Ex4-2

Solution to Chapter 4, Exercise 2
Question: Let fbe a simple function, taking its distinct values on disjoint sets E_1, \cdots, E_N. Show that fis measurable if and only if E_1,\cdots,E_Nare measurable.
Solution: Let the distinct values be a_1,\cdots,a_Nso that
f(x)=\sum_{k=1}^N a_k\chi_{E_k}(x)
on E=\bigcup_{k=1}^N E_k.
Suppose fis measurable. Then for any a, \{f \geq a\}and \{f>a\}are measurable. Then
\{f=a\}=\{f \geq a\}-\{f>a\}
is measurable for any a. Since E_k=\{f=a_k\}, each E_kis measurable.
Conversely, suppose E_1, \cdots, E_Nare measurable. Then for any a, if a \geq \max\{a_1,\cdots,a_N\}, then
\{f>a\}=\emptyset
is measurable so fis measurable. If a<\max\{a_1,\cdots,a_N\}, then
\{f>a\}=\bigcup_{k~:~a_k>a}E_k
is a finite union of measurable sets so it is measurable. Thus, fis measurable.