measure on sphere In a set of notes I am using (physics related), it states that "we denote $$M = \{ (x,y,z)| x^2 + y^2 + z^2 = 1 \}$$ and $L^2(M)$ the space of square integrable functions on $M$ with measure $$\Omega = (\theta, \phi),~~0 \leq \theta \leq \pi, ~0 \leq \phi \leq 2 \pi$$ and $$d \Omega = \sin \theta d \theta d \phi."$$ Is it clear how this defines a measure? As I know a measure is defined as a function from a $\sigma-$algebra to the real line, which satisfies the properties as in the link. 
 A: Here $M$ is parameterized via the map $\varphi:\Omega\to M$ defined by
$$\varphi(\theta,\phi)=\left(\begin{matrix}\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta\end{matrix}\right).$$
When we say that $d\Omega=\sin\theta~d\theta~d\phi$ is a measure, we really mean that it is the "density" of a measure. That is, for a measurable set $U\subseteq\Omega$ (respect to the Borel or Lebesgue $\sigma$-algebra) we have the measure of $U$ given by $d\Omega$ is 
$$\Omega(U):=\int_{U}d\Omega=\int_{U}\sin\theta~d\theta~d\phi=\int_{U}\sin\theta~d\mathcal{L}^2(\theta,\phi),$$
where $\mathcal{L}^2$ is the usual Lebesgue measure on $\mathbb{R}^2$. This is all well defined since $\sin\theta$ is a measurable function with respect to $\mathcal{L}^2$. It shouldn't be hard to check that $\Omega$ defined this way is a measure.
Now if we have an integrable function $f:\Omega\to\mathbb{R}$ and a measurable set $U\subseteq\Omega$ one can check that  $$\int_{U}f(\theta,\phi)~d\Omega=\int_{U}f(\theta,\phi)\sin\theta~d\theta~d\phi,$$
which in some sense justifies the notation $d\Omega$. 
