1054 CHAPTER 29. THE AREA FORMULA

Therefore, d (f,Ω,B) ̸= 0 and so for every z ∈ B, it follows z ∈ f(Ω) . Thus B ⊆ f(Ω) .On the other hand, f(Ω) cannot have points in both U and B because it is a connectedset. Therefore f(Ω) ⊆ B and this shows B = f(Ω). Thus d (f,Ω,B) = d (f,Ω,y) for eachy ∈ B and the above formula shows this equals either 1 or −1 because the degree is aninteger. In the case where n = 1, the argument is similar but here you have 3 componentsin R1 \ f(∂Ω) so there are more terms in the above sum although two of them give 0. Thisproves the proposition.

The following is a version of the area formula.

Lemma 29.12.4 Let h∈W 1,p (Rn;Rn) , p> n where h is one to one, h(∂Ω) ,∂Ω have mea-sure zero for Ω a bounded open connected set in Rn. Then h(∂Ω)C has two components,three if n = 1, and for y ∈ h(Ω) , and f ∈Cc (Rn) .∫

h(Ω)f (y)dy =

∫Ω

|det(Dh(x))| f (h(x))dx

If O is an open set, it is also true that∫h(Ω)

XO (y)dy =∫

Ω

|det(Dh(x))|XO (h(x))dx

Also if f is any nonnegative Borel measurable function∫h(Ω)

f (y)dy =∫

Ω

|det(Dh(x))| f (h(x))dx

Proof: Consider the first claim. Let δ be such that B(x1,δ ) ⊆ Ω and let{

f j (y)}∞

j=1be nonnegative, increasing in j and converging pointwise to Xh(B(x1,δ )) (y). This can bedone because h(B(x1,δ )) is an open bounded set thanks to invariance of domain, Theorem23.4.3. By Proposition 29.12.3, d (y,Ω,h) either equals 1 or −1. Suppose it equals −1.Then from Lemma 29.11.2∫

h(Ω)f j (y)dy =−

∫Ω

det(Dh(x)) f j (h(x))dx

The integrand on the right is uniformly integrable thanks to the fact the f j are bounded anddet(Dh(x)) is in Lr (Ω) for some r > 1. Therefore, by the Vitali convergence theorem andthe monotone convergence theorem,∫

h(Ω)Xh(B(x1,δ )) (y)dy =−

∫Ω

det(Dh(x))XB(x1,δ ) (x)dx

som(h(B(x1,δ )))

1m(B(x1,δ ))

=− 1m(B(x1,δ ))

∫B(x1,δ )

det(Dh(x))dx

If x1 is a Lebesgue point of det(Dh(x)) , then you can pass to the limit as δ → 0 andconclude

−det(Dh(x1))≥ 0