Lebesgue integral - definition of the domain of the simple function.

Question

The Lebesgue integral is defined as,

$$\int f \, d\mu = \text{sup}\, \Big\{ \sum_{z\in s(M)} z\,\mu \,\big(\text{pr_im}_s(\{z\})\big) \Big\}$$ or the supremum of the sum of the areas under the curve of the simple function $s$.

The question is, where in the definition of a simple function is it made clear that there is no overlap in the measures over which the sums are taken to prevent overcounting.

Graphically,

I guess the reason has to be in the definition of the domain of $z$: the measure $M$ as a topology space with no overlapping intervals?

Answer 1

The definition of a simple function includes the fact that a simple function is a function. A function has just one value for each argument.

Answer 2

If $f(a)\ne f(b)$, then $f^{-1}\{f(a)\}\cap f^{-1}\{f(b)\}=\emptyset$. That is, if a function, being a function, maps $a$ to $f(a),$ it cannot also map $a$ to $f(b)$.