The updated version can be found [here](https://mathoverflow.net/questions/223203/can-we-represent-a-n-1-rectifiable-set-locally-as-a-graph-with-some-price#223203).


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!



The partition of $N-1$ rectifiable set
The answer is "no".


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. 


The latter does not hold in general.
Say take a logarithmic spiral. 


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.


