Let $\Omega\subset\mathbb{R}^N$. For compact $K\subset \Omega$ we can define the $p$-capacity, $p\in (1,\infty)$ as the number $$\operatorname{cap}_p(K)=\inf \int_\Omega |\nabla u|^p$$
where the infimum is taken over all $C_0^\infty(\Omega)$ with $u\ge 1$ in $K$. If $U\subset\Omega$ is open, set $$\operatorname{cap}_p(U)=\sup_{K\subset U}\operatorname{cap}_p (K)$$
where $K$ is compact.
My question is: how to find a open set $U\subset \Omega$ with $\overline{U}\subset \Omega$ such that $$\operatorname {cap}_p(U)\ne\operatorname {cap}_p(\overline {U})$$
I am aware of some equivalent definitions of capacity, for example, $u$ can be chosen in $W_0^{1,p}(\Omega)$ and $u\ge 1$ a.e. in $K$, or $u=1$ in $K$ or $u=1$ in a neighbourhood of $K$, so feel free to choose any of them.
I think I do not understood quite well what capacity measure, so I was unable to find such example. I was thinking in take some $U$ such that it's boundary has positive measure, however, this does not seem to change the value of the capacity, since the derivative of $u$ will be zero on the boundary of $U$.
In the (most interesting) case $p < n$, you can argue as follows. For arbitrary $x \in \Omega$, you have $\mathrm{cap}_p(U_r(x)) \to 0$ as $r \to 0$, where $U_r(x)$ is the open ball with center $x$ and radius $r$.
Fix a closed set $K \subset \Omega$ and set $k = \mathrm{cap}_p(K)$. Now, fix a countable set $M = \{x_1, x_2, \ldots\} \subset \Omega$ with $K \subset \overline M$. For each $n \in \mathbb{N}$, choose $r_n$, such that $\mathrm{cap}_p(U_{r_n}(x_n)) \le k \, 2^{-n-1}$. Set $U = \bigcup_{n=1}^\infty U_{r_n}(x_n)$.
By subadditivity of the capacity, we get $\mathrm{cap}_p(U) \le k/2$ and, since $K \subset \overline U$, we have $\mathrm{cap}_p(\overline U) \ge k$.
If you drop the requirement of $U$ being open, you can simply set $U = M$. Then $\mathrm{cap}_p(U) = 0$, whereas $\mathrm{cap}_p(\overline U) \ge k$.