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Let $S\subset \mathbb R^N$ be a $\mathcal H^{N-1}$-rectifiable set. Then we know that there exist countably many Lipschitz $N-1$-graphs $\Gamma_i\subset \mathbb R^N$ such that $$ \mathcal H^{N-1}\left(S\setminus \bigcup_{i=1}^\infty \Gamma_i\right)=0 $$ Without lose of generality, we assume that $$ \mathcal H^{N-1}(\Gamma_i\cap \Gamma_j)=0\text{ if }i\neq j $$ Hence, for $\mathcal H^{N-1}$ a.e. $x\in S$, there exists only one $\Gamma_i$ such that $x\in\Gamma_i$, and let us name this $\Gamma_i$ by $\Gamma_x$.

My question: would it possible to hold that $$ \lim_{r\to 0}\frac{\mathcal H^{N-1}(\Gamma_x\cap B(x,r))}{r^{N-1}}=1 $$ for $\mathcal H^{N-1}$ a.e. $x\in S$?

Thank you!

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  • $\begingroup$ Maybe I am misunderstanding you again, but: once you fix $x \in S$, $\Gamma_x$ is just a Lipschitz graph which has density $1$ at almost all its points, which is the statement you want to hold. If your assumptions above about mutual intersections are correct, I don't see why it shouldn't, but I could have missed something, if so, please let me know. $\endgroup$ Jun 2, 2016 at 14:36
  • $\begingroup$ @SilviaGhinassi I think you understand me perfectly. My advisor keep asking me why $\mathcal H^{N-1}$ a.e. hold and it confuse me a lot... By the way, do you think there is any problem for mutual intersections assumption? $\endgroup$
    – JumpJump
    Jun 2, 2016 at 14:47
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    $\begingroup$ Sorry I don;t understand your notation. Is B(x,r) an N-dimensional ball? $\endgroup$ Jun 2, 2016 at 16:48
  • $\begingroup$ @PieroD'Ancona ah sorry, fixed! $\endgroup$
    – JumpJump
    Jun 2, 2016 at 17:18

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