Kenneth Kuttler

460 CHAPTER 17. THE AREA FORMULA

so, XH (h(x))det(U (x))−XF (h(x))det(U (x)) = 0 a.e. By completeness of Lebesguemeasure, x→XE (h(x))det(U (x))XEk (x) is Lebesgue measurable and∫

h(Ek)XE (y)dH n =

∫h(Ek)

XF (y)dH n =∫

XF (h(x))det(U (x))dx

=∫

XE (h(x))det(U (x))dx

Using the above argument, we can add these over k and obtain∫h(G)

#(y)XE (y)dH n =∫h(A)

#(y)XE (y)dH n =∫h(A+)

#(y)XE (y)dH n

=∫

A+XE (h(x))det(U (x))dx

=∫

AXE (h(x))det(U (x))dx

=∫

GXE (h(x))det(U (x))dx (17.13)

because G\A has measure zero and so does h(G\A) and det(U (x)) = 0 on A\A+. Also,from 17.10, h(A\A+) has measure zero. ■

Note that H n (G\A+) = 0 so it suffices to let #(y) simply be the number of points inh−1 (y). This has almost shown the following theorem.

Definition 17.3.2 To save on notation, I will denote det(U (x)) as J∗ (x).

Theorem 17.3.3 Suppose h : G→ Rm is Lipschitz, G some open set, and let A bethe Borel set on which Dh exists with mp (G\A) = 0.(Rademacher’s theorem). Then ifg≥ 0 and is H n measurable,∫

h(G)#(y)g(y)dH n =

∫G

g(h(x))J∗ (x)dmn.

and everything makes sense where here #(y) is defined as the number of times h hits yfrom points in A+ or G. If h is one to one on A+, we can replace #(y) with 1.

Proof: From 17.13 one can assert this holds for H n measurable simple functionsand then, passing to a limit with monotone convergence theorem, one obtains the abovetheorem. ■

Next is an interesting version of the chain rule for Lipschitz maps. The proof of thistheorem is based on the following lemma.

Lemma 17.3.4 If h : Rn→ Rn is Lipschitz, then if h(x) = 0 for all x ∈ A, then

det(Dh(x)) = 0 a.e.x ∈ A

Proof: By the area formula, 0 =∫{0} #(y)dy =

∫A |det(Dh(x))|dx, and so it follows

that det(Dh(x)) = 0 a.e. ■

Theorem 17.3.5 Let f , g be Lipschitz mappings from Rn to Rn with g (f (x)) = xon A, a measurable set. Then for a.e. x ∈ A, Dg (f (x)), Df (x), and D(g ◦f)(x) allexist and I = D(g ◦f)(x) = Dg (f (x))Df (x) .

460 CHAPTER 17. THE AREA FORMULAso, 2p (h(ax)) det (U (a)) — 2 (h(ax)) det (U (x)) =0a.e. By completeness of Lebesguemeasure, x + Zz (h(a)) det (U (x)) Ze, (a) is Lebesgue measurable andKel(ydH” = Xe (y dt" = | Mp (h(w)) det(U (w)) dxh(E) h(E) EXz (h(a)) det (U (x)) dxExUsing the above argument, we can add these over & and obtain[ gtoremae = | te) rewan=[ #y) %e(yare= | 2%e(h(w))det(U (@)) dx- [sens )) det (U (a)) dx- [ (ne )) det (U (x)) dx (17.13)because G\A has measure zero and so does h (G\ A) and det (U (x)) =0 on A\A™. Also,from 17.10, h(A\ AT) has measure zero. lilNote that .#" (G\ A‘) = 0 so it suffices to let #(y) simply be the number of points inh~!(y). This has almost shown the following theorem.Definition 17.3.2 10 save on notation, I will denote det (U (x)) as J. (a).Theorem 17.3.3 Suppose h : G+ R"” is Lipschitz, G some open set, and let A bethe Borel set on which Dh exists with mp(G\ A) = 0.(Rademacher’s theorem). Then ifg>O0and is #" measurable,hig #(y)g(y)d HO" = [she )) Ja (a) dim.and everything makes sense where here #(y) is defined as the number of times h hits yfrom points in A* or G. If h is one to one on A‘, we can replace #(y) with 1.Proof: From 17.13 one can assert this holds for #” measurable simple functionsand then, passing to a limit with monotone convergence theorem, one obtains the abovetheorem.Next is an interesting version of the chain rule for Lipschitz maps. The proof of thistheorem is based on the following lemma.Lemma 17.3.4 [fh :R” — R" is Lipschitz, then if h(a) = 0 for all x € A, thendet (Dh(x))=Oa.e.aEAProof: By the area formula, 0 = fro, #(y)dy = J, |det (Dh (a))| dx, and so it followsthat det (Dh (x)) =Oa.e.Theorem 17.3.5 Let f, g be Lipschitz mappings from R" to R" with g (f(@))=aon A, a measurable set. Then for ae. x € A, Dg(f (x)), Df (a), and D(go f) (x) allexist and I = D(go f)(a) =Dg(f (x))Df (x).