# For what reason, the surface measure represents the surface area?

Let $||\cdot||$ be a norm on $\mathbb{R}^n$ (It's an arbitrary norm, not 2-norm)

Define $S^{n-1}=\{x\in\mathbb{R}^n : ||x||=1\}$

Let $\mu$ be the n-dimensional Lebesgue measure.

Define $\sigma(E)=n\mu((0,1]\cdot E), \forall E\in\mathscr{B}_{S^{n-1}}$

This measure $\sigma_{n-1}:\mathscr{B}_{S^{n-1}}\rightarrow [0,\infty]$ is said to be the surface measure on $S^{n-1}$.

(It has a property that $\mu(\Phi^{-1}(A\times B))=(\int_A r^{n-1}dr)\cdot\sigma_{n-1}(B)$ where $\Phi:\mathbb{R}^n\setminus\{0\}\rightarrow (0,\infty)\times S^{n-1}:x\mapsto (||x||,\frac{x}{||x||})$ is a homeomorphism and $A\in\mathscr{B}_{(0,\infty)}$ and $B\in\mathscr{B}_{S^{n-1}}$)

I'm curious to know, for what reason this $\sigma_{n-1}$ represents the usual sense of the area of a surface, and why is it defined in a such way? why $n$ is multiplied? It may have a geometrical meaning.

For example, when $||\cdot||$ is the max-norm, $\sigma_2(S^2)=24$. This is the "intuitive" area of the surface of a cube of length 2.

Another example, when $||\cdot||$ is the 2-norm, $\sigma_2(S^2)=4\pi$. This is also the usual sense of the area of the surface of a ball.

I wonder why this result is not really surprising, but rather it has to have these values.

I have another question whether there is a way to expand this idea of measuring the surface of a given measurable set,(e.g. area of the surface of a star-shaped subset, which cannot be represented as $S^{n-1}$) but i think it deserves another post..

Let $V(r)$ be the measure of the ball of radius $r$. The surface area of this ball can be intuitively understood (or even defined) as $V'(r) = \lim_{h\to 0 }(V(r+h)-V(r))/h$. Think of covering $V(r)$ by a coat of paint with thickness $h$. The volume of paint used, divided by $h$, gives the surface area.
Since $V(r)=r^n V(1)$, it follows that $V'(r)=nr^{n-1}V(1)$ and in particular $V'(r)=nV(1)$. This explains the factor of $n$.
• @Number8 There are two definitions in the wiki. One is for the case when $A$ is the "solid" set whose surface (boundary) we want to measure. Then you argue as above: paint $A$ with thickness $h$, etc. The other is for the case when $A$ is itself the surface we want to measure, e.g., a sphere. When we paint all of it, both the inside and outside get covered by layer $h$, so the total thickness is $2h$. – user127096 Mar 6 '14 at 0:03