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!

thetangent vector $\nu(x)$? There should be an hyperplane tangent to $x \in \Gamma$, or am I misunderstanding? $\endgroup$normalvector to the tangent hyperplane of $x$,notthe tangent vector. Which makes $T_x$ the tangent hyperplane. $\endgroup$3more comments