Lebesgue Integration and Intersection in simple functions. In my previous question, I encountered a statement regarding the simple functions in Lebesgue integration which I would like to clarify:
Let $f$ be a simple function, i.e. $f(x)=\sum_{i=1}^{n}f_i 1_{F_i}(x)$, where $\{ f_i\}_{i=1}^n\subset \mathbb{R}$ and $F_i \in \mathcal{F}$ is a finite separation of $X$ into measurable subsets of $X$. Then
$$\int_X f(x)\mu(dx)=\sum_{i=1}^{n}f_i \cdot\mu(F_i)$$
Let us note that for the integral it does not matter whether they intersect or not. 
If $F_i$s never intersect then 
$$F_i=\{x\in X| f(x)=f_i \},$$
however this does not hold if they intersect. In this situation $f$ will not be one of these numbers $f_1,f_2,..,f_n$, but sometimes it will be the sum of some of them. Like the simple function in this picture(below):
https://en.wikipedia.org/wiki/Lebesgue_integration#/media/File:RandLintegrals.png
In the definition that I have found it is not stated that the $F_i$s in a simple function cannot intersect and I suspect that we simply put one simple function over another like in the problem of Hanoi Tower or like in the aforementioned picture.
So the final question is: can $F_i$s intersect or not?
 A: They can intersect; This is because if they do intersect, you can always rewrite it in a form where they don't intersect. That is, if $f = ∑f_i \mathbf{1}_{F_i}$, we can always find $G_i$ disjoint and numbers $g_i$ such that $$ f = ∑g_i \mathbf{1}_{G_i}$$
The uniqueness of the integral i.e. $∑f_i \mu(F_i) = ∑g_i \mu(G_i)$ does follow but it is a little fiddly. I have scribbled it down in my notes but forgot to note where the source is (probably this website).

Define first the integral of $f := ∑c_i \mathbf{1}_{C_i}$ as $∫ f := ∑c_i \mu(C_i) $, where the $C_i$ must be disjoint. We aim to prove that this formula holds when we have an arbitrary decomposition $f=∑b_i \mathbf{1}_{B_i}$, where the $B_i$ are not disjoint.
Let $[k]:=\{1,2,…,k\}$. 
Let the measurable sets $\{C_i\}_{i\in [m]}$ be the unique 
coarsest partition of $\bigcup B_j$ such that for any $j\in [N]$, 
we can write each nonempty $B_j$ as a disjoint union of $C_i$. 
For each $i$, let $I(i)$ be the unique indexing set 
such that $j\in I(i)$ iff $C_i \subset B_j$. 
Thus by construction,
    $\newcommand{\one}{\mathbf{1}}$ $$ \bigsqcup_{i : j\in I(i)} C_i = B_j$$
And also 
    $$f
    = \sum_{j=1}^N b_j \one_{B_j} 
    = \sum_{i=1}^M \left(\sum_{j\in I(i)} b_j \right)\one_{C_i}
    $$
Then $$ \int f d\mu  
    = \sum_{i=1}^M \left(\sum_{j\in I(i)}b_j\right) \mu(C_i) 
    = \sum_{j=1}^N b_j \sum_{i:j\in I(i)} \mu(C_i) = \sum_{j=1}^N b_j \mu(B_j) $$
