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