28.4. THE PROPER VALUE OF β (n) 1003

28.4 The Proper Value Of β (n)I will show that the proper determination of β (n) is α (n), the volume of the unit ball.Since β (n) has been adjusted such that k = 1, mn (B(0,1)) = H n (B(0,1)). There exists acovering of B(0,1) of sets of radii less than δ ,{Ci}∞

i=1 such that

H nδ(B(0,1))+ ε > ∑

iβ (n)r (Ci)

n

Then by Theorem 28.3.5, the isodiametric inequality,

H nδ(B(0,1))+ ε > ∑

iβ (n)r (Ci)

n =β (n)α (n) ∑

iα (n)r

(Ci)n

≥ β (n)α (n) ∑

imn(Ci)≥ β (n)

α (n)mn (B(0,1)) =

β (n)α (n)

H n (B(0,1))

Now taking the limit as δ → 0,

H n (B(0,1))+ ε ≥ β (n)α (n)

H n (B(0,1))

and since ε > 0 is arbitrary, this shows α (n)≥ β (n).By the Vitali covering theorem, there exists a sequence of disjoint balls, {Bi} such that

B(0,1) = (∪∞i=1Bi)∪N

where mn (N) = 0. Then H nδ(N) = 0 can be concluded because H n

δ≤H n and Lemma

28.2.2. Using mn (B(0,1)) = H n (B(0,1)) again,

H nδ(B(0,1)) = H n

δ(∪iBi)≤

∞

∑i=1

β (n)r (Bi)n

=β (n)α (n)

∞

∑i=1

α (n)r (Bi)n =

β (n)α (n)

∞

∑i=1

mn (Bi)

=β (n)α (n)

mn (∪iBi) =β (n)α (n)

mn (B(0,1)) =β (n)α (n)

H n (B(0,1))

which implies α (n) ≤ β (n) and so the two are equal. This proves that if α (n) = β (n) ,then the H n = mn on the measurable sets of Rn.

This gives another way to think of Lebesgue measure which is a particularly nice waybecause it is coordinate free, depending only on the notion of distance.

For s< n, note that H s is not a Radon measure because it will not generally be finite oncompact sets. For example, let n= 2 and consider H 1(L) where L is a line segment joining(0,0) to (1,0). Then H 1(L) is no smaller than H 1(L) when L is considered a subset ofR1,n = 1. Thus by what was just shown, H 1(L) ≥ 1. Hence H 1([0,1]× [0,1]) = ∞.The situation is this: L is a one-dimensional object inside R2 and H 1 is giving a one-dimensional measure of this object. In fact, Hausdorff measures can make such heuristicremarks as these precise. Define the Hausdorff dimension of a set, A, as

dim(A) = inf{s : H s(A) = 0}