it can be assumed the constant k is 1.

The exact determination of β (p) is more technical.

28.3 Technical ConsiderationsLet α(n) be the volume of the unit ball in Rn. Thus the volume of B(0,r) in Rn is α(n)rn

from the change of variables formula. There is a very important and interesting inequalityknown as the isodiametric inequality which says that if A is any set in Rn, then

m(A)≤ α(n)(2−1diam(A))n = α (n)r (A)n .