35.6. CHANGE OF VARIABLES FORMULA LIPSCHITZ MAPS 1247

and so h(T ) is mn measurable. Therefore, h(Ak) is mn measurable because mn is a completemeasure and this exhibits h(Ak) between two mn measurable sets whose difference hasmeasure 0. Now

h(A) = ∪∞k=1h(Ak)

so h(A) is also mn measurable and this proves the lemma.The following lemma, depending on the Brouwer fixed point theorem and found in

Rudin [113], will be important for the following arguments. The idea is that if a continuousfunction mapping a ball in Rk to Rk doesn’t move any point very much, then the image ofthe ball must contain a slightly smaller ball.

Lemma 35.6.9 Let B = B(0,r), a ball in Rk and let F : B→Rk be continuous and supposefor some ε < 1,

|F(v)−v|< εr

for all v ∈ B. ThenF(B)⊇ B(0,r (1− ε)).

Proof: Suppose a ∈ B(0,r (1− ε))\F(B)

and let

G(v)≡ r (a−F(v))|a−F(v)|

.

Then by the Brouwer fixed point theorem, G(v) = v for some v ∈ B. Using the formula forG, it follows |v|= r. Taking the inner product with v,

(G(v) ,v) = |v|2 = r2 =r

|a−F(v)|(a−F(v) ,v)

=r

|a−F(v)|(a−v+v−F(v) ,v)

=r

|a−F(v)|[(a−v,v)+(v−F(v) ,v)]

=r

|a−F(v)|

[(a,v)−|v|2+(v−F(v) ,v)

]≤ r|a−F(v)|

[r2 (1− ε)− r2+r2

ε]= 0,

a contradiction. Therefore, B(0,r (1− ε))\F(B)= /0 and this proves the lemma.

Now let Ω be a Lebesgue measurable set and suppose h : Rn→ Rn is Lipschitz contin-uous and one to one on Ω. Let

N ≡ {x ∈Ω : Dh(x) does not exist} (35.6.11)

S≡{

x ∈Ω\N : Dh(x)−1 does not exist}

(35.6.12)

Lemma 35.6.10 Let x ∈ Ω \ (S∪N). Then if ε ∈ (0,1) the following hold for all r smallenough.

mn

(h(

B(x,r)))≥ mn (Dh(x)B(0,r (1− ε))), (35.6.13)