29.3. A DECOMPOSITION 1013

By regularity, we can and will assume A is a Borel set. Of course this is automatic if his Lipschitz. I have in mind the assumption that h is Lipschitz. This makes things veryconvenient because then h(E) is H n Hausdorff measurable whenever E is n dimensionalLebesgue measurable. However, there are interesting things which don’t depend on Lips-chitz continuity. Initially, I will only assume that h is continuous on G and differentiableon A.

For x ∈ A, let Dh(x)≡ R(x)U (x) where R(x) preserves lengths and

U (x)≡(Dh(x)∗Dh(x)

)1/2

Let A+ denote those points of A for which U (x)−1 exists. Thus this is a measurable subsetof A.

Let B be a Borel measurable subset of A+ and let b ∈ B. Let S be a countable densesubset of the space of symmetric invertible matrices and let C be a countable dense subsetof B. The idea is to decompose B into countably many Borel sets E on which h is one toone and Lipschitz with h−1 Lipschitz on h(E) . This will be done by establishing 29.3.10given below where T is an invertible symmetric transformation.

Let ε be a small number. Since U (b) is invertible, Lemma 29.2.2 implies o(a−b) =o(U (b)(a−b)) and so

|h(a)−h(b)−Dh(b)(a−b)|< ε |U (b)(a−b)| (29.3.5)

provided that a ∈ B(b, 2

i

)for i sufficiently large. By Lemma 29.2.1,

|h(a)−h(b)−Dh(b)(a−b)|< ε |T (a−b)| (29.3.6)

where U (b) is replaced by another linear one to one and onto symmetric mapping T pro-vided T is sufficiently close to U (b).

Now let c ∈ C be close enough to b that b ∈ B(c, 1

i

). Thus b ∈ E (T,c, i) where for

i ∈ N,c ∈ C , T ∈S , E (T,c, i) consists of those b ∈ B(c, 1

i

)such that for all a ∈ B

(b, 2

i

),

29.3.6 holds and also

infv̸=0

|Dh(b)v||T v|

= infv̸=0

|U (b)v||T v|

> 1− ε, (29.3.7)

supv̸=0

|Dh(b)v||T v|

= supv̸=0

|U (b)v||T v|

< 1+ ε (29.3.8)

It follows then from the above inequalities and 29.3.6 that for all a ∈ B(b, 2

i

),

|h(a)−h(b)| ≤ (1+2ε) |T (a−b)||h(a)−h(b)| ≥ (1−2ε) |T (a−b)| (29.3.9)

and so

(1−2ε) |T (a−b)| ≤ |h(a)−h(b)| ≤ (1+2ε) |T (a−b)| (29.3.10)

Then if a,b ∈ E (T,c, i) , 29.3.10 holds for these two a,b because |a−b|< 2/i.