17.2. COMPARISON THEOREMS 455

Corollary 17.1.7 Let T ⊆ Rm. Then H n (T )≥H n (R∗T ) .

Hausdorff measure makes possible a unified development of p dimensional area. Asin the case of Lebesgue measure, the first step in this is to understand basic considerationsrelated to linear transformations.

Lemma 17.1.8 Let R ∈L (Rp,Rm), p≤m, and R∗R = I. Then if A⊆Rp, H p(RA) =H p(A). In fact, if P : Rp→ Rm satisfies |Px−Py|= |x−y| , then H p (PA) = H p (A) .

Proof: Now let P be an arbitrary mapping which preserves lengths, like R, and let A bebounded so P(A) is also bounded. Then P(A) ⊆ ∪∞

j=1C j, r(C j) < δ , and H pδ(PA)+ ε >

∑∞j=1 α(p)(r(C j))

p.Since P preserves lengths, it follows P is one to one and P−1 is one toone on P(Rp) and P−1 also preserves lengths on P(Rp) . Replacing each C j with C j∩(PA),

H pδ(PA)+ ε >

∞

∑j=1

α(p)r(C j ∩ (PA))p =∞

∑j=1

α(p)r(P−1 (C j ∩ (PA))

)p ≥H pδ(A).

Thus H pδ(PA) ≥H p

δ(A). Similarly H p

δ(P−1 (PA)) ≥H p

δ(PA) so H p

δ(A) ≥H p

δ(PA).

Letting δ → 0 yields the desired conclusion in the case where A is bounded. For the generalcase, let Ar = A∩B(0,r). Then H p(PAr) = H p(Ar). Now let r→ ∞. ■

Lemma 17.1.9 Let F ∈ L (Rp,Rm), p ≤ m, and let F = RU where R and U are de-scribed in Theorem 1.5.5 on Page 23. Then if A⊆ Rp is Lebesgue measurable,

H p(FA) = det(U)mp(A).

Proof: Using Theorem 11.7.4 on Page 330 and Theorem 16.7.4,

H p(FA) = H p(RUA) = H p(UA) = mp(UA) = det(U)mp(A). ■

Definition 17.1.10 Define J to equal det(U). Thus

J = det((F∗F)1/2) = (det(F∗F))1/2.

This is the essential idea for the area formula, but in the area formula, we must considerh : Rp→ Rm for h nonlinear and so h(U) is not a subspace.

17.2 Comparison TheoremsFirst is a simple lemma which is fairly interesting which involves comparison of two lin-ear transformations. These are Lemmas 5.3.2 and 5.3.1 which follows from fundamentalproperties of the operator norm. I am stating them here for convenience.

Lemma 17.2.1 Suppose S,T are linear, defined on a finite dimensional normed linearspace, S−1 exists, and let δ ∈ (0,1). Then whenever ∥S−T∥ is small enough, it followsthat

|Tv||Sv|

∈ (1−δ ,1+δ ) (17.1)

for all v ̸= 0. Similarly if T−1 exists and ∥S−T∥ is small enough,

|Tv||Sv|

∈ (1−δ ,1+δ )