1032 CHAPTER 29. THE AREA FORMULA

Theorem 29.8.2 Let f, g be Lipschitz mappings from Rn to Rn with g(f(x)) = x on A,a measurable set. Then for a.e. x ∈ A, Dg(f(x)), Df(x), and D(g◦ f)(x) all exist andI = D(g◦ f)(x) = Dg(f(x))Df(x).

Proof: By Lemma 29.8.1 there is a set of measure zero N1 off which

det(D(g◦ f)(x)− I) = 0

and in particular D(g◦ f)(x) exists. Let N2 be the set of measure zero off which f is differ-entiable. Let M be the set of points in f(Rn \N2) where, g fails to be differentiable. Whatabout f−1 (M)? If x ∈ f−1 (M) then Dg(f(x)) fails to exist and so x is in the first excep-tional set N1 or else in N2 because D(g◦ f)(x) will fail to exist. Thus f−1 (M) is a set ofmeasure zero. So let x /∈ N1∪N2. Then for such x, D(g◦ f)(x) ,Dg(f(x)) ,Df(x) all existand I = Dg(f(x))Df(x).

You could give a generalization to the above by essentially repeating the argument.

Corollary 29.8.3 Suppose h is differentiable on A, a measurable set and that f,g are Lip-schitz with g(f(x)) = h(x) for x ∈ A. Then for a.e. x ∈ A,

Dh(x) = Dg(f(x))Df(x)

In other words, the chain rule holds off a set of measure zero.The Reynolds transport formula is an interesting application of the divergence theorem

which is a generalization of the formula for taking the derivative under an integral.

ddt

∫ b(t)

a(t)f (x, t)dx =

∫ b(t)

a(t)

∂ f∂ t

(x, t)dx+ f (b(t) , t)b′ (t)− f (a(t) , t)a′ (t)

First is an interesting lemma about the determinant. A p× p matrix can be thought ofas a vector in Cp2

. Just imagine stringing it out into one long list of numbers. In fact, away to give the norm of a matrix is just ∑i ∑ j

∣∣Ai j∣∣2 ≡ ∥A∥2. This is called the Frobenius

norm for a matrix. It makes no difference since all norms are equivalent, but this one isconvenient in what follows. Also recall that det maps p× p matrices to C. It makes senseto ask for the derivative of det on the set of invertible matrices, an open subset of Cp2

withthe norm measured as just described because A→ det(A) is continuous, so the set wheredet(A) ̸= 0 would be an open set. Recall from linear algebra that the sum of the entrieson the main diagonal satisfies trace(AB) = trace(BA) whenever both products make sense.Indeed, trace(AB)≡ ∑i ∑ j Ai jB ji = trace(BA)

This next lemma is a very interesting observation about the determinant of a matrixadded to the identity.

Lemma 29.8.4 det(I +U) = 1+ trace(U)+ o(U) where o(U) is defined in terms of theFrobenius norm for p× p matrices.

Proof: This is obvious if p= 1 or 2. Assume true for n−1. Then for U an n×n, expandthe matrix along the last column and use induction on the cofactor of 1+Unn.