1022 CHAPTER 29. THE AREA FORMULA

Now define

n(y) =∞

∑i=1

Xh(Fi) (y).

By Lemma 29.1.2, h(Fi) is H n measurable and so n is a H n measurable function. Foreach y ∈ B, n(y) gives the number of elements in h−1 (y)∩B. From 29.6.26,∫

h(Rn)n(y)g(y)dH n =

∫B

g(h(x))J∗ (x)dm. (29.6.27)

Now define

#(y)≡ number of elements in h−1 (y).

Theorem 29.6.4 Let h : Rn→ Rm be Lipschitz. Then the function y→ #(y) is H n mea-surable and if

g : h(Rn)→ [0,∞]

is H n measurable, then∫h(Rn)

g(y)#(y)dH n =∫Rn

g(h(x))J∗ (x)dm.

Proof: If y /∈ h(S∪N), then n(y) = #(y). By 29.6.25

H n (h(S∪N)) = 0

and so n(y) = #(y) a.e. Since H n is a complete measure, #(·) is H n measurable. Letting

G≡ h(Rn)\h(S∪N),

29.6.27 implies∫h(Rn)

g(y)#(y)dH n =∫

Gg(y)n(y)dH n =

∫B

g(h(x))J∗ (x)dm

=∫Rn

g(h(x))J∗ (x)dmn.

Note that the same argument would hold if h : G→Rm is continuous and if A is the setwhere h is differentiable and H n (h(G\A)) = 0, then for g as above,∫

h(A)g(y)#(y)dH n =

∫A

g(h(x))J∗ (x)dmn

The details are left to the reader.