1014 CHAPTER 29. THE AREA FORMULA
•b•c
B(c, 1i )
B(b, 2i )
Note that this proves that on E (T,c, i) the function h is one to one and T is a closeapproximation to U (b) for each b ∈ E (T,c, i). It also shows the Lipschitz continuity of hand h−1 on E by comparison with T . What has just been shown is a very interesting resultfor its own sake. It is summarized in the following lemma.
29.4 Estimates and a LimitLemma 29.4.1 Let h be differentiable on A⊆G and let A+ consist of those points x wheredet(Dh(x)∗Dh(x)
)> 0. Then if B is any Borel subset of A+, there is a disjoint sequence
of Borel sets and invertible symmetric transformations Tk, {(Ek,Tk)} ,∪kEk = B such thath is Lipschitz on Ek and h−1 is Lipschitz on h(E (T,c, i)). Also for any b ∈ Ek,29.3.7 and29.3.8 both hold. Also, for b ∈ Ek
(1− ε) |Tkv|< |Dh(b)v|= |U (b)v|< (1+ ε) |Tkv| (29.4.11)
One can also conclude that for b ∈ Ek,
(1− ε)−n |det(Tk)| ≤ det(U (b))≤ (1+ ε)n |det(Tk)| (29.4.12)
Proof: It follows from 29.3.10 that for x,y ∈ T (E (T,c, i))∣∣h(T−1 (x))−h
(T−1 (y)
)∣∣≤ (1+2ε) |x−y| (29.4.13)
and for x,y in h(E (T,c, i)) ,∣∣T (h−1 (x))−T
(h−1 (y)
)∣∣≤ 1(1−2ε)
|x−y| (29.4.14)
The symbol h−1 refers to the restriction to h(E (T,c, i)) of the inverse image of h. Thus,on this set, h−1 is actually a function even though h might not be one to one. This alsoshows that h−1 is Lipschitz on h(E (T,c, i)) and h is Lipschitz on E (T,c, i). Indeed, from29.4.13, letting T−1 (x) = a and T−1 (y) = b,
|h(a)−h(b)| ≤ (1+2ε) |T (a)−T (b)| ≤ (1+2ε)∥T∥|a−b| (29.4.15)
and using the fact that T is one to one, there is δ > 0 such that |T z| ≥ δ |z| so 29.4.14implies that ∣∣h−1 (x)−h−1 (y)
∣∣≤ 1δ (1−2ε)
|x−y| (29.4.16)
Now let (Ek,Tk) result from a disjoint union of measurable subsets of the countablymany E (T,c, i) such that B = ∪kEk. Thus the above Lipschitz conditions 29.4.13 and