1018 CHAPTER 29. THE AREA FORMULA

29.5 The Area FormulaAt this point, I will begin assuming h is Lipschitz continuous to avoid the fuss with whethersets are appropriately measurable and to ensure that the measure considered below is ab-solutely continuous without any heroics. Let h be Lipschitz continuous on G, an openset containing A, the set where h is differentiable. Suppose also that h is one to one onA. Then H n (h(G\A)) = 0 because mn (G\A) = 0 by Rademacher’s theorem. HenceLemma 29.1.2 applies.

Lemma 29.5.1 Let h be Lipschitz. Let h be one to one and differentiable on A withmn (G\A) = 0. If N ⊆ G has measure zero, then h(N) has H n measure zero and ifE is Lebesgue measurable subset of G, then h(E) is H n measurable subset of Rm. Ifν (E)≡H n (h(E)) , then ν ≪ mn.

Proof: Lemma 29.1.2 implies h(N) = 0 if mn (N) = 0.Also from this lemma, h(E) isH n measurable if E is. Is ν a measure? Suppose {Ei} are disjoint Lebesgue measurablesubsets of G. Then for A the set where Dh exists as above,

ν (∪iEi) ≡ H n (h(∪iEi))≤H n (h(∪iEi∩A)∪h(G\A))

≤ ∑i

H n (h(Ei∩A))+H n (h(G\A)) = ∑i

H n (h(Ei∩A)) = ∑i

ν (Ei)

Thus ν ≪ mn.It follows from the Radon Nikodym theorem for Radon measures, that

ν (E)≡H n (h(E)) =∫

EDmnνdmn

but Dmnν ≡ limr→0H n(h(B(x,r)))

mn(x,r) = limr→0H n(h(B(x,r))∩A)

mn(x,r) = J∗ (x) for a.e. x from Theorem29.4.3. Also, ν is finite on closed balls so it is regular thanks to Corollary 11.6.8. Thisshows from the Radon Nikodym theorem, Theorem 31.3.5 that∫

Xh(E)dH n =∫

EDmnνdmn =

∫XE (x)J∗dmn

Note also that, since AC has measure zero,∫h(A)

Xh(E)dH n =∫

AXE (x)J∗dmn

Now let F be a Borel set in Rm. Recall this implies F is H n measurable. Then∫h(A)

XF (y)dH n =∫

XF∩h(A) (y)dH n = H n (h(h−1 (F)∩A))

= ν(h−1 (F)

)=∫

XA∩h−1(F) (x)J∗ (x)dmn

=∫

AXF (h(x))J∗ (x)dmn. (29.5.22)