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Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $||g_\epsilon - g_0||_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $||\nabla(g_\epsilon - g_0)||_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 \cdot d n - \int_{g_0^{-1}(\mu)} \nabla g_0 \cdot d n \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ? Also we assume that for any $x \in g_0^{-1}(\mu)$, $\nabla g_0 (x)\neq 0$

Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $\|g_\epsilon - g_0\|_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $\|\nabla(g_\epsilon - g_0)\|_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 \cdot d n - \int_{g_0^{-1}(\mu)} \nabla g_0 \cdot d n \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ? Also we assume that for any $x \in g_0^{-1}(\mu)$, $\nabla g_0 (x)\neq 0$

Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $||g_\epsilon - g_0||_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $||\nabla(g_\epsilon - g_0)||_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 \cdot d n - \int_{g_0^{-1}(\mu)} \nabla g_0 \cdot d n \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ?

Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $||g_\epsilon - g_0||_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $||\nabla(g_\epsilon - g_0)||_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 \cdot d n - \int_{g_0^{-1}(\mu)} \nabla g_0 \cdot d n \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ? Also we assume that for any $x \in g_0^{-1}(\mu)$, $\nabla g_0 (x)\neq 0$

Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $||g_\epsilon - g_0||_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $||\nabla(g_\epsilon - g_0)||_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 - \int_{g_0^{-1}(\mu)} \nabla g_0 \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ?

Let $g_\epsilon : K \subset \mathbb{R}^d \rightarrow \mathbb{R}$ (more regularity can be assumed if necessary) be defined on a compact set (with regular boundary) $K \subset \mathbb{R}^d$, and the familly $g_\epsilon$ is indexed by $\epsilon \in [0,1]$. $||g_\epsilon - g_0||_{L^\infty(K)} \le c \epsilon$ where $c$ does not depend on $\epsilon$, and $||\nabla(g_\epsilon - g_0)||_{L^\infty(K)} \le c \epsilon$. How to prove that \begin{align*} \left| \int_{g_\epsilon^{-1}(\mu)} \nabla g_0 \cdot d n - \int_{g_0^{-1}(\mu)} \nabla g_0 \cdot d n \right| \le c \epsilon \end{align*} where $c$ does not depend on $\epsilon$ ?