17.5. THE COAREA FORMULA 463

Now let ic ∈Λ(n,n−m) consist of the remaining indices taken in order where i∈Λ(n,m) .For i= (i1, · · · , im), define xi ≡ (xi1 , ...,xim) and xic to be the other components of x taken

in order. Then let f i (x)≡(f (x)xic

). Thus there are C (n,n−m) =C (n,m) different f i

featuring C (n,m) different xi,xic .

Example 17.5.2 Say f : R4→ R2. Here are some examples for f i:f1 (x1,x2,x3,x4)f2 (x1,x2,x3,x4)

x2x4

 ,

f1 (x1,x2,x3,x4)f2 (x1,x2,x3,x4)

x1x2

 ,

f1 (x1,x2,x3,x4)f2 (x1,x2,x3,x4)

x3x4

Suppose first that xic =

(xm+1 · · · xn

)T so

f i (x) =

(f (x)xic

), Df (x) =

(Dxi

f (x) Dxicf (x)

0 I

)and so from row operations, detDf i (x) = detDxi

f (x) . It is similar in the general caseexcept one might have a sign change which is not important in what follows. So

detDf i (x) = detDxif (x) . (17.16)

Earlier with the area formula, we integrated J∗ (x) ≡ det(Df (x)∗Df (x)

)1/2. With

the coarea formula, we integrate J∗ (x) ≡ det(Df (x)Df (x)∗

)1/2. This proof involvesdoing this integration and seeing what happens. In case n = m the claim of the theoremwill follow from the area formula because H 0 (E) is the number of elements of E, so onecan assume if desired that in what follows n > m although the argument does include thiscase.

Theorem 17.5.3 Let f : Rn → Rm where n ≥ m be a Lipschitz map. Let A beLebesgue measurable. Then the following formula holds along with all measurability as-sertions needed for it to make sense.∫

RmH n−m (A∩f−1 (y)

)dy =

∫f(A)

H n−m (A∩f−1 (y))

dy =∫

AJ∗ (x)dx (17.17)

where J∗ (x)≡ det(Df (x)Df (x)∗

)1/2.

Proof: First assume A is Borel, f differentiable on A. Now note that

det(Df (x)Df (x)∗

)= ∑

i∈Λ(n,m)

det(Df i (x)

)2

by the Binet Cauchy theorem and 17.16.

Lemma 17.5.4 Suppose A is a measurable nonempty set and det(Df (x)Df (x)∗

)> 0

for all x ∈ A. Then there exist disjoint, measurable Ai, one for each i ∈ Λ(n,m) such thatfor all j ̸= i,det

(Df j (x)

)2= 0 for x ∈ Ai.

17.5. THE COAREA FORMULA 463Now let i, € A(n,n —m) consist of the remaining indices taken in order where ¢ € A(n,m).For 7 = (i1,-++ ,im), define x; = (xj, ,..-,%;,,) and x4, to be the other components of « takenin order. Then let f* (2) = ( Ft) ) Thus there are C (n,n —m) =C (n,m) different f*Uefeaturing C(n,m) different x;,x;,.Example 17.5.2 Say f : R+ — R?. Here are some examples for f’:fi (%1,.42,%3,%4) Si (1,%2,%3,%4) Si (%1,%2,%3,.%4)fa (*1,.X25%3,%4) fa (*1,%2,%3,%4) fa (%1,X2,%3,%4)x2 , x] , x3x4 x2 x4TSuppose first that 7; =( Xm41 -*+ Xn ) soi x Dz, f (a) Daz, f(xand so from row operations, detDf* (x) = detDz, f (x). It is similar in the general caseexcept one might have a sign change which is not important in what follows. Sodet Df* (x) = detDy, f (x). (17.16)Earlier with the area formula, we integrated J, (x) = det (Df (x)* Df («)) 12 | Withthe coarea formula, we integrate J* (a) = det (Df (x) Df (w)*)"/ *. This proof involvesdoing this integration and seeing what happens. In case n = m the claim of the theoremwill follow from the area formula because .#° (E) is the number of elements of E, so onecan assume if desired that in what follows n > m although the argument does include thiscase.Theorem 17.5.3 Le: f : R" > R"” where n> m be a Lipschitz map. Let A beLebesgue measurable. Then the following formula holds along with all measurability as-sertions needed for it to make sense.n—m —l _ n—m -1 _ i[ ve" (aos w)dy= | (anf (w))ay= [J (w)dx (17.17)where J* (a) = det (Df (a) Df (x)*) 2Proof: First assume A is Borel, f differentiable on A. Now note that2det (Df (a)Df(x)*)= ) det(Df*(zx))4€A(n,m)by the Binet Cauchy theorem and 17.16.Lemma 17.5.4 Suppose A is a measurable nonempty set and det (Df (x) Df (x)*) >0for all x € A. Then there exist disjoint, measurable A;, one for each 4 © A(n,m) such thatfor all j #%,det (Df4 (w))” = 0 for @ € Ay.