1028 CHAPTER 29. THE AREA FORMULA

+∫

Ai

∫ gi(x1,··· ,xp−1)

−∞

Dp (ψ i f (x1, · · · ,xp−1,xp))wpdxpdx1 · · ·dxp−1

Doing the integrals using the observation that gi, j (x1, · · · ,xp−1) does not depend on xp, thisreduces further to∫

Ai

(ψ i f )(x1, · · · ,xp−1,xp)Ni (x1, · · · ,xp−1,gi (x1, · · · ,xp−1)) ·wdmp−1 (29.7.33)

where Ni (x1, · · · ,xp−1,gi (x1, · · · ,xp−1)) is given by

(−gi,1 (x1, · · · ,xp−1) ,−gi,2 (x1, · · · ,xp−1) , · · · ,−gi,p−1 (x1, · · · ,xp−1) ,1) . (29.7.34)

At this point I need a technical lemma which will allow the use of the area formula. Thepart of the boundary of U which is contained in Qi is the image of the map, hi (x1, · · · ,xp−1)given by (x1, · · · ,xp−1,gi (x1, · · · ,xp−1)) for (x1, · · · ,xp−1) ∈ Ai. I need a formula for

det(Dhi (x1, · · · ,xp−1)

∗Dhi (x1, · · · ,xp−1))1/2

.

To avoid interupting the argument, I will state the lemma here and prove it later.

Lemma 29.7.9

det(Dhi (x1, · · · ,xp−1)

∗Dhi (x1, · · · ,xp−1))1/2

=

√√√√1+p−1

∑j−1

gi, j (x1, · · · ,xp−1)2 ≡ J∗i (x1, · · · ,xp−1) .

Fory = (x1, · · · ,xp−1,gi (x1, · · · ,xp−1)) ∈ ∂U ∩Qi

and n defined by

ni (y) =1

J∗i (x1, · · · ,xp−1)Ni (y)

it follows from the description of J∗i (x1, · · · ,xp−1) given in the above lemma, that ni is aunit vector. All components of ni are continuous functions of limits of continuous func-tions. Therefore, ni is Borel measurable and so it is H p−1 measurable. Now 29.7.33reduces to ∫

Ai

(ψ i f )(x1, · · · ,xp−1,gi (x1, · · · ,xp−1))×

ni (x1, · · · ,xp−1,gi (x1, · · · ,xp−1)) ·wJ∗i (x1, · · · ,xp−1)dmp−1.

By the area formula this equals∫h(Ai)

ψ i f (y)ni (y) ·wdH p−1.