The definition of $p$ capacity of a set $A\subset\mathbb{R}^n$

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I am having a bit of difficult understanding the definition of the $p$-capacity of a set $A\subset\mathbb{R}^n$ and I was wondering if anyone would be able to clarify whether I have the right idea or not.

This is what I understand so far. Formerly the definition of the $p$-capacity of a set is presented as follows.

Fix $1\leq p<n$. Define, \begin{equation} K^p\equiv\{f:\mathbb{R^n}\rightarrow\mathbb{R}\ \vert\ f\geq 0, f\in L^{p^{\ast}}(\mathbb{R}^n), Df\in L^{p}(\mathbb{R}^n;\mathbb{R}^n)\}. \end{equation} If $A\subset\mathbb{R}^n$ we define the quantity \begin{equation} \text{Cap}_p(A) \equiv \inf\left\{\int_{\mathbb{R}^n}\vert Df\vert^p\text{ d}x\ \middle|\ f\in K^p, A\subset\text{int}\{f\geq 1\}\right\} \end{equation} as the $p$-capacity of $A$ (denoted by Cap$_p(A)$).

My thought process on this is as follows:

  1. Pick an $f\in K^p$, and look at int$\{f\geq 1\}$.
  2. At this point, we ask ourselves : is $A\subset\text{int}\{f\geq 1\}$? If not then repeat step 1.
  3. Compute the quantity $\int_{\mathbb{R}^n}\vert Df\vert^p\text{ d}x$
  4. Save it in a list $L$.
  5. Repeat the process until all possible $f$ from $K^p$ have been considered.

Taking the 'smallest' number from the list $L$ defines the $p$-capacity of $A$. Is this a correct way to think about the definition of Cap$_p(A)$? I am aware of the distinction between minimum and infimum which is why I 've put 'smallest' in inverted commas.

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Basically, you are right. However, thinking of a list might be misleading, since there are infinitely many functions in $K^p$.