Why the Jauge $p:\mathbb R^n\to \mathbb R$ of a set $C\subset \mathbb R^n$ is defined as $$p(v)=\inf\{r>0\mid v/r\in C\}$$ and not as $$p(v)=\inf\{r>0\mid rv\notin C\} \quad \text{or}\quad p(v)=\sup\{r>0\mid rv\in C\}\ \ ?$$
Because both explain the same concept as : the first time we enter in $C$ or go out of $C$, and it looks more easy to work with $rv$ than with $\frac{v}{r}$. So I was wondering what is the motivation to use $\frac{v}{r}$ instead of $rv$.
For $p(v)=\inf\{r>0\mid rv\notin C\}$, if $\mathbb R\setminus C$ contains a neighborhood of $0$, then $p(v) = 0$ for all $v$, which is not very instructive.
For fixed $v$, let $P = \{r>0\mid v/r \in C\}$ and $P^* = \{r>0\mid rv\in C\}$. If $r>0$, we have
$$r \in P^* \iff rv \in C \iff \frac {v}{1/r} \in C \iff \frac 1r \in P$$
It follows that
$$\sup P^* = \frac1{\inf P},$$
where the 'equality' $\infty = 1/0$ is implicit. Hence, with $p^*(v) = \sup\{r>0\mid rv\in C\}$, we'd have $p^*(v) = 1/p(v)$.
I guess what you should ask yourself at this point is: do you prefer to work with infinites, or with $0$s?
If you want a more prosaic answer (but probably also more correct), I'd just say that people have worked with $p(v)$ and found it to have nice properties that might not be immediately or as easily intuitive (at least notationally) when working with $1/p(v)$. In particular, $p$ can be equivalently written as
$$p(v) = \inf \{r>0\mid v \in rC\},$$
which may now remind you of Minkowski functionals. Under certain conditions $($on $C)$, these functionals have very nice properties: they are a norm for which $C$ is a ball!
