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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 $\mu(\partial F)=0$?

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  • $\begingroup$ @NateEldredge Apologies for the typo, the $\mu$ in (1) was meant to be $\mu_c$. I have fixed this. $\endgroup$ – Nirav Jan 7 '18 at 2:19
  • $\begingroup$ @NateEldredge The only other information that we have about $\mu_m$ is that each one is diffuse with respect to $p\text{-cap}$. $\endgroup$ – Nirav Jan 7 '18 at 2:28
  • $\begingroup$ Ok. I don't think that places any restrictions on $\mu$ since every finite measure is a weak limit of smooth measures (e.g. by convolution). $\endgroup$ – Nate Eldredge Jan 7 '18 at 2:54
  • $\begingroup$ Accidentally deleted my earlier comment, but doesn't the same argument still work? If $F$ has zero capacity then it has empty interior, so $\mu(\partial F) = \mu(\overline{F}) \ge \mu(F) \ge \mu_c(F) > 0$. $\endgroup$ – Nate Eldredge Jan 7 '18 at 2:56
  • $\begingroup$ @NateEldredge Oh yeah, because $\mu_c(F)=\mu_c(\mathbb{R}^2)>0$. Thanks. $\endgroup$ – Nirav Jan 7 '18 at 3:10

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