There is this formula that relates the volume of a unit $n$-ball with that of the surface area of the unit $n-1$-sphere. Namely:
$$\sigma(S^{n-1}) = n \cdot m_n(B^n)$$
where $S^{n-1}$ is the $n-1$-sphere, $\sigma(A)$ is the surface area of $A$, $m_n$ is the $n$-dimensional Lebesgue measure and $B^n$ is the unit $n$-ball.
Now, my problem is when $n = 1$. In this case we have the $1$-ball which is the segment $(-1,1)$ in $\mathbb{R}$, and $S^0$ = $\{-1 ,1\}$. Is it possible to calculate the surface area of $S^0$ directly if it is just two points? (By directly I mean, for example in the case of $n = 2$ there is a formula for the surface area of a $1$-sphere, namely the formula for the perimeter of a circle). It can't be $0$ since we obtained from the formula that $n \cdot m_1(B^1) = 1 \cdot m((-1,1)) = 2$. The only way I can think it would work is if the surface area is the number of points in the set. Why? If this is the case, does this mean that the $0$-dimensional Lebesgue measure $m_0$ is the counting measure?