29.9. THE COAREA FORMULA 1037

Thus µ(∩∞

n=1∪m≥n [gkm ≥ m−1])= 0. Therefore, for ω /∈ ∩∞

n=1 ∪m≥n [gkm ≥ m−1], a setof measure zero, for all m large enough, [gkm < m−1] and so gkm (ω)→ 0 a.e. ω. Sincefkm (ω)≤ gkm (ω), this proves the lemma.

It might help a little before proceeding further to recall the concept of a level surfaceof a function of n variables. If f : U ⊆ Rn→ R, such a level surface is of the form f−1 (y)and we would expect it to be an n−1 dimensional thing in some sense. In the next lemma,consider a more general construction in which the function has values in Rm,m≤ n. In thismore general case, one would expect f−1 (y) to be something which is in some sense n−mdimensional. As earlier, sets will not be assumed measurable and H k will refer to an outermeasure.

Lemma 29.9.4 Let A⊆ Rp and let f : Rp→ Rm be Lipschitz. Then∫ ∗Rm

H s (A∩ f−1 (y))

dH m ≤ β (s)β (m)

β (s+m)(Lip(f))m H s+m (A).

Proof: The formula is obvious if H s+m (A) = ∞ so assume H s+m (A)< ∞. The diam-eter of the closure of a set is the same as the diameter of the set and so one can assume

A⊆ ∪∞i=1B j

i , r(

B ji

)≤ j−1, B j

i is closed,

and

H s+mj−1 (A)+ j−1 ≥

∑i=1

β (s+m)(

r(

B ji

))s+m(29.9.46)

Now define g ji (y) ≡ β (s)

(r(

B ji

))sX

f(

B ji

) (y). If f−1 (y) /∈ B ji , this indicator function

Xf(

B ji

) just gives 0. If f−1 (y) ∈ B ji then y ∈ B j

i . Thus

H sj−1

(A∩ f−1 (y)

)≤

∑i=1

β (s)(

r(

B ji

))sX

f(

B ji

) (y) = ∞

∑i=1

g ji (y),

a Borel measurable function. It follows,∫ ∗Rm

H s (A∩ f−1 (y))

dH m =∫ ∗Rm

limj→∞

H sj−1

(A∩ f−1 (y)

)dH m

≤∫ ∗Rm

lim infj→∞

∑i=1

g ji (y)dH m.

By Borel measurability of the integrand, the last term is no more than∫Rm

lim infj→∞

∑i=1

g ji (y)dH m

By Fatou’s lemma,

≤ lim infj→∞

∫Rm

∑i=1

g ji (y)dH m = lim inf

j→∞

∑i=1

β (s)(

r(

B ji

))s ∫Rm

Xf(

B ji

) (y)dH m

29.9. THE COAREA FORMULA 1037Thus { (_| Um>n [Sk_ 2 ')]) =0. Therefore, for @ EM, Umen [Sim 2m |], a setof measure zero, for all m large enough, [g;,, < m~'] and so gi,,(@) + 0 ae. @. SinceFn (@) < 8x, (@), this proves the lemma.It might help a little before proceeding further to recall the concept of a level surfaceof a function of n variables. If f : U CR" > R, such a level surface is of the form f~! (y)and we would expect it to be an n — | dimensional thing in some sense. In the next lemma,consider a more general construction in which the function has values in R”,m <n. In thismore general case, one would expect f~! (y) to be something which is in some sense n—mdimensional. As earlier, sets will not be assumed measurable and #* will refer to an outermeasure.Lemma 29.9.4 Let A C R? and let f : R? — R” be Lipschitz. ThenUs mm BUS)BOM) am ssn[Le ane! y)azen < BESS ip ny” sem (A)Proof: The formula is obvious if #°+™ (A) = 0 so assume #51 (A) < 00. The diam-eter of the closure of a set is the same as the diameter of the set and so one can assumeAC U2, B!, r(B/) < 7, BI is closed,and ooHAYA FYB stm) (-(s/))" (29.9.46)Now define g/(y) = B(s) (r (s/)) Xu! (y). If f-'(y) ¢ Bi, this indicator function4(n/) just gives 0. If f-! (y) € BI then y € Bi. ThusMs2}. (ANE"(y)) SVB) (+(B/)) Ziq = Yel (y).a Borel measurable function. It follows,[. ze (ant (yas = | tim.762, (Ane (y))azenRn Rm” jJrow* oO< | lim inf ¥ gi (y)a.0™R”™ Je i=1By Borel measurability of the integrand, the last term is no more than[ lim inf Y" g/ (y) de"Rn 1JeBy Fatou’s lemma,< lim inf [ ; y ai (y)d.70" = lim ink y B(s)(r(B})). [ » Ae(a)) 9420"jee