There is this definition I couldn't quite get my head around. It is the definition of the integral in regard to the simple functions. The integral of the measurable positive function
$f:X \rightarrow \mathbb{R}$, with (X,$\mathcal{A}$,$\mu$) being the measure space is defined as:
$\int f d\mu=sup\{\int\phi\ d\mu|0\leq\phi\leq f,\phi $ is a simple function$\}$.
While simple functions have the form $\phi = \sum_{i=1}^Nc_i\chi_{E_i}$,
where $c_i\geq0$ and $E_i \in \mathcal{A}$ and its integral the form:
$\int\phi\ d\mu = \sum_{i=1}^Nc_i\mu({E_i})$.
I do believe I get the point of the definition, not to be paraphrasing, it aims to find $\phi \leq f$ that composes the highest sum. So the aim is to approximate f with simple functions.
But I did not understand what domain is considered for the supremum.
Are we just considering the domain of simple functions ? Because if that was the case , the supremum of the set of the possible simple functions would again be a simple function. But I found that highly unlikely because in that case , every integral should be able to be written as a finite sum.
My thought was that the idea is that also the limits of the simple functions are considered in that domain. Since another definition I found was one where the limit of the integral of simple functions was considered to be the integral of f.
Is that the reasoning behind this supremum definition ?
The supremum of a set need not be a member of the set. For example:
Similarly, $$\sup \left\{\sum_{i=1}^n \frac 1n\sqrt{1 -\left(\frac in\right)^2}\,\middle|\, n \in \Bbb N\right\} = \frac \pi 2$$ even though every element of that set is $< \frac \pi 2$. A fact that remains true when you extend the set to include all integrals over simple functions under $y = \sqrt{1 - x^2}$ on $[0,1]$.