Given a measure on a measurable space and a function to integrate, what is the correspondent Riemann integral? I'm having troubles understanding the connection between the Riemann and Lebesgue integral and the role that the measure function plays in this.
Let's say I have a measurable space and a measure $\mu$ on it. Then I can integrale a function $f$ and the integral with respect to the measure is defined as
$$
\int_X f d\mu = sup \left\{  \int_X s d\mu : 0 \leq s \leq f, s \:\: \text{simple}  \right\}
$$
So taht I should find a simple function that approaches my $f$, integrate the simple function, and then take the sup.
Now I'd like instead to take an alternative way and use the Riemann integral to evaluate $
\int_X f d\mu$.
What I'm having troubles with is, given the measure $\mu$, what are the variables I have to integrate with respect to in the Riemann integral.
So, given the measure $\mu$, how do I explicitly continue the equation 
$$
\int_X f d\mu = ...
$$
passing to the Riemann integral?
Alternatively, as an example, let's say that I have the usual integral "with the measure on a sphere": 
$$
\int_{S^2} f(r, \theta, \phi) r^2 \sin(\theta) dr d\theta d\phi
$$
What I always hear is that 
$$
r^2 \sin(\theta) dr d\theta d\phi
$$
 is the measure. But what exactly does this mean? I know that a measure takes a set as input and gives a positive number as a result. Now, it's evident that $r^2 \sin(\theta) dr d\theta d\phi$ cannot take a set as an argument. 
So which exactly is the measure $\mu$ and how can I formally complete the equation
$$
\int_{S^2} f d\mu = ... = \int_{S^2} f(r, \theta, \phi) r^2 \sin(\theta) dr d\theta d\phi
$$ 
Thanks in advance!
 A: $1.$ If  $f$ is Riemann integrable, then it's Lebesgue integrable, as well, wrt the Lebesgue measure. 
$2.$ Since you're only integrating over the sphere, you just perform a change of variables (formula holds for both integrals). 
$3.$ $drd\theta d\phi$
is a measure in the same way as $dxdydz$ is. 
So, say we have a Riemann integrable function. Then, for bounded $X\subseteq \mathbb{R}^n$ $$\int\limits_X fdm=\int\limits_X f dx,$$ where $m$ denotes the Lebesgue measure (and $x$ the Jordan). If we stick with Lebesgue measure (you may be more comfortable), the statement $dm(X_j)$ does not make sense, as $dm$ is not a measure; $m$ is. However, you do calculate the measure of sets when integrating simple functions, which influences the integral of an arbitrary integrable function. So, the integral notation naturally reflects this. 
Indeed, if $s=\sum\limits_{j=1}^n c_j\chi_{X_j},$ where $X_j\subseteq X$ Lebesgue measurable, then $$\int\limits_X s \, dm = \sum\limits_{j=1}^n c_jm(X_j).$$
For your original question, if $\mu$ is the Lebesgue measure, then $$\int\limits_{S^2} f\, d\mu=\int\limits_{S^2} f(x,y,z)\, dxdydz=\int\limits_0^\pi\int\limits_{0}^{2\pi} f(\theta,\phi)\, \sin\phi\, d\theta d\phi.$$ If you want to integrate over the entire ball, then
$$\int\limits_{B_1(0)} f\, d\mu=\int\limits_{B_1(0)} f(x,y,z)\, dxdydz=\int\limits_0^\pi\int\limits_{0}^{2\pi}\int\limits_0^1 f(r,\theta,\phi)\, r^2\sin\phi\, dr d\theta d\phi.$$ 
