29.12. THE CASE OF W 1,p 1053

This is because spt( f ) is at a positive distance from h(∂Ω)C. Therefore, for all ε smallenough, ∫

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

=∫

Ω

detDh(x)∫

f (h(x)− εu)φ (u)dudx

Using the uniform continuity of f , you can now pass to a limit as ε → 0 and obtain, usingthe fact that 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 29.12.2 Let f ∈Cc

(h(∂Ω)C

)and let h ∈W 1,p (Rn;Rn) , p > n, h(∂Ω) has mea-

sure zero for Ω a bounded open set. Then everything is measurable which needs to beand ∫

f (y)d (y,Ω,h)dy =∫

Ω

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

Note that h is not necessarily one to one. The difficult issue is handling d (y,Ω,h) whichhas integer values constant on each component of h(∂Ω)C and the difficulty arrises in notknowing how many components there are. What if there are infinitely many, for example,and what if the degree changes sign. If this happens, it is hard to exploit convergencetheorems to get generalizations of f ∈ Cc

(h(∂Ω)C

). One way around this is to insist h

be one to one and that Ω be connected having a boundary which separates Rn into twocomponents, three if n = 1. That way, you can use the Jordan separation theorem and asserth(∂Ω) also separates Rn into the same number of components with h(Ω) being the onlyone on which the degree is nonzero.

First recall the following proposition.

Proposition 29.12.3 Let Ω be an open connected bounded set in Rn,n≥ 1 such that Rn \∂Ω consists of two, three if n = 1, connected components. Let f ∈C

(Ω;Rn

)be continuous

and one to one. Then f(Ω) is the bounded component of Rn \ f(∂Ω) and for y ∈ f(Ω) ,d (f,Ω,y) either equals 1 or −1.

Proof: First suppose n ≥ 2. By the Jordan separation theorem, Rn \ f(∂Ω) consistsof two components, a bounded component B and an unbounded component U . Using theTietze extention theorem, there exists g defined on Rn such that g = f−1 on f

(Ω). Thus on

∂Ω,g◦ f = id. It follows from this and the product formula that

1 = d (id,Ω,g(y)) = d (g◦ f,Ω,g(y))= d (g,B,g(y))d (f,Ω,B)+d (f,Ω,U)d (g,U,g(y))= d (g,B,g(y))d (f,Ω,B)