23.1. SARD’S LEMMA AND APPROXIMATION 759

Definition 23.1.10 Let g ∈ C∞(Ω;Rp

)where Ω is a bounded open set. Also let φ ε be a

mollifier.

φ ε ∈C∞c (B(0,ε)) , φ ε ≥ 0,

∫φ ε dx = 1.

The idea is that ε will converge to 0 to get suitable approximations.

First, here is a technical lemma which will be used to identify the degree with an inte-gral.

Lemma 23.1.11 Let y /∈ g(∂Ω) for g ∈C∞(Ω;Rp

). Also suppose y is a regular value of

g. Then for all positive ε small enough,∫Ω

φ ε (g(x)−y)detDg(x)dx = ∑{

sgn(detDg(x)) : x ∈ g−1 (y)}

Proof: First note that the sum is finite from Lemma 23.1.5. It only remains to verifythe equation.

I need to show the left side of this equation is constant for ε small enough and equals theright side. By what was just shown, there are finitely many points, {xi}m

i=1 = g−1 (y). By theinverse function theorem, there exist disjoint open sets Ui with xi ∈Ui, such that g is one toone on Ui with det(Dg(x)) having constant sign on Ui and g(Ui) is an open set containingy. Then let ε be small enough that B(y,ε) ⊆ ∩m

i=1g(Ui) . Also, y /∈ g(Ω\(∪n

i=1Ui))

, acompact set. Let ε be still smaller, if necessary, so that B(y,ε)∩g

(Ω\(∪n

i=1Ui))

= /0 andlet Vi ≡ g−1 (B(y,ε))∩Ui.

g(U2)g(U3)

g(U1)yε

x1

x2

x3V1

V2

V3

Therefore, for any ε this small,∫Ω

φ ε (g(x)−y)detDg(x)dx =m

∑i=1

∫Vi

φ ε (g(x)−y)detDg(x)dx

The reason for this is as follows. The integrand on the left is nonzero only if g(x)−y ∈ B(0,ε) which occurs only if g(x) ∈ B(y,ε) which is the same as x ∈ g−1 (B(y,ε)).Therefore, the integrand is nonzero only if x is contained in exactly one of the disjoint sets,Vi. Now using the change of variables theorem, (z = g(x)−y,g−1 (y+ z) = x.)

=m

∑i=1

∫g(Vi)−y

φ ε (z)detDg(g−1 (y+ z)

)∣∣detDg−1 (y+ z)∣∣dz

23.1. SARD’S LEMMA AND APPROXIMATION 759Definition 23.1.10 Let g < C” (Q:R? ) where Q is a bounded open set. Also let @, be amollifier.. €C?(B(0,£)), 6. >0, | ecax= 1.The idea is that € will converge to 0 to get suitable approximations.First, here is a technical lemma which will be used to identify the degree with an inte-gral.Lemma 23.1.11 Let y ¢ g(0Q) for g € C® (Q;R”). Also suppose y is a regular value ofg. Then for all positive € small enough,[gelees y) det Dg (x) dx = )° {sgn (det Dg(x)) : x € g”! (y)}Proof: First note that the sum is finite from Lemma 23.1.5. It only remains to verifythe equation.I need to show the left side of this equation is constant for € small enough and equals theright side. By what was just shown, there are finitely many points, {x;}/"_, = g'(y). By theinverse function theorem, there exist disjoint open sets U; with x; € Uj, such that g is one toone on U; with det (Dg (x)) having constant sign on U; and g(U;) is an open set containingy. Then let € be small enough that B(y,e) C M%,g(U;). Also, y ¢ 8(2\ (Uj \Ui)),acompact set. Let € be still smaller, if necessary, so that B(y, €) Ng (Q\ (UL Ui \) = @ andlet Vj =g°! (B(y,€)) Uj.g(U3) g(U2)g(U1)Therefore, for any € this small,[geles y) det Dg (x yav=¥ | 6. (g(x) —y) detDg (x) dxThe reason for this is as follows. The integrand on the left is nonzero only if g(x) —y € B(0,€) which occurs only if g(x) € B(y,€) which is the same as x € g! (B(y,€)).Therefore, the integrand is nonzero only if x is contained in exactly one of the disjoint sets,V;. Now using the change of variables theorem, (z = g(x) —y,g-! (y+z) =x.)- » hey 5 (2) detDg(g' (y +2) |detDg”' (y+ 2)| dz