29.9. THE COAREA FORMULA 1039

Proof: If not, then there exists a sequence {zk} such that zk → y, and xk ∈ (A ∩f−1 (zk))\∪∞

i=1Si. Thus f(xk) = zk. Taking a subsequence still denoted by k we can have

zk→ y,xk→ x ∈ A\∪∞i=1Si.

Hence f(x) = limk→∞ f(xk) = limk→∞ zk = y, so x ∈ f−1 (y)∩A \∪∞i=1Si contrary to the

assumption that A∩ f−1 (y)⊆ ∪∞i=1Si.

It follows from this claim that whenever z is close enough to y,

H n−mδ

(A∩ f−1 (z)

)< t.

Thus ifUδ ≡ {z : H n−m

δ

(A∩ f−1 (z)

)< t +δ},

then Uδ is open. Hence, letting δ i→ 0+,

{z : H n−m (A∩ f−1 (z))≤ t}= ∩∞

i=1Uδ i = Borel set.

Thus, if A is compact, then for each y ∈ Rm, A∩ f−1 (y) is H n−m measurable and also thefunction

y→H n−m (A∩ f−1 (y))

is a Borel measurable function, hence Lebesgue measurable.Let A be Lebesgue measurable not just compact. Then by regularity, there exists F ⊆ A

where F is the countable union of compact sets and mm (A\F) = 0. Say F = ∪kFk,Fk+1 ⊇Fk and each Fk is compact. Then H n−m

(F ∩ f−1 (y)

)= limn→∞ H n−m

(Fn∩ f−1 (y)

)so

y→H n−m(F ∩ f−1 (y)

)is Lebesgue measurable.∫ ∗

RmH n−m ((A\F)∩ f−1 (y)

)dH n−m ≤Cm,nH

n (A\F) =Cm,nmn (A\F) = 0

From Lemma 29.9.3 H n−m((A\F)∩ f−1 (y)

)= 0 for H n−m a.e. y. Hence, regarding

H n−m as an outer measure,

H n−m (F ∩ f−1 (y))≤ H n−m (A∩ f−1 (y)

)≤ H n−m ((A\F)∩ f−1 (y)

)+H n−m (F ∩ f−1 (y)

)= H n−m (F ∩ f−1 (y)

)and so y→H n−m

(A∩ f−1 (y)

)= H n−m

(F ∩ f−1 (y)

)is Lebesgue measurable.

With this lemma proved, it is possible to obtain the following useful inequality whichwill be used repeatedly.

Lemma 29.9.6 If A⊆ Rn is Lebesgue measurable, then∫Rm

H n−m (A∩ f−1 (y))

dy

≤C (n,m)(Lip(f))m mn (A), C (n,m) =β (n−m)β (m)

β (n)