Spherical coordinates and Lebesgue integral I found a textbook that says that $r \in [0,\infty), \theta \in [0,\pi], \phi \in [-\pi,\pi]$ are the spherical coordinates.
Then he goes on with: This corresponds to a unitary map $L^2(\mathbb{R}^3) \rightarrow L^2((0,\infty),r^2dr) \otimes L^2((0,\pi),\sin(\theta) d \theta) \otimes L^2((0,2\pi),d\phi)$.
I was wondering whether there is any point in using open intervals now?  Or to say it differently: If I had used the intervals as they are given by the transform would this make any difference?
 A: A very important property of a transformation is bijectivity, i.e. if we consider a map $T: U \to V$, $U \subseteq \mathbb{R}^d$, $V \subseteq \mathbb{R}^n$, then for any $v \in V$ there should exist exactly one $u \in U$ such that $T(u)=v$. For the spherical coordinates we see that for
$$V := \mathbb{R}^3 \qquad \quad U_1 := [0,\infty) \times [0,\pi] \times [-\pi,\pi]$$
the transformation is not bijective (it is not injective). On the other hand, if we consider
$$V := \mathbb{R}^3 \qquad \quad U_2 := (0,\infty) \times (0,\pi) \times (-\pi,\pi)$$
then the transformation is still not bijective (it is not surjective). So what is the advantage of this setting? It is a differentiable map! This property is of importance if we want to apply the transformulation formula (i.e. make a change of variables):
$$\int_V f(v) \, dv = \int_{U_2} f(T(u)) \cdot |\text{det} \, (DT(u))| \, du.$$
Since $\mathbb{R}^3 \backslash T(U_2)$ (i.e. the set which is not "covered" by our transformation) is a Lebesgue null set, we may indeed apply the transformation formula.
