248 CHAPTER 10. CHANGE OF VARIABLES

10.3 Mappings Which are Not One to OneNow suppose h : U → V = h(U) and h is only C1, not necessarily one to one. Note that Iam using C1, not just differentiable. This makes it convenient to use the inverse functiontheorem. You can get more generality if you work harder. For

U+ ≡ {x ∈U : |detDh(x)|> 0}

and Z the set where |detDh(x)| = 0, Lemma 10.2.3 implies mp(h(Z)) = 0. For x ∈U+,the inverse function theorem implies there exists an open set Bx ⊆U+, such that h is one toone on Bx.

Let {Bi} be a countable subset of {Bx}x∈U+ such that U+ = ∪∞i=1Bi. Let E1 = B1. If

E1, · · · ,Ek have been chosen, Ek+1 = Bk+1 \∪ki=1Ei. Thus

∪∞i=1Ei =U+, h is one to one on Ei, Ei∩E j = /0,

and each Ei is a Borel set contained in the open set Bi. Now define

n(y)≡∞

∑i=1

Xh(Ei)(y)+Xh(Z)(y).

The sets h(Ei) ,h(Z) are measurable by Proposition 10.1.2. Thus n(·) is measurable.

Lemma 10.3.1 Let F ⊆ h(U) be measurable. Then∫h(U)

n(y)XF(y)dmp =∫

UXF(h(x))|detDh(x)|dmp.

Proof: Using Lemma 10.2.3 and the Monotone Convergence Theorem

∫h(U)

n(y)XF(y)dmp =∫

h(U)

 ∞

∑i=1

Xh(Ei)(y)+

mp(h(Z))=0︷ ︸︸ ︷Xh(Z)(y)

XF(y)dmp

=∞

∑i=1

∫h(U)

Xh(Ei)(y)XF(y)dmp

=∞

∑i=1

∫h(Bi)

Xh(Ei)(y)XF(y)dmp =∞

∑i=1

∫Bi

XEi(x)XF(h(x))|detDh(x)|dmp

=∞

∑i=1

∫U

XEi(x)XF(h(x))|detDh(x)|dmp

=∫

U

∞

∑i=1

XEi(x)XF(h(x))|detDh(x)|dmp

=∫

U+

XF(h(x))|detDh(x)|dmp =∫

UXF(h(x))|detDh(x)|dmp. ■

Definition 10.3.2 For y ∈ h(U), define a function, #, according to the formula

#(y)≡ number of elements in h−1(y).