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This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (updated) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can 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}

<!-The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$-->

By @Anton Petrunin's answer in the linked post, this question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap S(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap S(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!

This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (updated) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can 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}

<!-The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$-->

By @Anton Petrunin's answer in the linked post, this question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap S(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap S(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!

This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (updated) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can 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}

<!-The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$-->

By @Anton Petrunin's answer in the linked post, this question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap B(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap B(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!

This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (updated) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can 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}

<!-The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$-->

By @Anton Petrunin's answer in the linked post, this question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap S(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap S(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!

This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (update) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can choose a ball(cube) $B$ centered at $x_0$ with ridus $r>0$ 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}

The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$

By @Anton Petrunin's answer in the linked post. This question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap B(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap B(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!

This is a follow up discussion for this post.

Let me copy part of the definition here.

Let $S\subset \mathbb R^N$ be a $N-1$ rectifiable set with $\mathcal H^{N-1}(S)<\infty$.

My (updated) question:

Can we find, for any $\epsilon>0$ be given, a set $S_0\subset S$ with $\mathcal H^{N-1}(S\setminus S_0)<\epsilon$ so that for each $x_0\in S_0$, we can 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}

<!-The last line is not readable, I repeat it here: $B_\nu = \{x\in\pi_\nu:\,B_x\neq \varnothing\}.$-->

By @Anton Petrunin's answer in the linked post, this question is equivalent to asking can we represent this curve locally as a graph.

Little background: Here my set $S$ is the jump set of a $BV$ function in $\mathbb R^N$.

My try: We know that $\mathcal H^{N-1}$ a.e. $x\in S$ has density $1$. (I don't know whether this is hold for general rectifiable curve but it does hold in my case). That is, for a.e. $x\in S$ we have $$ \lim_{r\to 0} \frac{\mathcal H^{N-1}(S\cap B(x,r))}{S_{N}r^{N-1}} = 1 $$ where $S_N$ denote the measure of unit sphere in $\mathbb R^{N}$.

Therefore, we have, for $r$ small enough, $$ \mathcal H^{N-1}(S\cap B(x,r))\geq {S_{N}r^{N-1}}(1-\epsilon) $$

My idea is that as $r$ getting smaller, density $1$ makes sure that $S\cap B(x,r)$ can not bending too much other wise the limit is not $1$. But I can't not finish my argument from here... Please advise!