1044 CHAPTER 29. THE AREA FORMULA
Then Dkε (x,z) =(
Df(x) εI)=(
UR εI)
where the dependence of U and R onx has been suppressed. Here RR∗ = I and U is a non-negative symmetric transformation.Thus
(J∗kε)2 = det
(UR εI
)( R∗UεI
)= det
(U2 + ε
2I)
= det(Q∗DQQ∗DQ+ ε
2I)= det
(D2 + ε
2I)
=m
∏i=1
(λ
2i + ε
2)∈ [ε2m,C2
ε2] (29.9.58)
since one of the λ i equals 0 due to det(U) = 0. All the eigenvalues of U must be boundedindependent of x, since ∥Df(x)∥ is bounded independent of x due to the assumption that fis Lipschitz. Since the corresponding S = /0, the first part of the argument implies
εCmn+m
(A×B(0,1)
)≥∫
A×B(0,1)|J∗kε |dmn+m
=∫Rm
H n(
k−1ε (y)∩A×B(0,1)
)dy (29.9.59)
Now it is clear that k−1ε (y)⊇ f−1 (y) . Indeed, if f(x) = y, then f(x)+ ε0 = y.
Note that A⊆ Rn and B(0,1) ∈ Rm. By Lemma 29.9.4, and what was just noted,
H n(
k−1ε (y)∩A×B(0,1)
)≥
Cnm1
(Lip(p))m
∫Rm
H n−m(
k−1ε (y)∩p−1 (w)∩A×B(0,1)
)dw
Therefore, from 29.9.59,εCmn+m
(A×B(0,1)
)≥
Cnm
∫Rm
∫Rm
H n−m(
k−1ε (y)∩p−1 (w)∩A×B(0,1)
)dwdy (29.9.60)
≥Cnm
∫Rm
∫Rm
H n−m(
f−1 (y)∩p−1 (w)∩A×B(0,1))
dwdy
The inside set is f−1 (y)∩p−1 (w)∩A×B(0,1). That is,{(x,w) ∈ A×B(0,1) : f(x) = y
}thus the set inside H n−m is
(f−1 (y)∩A
)×B(0,1), then continuing the chain of inequali-
ties,
≥ Cnm
∫Rm
∫Rm
H n−m((
f−1 (y)∩A)×B(0,1)
)dwdy
≥ Cnm
∫Rm
∫B(0,1)
H n−m (f−1 (y)∩A)
dwdy