Let $\mathscr{H}^m$ be the $m$ dimensional Hausdorff measure in $\mathbb{R}^n$, $m\leq n$. Is it true that for $\mathscr{H}^m$-almost every point $p$ on a Lipschitz manifold $M$ of dimension $m$ embedded in $\mathbb{R}^n$ there is a neighborhood $B\_\epsilon(p)\subset \mathbb{R}^n$ such that $M\cap B\_\epsilon(p)=graph(u)\cap B\_\epsilon(p)$ for some
$$
u:p+T\_pM\to p+(T\_pM)^\perp
$$
defined on the (approximate) tangent space of $p$ (which exists for a.e. $p$) with $Lip(u)<\epsilon$?


Please note that the question is trivial if we ask $u$ only to be a Lipschitz function without the requirement that it satisfies $Lip(u)<\epsilon$.


Edit: Since a Lipschitz manifold can be defined in several ways not all equivalent to each other, here we can just consider the simplest case of $M=graph(f)$ for some Lipschitz function $f:\mathbb{R}^m\to\mathbb{R}^{n-m}$. If it helps, we can also set $m=1$ and $n=2$, so that $f:\mathbb{R}\to\mathbb{R}$.



$(1+\epsilon)$-bilipschitz parametrization of Lipschitz manifold
The answer to my question is no, and a counter-example is provided by the staircase function of the fat-Cantor set, which is a Lipschitz function $f:[0,1]\to\mathbb{R}$ with the property that $\{f'=0\}$ is dense and $\{f'\geq1\}$ has positive measure. Therefore, for $\epsilon>0$ small enough, a function $u$ as in the body of the question cannot exist around all points $(x,f(x))$ with $x\in\{f'\geq1\}$. This set has positive $\mathscr{H}^1$-measure (in fact, at least $\sqrt{2}\mathscr{L}^1(\{f'\geq1\})$, since for all measurable $A\subset [0,1]$ it holds $\mathscr{H}^1(G(A))=\int\_{[0,1]}1\_A \sqrt{1+|f'(x)|^2}dx$, where $G:[0,1]\to[0,1]\times\mathbb{R}$ is given by $G(x)=(x,f(x))$.


See also this question <https://math.stackexchange.com/questions/3380703/where-is-a-fat-cantor-staircase-differentiable>.


