Multivariable Advanced Calculus

284 BROUWER DEGREEFor y ∈ S, let z ∈ B (y, ε) where ε < ε 0 and suppose x ∈ ∂Ω, and k, m ≥ M. Thenfor t ∈ [0, 1] ,|(1 − t) h m (x) + h k (x) t − z| ≥ |h m (x) − z| − t |h k (x) − h m (x)|> 2ε 0 − t2ε 0 ≥ 0showing that for each y ∈ S, B (y, ε) ∩ ((1 − t) h m + th k ) (∂Ω) = ∅. By Lemma 10.2.5,for all y ∈ S, ∫ϕ ε (h m (x) − y) det (Dh m (x)) dx =Ω∫Ωϕ ε (h k (x) − y) det (Dh k (x)) dx (10.24)for all k, m ≥ M. By this lemma again, which says that for small enough ε the integralis constant and the definition of the degree in Definition 10.2.4,∫d (y,Ω, h m ) = ϕ ε (h m (x) − y) det (Dh m (x)) dx (10.25)for all ε small enough. For x ∈ ∂Ω, y ∈ S, and t ∈ [0, 1],Ω|(1 − t) h (x) + h m (x) t − y| ≥ |h (x) − y| − t |h (x) − h m (x)|> 3ε 0 − t2ε 0 > 0and so by Theorem 10.2.9, the part about homotopy, for each y ∈ S,∫Ωd (y,Ω, h) = d (y,Ω, h m ) =ϕ ε (h m (x) − y) det (Dh m (x)) dxwhenever ε is small enough. Fix such an ε < ε 0 and use 10.24 to conclude the right sideof the above equation is independent of m > M. Now from the uniform convergencenoted above,∫d (y,Ω, h) = lim ϕ ε (h m (x) − y) det (Dh m (x)) dxm→∞Ω∫= ϕ ε (h (x) − y) det (Dh (x)) dx.ΩThis proves the proposition. The next lemma is quite interesting. It says a C 1 function maps sets of measurezero to sets of measure zero. This was proved earlier in Lemma 9.8.1 but I am statinga special case here for convenience.Lemma 10.7.2 Let h ∈ C 1 (R n ; R n ) and h vanishes off a bounded set. Let m n (A) =0. Then h (A) also has measure zero.Next is an interesting change of variables theorem. Let Ω be a bounded open setwith the property that ∂Ω has measure zero and let h be C 1 and vanish off a boundedset. Then from Lemma 10.7.2, h (∂Ω) also has measure zero.Now suppose f ∈ C c(h (∂Ω) C) . By compactness, there are finitely many componentsof h (∂Ω) C which have nonempty intersection with spt (f). From the Propositionabove,∫∫f (y) d (y, Ω, h) dy = f (y) lim ϕ ε (h (x) − y) det Dh (x) dxdyε→0∫Ω