29.2. COMPARISON THEOREMS 1011

The second set is Borel and the first has H n measure zero. By completeness of H n, h(S)is H n measurable.

By Theorem 5.9.6 on Page 94, when Dh(x) exists,

Dh(x) = R(x)U (x)

where (U (x)u,v) = (U (x)v,u) ,(U (x)u,u)≥ 0 and R∗R = I so R preserves lengths. Thisconvention will be used in what follows.

Lemma 29.1.3 In this situation where R∗R = I, |R∗u| ≤ |u|.

Proof: First note that

(u−RR∗u,RR∗ u) = (u,RR∗ u)−|RR∗u|2

= |R∗u|2−|R∗u|2 = 0,

and so

|u|2 = |u−RR∗u+RR∗u|2

= |u−RR∗u|2 + |RR∗u|2

= |u−RR∗u|2 + |R∗u|2.

Then the following corollary follows from Lemma 29.1.3.

Corollary 29.1.4 Let T ⊆ Rm. Then

H n (T )≥H n (R∗T ).

29.2 Comparison TheoremsFirst is a simple lemma which is fairly interesting which involves comparison of two lineartransformations.

Lemma 29.2.1 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

|T v||Sv|∈ (1−δ ,1+δ ) (29.2.3)

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

|T v||Sv|∈ (1−δ ,1+δ )

Proof: Say S−1 exists. Then v→ |Sv| is a norm. Then by equivalence of norms,Theorem 8.4.9, there exists η > 0 such that for all v, |Sv| ≥ η |v| . Say ∥T −S∥< r < δη

|Sv|−∥T −S∥|v||Sv|

≤ |T v||Sv|

=|[S+(T −S)]v|

|Sv|≤ |Sv|+∥T −S∥|v|

|Sv|