29.9. THE COAREA FORMULA 1041

Proof: First note that det(Df(x)Df(x)∗

)=∑i∈Λ(n,m) det

(Dfi (x)

)2 by the Binet Cauchytheorem. Let S ≡ {x : J∗ (x) = 0}. For each i, fi ({x : det

(Dfi (x)

)= 0})

has measurezero due to Sard’s theorem and so it will follow from the argument presented below thatSi ≡ fi ({x : det

(Dfi (x)

)= 0})

has measure zero. Thus Si \S can be neglected. For N ≡{x : Df(x) does not exist} ,mn (N) = 0 by Rademacher’s theorem.Thus in what follows, wecan always assume that either Dfi (x) does not exist or det

(Dfi (x)

)exists and is not 0. This

will be clear from the argument. Let A be a closed subset of Rn \ {S∪N}. By Lemma

29.4.1, there exist disjoint Borel measurable sets{

F ij

}∞

j=1such that fi is one to one on F i

j ,(fi)−1 is Lipschitz on fi

(F i

j

), and

∪∞j=1F i

j ={

x : Dfi (x) exists anddetDfi (x) ̸= 0}.

If x ∈ Rn \{S∪N}, it follows x ∈ F ij for some i and j. Hence ∪i, jF i

j ⊇ A.

Now let{

E ij

}be measurable sets such that E i

j ⊆ F ik for some k, the sets E i

j are disjoint,

and their union coincides with ∪i, jF ij . Let g :Rn→Rn be a Lipschitz function which equals(

fi)−1 on fi(

E ij

). I am supressing the dependence on i. Then for any x ∈ E i

j,g(fi (x)

)= x.

In particular, gic(fi (x)

)= xic where

gi (y)≡(

gi1 (y) · · · gim (y))T

for i≡ (i1, · · · , im) with gic (y) defined similarly and x ∈ E ij, with

y≡(

y1y2

)≡(

f(x)xic

)≡ fi (x) ∈ fi

(E i

j

),

xi = gi

(fi (x)

), y2 ≡ xic = gic

(fi (x)

)(29.9.49)

Then, by definition, ∫A

J∗ (x)dx≡∫

Adet(Df(x)Df(x)∗

)1/2 dx (29.9.50)

First, using Theorem 29.8.2, and the fact that Lipschitz mappings take sets of measure zeroto sets of measure zero, replace E i

j with Ẽ ij ⊆ E i

j such that E ij \ Ẽ i

j has measure zero and

Dfi (g(y))Dg(y) = I, |det(Dg(y))|=∣∣∣detDfi (g(y))

∣∣∣−1(29.9.51)

on fi(

Ẽ ij

). Changing the variables using the area formula and 29.9.51, the expression in

29.9.50 equals∫A

J∗ (x)dx =∞

∑j=1

∑i ∈Λ(n,m)

∫Ẽ i

j∩A

(det(Df(x)Df(x)∗

))1/2 dx

=∞

∑j=1

∑i ∈Λ(n,m)

∫E i

j∩A

(det(Df(x)Df(x)∗

))1/2 dx

29.9. THE COAREA FORMULA 1041Proof: First note that det (Df (x) Df (x)*) = Yj- A(nym) det (Df (x))° by the Binet Cauchytheorem. Let S = {x:J*(x) =O}. For each i, fi({x: det (Dfi(x)) =0}) has measurezero due to Sard’s theorem and so it will follow from the argument presented below thatSi=fi({x: det (Dfi(x)) =0}) has measure zero. Thus S! \ S can be neglected. For N ={x : Df (x) does not exist} ,m, (N) =0 by Rademacher’s theorem.Thus in what follows, wecan always assume that either Df' (x) does not exist or det (Df' (x)) exists and is not 0. Thiswill be clear from the argument. Let A be a closed subset of R” \ {SUN}. By Lemma29.4.1, there exist disjoint Borel measurable sets {Fi iy ; such that fi is one to one on Fi,jc(f') lis Lipschitz on fi (Fi) , andUF} = {x : Dfi (x) exists and det Dfi (x) 4 of .Ifx € R"\ {SUN}, it follows x € F} for some i and j. Hence Uj jF} 2 A.Now let {ei} be measurable sets such that Et Cc Fi for some k, the sets Ek are disjoint,and their union coincides with Uj, ;F i: Let g: R” — R" be a Lipschitz function which equals(f') ~ on fi (zi). I am supressing the dependence on i. Then for any x € E},g (ff (x)) =x.In particular, gj, (fi (x)) = xj, wheregi(y)=( en (y) - gin (y) )”for i= (i1,--* ,im) with gj, (y) defined similarly and x € E', with_(y \_( €®) \_g¢ i (pi=(3)- (82 otnertXj = Qj (f (x)) ,¥2 =X, = gi, (f (x)) (29.9.49)Then, by definition,| J*(x)dx= / det (Df (x) Df (x)*)/” dx (29.9.50)A JAFirst, using Theorem 29.8.2, and the fact that Lipschitz mappings take sets of measure zeroto sets of measure zero, replace Ej with E¥ C E¥ such that E¥ \E; has measure zero andpf (g(y)) Dg(y) =1, |det (Dg (y ))| = [de ( g(y )) (29.9.51)on fi (21). Changing the variables using the area formula and 29.9.51, the expression in29.9.50 equals|e (x)dx =41syf (det (Df (x ) Df (x)*)) 7 axBinaiIlfayi €A(n,m)IP18© es! (det (DE (x) Df(x)*))"/? dxi €A(n,m) Ejnll