334 CHAPTER 11. REGULAR MEASURES

Proof: Since the Ei are disjoint and h is one to one. λ (∪iEi) ≡ mp (h(∪iEi∩H)) =

∑i mp (h(Ei∩H)) = ∑i λ (Ei). If mp (E) = 0, then λ (E)≡mp (h(E ∩H)) = 0 because ofLemma 11.8.1. ■

Since λ ≪mp, it follows from the Radon Nikodym theorem of Corollary 10.13.14 thatthere exists g ∈ L1

loc (U) such that for F a measurable subset of U,

λ (F) = mp (h(F ∩H)) =∫

Fgdmp (11.11)

where g = 0 off H. To see that this corollary applies, note that both λ and mp are finite oncompact sets and that every open set is a countable union of compact sets.

Now let F be a Borel set so that h−1 (F)∩H is measurable and plays the role of F inthe above. Then

λ(h−1 (F)

)≡ mp

(h(h−1 (F)∩H

))=∫

UXh−1(F)∩H (x)g(x)dmp (x) =

∫H

XF (h(x))g(x)dmp (x)

Thus also for s a Borel measurable nonnegative simple function,∫h(H)

s(y)dmp (y) =∫

Hs(h(x))(x)g(x)dmp (x)

Using a sequence of nonnegative simple functions to approximate a nonnegative Borelmeasurable f , we obtain from the monotone convergence theorem that∫

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

∫H

f (h(x))(x)g(x)dmp (x)

If f is only Lebesgue measurable, then there are nonnegative Borel measurable functionsk, l such that k (y)≤ f (y)≤ l (y) with equality holding off a set of mp measure zero. Thenk (h(x))g(x) ≤ f (h(x))g(x) ≤ l (h(x))g(x) and the two on the ends are Lebesguemeasurable which forces the function in the center to also be Lebesgue measurable bycompleteness of Lebesgue measure because∫

Hl (h(x))g(x)− k (h(x))g(x)dmp =

∫h(H)

l (y)dmp−∫h(H)

k (y)dmp

=∫h(H)

f (y)dmp−∫h(H)

f (y)dmp = 0

Thus l (h(x))g(x)− k (h(x))g(x) = 0 a.e. Then for f nonnegative and Lebesgue mea-surable, ∫

Hf (h(x))g(x)dmp =

∫h(H)

f (y)dmp.

This shows the following lemma.

Lemma 11.9.4 Let h : U→h(U) be continuous, U open, and let H ⊆U be measurableand h is one to one and differentiable on H. Then there exists nonnegative measurableg ∈ L1

loc such that whenever f is nonnegative and Lebesgue measurable,∫h(H)

f (y)dmp =∫

Hf (h(x))g(x)dmp

where all necessary measurability is obtained.

334 CHAPTER 11. REGULAR MEASURESProof: Since the £; are disjoint and h is one to one. A (UjE;) = mp (h (UjE;NA)) =Limp (R(E;NA)) = YA (Ej). If mp (E) = 0, then A (E) = m, (h(E MA)) = 0 because ofLemma 11.8.1. HfSince A < mp, it follows from the Radon Nikodym theorem of Corollary 10.13.14 thatthere exists g € L},.(U) such that for F a measurable subset of U,2 (F) =mp(h(FAH)) )= | sam, (11.11)where g = 0 off H. To see that this corollary applies, note that both A and my are finite oncompact sets and that every open set is a countable union of compact sets.Now let F be a Borel set so that h~! (F ) AH is measurable and plays the role of F inthe above. ThenA (hk! (F)) =m, " (h! (F) mu_ [ Burry (@)8 (@) dnp (x =| 2p (h(a) g (aw) dmy (x)Thus also for s a Borel measurable nonnegative simple function,I py an) = [, sea) (@)(@) dim ()Using a sequence of nonnegative simple functions to approximate a nonnegative Borelmeasurable f, we obtain from the monotone convergence theorem that[hf Damo) = [, Fa) (@) (a) Amp)If f is only Lebesgue measurable, then there are nonnegative Borel measurable functionsk,l such that k(y) < f(y) </(y) with equality holding off a set of m, measure zero. Thenk(h(x)) g(x) < f(h(x)) g(x) < 1(h(ax)) g(x) and the two on the ends are Lebesguemeasurable which forces the function in the center to also be Lebesgue measurable bycompleteness of Lebesgue measure because[1 (@))8(@) —k(h(@)) g(@)dm,| Ce CO[hg foarre~ Ji yf am =0Thus /(h(a)) g(a) —k(h(ax)) g(x) =O ae. Then for f nonnegative and Lebesgue mea-surable,[fo @ala)amy = J. foamThis shows the following lemma.Lemma 11.9.4 Leth:U sh (U) be continuous, U open, and let H CU be measurableand h is one to one and differentiable on H. Then there exists nonnegative measurablegE Li . such that whenever f is nonnegative and Lebesgue measurable,how f(y Jdmy = | faa «)dmywhere all necessary measurability is obtained.