1010 CHAPTER 29. THE AREA FORMULA

In the first version of the area formula h will be a Lipschitz function,

|h(x)−h(y)| ≤ K |x−y|

defined on Rn. This is no loss of generality because of Theorem 29.0.1.The following lemma states that Lipschitz maps take sets of measure zero to sets of

measure zero. It also gives a convenient estimate. It involves the consideration of H n asan outer measure. Thus it is not necessary to know the set B is measurable.

Lemma 29.1.1 If h is Lipschitz with Lipschitz constant K then

H n (h(B))≤ KnH n (B)

Also, if T is a set in Rn, mn (T ) = 0, then H n (h(T )) = 0. It is not necessary that h be oneto one.

Proof: Let {Ci}∞

i=1 cover B with each having diameter less than δ and let this cover besuch that

∑i

β (n)12

diam(Ci)n < H n

δ(B)+ ε

Then {h(Ci)} covers h(B) and each set has diameter no more than Kδ . Then

H nKδ

(h(B)) ≤ ∑i

β (n)(

12

diam(h(Ci))

)n

≤ Kn∑

iβ (n)

(12

diam(Ci)

)n

≤ Kn (H nδ(B)+ ε

)Since ε is arbitrary, this shows that

H nKδ

(h(B))≤ KnH nδ(B)

Now take a limit as δ → 0. The second claim follows from mn = H n on Lebesgue mea-surable sets of Rn.

Lemma 29.1.2 If S is a Lebesgue measurable set and h is Lipschitz then h(S) is H n

measurable. Also, if h is Lipschitz with constant K,

H n (h(S))≤ Knmn (S)

It is not necessary that h be one to one.

Proof: The estimate follows from Lemma 29.1.1 and the observation that, as shownbefore, Theorem 28.2.4, if S is Lebesgue measurable in Rn, then H n (S) = mn (S). Theestimate also shows that h maps sets of Lebesgue measure zero to sets of H n measure zero.Why is h(S)H n measurable if S is Lebesgue measurable? This follows from completenessof H n. Indeed, let F be Fσ and contained in S with mn (S\F) = 0. Then

h(S) = h(S\F)∪h(F)