# Decomposition of rectifiable curves in $\mathbb R^2$

Let $$\gamma:[0,1]\to \mathbb{R}^2$$ be a rectifiable curve and $$\Gamma=\gamma[0,1]$$ be its image. Is it possible to cover $$\Gamma$$ by a countable collection of sets $$N,R_1,R_2,\dots$$ such that $$N$$ has vanishing 1-dimensional Hausdorff measure and each $$R_i$$ admits a bi-Lipschitz embedding into $$\mathbb R$$?

If yes, is there an analogue of this statement for $$k$$-rectifiable sets in $$\mathbb{R}^n$$?

• Once you have a Lipschitz map into part of your rectifiable set, you can ignore the set where the (metric) derivative is non-injective because by a Sard-type theorem the image of this set has zero Hausdorff measure. On the rest of the domain the map is locally one-to-one, and indeed bi-Lipschitz because locally you have positive lower bound on modulus of derivative. (By metric derivative I am talking about Kirchheim's differentiability result from 1994.) Sep 23, 2022 at 12:18

Regardless of its dimension, a $$k$$-rectifiable set $$M \subset \mathbf{R}^n$$ can be covered by a collection $$(N_j \mid j \geq 0 )$$ of sets, with $$\mathcal{H}^k(N_0) = 0$$ and where the $$N_j$$, $$j \geq 1$$ are $$k$$-dimensional embedded $$C^1$$ submanifolds of $$\mathbf{R}^n$$. Here by 'covered' I mean that $$M \subset \cup_{j \geq 0} N_j$$.