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Why is the Hausdorff measure of this set is zero?

Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set, and let $\phi:\Omega\to\mathbb{R}^N$ be a $C^1$ function with the property that $\phi^{-1}(0)\neq\emptyset$, and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$.

How can we prove or disprove that:

$$\mathcal{H}^{N-1}\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )=0$$

It is well-known that $\mathcal{L}^N(\phi^{-1}(0))=\mathcal{H}^N(\phi^{-1}(0))=0$ (Lebesgue measure of the zero level set is null).

I denote by $\mathcal{H}^{N-1}$ the $N-1$ dimensional Hausdorff measure on $\mathbb{R}^N$.

Why Hausdorff measure of this set is zero?

Let $\Omega\subseteq\mathbb{R}^N$ an open and bounded set, and let $\phi:\Omega\to\mathbb{R}^N$ be a $C^1$ function with the property that $\phi^{-1}(0)\neq\emptyset$, and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$.

How can we prove or disprove that:

$$\mathcal{H}^{N-1}\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )=0$$

It is well-known that $\mathcal{L}^N(\phi^{-1}(0))=\mathcal{H}^N(\phi^{-1}(0))=0$ (Lebesgue measure of the zero level set is null).

I denote $\mathcal{H}^{N-1}$ the $N-1$ dimensional Hausdorff measure on $\mathbb{R}^N$.

Why is the Hausdorff measure of this set zero?

Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set, and let $\phi:\Omega\to\mathbb{R}^N$ be a $C^1$ function with the property that $\phi^{-1}(0)\neq\emptyset$, and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$.

How can we prove or disprove that:

$$\mathcal{H}^{N-1}\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )=0$$

It is well-known that $\mathcal{L}^N(\phi^{-1}(0))=\mathcal{H}^N(\phi^{-1}(0))=0$ (Lebesgue measure of the zero level set is null).

I denote by $\mathcal{H}^{N-1}$ the $N-1$ dimensional Hausdorff measure on $\mathbb{R}^N$.

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Martin Sleziak
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Why Hausdorff measure of this set is zero?

Let $\Omega\subseteq\mathbb{R}^N$ an open and bounded set, and let $\phi:\Omega\to\mathbb{R}^N$ be a $C^1$ function with the property that $\phi^{-1}(0)\neq\emptyset$, and $\nabla\phi(x)\neq 0,\ \forall\ x\in \phi^{-1}(0)$.

How can we prove or disprove that:

$$\mathcal{H}^{N-1}\left (\overline{\phi^{-1}(0)}\setminus\phi^{-1}(0)\right )=0$$

It is well-known that $\mathcal{L}^N(\phi^{-1}(0))=\mathcal{H}^N(\phi^{-1}(0))=0$ (Lebesgue measure of the zero level set is null).

I denote $\mathcal{H}^{N-1}$ the $N-1$ dimensional Hausdorff measure on $\mathbb{R}^N$.