upper G left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis"/> for which 
Any finite set
is called a division of
Definition 1.3: A function
It is clear that
Theorem 1.4: Let and consider a function . The assertions below are equivalent:
1 is the uniform limit of step functions , with ;
2 ;
3 given , there exists a division such that
Proof. We will prove (i)
(ii) Note that for all . We need to show that , see Remark 1.2. Let . We will only prove that exists, because the existence of follows analogously. Consider a sequence in such that , that is, , for every , and converges to as . Consider the sequence of step functions from to such that uniformly as . Then, given , there exists such that , for all . In addition, since is a step function, there exists such that , for all . Therefore, for , we haveThen, once is a Banach space, exists.
(iii) Let be given. Since , it follows that (see Remark 1.2). Thus, for every , there exists such thatSimilarly, there are such thatNotice that the set of intervals is an open cover of the interval and, hence, there is a division of , with , such that is a finite subcover of for and, moreover,
(i). Given , let , , be a division of such thatand , . Definewhere denotes the characteristic function of a measurable set . Note that for all and all . Moreover, is a sequence of step functions which converge uniformly to , as .
It is a consequence of Theorem 1.4 (with
