10.2. CHANGE OF VARIABLES NONLINEAR MAPS 245

and so, using Theorem 10.1.4 again,

mp (h(B(x,r)))≥ |det(Dh(x))|mp (B(0,r (1− ε)))

for r small enough. Thus, since

mp (h(B(x,r))) = mp (h(B(x,r)∩H)) = λ (B(x,r)) =∫

B(x,r)f (x)dmp,

it follows that

|det(Dh(x))|mp (B(0,r (1− ε))) ≤∫

B(x,r)f (x)dmp

≤ |det(Dh(x))|mp (B(0,r (1+ ε)))

for r small enough. Now divide by mp (B(x,r)) and use the fundamental theorem of calcu-lus Corollary 9.5.3 to find that for ε small ε > 0,

|det(Dh(x))|(1− ε)p ≤ f (x)≤ |det(Dh(x))|(1+ ε)p a.e.

Letting εk → 0 and picking a set of measure zero for each εk, it follows that off a set ofmeasure zero f (x) = |det(Dh(x))| .

If Dh(x)−1 does not exist, then you have

h(B(x,r))−h(x)⊆ Dh(x)B(0,r)+B(0,εr)

and Dh(x) maps into a bounded subset of a p−1 dimensional subspace. Therefore, usingProposition 10.1.5, the right side has measure no more than an expression of the form Crpε,C depending on Dh(x). Therefore, in this case,

limr→0

1mp (B(x,r))

∫B(x,r)

f (x)dmp = limr→0

mp (h(B(x,r)))mp (B(x,r))

≤ εCrp

α prp

and since ε is arbitrary, this shows that for a.e. x, such that Dh(x)−1 does not exist, f (x) =0 = |det(Dh(x))| in this case also. Therefore, whenever E is a Lebesgue measurable setE ⊆ H,

mp (h(E)) =∫

h(H)Xh(E) (y)dmp =

∫U

XE (x) |det(Dh(x))|dmp

=∫

HXE (x) |det(Dh(x))|dmp

The difficulty here is that the inverse image of a Lebesgue measurable set might not bemeasurable. However, there is no problem with the inverse image of a Borel set. Let F bea Borel subset of the measurable set h(H) . Then h−1 (F) is measurable. Indeed, h−1 (F)is open if F is open. If B consists of the sets h−1 (F) were F is Borel, B is a σ algebrawhich contains the open sets. Thus B contains the Borel sets. Then from what was justshown, ∫

h(H)Xh(h−1(F)) (y)dmp =

∫H

Xh−1(F) (x) |det(Dh(x))|dmp

and rewriting this gives∫h(H)

XF (y)dmp =∫

HXF (h(x)) |det(Dh(x))|dmp