Let $\mu_m$ be a sequence of finite Radon measures on $\mathbb{R}^2$ that converge weakly to the finite Radon measure $\mu$, that is, \begin{equation} \lim_{m\rightarrow\infty}\int_{\mathbb{R}^2}\varphi\ \mathrm{d}\mu_m=\int_{\mathbb{R}^2} \varphi\ \mathrm{d}\mu\quad (\varphi\in C_c(\mathbb{R}^2)). \end{equation}

Fix $p\in [1, 2)$ and denote the $p$-capacity of a compact set $K$ as $p$-$\text{cap}(K)$, i.e., \begin{equation} p\text{-cap}(K)\equiv\left\{\int_{\mathbb{R}^2}|D\varphi|^p\ \mathrm{d}x\ \Big|\ \varphi\in C_c^{\infty}(\mathbb{R}^2),\ \varphi\geq 0\text{ and } \varphi=1\text{ on }K\right\}. \end{equation}If $U$ is open, then \begin{equation} p\text{-cap}(U)\equiv\sup\{p\text{-cap}(K)\ |\ K\subset U\text{ and } K \text{ is compact}\}, \end{equation}and finally, \begin{equation} p\text{-cap}(A)\equiv\inf\{p\text{-cap}(U)\ |\ A\subset U\text{ and } U \text{ is open}\}\quad(A\subset\mathbb{R}^2). \end{equation}

We say that $\mu$ is diffuse with respect to $p\text{-cap}$ if \begin{equation} p\text{-cap}(A)=0\implies \mu(A)=0, \end{equation}and $\mu$ is concentrated with respect to $p\text{-cap}$ if there is a Borel $A\subset\mathbb{R}^2$ such that \begin{equation} p\text{-cap}(A)=\mu(\mathbb{R}^2\setminus A)=0. \end{equation}

The measure $\mu$ can be decomposed into two: \begin{equation} \mu\equiv \mu_d+\mu_c. \end{equation}Here $\mu_d$ is a Radon measure that is diffuse with respect to $p\text{-cap}$ and $\mu_c$ is a Radon measure that is concentrated with respect to $p\text{-cap}$. If $\mu_c(\mathbb{R}^2)\neq 0$, there exists a Borel set $F$ such that \begin{equation}\tag 1 p\text{-cap}(F)=\mu_c(\mathbb{R}^2\setminus F)=0. \end{equation}

Question: Does there exist a Borel set $F$ satisfying (1) and in addition $\mu(\partial F)=0$?

Fix $p\in [1, 2)$ and denote the $p$-capacity of a compact set $K$ as $p$-$\text{cap}(K)$, i.e., \begin{equation} p\text{-cap}(K)\equiv\left\{\int_{\mathbb{R}^2}|D\varphi|^p\ \mathrm{d}x\ \Big|\ \varphi\in C_c^{\infty}(\mathbb{R}^2),\ \varphi\geq 0\text{ and } \varphi=1\text{ on }K\right\}. \end{equation}If $U$ is open, then \begin{equation} p\text{-cap}(U)\equiv\sup\{p\text{-cap}(K)\ |\ K\subset U\text{ and } K \text{ is compact}\}, \end{equation}and finally, \begin{equation} p\text{-cap}(A)\equiv\inf\{p\text{-cap}(U)\ |\ A\subset U\text{ and } U \text{ is open}\}\quad(A\subset\mathbb{R}^2). \end{equation}

We say that a Radon measure $\mu$ is diffuse with respect to $p\text{-cap}$ if \begin{equation} p\text{-cap}(A)=0\implies \mu(A)=0, \end{equation}and $\mu$ is concentrated with respect to $p\text{-cap}$ if there is a Borel $A\subset\mathbb{R}^2$ such that \begin{equation} p\text{-cap}(A)=\mu(\mathbb{R}^2\setminus A)=0. \end{equation}

Let $\mu_m$ be a sequence of finite Radon measures on $\mathbb{R}^2$ that converge weakly to the finite Radon measure $\mu$, that is, \begin{equation} \lim_{m\rightarrow\infty}\int_{\mathbb{R}^2}\varphi\ \mathrm{d}\mu_m=\int_{\mathbb{R}^2} \varphi\ \mathrm{d}\mu\quad (\varphi\in C_c(\mathbb{R}^2)). \end{equation}Assume also that $\mu_m$ is diffuse with respect to $p\text{-cap}$ for each $m\in \mathbb{N}$. The measure $\mu$ can be decomposed into two: \begin{equation} \mu\equiv \mu_d+\mu_c. \end{equation}Here $\mu_d$ is a Radon measure that is diffuse with respect to $p\text{-cap}$ and $\mu_c$ is a Radon measure that is concentrated with respect to $p\text{-cap}$. If $\mu_c(\mathbb{R}^2)\neq 0$, there exists a Borel set $F$ such that \begin{equation}\tag 1 p\text{-cap}(F)=\mu_c(\mathbb{R}^2\setminus F)=0. \end{equation}

Question: Does there exist a Borel set $F$ satisfying (1) and in addition $\mu(\partial F)=0$?

