246 CHAPTER 10. CHANGE OF VARIABLES

and all needed measurability holds. Now for E an arbitrary measurable subset of H letF ⊆ E ⊆ G where F is Fσ and G is Gδ and mp (G\F) = 0. Without loss of generality,assume G⊆U . Then∫

H(XG (h(x)) |det(Dh(x))|−XF (h(x)) |det(Dh(x))|)dmp = 0

because the integral of each function in the difference equals∫h(H)

XE (y)dmp =∫

h(H)XF (y)dmp =

∫h(H)

XG (y)dmp

and so the integrands are equal off a set of measure zero. Furthermore,

XE (h(x)) |det(Dh(x))| ∈ [XF (h(x)) |det(Dh(x))| ,XG (h(x)) |det(Dh(x))|]

and so XE (h(x)) |det(Dh(x))| equals a measurable function a.e. By completeness of themeasure mp it follows that x→XE (h(x)) |det(Dh(x))| is also measurable. I am not sayingthat x→XE (h(x)) is measurable. In fact this might not be so because it is Xh−1(E) (x) andthe inverse image of a measurable set is not necessarily measurable. For a well knownexample, see Problem 26 on Page 156. The thing which is measurable is the product in theintegrand. Therefore, for E a measurable subset of H∫

h(H)XE (y)dmp =

∫H

XE (h(x)) |det(Dh(x))|dmp (10.3)

The following theorem gives a change of variables formula.

Theorem 10.2.2 Let U ⊆ Rp be open, h : U → Rp continuous, and

mp (h(U \H)) = 0

where H ⊆U and H is Lebesgue measurable. Suppose also that h is differentiable on Hand is one to one on H. Then h(H) is Lebesgue measurable and if g ≥ 0 is Lebesguemeasurable, then ∫

h(H)g(y)dmp =

∫H

g(h(x)) |det(Dh(x))|dmp (10.4)

and all needed measurability holds.

Proof: Formula 10.3 implies that 10.4 holds for any nonnegative simple function s.Then for g nonnegative and measurable, it is the pointwise increasing limit of such simplefunctions. Therefore, 10.4 follows from the monotone convergence theorem. ■

Note that the above theorem holds if H = U . One might wonder why the fuss overhaving a separate H on which h is differentiable. One reason for this is Rademacher’stheorem which states that every Lipshitz continuous function is differentiable a.e. Thus ifyou have a Lipshitz function defined on U an open set, then if you let H be the set wherethis function is differentiable, it will follow that U \H has measure zero and so, by Problem9 on Page 212, you also have h(U \H) has measure zero. Thus the above theorem is atleast as good as what is needed to give a change of variables formula for transformationswhich are only Lipschitz continuous.

Next is a significant result called Sard’s lemma. In the proof, it does not matter whichnorm you use in defining balls but it may be easiest to consider the norm

∥x∥ ≡max{|xi| , i = 1, · · · , p}