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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?

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Yes, it is the standard "surface measure" on $S^{n-1}$.

In order to prove it, it is enough to check that it gives the expected measure on some simple "surface elements" of your choosing (as long as they generate the Borel algebra). You will find the expression of the spherical volume element (in spherical coordinates) here (we are talking about the standard Euclidean sphere here).

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: $$\mu_n((0,1]A) = \int_0^1 \sigma_r(rA)\,dr = \int_0^1 r^{n-1} \sigma_1(A)\,dr = {\sigma_1(A) \over n}~.$$

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