17.7. INTEGRATION AND THE DEGREE 469

So letting ε → 0,∫

f (z)d (hm,Ω,z)dz =∫

f (z)det(Dhm (z))dz. Next suppose

f ∈Cc

(h(∂Ω)C

)Then spt( f ) is covered by finitely many of such balls like the above, {Bi}m

i=1 with theproperty that if g ∈Cc (Bi) , then∫

g(z)d (hm,Ω,z)dz =∫

Ω

g(hm (x))det(Dhm (x))dx

Now let{

ψ j

}be a partition of unity on spt f with sptψ i ⊆ Bi. Then

∫f (z)d (hm,Ω,z)dz =

∫ m

∑i=1

ψ i (z) f (z)d (hm,Ω,z)dz

=m

∑i=1

∫ψ i (z) f (z)d (hm,Ω,z)dz =

m

∑i=1

∫Ω

ψ i (hm (x)) f (hm (x))det(Dhm (x))dx

=∫

Ω

m

∑i=1

ψ i (hm (x)) f (hm (x))det(Dhm (x))dx =∫

Ω

f (hm (x))det(Dhm (x))dx

If h is Lipschitz, then limm→∞hm (x) = h(x) uniformly and also for a.e. x, Dhm (x)→Dh(x) and so, since everything is bounded, we can apply the dominated convergencetheorem and conclude that∫

f (z)d (h,Ω,z)dz =∫

Ω

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

This proves the following interesting proposition.

Proposition 17.7.4 Let f ∈Cc

(h(∂Ω)C

)and let h be Lipschitz on Rn. Then

∫f (y)d (h,Ω,y)dy =

∫Ω

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

Note that d (h,Ω,y) = 0 if y /∈ h(Ω) so the integral on the left is taken over h(Ω).Recall that the area formula gives the formula∫

h(Ω)f (y)#(y)dy =

∫Ω

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

You could probably say more. For example, the degree is constant on components ofh(dΩ)C and so considering these components, you maybe could use the Riesz represen-tation theorem for positive linear functionals to get this formula for more general f . SayΩ is open and connected, for example, and suppose d (h,Ω,y) is 3 on y ∈ h(Ω) . Thenyou could let Λ f ≡

∫f (y)3dy =

∫Ω

f (h(x))det(Dh(x))dx and this would be a positive

linear functional on Cc

(h(∂Ω)C

).