1042 CHAPTER 29. THE AREA FORMULA

=∞

∑j=1

∑i ∈Λ(n,m)

∫fi(

E ij∩A

) (det(Df(g(y))Df(g(y))∗

))1/2∣∣∣detDfi (g(y))

∣∣∣−1dy (29.9.52)

Recall everything is Borel measurable. I will consider one of the integrals in the sum. Forconvenience, replace E i

j with a compact set, Kij contained in it to obtain Borel measurability

in what follows. ∫Ki

j∩Adet(Df(x)Df(x)∗

)1/2 dx (29.9.53)

=∫

fi(

Kij∩A

) det(Df(g(y))Df(g(y))∗

)1/2|Dg(y)|dy

=∫

fi(

Kij∩A

) det(Df(g(y))Df(g(y))∗

)1/2∣∣∣detDfi (g(y))

∣∣∣−1dy

Now(

y1y2

)= fi (x) =

(f(x)xic

)for x ∈ Ki

j ∩A if and only if x is also in f−1 (y1) which

recall is a vector in Rn. Therefore, by 29.9.47, the above equals the following iteratedintegral.

=∫Rm

∫f−1(y1)∩Ki

j∩Adet(Df(g(y))Df(g(y))∗

)1/2∣∣detDfxi (g(y))∣∣−1 dy2dy1 (29.9.54)

where y1 = f(x) and y2 = xic . Since y1 is fixed in the inner integral of 29.9.54, and y1 =f(g(y)), and by definition gic

(fi (x)

)= y2, one can take the partial derivative of y1 =

f(g(y)) with respect to y2 to obtain

0 = Dxi f(g(y))Dy2gi (y)+Dxicf(g(y))Dy2gic (y)

= Dxi f(g(y))Dy2gi (y)+Dxicf(g(y)). (29.9.55)

Now consider the inner integral in 29.9.54 in which y1 is fixed. The integrand equals

det[(

Dxi f(g(y)) Dxicf(g(y))

)( Dxi f(g(y))∗

Dxicf(g(y))∗

)]1/2 ∣∣detDfxi (g(y))∣∣−1

. (29.9.56)

Let A ≡ Dxi f(g(y)) so A is m×m and B ≡ Dy2gi (y) an m× (n−m) , and using 29.9.55,29.9.56 is of the form

det[(

A −AB)( A∗

−B∗A∗

)]1/2

|detA|−1

= det [AA∗+ABB∗A∗]1/2 |detA|−1

= det [A(I +BB∗)A∗]1/2 |detA|−1 = det(I +BB∗)1/2

which, by Corollary 29.9.2, equals det(I +B∗B)1/2. (Note the size of the identity changesin these two expressions.) Since B = Dy2gi (y) and Dy2gic (y) = I, the above reduces to

det(I +B∗B)1/2 = det[(

B∗ I)( B

I

)]1/2

=