Let $\Gamma\subset \mathbb R^N$ be a $N-1$ rectifiable curve such that $\mathcal H^{N-1}(\Gamma)<\infty$. I am wondering that would it be possible to partition it into countably many connection pieces, up to $\epsilon>0$ error? (where $\epsilon>0$ is given)

That is, would it be possible to find a set $\{\Gamma_n\}_{n=1}^\infty$ such that each $\Gamma_n\subset \Gamma$ is connected, and $$ \mathcal H^{N-1}\left(\Gamma\setminus \bigcup_{n=1}\Gamma_n\right)<\epsilon. $$

My try: I simply think there can not be uncountably many connected pieces with positive $\mathcal H^{N-1}$ measure, otherwise $$ \sum_{n=1}^{\infty} \mathcal H^{N-1}(\Gamma_n) = \infty, $$ which contradicts to the fact that $\Gamma$ has finite measure.

Is it my argument correct?


1 Answer 1


By definition, if $\Gamma$ is rectifiable, there exist countably many Lipschitz maps $f_i \colon \mathbb R^{N-1} \to \mathbb R^N$ such that $$ \mathcal{H}^{N-1}\left(\Gamma \smallsetminus \bigcup_{i=1}^{\infty} f_i(\mathbb R^{N-1})\right)=0. $$ Since each $f_i$ is in particular continuous, the continuous image of a connected set is connected, so by setting $\Gamma_i=f_i(\mathbb R^{N-1})$ you get the desired result, with what is left having $0$ measure.

Note that this holds without assuming that $\Gamma$ has finite $\mathcal{H}^{N-1}$ measure, as it follows directly from the definition of rectifiability.

Edit: what I've said is not precisely an answer to your question, my apologies.

In the definition of the rectifiability, we don't require $\Gamma_i=f_i(\mathbb R^{N-1})$ to be contained in $\Gamma$, but just to cover it. With your additional requirement (that the connected sets are contained in $\Gamma$) then I believe this is not possible, even in the case where $\mathcal{H}^{N-1}(\Gamma)$ is finite.

Take for, instance $\Gamma=([0,1]\smallsetminus \mathbb Q)\times \{0\}$. Then $\Gamma$ is $1$-rectifiable (according to the usual definition, because it's covered by a line segment), but it has uncountably many connected components. The same example works if you take away from it a set of small measure.

So, the answer to your specific question is, no, you can't. But if you allow the $\Gamma_i$ to cover your set $\Gamma$, then it is immediate by the definition.

  • $\begingroup$ Thank you! Can you provide a reference for what you are saying? $\endgroup$
    – spatially
    Nov 19, 2015 at 0:43
  • $\begingroup$ A good reference for rectifiability and related topics is Pertti Mattila's "Geometry of sets and measures in Euclidean spaces". Also the definition is here. Note that in Federer's book (look at Wikipedia references) what we call rectifiability is referred to as "countable rectifiability". $\endgroup$ Nov 19, 2015 at 0:44
  • $\begingroup$ Thank you! One more thing. Do you think my argument makes sense? $\endgroup$
    – spatially
    Nov 19, 2015 at 0:47
  • $\begingroup$ I don't think that would work. The reason is that, as I mentioned, the result holds regardless of the finiteness of the measure, so your argument would not lead to a contradiction. But there is no problem, since the definition is the only argument needed. Also, as I mentioned to you before, unless $N-1=1$ and $\Gamma$ is connected itself, the correct terminology is rectifiable set, not curve. $\endgroup$ Nov 19, 2015 at 0:51
  • $\begingroup$ @tankonetoone please look at my edit. I didn't see the condition $\Gamma_n \subset \Gamma$ in your question before. I hope my edit makes it clear. $\endgroup$ Nov 19, 2015 at 4:27

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