Kenneth Kuttler

338 CHAPTER 11. REGULAR MEASURES

and one of those points c and look for all b, if any, which do the right things. There arecountably many of these pieces of B being denoted as E (T,c, i).

The union of these E (T,c, i) is all of B because if b ∈ B,

|h(a)−h(b)−Dh(b)(a−b)|< ε |Dh(b)(a−b)| (11.14)

whenever a ∈ B(b, 2

)provided i is sufficiently large. Thus also, by Lemma 5.3.1, there

is T ∈S such that the above holds for Dh(b) replaced with T and a ∈ B(b, 2

)and also

11.13. Thus b ∈ E (T,c, i), so indeed the union of these sets is B.Now let a,b ∈ E (T,c, i) . Since a,b ∈ E (T,c, i) , a,b are within 1/i of c and so a is

within 2/i of b and so 11.12 holds because of the definition of E (T,c, i). Therefore, from11.12 and the inequalities which follow, 11.13,

(1−3ε) |T (a−b)| ≤ |h(a)−h(b)| ≤ (1+3ε) |T (a−b)| (11.15)

Indeed from 11.14 and 11.13

|h(a)−h(b)| ≤ (1+ ε) |Dh(b)(a−b)| ≤ (1+ ε)2 |T (a−b)| ≤ (1+3ε) |T (a−b)|

The bottom inequality is similar. Thus h is one to one on E (T,c, i). Now enumerate theseBorel sets {Ei}∞

i=1. Let F1 = E1 and if F1, ...,Fm have been chosen, let Fm+1 ≡ Em+1 \(∪m

i=1Fi). ■

Theorem 11.10.2 Let U be an open set and let h : U → h(U) be continuous anddifferentiable on the measurable H ⊆ U such that h(U \H) has measure zero. Then iff ≥ 0 is Lebesgue measurable,∫

h(H)#(y) f (y)dmp =

∫H

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

where #(y) is the number of elements of h−1 (y) in U.

Proof: Let {Fi} be the Borel sets of Lemma 11.10.1 whose union equals H \Z whereZ is the set where Dh(x) exists but is not invertible and h one to one on each Fi. Thus forf Lebesgue measurable,

∫h(H\Z)Xh(Fi) f (y)dmp =

∫Fi

f (h(x)) |det(Dh(x))|dmp. Letn(y)≡ ∑i Xh(Fi) (y). Then, adding these yields,∫

h(H\Z)n(y) f (y)dmp =

∫H\Z

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

Now #(y) = n(y) except for h(U \H)∪h(Z) which is a set of mp measure zero byassumption and by Sard’s Lemma, Lemma 11.8.3 for h(Z). Therefore,∫

h(H)#(y) f (y)dmp =

∫h(H\Z)

#(y) f (y)dmp =∫

H\Zf (h(x)) |det(Dh(x))|dmp

=∫

Hf (h(x)) |det(Dh(x))|dmp ■

h is one to one when #(y)= 1 and in this case we get the usual change of variables formula.

338 CHAPTER 11. REGULAR MEASURESand one of those points c and look for all b, if any, which do the right things. There arecountably many of these pieces of B being denoted as E (T,c,i).The union of these E (T,c,i) is all of B because if b € B,|h (a) —h(b) —Dh(b) (a—b)| < €|Dh(b) (a—b)| (11.14)whenever a € B (0, 2) provided i is sufficiently large. Thus also, by Lemma 5.3.1, thereis T € Y such that the above holds for Dh (b) replaced with T and a € B (0, 2) and also11.13. Thus b € E (T,c,7), so indeed the union of these sets is B.Now let a,b € E(T,c,i). Since a,b € E(T,c,i), a,b are within 1/i of c and so a iswithin 2/i of b and so 11.12 holds because of the definition of E (T,c,i). Therefore, from11.12 and the inequalities which follow, 11.13,(1 —3¢)|T (a—b)| < |h(a) —h(b)| < (1+3e) |T (a—5)| (11.15)Indeed from 11.14 and 11.13|h (a) —h(b)| < (1+¢) |Dh(b) (a—b)| < (1+)? |T (a—b)| < (1+3e) |T (a—b)|The bottom inequality is similar. Thus h is one to one on E (T,c,i). Now enumerate theseBorel sets {E;};,. Let Fj = E; and if F,,..., Fj, have been chosen, let Fn41 = Em+1 \(umf).Theorem 11.10.2 Let U be an open set and let h: U —> h(U) be continuous anddifferentiable on the measurable H CU such that h(U \H) has measure zero. Then iff = 0 is Lebesgue measurable,in H) #(y) f(y)dmy =f f (h(a)) |det (Dh (a))|dm,where #(y) is the number of elements of h~' (y) in U.Proof: Let {F;} be the Borel sets of Lemma 11.10.1 whose union equals H \ Z whereZ is the set where Dh (x) exists but is not invertible and h one to one on each F;. Thus forf Lebesgue measurable, fh,(4\z) Zam (y)dmp = Jy, f (h(x)) |det (Dh (x))|dmp. Letn(y) =Yi Zac) (y)- Then, adding these yields,| [ ing DF wan = [ _F((@))|det(Dh (a) dm,Now #(y) =n(y) except for h(U\H) Uh(Z) which is a set of mp measure zero byassumption and by Sard’s Lemma, Lemma 11.8.3 for h (Z). Therefore,[tsar = Lanat ore )diny = [| F0o(@))det(Dh (x) amy[fre )) \det(Dh (a))|dm,his one to one when #(y) = | and in this case we get the usual change of variables formula.