1030 CHAPTER 29. THE AREA FORMULA

which by the first part of the argument given above equals∫W

ψ i f ni ·wdH p−1 =∫

Wf w ·nidH p−1.

Thus for all f ∈C1c (O) , ∫

Wf w ·ndH p−1 =

∫W

f w ·nidH p−1 (29.7.36)

Since C1c (O) is dense in Cc (O) , the above equation is also true for all f ∈ Cc (O). Now

letting h ∈ Cc (W ) , the Tietze extension theorem implies there exists f1 ∈ C(O)

whoserestriction to W equals h. Let f be defined by

f1 (x)dist(x,OC

)dist(x,spt(h))+dist(x,OC)

= f (x) .

Then f = h on W and so this has shown that for all h ∈ Cc (W ) , 29.7.36 holds for h inplace of f . But as observed earlier, H p−1 is outer and inner regular on ∂U and so Cc (W )is dense in L1

(W,H p−1

)which implies w ·n(y) = w ·ni (y) for a.e. y. Considering a

countable dense subset of the unit sphere as above, this implies n(y) = ni (y) a.e. y. Thisproves |n(y)|= 1 a.e. and in fact n(y) can be computed by using the formula for ni (y).

It remains to prove Lemma 29.7.9.Proof of Lemma 29.7.9: Let h(x) = (x1, · · · ,xp−1,g(x1, · · · ,xp−1))

T

Dh(x) =

1 0...

. . ....

0 1g,x1 · · · g,xp−1

Then,

J∗ (x) =(det(Dh(x)∗Dh(x)

))1/2.

Therefore, J∗ (x) is the square root of the determinant of the following (p−1)× (p−1)matrix. 

1+(g,x1)2 g,x1 g,x2 · · · g,x1 g,xp−1

g,x2g,x1 1+(g,x2)2 · · · g,x2g,xp−1

.... . .

...g,xp−1g,x1 g,xp−1g,x2 · · · 1+(g,xp−1)

2

 . (29.7.37)

By Lemma 29.7.6, this determinant is 1+∑p−1i=1 (g,xi (x))

2 .Now Lemma 29.7.8 implies the divergence theorem.

Theorem 29.7.10 Let U be a bounded open set with a Lipschitz boundary which lies onone side of its boundary. Then if f ∈C1

c (Rp) ,∫U

f,k (x)dmp =∫

∂Uf nkdH p−1 (29.7.38)