1072 CHAPTER 30. INTEGRATION OF DIFFERENTIAL FORMS

Next let I always denote an increasing list of indices. Note that S j ◦S−1i maps the open

set A to an open set which therefore has positive Lebesgue measure. It follows from thearea formula that

det(D(S j ◦S−1

i))

= det

n×m︷ ︸︸ ︷

D(S j(S−1

i (u))) m×n︷ ︸︸ ︷

DS−1i (u)

 (30.4.20)

must be nonzero on a set of positive measure. It follows that at least some

∂(xi1 · · ·xin−1

)∂ (u2, · · · ,un)

must be nonzero since by the Binet Cauchy theorem, the above determinant in 30.4.20 isthe sum of products of these multiplied by other determinants which come from deletingcorresponding columns in the matrix for D

(S j(S−1

i (u)))

. It follows that

∑I

(∂(xi1 · · ·xin−1

)∂ (u2, · · · ,un)

)2

is positive on a set of positive measure. Let

limp→∞

aI p ◦S−1j =

∂(xi1 · · ·xin−1

)∂ (u2, · · · ,un)

in L2(S j ◦S−1

i (A))

for each I = (i1, · · · , in−1) . Replacing aI ◦S−1j with aI p ◦S−1

j in 30.4.19and passing to the limit, it follows

0 = limp→∞

∫S j◦S−1

i (A)∑

IaI p ◦S−1

j (u1)∂(xi1 · · ·xin−1

)∂ (u2, · · · ,un)

du1

=∫

S j◦S−1i (A)

∑I

(∂(xi1 · · ·xin−1

)∂ (u2, · · · ,un)

)2

du1 > 0

a contradiction. Therefore, d(u1,A,S j ◦S−1

i

)= 1 and this shows the atlas is an oriented

atlas for ∂Ω. This has proved a general Stokes theorem.

Theorem 30.4.1 Let Ω be an oriented Lipschitz manifold and let

ω = ∑I

aI (x)dxi1 ∧·· ·∧dxin−1 .

where each aI is C1(Ω). For

{U j,R j

}pj=0 an oriented atlas for Ω where R j (U j) is a

relatively open set in{u ∈ Rn : u1 ≤ 0} ,