248 CHAPTER 10. REGULAR MEASURES

Lemma 10.5.7 For a.e. x ∈ H,g(x) = |detDh(x)|.

Proof: First consider x such that |det(Dh(x))| ̸= 0. Then by Lemmas 10.5.5 and 10.5.6

limr→0

mp (h(B(x,r)∩H))

mp (B(x,r))= lim

r→0

mp (h(B(x,r)∩H))

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

mp (B(x,r))

=g(x)

|detDh(x)||detDh(x)|= g(x)

for a.e. x where |det(Dh(x))| ̸= 0.If |detDh(x)|= 0 then for r small enough,

1mp (B(x,r))

∫B(x,r)

gdmp =mp (h(B(x,r)∩H))

mp (B(x,r))

≤mp (h(x)+Dh(x)B(0,r)+B(0,εr))

mp (B(x,r))=

mp (Dh(x)B(0,r)+B(0,εr))mp (B(x,r))

Now Dh(x)B(0,r) + B(0,εr) has finite diameter and lies in a p− 1 dimensional sub-set. Therefore, from Theorem 10.3.4 on linear mappings, there is an orthogonal matrix Qpreserving all distances such that

|detQ|mp (Dh(x)B(0,r)+B(0,εr)) = mp (QDh(x)B(0,r)+B(0,εr))

where QDh(x)B(0,r) lies in a ball in Rp−1 of some radius r̂ = ∥Dh(x)∥r,. Thus the seton the right side is contained in a cylinder of radius r̂+ εr and height 2rε so its measure isno more than α p−1 (r̂+ rε)p−1 2εr for α p−1 = mp−1 (B(0,1)) . Thus,

1mp (B(x,r))

∫B(x,r)

gdmp ≤(∥Dh(x)∥+1)p

α p−1 (r+ rε)p−1 2εrα prp

= 2(∥Dh(x)∥+1)p α p−1

α p(1+ ε)p−1

ε

Since ε is arbitrary, for every Lebesgue point where |detDh(x)| = 0, it follows g = 0 =|detDh(x)| . ■

Here is the change of variables formula which follows from Lemma 10.5.4 now that ghas been identified.

Theorem 10.5.8 Let U be an open set and let h : U→h(U) be continuous and oneto one and differentiable on the measurable H ⊆U. Then if f ≥ 0 is Lebesgue measurable,∫

h(H)f (y)dmp =

∫H

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

10.6 Mappings Which are Not One to OneNow suppose h : U → V = h(U) and h is only C1, not necessarily one to one. Note thatI am using C1, not just differentiable. This makes it convenient to use the inverse functiontheorem. You can get more generality if you work harder. See my book “Real and AbstractAnalysis” for example. For

U+ ≡ {x ∈U : |detDh(x)|> 0}