17.5. THE COAREA FORMULA 465

From Theorem 17.5.1, 17.23 equals det(I +B∗B)1/2. Note how the size of the matriceschanges. Since B = Dy2g

ii (y) and Dy2g

iic(y) = I, the above reduces to

det(I +B∗B)1/2 = det[(

B∗ I)( B

I

)]1/2

=

det[(

Dy2gii (y)

∗ Dy2giic(y)∗

)( Dy2gii (y)

Dy2giic(y)

)]1/2

= det(Dy2g

i (y)∗Dy2gi (y)

)1/2

Therefore, from area formula and the above simplification of 17.24 and 17.20, 17.23∫XEi∩A (x)det

(Dxi

f (x)Dxif (x)∗

)1/2 dx

=∫Rm

∫f−1(y1)∩Ei∩A

det(Dy2g

i (y)∗Dy2gi (y)

)1/2dy2dy1

=∫Rm

H n−m (f−1 (y1)∩Ei∩A)

dy1 (17.25)

Note how this also shows that y1→H n−m(f−1 (y1)∩Ei∩A

)is measurable since it

equals the inner integral in an iterated integral having Borel integrand.Now suppose that A = A+ ≡ {x ∈ A : J∗ (x)> 0} , and A is a Borel set. Let Ai be as in

Lemma 17.5.4. Lemma 17.3.1 says there are disjoint Borel sets{

Eij

}∞

j=1whose union is

Ai on which the conditions for Ei in the above argument hold. Adding the above in 17.25over j, for Ei replaced with Ei

j and using Lemma 17.5.4,∫XAi

(x)det(Df i (x)Df i (x)∗

)1/2dx

=∫

XAi(x)det

(Df (x)Df (x)∗

)1/2 dx =∫Rm

H n−m (f−1 (y1)∩Ai

)dy1

Now add the above over all Ai to obtain 17.17.What if A\A+ ̸= /0? Then consider (A\A+)×Rm ≡ Â as the new A and

f̂ (x1, ...,xn+m)≡(f (x1, ...,xn) εxn+1e1 · · · εxn+mem

)Thus Df̂ (x)Df̂ (x)∗ = Df (x)Df (x)∗+ ε2mI and so if ε > 0, det

(Df̂ (x)Df̂ (x)∗

)≡

J∗ε (x) ̸= 0 since Df̂ (x)Df̂ (x)∗ has positive eigenvalues at least ε2m. Then from what wasdone above, letting Ê be a bounded Borel set in Rn+m and E the corresponding boundedset in Rn,∫

XÂ∩Ê (x)J∗ε (x)dmn+m =∫Rm

H n−m(f̂−1

(y1)∩ Â∩ Ê)

dy1

≥∫Rm

H n−m (f−1 (y1)∩(A\A+

)∩E)

dy1

where limε→0 J∗ε (x) = 0. This follows since projections decrease distance. Then by thedominated convergence theorem we can pass to a limit and find∫

RmH n−m (f−1 (y1)∩

(A\A+

)∩E)= 0.

17.5. THE COAREA FORMULA 465From Theorem 17.5.1, 17.23 equals det (J + B*B)!/ *. Note how the size of the matriceschanges. Since B = Dy, gi (y) and Dy, gi. (y) =I, the above reduces todet (U-+B°B)" = det|( BY 1) ( | "=i 1/2 ; ;det ( Dygily)” Dygi.(y)” )( pie iA )] = det (Dy,g' (y)* Dy,g' (y))”Therefore, from area formula and the above simplification of 17.24 and 17.20, 17.231/2/ Kring (w) det (De, f (w) Do, f (w)*) "7 dxdet (D,,g* (y)* Dy,g? dyxdI. Fvone as (Dy,9° (y) Dyg’ (y)) © dy2dy1= [ 2" (¢" (yn Et A) dy (17.25)R”Note how this also shows that y; > #7"~"" (f-! (y,)NE* MA) is measurable since itequals the inner integral in an iterated integral having Borel integrand.Now suppose that A = A* = {a € A: J* (a) > 0}, and A is a Borel set. Let A; be as inLemma 17.5.4. Lemma 17.3.1 says there are disjoint Borel sets {Es} - whose union is. J=A, on which the conditions for E* in the above argument hold. Adding the above in 17.25over j, for E* replaced with EX and using Lemma 17.5.4,[@) ) det (Df? (x) Df* (a)")\? ax= [%,@) ) det (Df (a DF (a))'Pax= [20 (F (ys) Aa) dyNow add the above over all A; to obtain 17.17.What if A \\A* 4 0? Then consider (A \ At) x R” =A as the new A andFf (x1, --:Xnim) = ( f (%1,.--,%n) EXn+1E1 +°* EXnt+mem )Thus Df (a) Df (a)* =Df (« ) Df (a)* + €"1 and so if € > 0, det (Df (w) Df (a)") =Jz (x) £0 since Df (x ) Df (a)* has positive eigenvalues at least €2”. Then from what wasdone above, letting E be a bounded Borel set in R’*” and E the corresponding boundedset in R”,im (e-} ns| Pive (x) dmn+m = rome (f (y)NANE) dy= fe" (Fy) (A\A*) ) dywhere limeg_,o JZ (x) = 0. This follows since projections decrease distance. Then by thedominated convergence theorem we can pass to a limit and find[eo Fn (A\ at) nE) =0.