Let $\mu_m$ be a sequence of finite Radon measures on $\mathbb{R}^2$ that converge weakly to the finite Radon measure $\mu$, that is, \begin{equation} \lim_{m\rightarrow\infty}\int_{\mathbb{R}^2}\varphi\ \mathrm{d}\mu_m=\int_{\mathbb{R}^2} \varphi\ \mathrm{d}\mu\quad (\varphi\in C_c(\mathbb{R}^2)). \end{equation}

Fix $p\in [1, 2)$ and denote the $p$-capacity of a compact set $K$ as $p$-$\text{cap}(K)$, i.e., \begin{equation} p\text{-cap}(K)\equiv\left\{\int_{\mathbb{R}^2}|D\varphi|^p\ \mathrm{d}x\ \Big|\ \varphi\in C_c^{\infty}(\mathbb{R}^2),\ \varphi\geq 0\text{ and } \varphi=1\text{ on }K\right\}. \end{equation}If $U$ is open, then \begin{equation} p\text{-cap}(U)\equiv\sup\{p\text{-cap}(K)\ |\ K\subset U\text{ and } K \text{ is compact}\}, \end{equation}and finally, \begin{equation} p\text{-cap}(A)\equiv\inf\{p\text{-cap}(U)\ |\ A\subset U\text{ and } U \text{ is open}\}\quad(A\subset\mathbb{R}^2). \end{equation}

We say that $\mu$ is diffuse with respect to $p\text{-cap}$ if \begin{equation} p\text{-cap}(A)=0\implies \mu(A)=0, \end{equation}and $\mu$ is concentrated with respect to $p\text{-cap}$ if there is a Borel $A\subset\mathbb{R}^2$ such that \begin{equation} p\text{-cap}(A)=\mu(\mathbb{R}^2\setminus A)=0. \end{equation}

The measure $\mu$ can be decomposed into two: \begin{equation} \mu\equiv \mu_d+\mu_c. \end{equation}Here $\mu_d$ is a Radon measure that is diffuse with respect to $p\text{-cap}$ and $\mu_c$ is a Radon measure that is concentrated with respect to $p\text{-cap}$. If $\mu_c(\mathbb{R}^2)\neq 0$, there exists a Borel set $F$ such that \begin{equation}\tag 1 p\text{-cap}(F)=\mu(\mathbb{R}^2\setminus F)=0. \end{equation}

Question: Does there exist a Borel set $F$ satisfying (1) and in addition $\mu(\partial F)=0$?

Let $\mu_m$ be a sequence of finite Radon measures on $\mathbb{R}^2$ that converge weakly to the finite Radon measure $\mu$, that is, \begin{equation} \lim_{m\rightarrow\infty}\int_{\mathbb{R}^2}\varphi\ \mathrm{d}\mu_m=\int_{\mathbb{R}^2} \varphi\ \mathrm{d}\mu\quad (\varphi\in C_c(\mathbb{R}^2)). \end{equation}

Fix $p\in [1, 2)$ and denote the $p$-capacity of a compact set $K$ as $p$-$\text{cap}(K)$, i.e., \begin{equation} p\text{-cap}(K)\equiv\left\{\int_{\mathbb{R}^2}|D\varphi|^p\ \mathrm{d}x\ \Big|\ \varphi\in C_c^{\infty}(\mathbb{R}^2),\ \varphi\geq 0\text{ and } \varphi=1\text{ on }K\right\}. \end{equation}If $U$ is open, then \begin{equation} p\text{-cap}(U)\equiv\sup\{p\text{-cap}(K)\ |\ K\subset U\text{ and } K \text{ is compact}\}, \end{equation}and finally, \begin{equation} p\text{-cap}(A)\equiv\inf\{p\text{-cap}(U)\ |\ A\subset U\text{ and } U \text{ is open}\}\quad(A\subset\mathbb{R}^2). \end{equation}

We say that $\mu$ is diffuse with respect to $p\text{-cap}$ if \begin{equation} p\text{-cap}(A)=0\implies \mu(A)=0, \end{equation}and $\mu$ is concentrated with respect to $p\text{-cap}$ if there is a Borel $A\subset\mathbb{R}^2$ such that \begin{equation} p\text{-cap}(A)=\mu(\mathbb{R}^2\setminus A)=0. \end{equation}

The measure $\mu$ can be decomposed into two: \begin{equation} \mu\equiv \mu_d+\mu_c. \end{equation}Here $\mu_d$ is a Radon measure that is diffuse with respect to $p\text{-cap}$ and $\mu_c$ is a Radon measure that is concentrated with respect to $p\text{-cap}$. If $\mu_c(\mathbb{R}^2)\neq 0$, there exists a Borel set $F$ such that \begin{equation}\tag 1 p\text{-cap}(F)=\mu_c(\mathbb{R}^2\setminus F)=0. \end{equation}

Question: Does there exist a Borel set $F$ satisfying (1) and in addition $\mu(\partial F)=0$?