> [!definition] > > Let $\real^n$ be the [[Euclidean Space|Euclidean space]], and $S^{n - 1} = \bracs{x \in \real^n: \abs{x} = 1}$ as the $2$-sphere. Define the **polar coordinate map** > $ > \Phi: (\real^n \setminus \bracs{0}) \to > (0, \infty) \times S^{n - 1} \quad > x \mapsto (\abs{x}, \wh x) \quad \wh x = \frac{x}{\abs{x}} > $ > that decomposes a point to its norm and direction, which is a [[Homeomorphism|homeomorphism]] whose inverse is $(r, \wh x) \mapsto r\wh x$. > > Let $m_*$ be the [[Pushforward of Vector Field|pushforward]] of the [[Lebesgue Measure|Lebesgue measure]] by $\Phi$, and $\rho = \rho_n$ be a [[Measure Space|measure]] on $(0, \infty)$ defined by $d\rho(r) = r^{n - 1}dr$. > [!definition] > > There exists a unique [[Borel Measure|Borel measure]] > $ > \sigma: \cb(S^{n - 1}) \to [0, \infty) \quad \sigma(E) = m(\Phi^{-1}((0, 1] \times E)) > $ > known as the **surface measure**, such that $m_* = \rho \times \sigma$. If $f$ is Borel measurable on $\real^n$ with $f \in L^+ \cup L^1$, then > $ > \int_{\real^n}fdx = \int_{(0, \infty)}\int_{S^{n - 1}}f(r\wh x) d\sigma(\wh x) d\rho(r) > $ > *Proof*. It's sufficient to show that the above holds for measurable sets, and the rest follows from a [[Monotone Class Argument]]. Let $E$ be a Borel set on $S^{n - 1}$, and let > $ > E_\alpha = \Psi^{-1}((0, 1] \times E) = \bracs{x: \abs{x} \le \alpha, \wh x \in E} > $ > Note that by scaling, $m(E_\alpha) = a^n m(E_1)$. Let $\phi(E) = \Phi^{-1}((0, 1] \times E)$, then $\phi$ maps Borel sets to Borel sets, while preserving unions, intersections, and complements. Define $\sigma = n \cdot m \circ \phi$, then $\sigma$ is also a Borel measure, and > $ > \begin{align*} > m(E_1) &= \frac{\sigma(E)}{n} = \sigma(E)\int_{(0, 1]}r^{n - 1}dr\\ > &= \int_{(0, 1]}\int_{E}r^{n - 1}d\sigma(\wh x)dr \\ > \end{align*} > $ > From here, we can extend this via > $ > \begin{align*} > m_*((a, b] \times E) &= m(E_b \setminus E_a) = \frac{b^n - a^n}{n}\sigma(E) \\ > &= \sigma(E)\int_a^br^{n - 1}dr = \rho \times \sigma((a, b] \times E) > \end{align*} > $ > Fix $E \in \cb(S^{n - 1})$, and let $\alg_E$ be the family of FDUs of sets of the form $(a, b] \times E$. As this is a $\pi$-system that generates the $\sigma$-algebra $\cm_E = \bracs{A \times E: A \in \cb_{(0, \infty)}}$, and $m_* = \rho \times \sigma$ on $\alg_E$, by [[Dynkin's Uniqueness Theorem]], $m_* = \rho \times \sigma$ on $\cm_E$. On the other hand, $\bigcup_{E \in \cb(S^{n - 1})}\cm_E$ are the rectangle sets that generate the Borel $\sigma$-algebra on $(0, \infty) \times S^{n - 1}$. By the uniqueness theorem again, $m_* = \rho \times \sigma$ on all Borel sets.