uniform measure on $S^2$ as a product measure Consider the unit sphere $S^2$, where every point $x\in S^2$ can be parametrized in spheric coordinates $x=(r=1,\theta, \phi)$ where $\theta \in [0,2\pi]$ and $\phi \in [-\pi/2,\pi/2]$, and denote the area measure on the sphere, i.e., a uniform measure w.r.t. the area, by $m_2$. **My Question:** can $m_2$ be parametrized as $m_2 = \mu _{\theta} \times \mu _{\phi} $, where $\mu _{\theta}$ and $\mu _{\phi}$ are measures on $[0,2\pi]$ and $[-\pi/2, \pi/2]$ respectively. _Remark:_ the product of Lebesgue measures on both spaces does not, inasmuch as I understand, produce the spheric area measure.

A famous theorem of Archimedes says that under the parametrization $$ (\theta, z) \mapsto(\sqrt{1 - z^{2}} \cos\theta, \sqrt{1 - z^{2}} \sin\theta, z),\qquad 0 \leq \theta \leq 2\pi,\quad -1 \leq z \leq 1, $$ the area element of the round sphere is $d\theta \times dz$. This gives you two natural ways to proceed:

* In the standard spherical parametrization, $z = \sin\phi$ and $\sqrt{1 - z^{2}} = \cos\phi$, the spherical measure is $d\theta \times \cos\phi\, d\phi$;

* In the parametrization $z = 2\phi/\pi$, the spherical area element is $(2/\pi)\, d\theta \times d\phi$.