1052 CHAPTER 29. THE AREA FORMULA

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) =∫

φ ε (hm (x)−y)det(Dhm (x))dx (29.12.74)

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 29.12.73 to conclude the right sideof the above equation is independent of m > M.

By 29.12.71, 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

=∫

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 and

let h ∈W 1,p (Rn). Also assumem(h(∂Ω)) = 0.

From Proposition 29.12.1, for y /∈ h(∂Ω),

d (y,Ω,h) = limε→0

∫Ω

φ ε (h(x)−y)detDh(x)dx,

showing that y→ d (y,Ω,h) is a measurable function since it is the limit of continuousfunctions off the set of measure zero h(∂Ω).

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, from Proposition 29.12.1 there exists an ε small enough that for all y ∈ spt( f ) ,

limε→0

∫Ω

φ ε (h(x)−y)detDh(x)dx =∫

φ ε (h(x)−y)detDh(x)dx

= d (y,Ω,h)

1052 CHAPTER 29. THE AREA FORMULAfor all k,m > M. By this lemma again, which says that for small enough € the integral isconstant and the definition of the degree in Definition 23.1.10,d(y,Q,h,,) =f, @¢ (By, (x) — y) det (Dh, (x)) dx (29.12.74)for all € small enough. For x € dQ, y € S, andt € [0,1],\(1—t)h(x) +h (x)t—y| > |h(x)—y|—t|h(x) — hy, (x)|> 3€9 —t2€) >0and so by Theorem 23.2.2, the part about homotopy, for each y € S,d(y,Q,h) =d(y,Q,h,,) =[Gln ( — y) det (Dh, (x)) dxwhenever € is small enough. Fix such an € < €9 and use 29.12.73 to conclude the right sideof the above equation is independent of m > M.By 29.12.71, there exists a subsequence still denoted by m such that Dh,, (x) ~ Dh (x)a.e. Since p > n, det(Dh,,) is bounded in L” (Q) for some r > | and so the integrands inthe following are uniformly integrable. By the Vitali convergence theorem, one can pass tothe limit as follows.d(y.Q.h) = lim hn ¢ (Hm (x) —y) det (Dh, (x)) dx- [gecn( y) det (Dh (x)) dx.This proves the proposition.Next is an interesting change of variables theorem. Let Q be a bounded open set andleth € W!? (R"). Also assumem(h(dQ)) =From Proposition 29.12.1, for y ¢ h(dQ),d(y,Q,h) = tim | $, (h(x) —y) detDh (x) dx,showing that y > d(y,Q,h) is a measurable function since it is the limit of continuousfunctions off the set of measure zero h(dQ).Now suppose f € C, (h (99)") . There are finitely many components of h(9Q)° whichhave nonempty intersection with spt (f). From the Proposition above,[fay.anay= [fty) tim | ¢, (h y) det Dh (x) dxdyActually, from Proposition 29.12.1 there exists an € small enough that for all y € spt(f),tim | (h(x) —y) detDh (x) dx -f[ bo( y) det Dh (x) dx= Soy