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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 $\mathbb{R}^n$?

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  • $\begingroup$ 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.) $\endgroup$
    – Behnam Esmayli
    Sep 23, 2022 at 12:18

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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 $M \subset \cup_{j \geq 0} N_j$.

This is Lemma 11.1 in Leon Simon's lecture notes on geometric measure theory.

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