Can a $N-1$ rectifiable set be partitioned into countably many connected pieces?

Question

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 $$ \mathcal H^{N-1}\left(\Gamma\setminus \bigcup_{n=1}\Gamma_n\right)<\epsilon. $$

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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?

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.

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

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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.