10.2. CHANGE OF VARIABLES NONLINEAR MAPS 247

Lemma 10.2.3 (Sard) Let U be an open set in Rp and let h : U → Rp be differentiable.Let

Z ≡ {x ∈U : detDh(x) = 0} .Then mp (h(Z)) = 0.

Proof: For convenience, assume the balls in the following argument come from ∥·∥∞

.First note that Z is a Borel set because h is continuous and so the component functionsof the Jacobian matrix are each Borel measurable. Hence the determinant is also Borelmeasurable.

Suppose that U is a bounded open set. Let ε > 0 be given. Also let V ⊇ Z with V ⊆Uopen, and

mp (Z)+ ε > mp (V ) .

Now let x ∈ Z. Then since h is differentiable at x, there exists δ x > 0 such that if r < δ x,then B(x,r)⊆V and also,

h(B(x,r))⊆ h(x)+Dh(x)(B(0,r))+B(0,rη) , η < 1.

Regard Dh(x) as an n×n matrix, the matrix of the linear transformation Dh(x) with respectto the usual coordinates. Since x∈ Z, it follows that there exists an invertible matrix A suchthat ADh(x) is in row reduced echelon form with a row of zeros on the bottom. Therefore,

mp (A(h(B(x,r))))≤ mp (ADh(x)(B(0,r))+AB(0,rη)) (10.5)

The diameter of ADh(x)(B(0,r)) is no larger than ∥A∥∥Dh(x)∥2r and it lies in Rp−1×{0} . The diameter of AB(0,rη) is no more than ∥A∥(2rη) .Therefore, the measure of theright side in 10.5 is no more than

[(∥A∥∥Dh(x)∥2r+∥A∥(2η))r]p−1 (rη)

≤ C (∥A∥ ,∥Dh(x)∥)(2r)pη

Hence from the change of variables formula for linear maps,

mp (h(B(x,r)))≤ ηC (∥A∥ ,∥Dh(x)∥)

|det(A)|mp (B(x,r))

Then letting δ x be still smaller if necessary, corresponding to sufficiently small η ,

mp (h(B(x,r)))≤ εmp (B(x,r))

The balls of this form constitute a Vitali cover of Z. Hence, by the Vitali covering theoremTheorem 8.6.6, there exists {Bi}∞

i=1 ,Bi = Bi (xi,ri) , a collection of disjoint balls, each ofwhich is contained in V, such that mp (h(Bi)) ≤ εmp (Bi) and mp (Z \∪iBi) = 0. Hencefrom Lemma 10.1.2,

mp (h(Z)\∪ih(Bi))≤ mp (h(Z \∪iBi)) = 0

Therefore,

mp (h(Z)) ≤ ∑i

mp (h(Bi))≤ ε ∑i

mp (Bi)

≤ ε (mp (V ))≤ ε (mp (Z)+ ε) .

Since ε is arbitrary, this shows mp (h(Z)) = 0. What if U is not bounded? Then considerZn = Z ∩B(0,n) . From what was just shown, h(Zn) has measure 0 and so it follows thath(Z) also does, being the countable union of sets of measure zero. ■