I have some confusion regarding Rudin RCA book, page number 56
Theorem 2.25 The Vitali – Caratheodory theorem
In the proof it is written that we see that there are measurable set $E_i$ (not necessarily disjoint) and constant $c_i >0$ such that $f(x)=\sum_{i=1}^{\infty}c_i\chi_{E_i}(x)$ for all $x \in X$
My confusion : why Rudin say that $E_i$ are not necessarily disjoint?
My thinking : If $E_i$ are not disjoint then it will contradicts the definition of simple function
If $E_1\cap E_2\cap …\cap E_n =\emptyset $ then $E_i= \{x : s(x)= c_i\} \implies f(x)=\sum_{i=1}^{\infty}c_i\chi_{E_i}(x) $
If $E_1\cap E_2\cap …\cap E_n \neq\emptyset $ then $E_i= \{x : s(x)\neq c_i\}\implies f(x)\neq\sum_{i=1}^{\infty}c_i\chi_{E_i}(x)$
Best Answer
He's not claiming that the sets cannot be disjoint, only that they may not be.
Indeed, there is no reason for that to be true: the sets $E_i$ come from the characteristic functions whose linear combinations you use to get the simple functions $t_n$ such that $f=\sum_{n=0}^\infty t_n$. You know absolutely nothing about $t_n$ besides the fact that they are non-negative, simple, and add up to $f$.
In most cases, it is impossible to choose $E_i$ to be disjoint: otherwise, each of their linear combinations could only have countably many distinct values. On the other hand, if you happen to have a function which does only have countably many distinct values, it is quite easy to find such $E_i$.