Benefits of alternative formulation of the integral (general measure) Let $(X,\mathcal{M},\mu)$ be a measure space and $f:X\to \mathbb{R}^+$ a measurable map. Set
$$F(t) = \mu (\{x\in X \mid f(x)>t\})$$
and define
$$\int_X f\,d\mu := \int_0^\infty F(t)\,dt,$$
where the integral on the right is the Riemann integral. Since $F$ is monotonically decreasing, we know that the Riemann integral always exists (assuming the limit is finite).
Apparently, this definition agrees with the typical construction of the integral, where we define the integral for simple functions and then approximate from below the integral on the left with integrals of simple functions taking the supremum.
My question is the following: What is the advantage of this alternate definition? I realize that one can piggyback on results derived from the Riemann integral, but since the main goal is always to prove convergence theorems, does it actually help there?
 A: You have probably been reading "Analysis" by Lieb & Loss, have you?
The main advantage that I see using that approach is that you can (seemingly) avoid a lot if overhead: You do not have to talk about simple functions (i.e. $\sum_{i=1}^n \alpha_i \chi_{A_i}$), you do not have to show that each measurable nonnegative function is the monotone limit of a sequence of simple functions and you do not have to show that the integral is well-defined on simple functions, etc. 
Thus, like the dark side of the force, it is "quicker, easier".
You can even prove the usual convergence theorems using this definition, as long as you believe/know/show that if $(f_n)$ is an increasing sequence of decreasing functions, then $\int f_n \, dx \to \int \lim f_n \, dx$ (the limit is Riemann integrable, because it is decreasing). 
But if you want to show things like additivity of the integral, $\int f+g \, d\mu =\int f \,d\mu +\int g\, d\mu$, this is not easy at all. The path taken in Lieb & Loss (hidden in an exercise) is to approximate $f$ and $g$ by simple functions, so in the end one does not really gain anything. 
Also, this new approach requires you to develop the Riemann integral beforehand, which I do not really like. 
