Show there's a disjoint partition over the sample space Been working on this problem - any suggestions would be greatly appreciated!
Let $e \in\varepsilon$ be a simple function over $(\Omega,A,P)$. Show that there is a disjoint partition $A_i, i = 1,\ldots,m,$ of $\Omega$ such that $e =  \sum\limits_{i=1}\alpha_i1_{A_i}$. 
 A: I guess your definition of a simple function is a function almost surely taking only finitely many values (I prefer say it is a simple random variable). Then take the $\alpha_i$ as the possible values of $e$ and define the events $A_i = \{e=\alpha_i\}$. 
A: The function $f:\Omega\to\mathbb{R}$ is simple if it has finite range. This is the definition given, for example, in Billingsley's probability book. If $\mathrm{Range}(f)=\{a_1,\dots,a_n\}$, and we define $A_i=\{\omega:f(\omega)=a_i\}$, for $i=1,\dots,n$, it is clear that we can represent a simple function $f$ as a sum of indicators
$$
  f(\omega) = \sum_{i=1}^n a_i I_{A_i}(\omega) \, .
$$
Note that this representation is not unique (you can break each $A_i$ in many sets). Also, there are books that prefer to define simple functions directly as sums of indicators. It is easy to check that the sets $A_i$ form a partition of $\Omega$. Given distincts $A_i$ and $A_j$, there is no $\omega$ in $A_i\cap A_j$, because $f$ is a function, and by definition every function assigns only one value to each of its domain points. Hence, the sets $A_i$ are pairwise disjoint. Also, it is not difficult to see that $\cup_{i=1}^n A_i = \Omega$. Just check both inclusions.
