Capacity and measure

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

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

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, $$\lim_{m\rightarrow\infty}\int_{\mathbb{R}^2}\varphi\ \mathrm{d}\mu_m=\int_{\mathbb{R}^2} \varphi\ \mathrm{d}\mu\quad (\varphi\in C_c(\mathbb{R}^2)).$$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: $$\mu\equiv \mu_d+\mu_c.$$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 $$\tag 1 p\text{-cap}(F)=\mu_c(\mathbb{R}^2\setminus F)=0.$$

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

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