16.9. THE PROPER VALUE OF β (p) 451

Proof: By definition, Pjx+e jx j ∈ S(A,ei) if and only if |xi| < 2−1m(APi(Pjx+e jx j)).Now xi ∈ APi(Pjx+e jx j) if and only if xi ∈ APi(Pjx+(−x je j)) by the assumption on A whichsays that A is symmetric in the e j direction. Hence Pjx+e jx j ∈ S(A,ei) if and only if|xi|< 2−1m(APi(Pjx+(−x j)e j)) if and only if Pj x+ (−x j)e j ∈ S(A,ei). ■

16.8.2 The Isodiametric InequalityThe next theorem is called the isodiametric inequality. It is the key result used to compareLebesgue and Hausdorff measures.

Theorem 16.8.6 Let A be any Lebesgue measurable set in Rp. Then it follows thatmp(A)≤ α(p)(r (A))p.

Proof: Suppose first that A is Borel. Let A1 = S(A,e1) and Ak = S(Ak−1,ek). Thenby Lemma 16.8.4, Ap is a Borel set, diam(Ap) ≤ diam(A) , mp(Ap) = mp(A) and Ap issymmetric. Thus x ∈ Ap if and only if −x ∈ Ap. It follows that Ap ⊆ B(0,r (Ap)). Ifx ∈ Ap \B(0,r (Ap)), then −x ∈ Ap \B(0,r (Ap)) and so diam(Ap) ≥ 2 |x| > diam(Ap).Therefore, there is no such x and mp(Ap) ≤ α(p)(r (Ap))

p ≤ α(p)(r (A))p. It remains toestablish this inequality for arbitrary measurable sets. Letting A be such a set, let {Kk} bean increasing sequence of compact subsets of A such that mp(A) = limk→∞ mp(Kk). Then

mp(A) = limk→∞

mp(Kk)≤ lim supk→∞

α(p)(r (Kk))p ≤ α(p)(r (A))p. ■

16.9 The Proper Value of β (p)I will show that the proper determination of β (p) is α (p), the volume of the unit ball. Sinceβ (p) has been adjusted such that l = 1 in Theorem 16.7.4, mp (B(0,1)) = H p (B(0,1)).There exists a covering of B(0,1) of sets of radii less than δ ,{Ci}∞

i=1 such that

H pδ(B(0,1))+ ε > ∑

iβ (p)r (Ci)

p

Then by Theorem 16.8.6, the isodiametric inequality,

H pδ(B(0,1))+ ε > ∑

iβ (p)r (Ci)

p =β (p)α (p) ∑

iα (p)r

(Ci)p

≥ β (p)α (p) ∑

imp(Ci)≥ β (p)

α (p)mp (B(0,1)) =

β (p)α (p)

H p (B(0,1))

Now taking the limit as δ → 0, H p (B(0,1))+ ε ≥ β (p)α(p)H

p (B(0,1)) and since ε > 0 isarbitrary, this shows α (p)≥ β (p).

By the Vitali covering theorem in Corollary 9.12.5, there exists a sequence of disjointballs of radius no more than δ , {Bi} such that B(0,1) = (∪∞

i=1Bi)∪N. where mp (N) =0. Then H p

δ(N) = 0 can be concluded because H p

δ≤H p and Lemma 16.7.3. Using

mp (B(0,1)) = H p (B(0,1)) again,

H pδ(B(0,1)) = H p

δ(∪iBi)≤

∞

∑i=1

β (p)r (Bi)p =

β (p)α (p)

∞

∑i=1

α (p)r (Bi)p