10.6. MAPPINGS WHICH ARE NOT ONE TO ONE 249

and Z the set where |detDh(x)| = 0, Lemma 10.4.3 implies mp(h(Z)) = 0. For x ∈U+,the inverse function theorem implies there exists an open set Bx ⊆U+, such that h is oneto one 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 10.4.1. Thus n(·) is measurable.

Lemma 10.6.1 Let F ⊆ h(U) be measurable. Then∫h(U)

n(y)XF(y)dmp =∫

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

Proof: Using Lemma 10.4.3 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 10.6.2 For y ∈ h(U), define a function, #, according to the formula

#(y)≡ number of elements in h−1(y).

Observe that#(y) = n(y) a.e. (10.7)

because n(y) = #(y) if y /∈h(Z), a set of measure 0. Therefore, # is a measurable functionbecause of completeness of Lebesgue measure.