Theorem

Let f : X → Y be Lipschitz, ℋ s ( X ) < ∞ and, for ℋ s -a.e. x ∈ X , suppose that

for some 0 < λ x ≤ 1 . Then there exists a countable Borel decomposition X = N ∪ ⋃ i X i with ℋ s ( N ) = 0 such that each f ∣ X i is bi-Lipschitz.

Definition

For 0 < κ < λ ≤ 1 and 0 < ξ < 1 , a function f : V ⊂ X → Y satisfies the condition D ( λ , κ , ξ ) on a set S ⊂ V if, for all x ∈ S and r < diam S ,

Lemma

Let 0 < κ < λ ≤ 1 , 0 < ξ < 1 and suppose f : V ⊂ X → Y satisfies D ( λ , κ , ξ ) on S ⊂ V . Let x , y ∈ S , set r = d ( x , y ) ∕ 4 and suppose ℋ s ( B ( f ( x ) , κ r ) ) ≥ ℋ s ( B ( f ( y ) , κ r ) ) . If

then

Proof

Suppose that (3) does not hold. Then by the triangle inequality,

Combining this with the definition of D ( λ , κ , ξ ) negates (2). Indeed, it gives

Proof of the Theorem

First let V ⊂ X be Borel and note that (1) holds for the function f ∣ V and for ℋ s -a.e. x ∈ V . Indeed, for ℋ s -a.e. x ∈ V , Θ * , s ( X ⧹ V , x ) = 0 (see [5, §2.10.18]) and for such an x ,

The first inequality uses ℋ s ( f ( A ) ) ≤ L s ℋ s ( A ) for any A ⊂ X , see [5, Corollary 2.10.11]. A consequence of (4) is

If ℋ s ( X ) = 0 , the result is immediate. Otherwise, let V 1 ⊂ X be a Borel subset with ℋ s ( V 1 ) > 0 . By the coarea formula (see [5, §2.10.25]), for ℋ s -a.e. y ∈ Y , card f − 1 ( y ) < ∞ . Hence, by (5), for ℋ s -a.e. x ∈ V 1 ,

Let V 2 ⊂ V 1 be a positive measure Borel set for which (6) holds for all x ∈ V 2 . By the Lusin-Novikov theorem (see [6, Exercise 18.14])^[1], there exists a Borel function g : f ( V 2 ) → V 2 such that V 3 ≔ g ( f ( V 2 ) ) is Borel and f ( g ( y ) ) = y for all y ∈ f ( V 2 ) . By (5), ℋ s ( f ( V 3 ) ) = ℋ s ( f ( V 2 ) ) > 0 and hence, since f is Lipschitz, ℋ s ( V 3 ) > 0 .

Note that if x ∈ X satisfies (1), then it also satisfies (1) for all 0 < λ ≤ λ x . For i ∈ N let 1 ≥ λ i ↘ 0 and define S i to be the set of x ∈ V 3 for which

Then, by (4), the S i monotonically increase to a full measure subset of V 3 . Therefore, there exist i ∈ N and S ′ ⊂ S i with ℋ s ( S ′ ) > 0 and diam S ′ ≤ λ i . For any 0 < r < λ i , setting r ′ = ( 1 − λ i 2 ) λ i r shows that f ∣ V 3 satisfies D ( λ i , ( 1 − λ i 2 ) λ i , ( 1 + λ i ) − s ) on S ′ . Since f ∣ V 3 is injective, (2) holds for all x , y ∈ S ′ and hence the lemma implies that f ∣ S ′ is bi-Lipschitz.

The bi-Lipschitz condition extends to the closure of S ′ . Hence, S ′ ¯ ∩ V 1 is a Borel subset of V 1 of positive measure on which f is bi-Lipschitz. Since V 1 is an arbitrary Borel subset of positive measure, the conclusion follows by exhaustion (e.g. as in [5, §3.2.14]).□