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I posted this question on MSE but I received no reply. So I repost this here for better luck. Thank you!

Let $\Gamma\subset \mathbb R^N$ be $\mathcal H^{N-1}$-rectifiable. Then we know that $\mathcal H^{N-1}$ a.e. $x\in \Gamma$ has density $1$. In particular, let $\nu(x)\in \mathcal S^{N-1}$ be the vector at $x\in \Gamma$ normal to $\Gamma$, and $Q(x,r)$ be a cube centered at $x$ with side length $r$ and two faces normal to $\nu(x)$, we have $$ \lim_{r\to 0}\frac{\mathcal H^{N-1}(Q(x,r)\cap \Gamma)}{r^{N-1}}=1 $$ for $\mathcal H^{N-1}$ a.e. $x\in\Gamma$.

My question: Fix $x\in \Gamma$ satisfies above. Let $T_x$ denote the hyperplane normal to $\nu(x)$ and passing through $x$. Then, do we have $$ \lim_{r\to 0}\frac{\mathcal H^{N-1}(\mathbb P_x[Q(x,r)\cap \Gamma])}{r^{N-1}}=1? $$

Where $\mathbb P_x$ is the projection operator which projects $x\in \Gamma$ onto the hyperplane $T_x$. For example, if $T_x = \{x_N=0\}$, then $\mathbb P_x(x)=(x_1,x_2,\ldots, x_{N-1},0)$ where $x=(x_1,\ldots, x_N)$.

PS: I know for instance that $$ \mathcal H^{N-1}\lfloor\left(\frac{\Gamma-x}{r}\right)\to \mathcal H^{N-1}\lfloor \mathbb P_x $$ weakly as $r\to 0$. I think I may conclude from this statement but still...feel there is still a gap...

Any help is really welcome! Thx!

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    $\begingroup$ Question: what is the tangent vector $\nu(x)$? There should be an hyperplane tangent to $x \in \Gamma$, or am I misunderstanding? $\endgroup$
    – Silvia Ghinassi
    Commented Jan 11, 2016 at 16:07
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    $\begingroup$ So it's the normal vector to the tangent hyperplane of $x$, not the tangent vector. Which makes $T_x$ the tangent hyperplane. $\endgroup$
    – Silvia Ghinassi
    Commented Jan 11, 2016 at 16:10
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    $\begingroup$ Although this argument shows that your statement is true for almost every point $x$, you cannot fix a point x as you say and only assume that the density and a weak tangent plane at $x$ exists. To see this consider the $1$-rectifiable set $\Gamma$ in $\mathbb{R}^2$ obtained as the union over all integers $n \in \mathbb{Z}$ of the vertical segments $\{\frac{1}{n}\} \times [0,\frac{1}{|n|(|n|+1)}]$. Then the horizontal axis is a weak tangent plane at $(0,0)$ (with the right density) but the projection of $\Gamma$ onto the horizontal axis has measure zero. $\endgroup$
    – rozu
    Commented Jan 19, 2016 at 11:02
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    $\begingroup$ @rozu: Thank you! So what I can have is that for this $\Gamma$, a.e. $x\in\Gamma$ has density 1 in normal sense and also in the sense as I defined in my post right? $\endgroup$
    – JumpJump
    Commented Jan 19, 2016 at 19:30
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    $\begingroup$ @tankonetoone: Yes I'm quite certain about that, following the sketch of a proof above (and you assume that $\Gamma$ has locally finite $\mathcal H^{N-1}$-measure). $\endgroup$
    – rozu
    Commented Jan 19, 2016 at 23:51

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