$\newcommand{\d}{\,\mathrm{d}}\newcommand{\b}{\mathscr{B}}\newcommand{\R}{\Bbb R^n}$I had never heard of surface measures before today! In investigating the integrability of $\|x\|^{-n}$, in the Euclidean and Lebesgue senses, I discovered the following this morning:
It would appear that $\sigma_r$, the surface measure on the $n$-sphere $S_r^{n-1}:=\{x\in\R:|x|=r\}$ - the unit sphere is just denoted $S^{n-1}$, with measure $\sigma$ - is characterised by: $$\begin{align}\tag{1}\forall A\in\b(\R),\quad\mu(A)&=\int_0^\infty\sigma_\rho(A\cap S_\rho^{n-1})\d\rho\\\tag{2}\forall A\in\b(S^{n-1}),\quad\sigma_r(r\cdot A)&=r^{n-1}\cdot\sigma(A)\end{align}$$Where $\b$ denotes the Borel algebra.
This gives the following formula, which I attempted a proof of - is the proof correct? Here, $(0,b]A$ denotes $\{ka:a\in A,k\in(0,b]\}$, a kind of smearing product. Perhaps there's a better name!
My derivation of statement $(3)$ (the statement is correct as corroborated online, but my derivation may well not be):
Letting $A\in\b(S^{n-1})$, we have $(0,r]A\in\b(\R)$. Then:$$\begin{align}\mu((0,r]A)&=\int_0^\infty\sigma_\rho((0,r]A\cap S_\rho^{n-1})\d\rho\\&=\int_0^\infty\sigma_\rho(\rho A\cap S_\rho^{n-1})\d\rho\\&=\int_0^r\rho^{n-1}\sigma(A)\d\rho\\&=n^{-1}r^n\cdot\sigma(A)\\\implies\sigma(A)&=nr^{-n}\cdot\mu((0,r]A)\\\tag{3}\sigma_r(r\cdot A)&=nr^{-1}\cdot\mu((0,r]A),\quad A\in\b(S^{n-1})\end{align}$$
I have of course gone from the definition above to the formula obtained just now, but $(3)$ is very familiar to me as it is essentially the same statement as what I have studied in the Gamma function derivations of surface area/volume in arbitrary dimensions - that is, I can prove it is true. I believe that if I simply start with the latter equality, I will get the definitions I used at the top, but those definitions were stolen from a fellow stack user, and I have never seen these things before, so I ask: I know how to show $(3)$ implies $(2)$, and I am pretty sure I can step back through the integration and show $(3)$ implies $(1)$, but I'd like confirmation on that; it would be nice to know if I have a rigorous justification of surface measure (on $n$-spheres). Furthermore, the definitions used the Borel algebra; what blocks us (if anything) from using the general sigma algebra of Lebesgue measurable sets?
The author of another stack post used the following formula:
For Borel measurable $f:\R\to[0,\infty]$, we have: $$\int_{\R} f\d\mu=\int_{S^{n-1}}\int_0^\infty \rho^{n-1}f(\rho\cdot\omega)\d\rho\d\sigma$$
I again attempted my own proof of this, but as I am learning measure theory somewhat erratically without a rigorous textbook I'd like a confirmation on this also, please.
It suffices to show that the result is true for characteristic functions, as the linearity of the Lebesgue integral shows that it holds for simples, and the monotone convergence theorem together with the simple approximation of measurable functions concludes it holds for all measurable $f:\Bbb R^n\to\Bbb R$. Let $A\in\mathscr{L}(\R)$ be measurable, and let $\chi$ denote the characteristic function on sets. I here assume definitions $1,2,3$ above are valid for arbitary $A$, and invoke Fubini-Torelli. Notationally, I have supposed $A\cap S_\rho^{n-1}=\rho\cdot\Omega$ for some $\Omega\in\b(S^{n-1})$, which is true! And likewise any $w\in S_\rho^{n-1}$ has the form $w=\rho\cdot\omega,\,\omega\in S^{n-1}$. $$\begin{align}\int_{\Bbb R^n}\chi_A\d\mu=\mu(A)&=\int_0^\infty\sigma_\rho(A\cap S_\rho^{n-1})\d\rho\\&=\int_0^\infty\sigma_\rho(\rho\cdot\Omega)\d\rho\\&=\int_0^\infty\rho^{n-1}\cdot\sigma(\Omega)\d\rho\\&=\int_0^\infty\rho^{n-1}\int_{S^{n-1}}\chi_{\Omega}(\omega)\d\sigma(\omega)\d\rho\\&=\int_{S^{n-1}}\int_0^\infty\rho^{n-1}\chi_{\Omega}(\omega)\d\rho\d\sigma(\omega)\\&=\int_{S^{n-1}}\int_0^\infty\rho^{n-1}\chi_A(\rho\cdot\omega)\d\rho\d\sigma(\omega)\quad\quad\blacksquare\end{align}$$Where the final equality was justified as $\omega\in\Omega\iff\rho\cdot\omega\in\rho\cdot\Omega=A\cap S_\rho^{n-1}\iff\rho\cdot\omega\in A$.
Is this right?
Many thanks.