1048 CHAPTER 29. THE AREA FORMULA

showing that for each y ∈ S, B(y,ε)∩ ((1− t)hm + thk)(∂Ω) = /0. By Lemma 23.1.11, forall y ∈ S, ∫

Ω

φ ε (hm (x)−y)det(Dhm (x))dx =∫Ω

φ ε (hk (x)−y)det(Dhk (x))dx (29.11.66)

for all k,m ≥M. By this lemma again, which says that for small enough ε the integral isconstant and the definition of the degree in Definition 23.1.10,

d (y,Ω,hm) =∫

Ω

φ ε (hm (x)−y)det(Dhm (x))dx (29.11.67)

for all ε small enough. For x ∈ ∂Ω, y ∈ S, and t ∈ [0,1],

|(1− t)h(x)+hm (x) t−y| ≥ |h(x)−y|− t |h(x)−hm (x)|> 3ε0− t2ε0 > 0

and so by Theorem 23.2.2, the part about homotopy, for each y ∈ S,

d (y,Ω,h) = d (y,Ω,hm) =∫Ω

φ ε (hm (x)−y)det(Dhm (x))dx

whenever ε is small enough. Fix such an ε < ε0 and use 29.11.66 to conclude the right sideof the above equation is independent of m > M.

By 29.11.64, there exists a subsequence still denoted by m such that Dhm (x)→Dh(x)a.e. Since p > n, det(Dhm) is bounded in Lr (Ω) for some r > 1 and so the integrands inthe following are uniformly integrable. By the Vitali convergence theorem, one can pass tothe limit as follows.

d (y,Ω,h) = limm→∞

∫Ω

φ ε (hm (x)−y)det(Dhm (x))dx

=∫

Ω

φ ε (h(x)−y)det(Dh(x))dx.

This proves the proposition.Next is an interesting change of variables theorem. Let Ω be a bounded open set with

the property that ∂Ω has measure zero and let h be Lipschitz continuous on Rn. Then fromLemma 29.1.1, h(∂Ω) also has measure zero.

Now suppose f ∈Cc

(h(∂Ω)C

). There are finitely many components of h(∂Ω)C which

have nonempty intersection with spt( f ). From the Proposition above,∫f (y)d (y,Ω,h)dy =

∫f (y) lim

ε→0

∫Ω

φ ε (h(x)−y)detDh(x)dxdy

Actually, there exists an ε small enough that for all y ∈ spt( f ) ,

limε→0

∫Ω

φ ε (h(x)−y)detDh(x)dx =∫

Ω

φ ε (h(x)−y)detDh(x)dx

= d (y,Ω,h)