29.7. THE DIVERGENCE THEOREM 1025

Therefore,(1+∑

pi=1 α2

i)

det(A(α1, · · · ,α p)) = det(B)2 = det(BT)2. However, using row

operations,

detBT = det

1 0 · · · 0 α10 1 0 α2...

. . ....

0 1 α p0 0 · · · 0 1+∑

pi=1 α2

i

= 1+p

∑i=1

α2i

and therefore, (1+

p

∑i=1

α2i

)det(A(α1, · · · ,α p)) =

(1+

p

∑i=1

α2i

)2

which shows det(A(α1, · · · ,α p)) =(1+∑

pi=1 α2

i).

Definition 29.7.7 A bounded open set, U ⊆ Rp is said to have a Lipschitz boundary andto lie on one side of its boundary if the following conditions hold. There exist open boxes,Q1, · · · ,QN ,

Qi =p

∏j=1

(ai

j,bij)

such that ∂U ≡U \U is contained in their union. Also, for each Qi, there exists k and aLipschitz function, gi such that U ∩Qi is of the form x : (x1, · · · ,xk−1,xk+1, · · · ,xp) ∈∏

k−1j=1

(ai

j,bij

)×

∏pj=k+1

(ai

j,bij

)and ai

k < xk < gi (x1, · · · ,xk−1,xk+1, · · · ,xp)

 (29.7.29)

or else of the form x : (x1, · · · ,xk−1,xk+1, · · · ,xp) ∈∏k−1j=1

(ai

j,bij

)×

∏pj=k+1

(ai

j,bij

)and gi (x1, · · · ,xk−1,xk+1, · · · ,xp)< xk < bi

j

 (29.7.30)

The function, gi has a derivative on Ai ⊆∏k−1j=1

(ai

j,bij

)×∏

pj=k+1

(ai

j,bij

)where

mp−1

(k−1

∏j=1

(ai

j,bij)×

p

∏j=k+1

(ai

j,bij)\Ai

)= 0.

Also, there exists an open set, Q0 such that Q0 ⊆ Q0 ⊆U and U ⊆ Q0∪Q1∪·· ·∪QN .

Note that since there are only finitely many Qi and each gi is Lipschitz, it follows froman application of Lemma 29.1.1 that H p−1 (∂U)< ∞. Also from Lemma 29.7.4 H p−1 isinner and outer regular on ∂U . In the following, dx will be used in place of dmp to conformwith more standard notation from calculus.