1016 CHAPTER 29. THE AREA FORMULA
≤ (1+2ε)n
(1−2ε)n
(1
1−2ε
)n
H n (h(B))
ε was arbitrary and so when h is one to one on B,
H n (h(B))≤∫
Bdet(U (x))dx≤H n (h(B))
Now B was completely arbitrary. Let it equal B(x,r)∩A+ where x ∈ A+. Then forx ∈ A+,
H n (h(B(x,r)∩A+))
=∫
B(x,r)XA+ (y)det(U (y))dy
Divide by mn (B(x,r)) and use the fundamental theorem of calculus. This yields that for xoff a set of mn measure zero,
limr→0
H n (h(B(x,r)∩A+))
mn (B(x,r))= XA+ (x)det(U (x)) (29.4.19)
This has proved the following lemma.
Lemma 29.4.2 Let h be continuous on G and differentiable on A ⊆ G and one to one onA+ which is as defined above. There is a set of measure zero N such that for x ∈ A+ \N,
limr→0+
H n (h(B(x,r)∩A+))
mn (B(x,r))= det(U (x))
The next theorem removes the assumption that U (x)−1 exists and replaces A+ withA. From now on J∗ (x) ≡ det(U (x)) . Also note that if F is measurable and a subset ofA+,h(Ek ∩F) is Hausdorff measurable because of the Lipschitz continuity of h on Ek.
Theorem 29.4.3 Let h : G ⊆ Rn→ Rm for n ≤ m,G an open set in Rn, and suppose h iscontinuous on G differentiable and one to one on A. Then for a.e. x ∈ A, the set in G whereDh(x) exists,
J∗ (x) = limr→0
H n (h(B(x,r)∩A))mn (B(x,r))
, (29.4.20)
where J∗ (x)≡ det(U (x)) = det(Dh(x)∗Dh(x)
)1/2.
Proof: The above argument shows that the conclusion of the theorem holds whenJ∗ (x) ̸= 0 at least with A replaced with A+. I will apply this to a modified function inwhich the corresponding U (x) always has an inverse. Let k : Rn→ Rm×Rn be defined as
k(x)≡(
h(x)εx
)in which dependence of k on ε is suppressed. Then Dk(x)∗Dk(x) = Dh(x)∗Dh(x)+ε2Inand so
J∗k(x)2 ≡ det(Dh(x)∗Dh(x)+ ε
2In)= det
(Q∗DQ+ ε
2In)> 0