29.7. THE DIVERGENCE THEOREM 1027
Letting xp = y+gi (x1, · · · ,xp−1) and changing the variable, this equals
=∫
Bi
∫ 0
−∞
p
∑j=1
D j (ψ i f )(x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)) ·
w jdydx1 · · ·dxp−1
=∫
Ai
∫ 0
−∞
p
∑j=1
D j (ψ i f )(x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)) ·
w jdydx1 · · ·dxp−1
Recall Ai is all of Bi except for the set of measure zero where the derivative does not exist.Also D j refers to the partial derivative taken with respect to the entry in the jth slot. In thepth slot is found not just xp but y+ gi (x1, · · · ,xp−1) so a differentiation with respect to x jwill not be the same as D j. In fact, it will introduce another term involving gi, j. Thus fromthe chain rule,
=∫
Ai
∫ 0
−∞
p−1
∑j=1
∂
∂x j(ψ i f (x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)))w j−
Dp (ψ i f )(x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)) ·gi, j (x1, · · · ,xp−1)w jdydx1 · · ·dxp−1
+∫
Ai
∫ 0
−∞
Dp (ψ i f )(x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1))wpdydx1 · · ·dxp−1 (29.7.32)
Consider the term∫Ai
∫ 0
−∞
p−1
∑j=1
∂
∂x j(ψ i f (x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)))w jdydx1 · · ·dxp−1
This equals
∫Bi
∫ 0
−∞
p−1
∑j=1
∂
∂x j(ψ i f (x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1)))w jdydx1 · · ·dxp−1,
and now interchanging the order of integration and using the fact that spt(ψ i) ⊆ Qi, itfollows this term equals zero. The reason this is valid is that
x j→ ψ i f (x1, · · · ,xp−1,y+gi (x1, · · · ,xp−1))
is the composition of Lipschitz functions and is therefore Lipschitz. Therefore, this func-tion can be recovered by integrating its derivative, Lemma 26.2.6.
Then, changing the variable back to xp it follows 29.7.32 reduces to
−∫
Ai
∫ gi(x1,··· ,xp−1)
−∞
∑p−1j=1 Dp (ψ i f )(x1, · · · ,xp−1,xp)
·gi, j (x1, · · · ,xp−1)w j
dxpdx1 · · ·dxp−1