# Definition of “the surface measure”?

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

Let $||\cdot||$ be a norm on $\mathbb{R}^n$.

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

I have proven that $\forall A\in\mathscr{B}_{S^{n-1}}, (0,1]A\in \mathscr{B}_{\mathbb{R}^n}$. ($\mathscr{B}$ denotes the Borel-algebra and $(0,1]A$ is defined as $\{rb:r\in(0,1] , b\in A\}$)

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

Then, $\sigma$ is a measure.

Is this "the surface measure" or the completion of $\sigma$ the surface measure?

Yes, it is the standard "surface measure" on $S^{n-1}$.
A perhaps more instrinsic/satisfying argument is that if you denote by $\sigma_r$ the measure on $S_r^{n-1}$ ($r$ is the radius), then
• "$\mu_ n = \int_{0}^{+\infty} \sigma_r\,dr$" in the sense that $\mu_n(A) = \int_0^{+\infty} \sigma_r(A \cap S^{n-1})\,dr$ for any $A \in \mathscr{B}_{\mathbb{R}^n}$.
• "$\sigma_r = r^{n-1} \sigma_1$" in the sense that $\sigma_r (rA) = r^{n-1} \sigma_1(A)$ for any $A \in \mathscr{B}_{S^{n-1}}$
Of course, these two claims would need to be justified, I'll let you think about it. Anyway, it follows that: $$\mu_n((0,1]A) = \int_0^1 \sigma_r(rA)\,dr = \int_0^1 r^{n-1} \sigma_1(A)\,dr = {\sigma_1(A) \over n}~.$$