Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma

For Borel $E\subseteq S^{n-1}$ put $E_r:=\{s\omega :\omega \in E,0<s\le r\}$. By [A1], each $E_r$ is Borel. If $(E_j)$ is pairwise disjoint, then the sets $(E_j{)}_1$ are pairwise disjoint and $({\bigcup }_j{E}_j{)}_1={\bigcup }_j(E_j{)}_1$, so countable additivity of ${\lambda }_n$ makes $\sigma$ a Borel measure. It is finite because $\sigma (S^{n-1})=n{\lambda }_n(\{x:0<\mathrm{\mid }x\mathrm{\mid }\le 1\})<\mathrm{\infty }$ by [L5].

For Borel $E\subseteq S^{n-1}$ and $0<a<b$, [L1] gives ${\lambda }_n(E_1)=\sigma (E)\mathrm{/}n$. The dilation $x\mapsto rx$ has determinant $r^n$, so [L2] gives ${\lambda }_n(E_r)=r^n{\lambda }_n(E_1)=\frac{r^n}{n}\sigma (E)\mathrm{.}$ Therefore $m^{\ast }((a,b]\times E)={\lambda }_n(E_b\setminus E_a)=\frac{b^n-a^n}{n}\sigma (E)=\sigma (E){\int }_a^b{r}^{n-1}dr\mathrm{.}$

Step 2.1 shows that $m^{\ast }$ and the product measure $r^{n-1}dr\times \sigma$ agree on the generating pi-system of [A2]. Both are sigma-finite there, so [L3] implies that they agree on every Borel subset of $(0,\mathrm{\infty })\times S^{n-1}$. Applying [L4] to this measure identity gives the stated polar-coordinate integral formula for every nonnegative Borel measurable $f$. If $\tilde{\sigma }$ is another Borel measure on $S^{n-1}$ with the same integral formula, apply that formula to ${\mathbf{1}}_{E_1}$, where $E_1=\{r\omega :\omega \in E,0<r\le 1\}$. Then ${\lambda }_n(E_1)={\int }_0^{\mathrm{\infty }}{\int }_{S^{n-1}}{\mathbf{1}}_{E_1}(r\omega )r^{n-1}d\tilde{\sigma }(\omega )dr=\tilde{\sigma }(E){\int }_0^1{r}^{n-1}dr=\frac{\tilde{\sigma }(E)}{n},$ so $\tilde{\sigma }(E)=n{\lambda }_n(E_1)=\sigma (E)$ by [L1]. Thus $\tilde{\sigma }=\sigma$, proving uniqueness.