28.4. THE PROPER VALUE OF β (n) 1007

Thus H nδ(PA)≥H n

δ(A).

Now let A⊆ ∪∞j=1C j,diam(C j)≤ δ , and

H nδ(A)+ ε ≥

∞

∑j=1

α(n)(r (C j))n

Then

H nδ(A)+ ε ≥

∞

∑j=1

α(n)(r (C j))n

=∞

∑j=1

α(n)(r (PC j))n ≥H n

δ(PA).

Hence H nδ(PA) = H n

δ(A). Letting δ → 0 yields the desired conclusion in the case where

A is bounded. For the general case, let Ar = A∩B(0,r). Then H n(PAr) = H n(Ar). Nowlet r→ ∞.

Lemma 28.4.5 Let F ∈L (Rn,Rm),n≤ m, and let F = RU where R and U are describedin Theorem 5.9.6 on Page 94. Then if A⊆ Rn is Lebesgue measurable,

H n(FA) = det(U)mn(A).

Proof: Using Theorem 13.5.7 on Page 354 and Theorem 28.2.4,

H n(FA) = H n(RUA)

= H n(UA) = mn(UA) = det(U)mn(A).

Definition 28.4.6 Define J to equal det(U). Thus

J = det((F∗F)1/2) = (det(F∗F))1/2.