29.11. INTEGRATION AND THE DEGREE 1049

This is because spt( f ) is at a positive distance from the compact set h(∂Ω)C. Therefore,for all ε small enough,∫

f (y)d (y,Ω,h)dy =∫ ∫

Ω

f (y)φ ε (h(x)−y)detDh(x)dxdy

=∫

Ω

detDh(x)∫

f (y)φ ε (h(x)−y)dydx

Using the uniform continuity of f , you can now pass to a limit and obtain using the factthat detDh(x) is in Lr (Rn) for some r > 1,∫

f (y)d (y,Ω,h)dy =∫

Ω

f (h(x))detDh(x)dx

This has proved the following interesting lemma.

Lemma 29.11.2 Let f ∈ Cc

(h(∂Ω)C

)for Ω a bounded open set and let h be Lipschitz

on Rn. Say ∂Ω has measure zero so that h(∂Ω) has measure zero. Then everything ismeasurable which needs to be and∫

f (y)d (y,Ω,h)dy =∫

Ω

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

Note that h is not necessarily one to one. Next is a simple corollary which replacesCc (Rn) with L1

loc (Rn) in the case that h is one to one. Also another assumption is made onthere being finitely many components.

Corollary 29.11.3 Let f ∈ L1loc (Rn) and let h be one to one and satisfy 29.11.62 - 29.11.63,

∂Ω has measure zero for Ω a bounded open set and h(∂Ω)C has finitely many components.Then everything is measurable which needs to be and∫

f (y)d (y,Ω,h)dy =∫

Ω

detDh(x) f (h(x))dx.

Proof: Since d (y,Ω,h) = 0 for all |y| large enough due to y /∈ h(Ω) for large y,there isno loss of generality in assuming f is in L1 (Rn). For all y /∈ h(∂Ω), a set of measure zero,d (y,Ω,h) is bounded by some constant, depending on the maximum of the degree on thevarious components of h(∂Ω)C. Then from Proposition 29.11.1∫

f (y)d (y,Ω,h)dy =∫

f (y) limε→0

∫Ω

φ ε (h(x)−y)detDh(x)dxdy (29.11.68)

This time, use the area formula to write∣∣∣∣∫Ω

φ ε (h(x)−y)detDh(x)dx∣∣∣∣≤ ∫Rn

φ ε (h(x)−y) |detDh(x)|dx

≤ K∫Rn

φ ε (z−y)dz < ∞