30.6. THE DIVERGENCE THEOREM 1077

where the hat means dxk is being left out. Here it is assumed Fk is the restriction to Ω of afunction in W 1,p (Rn) where p > 1.Then

dω (x) =n

∑k=1

n

∑j=1

∂Fk

∂x j(−1)k−1 dx j ∧dx1∧·· ·∧ d̂xk ∧·· ·∧dxn

=n

∑k=1

∂Fk

∂xkdx1∧·· ·∧dxk ∧·· ·∧dxn

≡ div(F)dx1∧·· ·∧dxk ∧·· ·∧dxn

The assertion between the first and second lines follows right away from properties ofdeterminants and the definition of the integral of the above wedge products in terms ofdeterminants. From Green’s theorem and the change of variables formula applied to theindividual terms in the description of

∫Ω

dω∫Ω

div(F)dx =

p

∑j=1

∫B j

n

∑k=1

(−1)k−1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)

(ψ jFk

)◦R−1

j (0,u2, · · · ,un)du1,

du1 short for du2du3 · · ·dun. Also, this shows the result on the right of the equal sign doesnot depend on the choice of partition of unity or on the atlas.

I want to write this in a more attractive manner which will give more insight in termsof the Hausdorff measure on ∂Ω. The above involves a particular partition of unity, thefunctions being the ψ i. Replace F in the above with ψsF. Next let

{η j}

be a partition ofunity η j ≺ O j such that ηs = 1 on sptψs. This partition of unity exists by Lemma 30.3.4.Then ∫

Ω

div(ψsF)dx =

p

∑j=1

∫B j

n

∑k=1

(−1)k−1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)

(η jψsFk

)◦R−1

j (0,u2, · · · ,un)du1

=∫

Bs

n

∑k=1

(−1)k−1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)(ψsFk)◦R−1

s (0,u2, · · · ,un)du1 (30.6.21)

because since ηs = 1 on sptψs, it follows all the other η j equal zero there. Consider thevector defined for u1 ∈ Rs (Ws)∩Rn

0 whose kth component is

(−1)k−1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)= (−1)k+1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)(30.6.22)

Suppose you dot this vector with a “tangent” vector ∂R−1s /∂ui. For each j this yields

∑j(−1)k+1 ∂ (x1, · · · x̂k · · · ,xn)

∂ (u2, · · · ,un)

∂xk

∂ui= 0