# The partition of $N-1$ rectifiable set

The updated version can be found here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My question, for each $x_0\in S$, would it be possible to choose a ball (cube) $B$ centered at $x_0$ with radius $r>0$ so that:

I can choose a direction vector $\nu$ so that each slice $B_x$, for $x\in B_\nu$, only intersect with $S\cap B$ once?

Here by slice $B_x$ please refer to the following definition:

\begin{align} \begin{cases} \pi_\nu = \{x\in\mathbb R^N:\,<x,\nu>=0\};\\ B_x=\{t\in R:\, x+t\nu\in B\}\,\,(x\in\pi_\nu);\\ B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}. \end{cases} \end{align}

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

Thank you!

• I don't understand the definitions of $B_x$ and $B_\nu$. Apparently, $B$ is centered at $x_0$, but $x+t\nu$ is centered at $0$, because so are $x$ and $\nu$? Maybe, you mean $x_0+x+t\nu$ in the definition of $B_x$? – Sebastian Goette Nov 10 '15 at 8:33
• @SebastianGoette No, $B_x$ is not a ball. It is only a slice, i.e., a 1-d line. – JumpJump Nov 10 '15 at 14:35

In fact the answer "yes" to your question would imply that any Lipschitz simple curve the plane $\mathbb{R}^2$ can be presented locally as graph.
One could also construct a $(1\pm\varepsilon)$-bi-Lipschitz curve which is not a graph in a neighborhood of any of its point — such construction can be done by recursive logarithmic twisting of straight line and passing to the limit.
• @AntonPetrunin sorry, I have one more question. About your last sentence: I don't see how to construct such a map in a biLip way. Also, this seems to contradict the fact that a rectifiable set can be written -up to a set of $\mathcal{H}^d$ measure $0$- as a union of Lip graphs. – Silvia Ghinassi Nov 11 '15 at 17:53
• @SilviaGhinassi Note that the unit-speed parametrization of logarithmic spiral is bi-Lipschitz, the constants can be made as close to 1 as you want by sending logarithm base to $\infty$. For the second question, I do not see a contradiction --- the logarithm spiral is a countable union of graphs. – Anton Petrunin Nov 11 '15 at 21:03
• Oh, my bad. Now I see where I was confused: the logarithmic spiral is chord-arc, hence bi-Lipschitz, and also the set of measure $0$ is allowed to be dense, so being a countable union of Lip graphs does not imply being locally a graph. Thanks, I learned a lot from this. – Silvia Ghinassi Nov 11 '15 at 21:40