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Let be a Banach space with norm . Here, we describe regulated functions , where , with , is a compact interval of the real line .

Definition 1.1: A function is called regulated, if the lateral limits

exist. The space of all regulated functions will be denoted by .

We denote the subspace of all continuous functions by and, by , we mean the subspace of regulated functions which are left‐continuous on . Then, the following inclusions clearly hold

Remark 1.2: Let . By , we mean the set of all elements for which for every . Thus, it is clear that and the range of belongs to . Note that, for a given , and do not necessarily belong to .

Any finite set of points in the closed interval such that

is called a division of . We write simply . Given a division of , its elements are usually denoted by , where, from now on, denotes the number of intervals in which is divided through the division . The set of all divisions of is denoted by .

Definition 1.3: A function is called a step function, if there is a division such that for each , , for all . We denote by the set of all step functions .

It is clear that Moreover, we have the following important result which is a general version of the result presented in [127, Theorem I.3.1].

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), (ii) (iii) and, then, (iii) (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 ) that the closure of is . Therefore, is a Banach space when equipped with the usual supremum norm

See also [127, Theorem I.3.6].

If denotes the Banach space of bounded functions from to , equipped with the supremum norm, then the inclusion

follows from Theorem 1.4, items (i) and (ii), taking the limit of step functions which are constant on each subinterval of continuity.

Recently, D. Franková established a fourth assertion equivalent to those assertions of Theorem 1.4 in the case where . See [97, Theorem 2.3]. One can note, however, that such result also holds for any open set . This is the content of the next lemma.

Lemma 1.5: Let and be a function. Then the assertions of Theorem 1.4 are also equivalent to the following assertion:

1 (iv) for every , there is a division such that

Proof. Note that condition (iii) from Theorem 1.4 implies condition (iv). Now, assume that condition (iv) holds. Given , there is a division such that , for all and According to [97, Theorem 2.3], take and consider a step function given by

Hence, .

The next result, borrowed from [7, Lemma 2.3], specifies the supremum of a function .

Proposition 1.6: Let . Then where either , for some , or , for some , or for some .

Proof. Let . Since , . By the definition of the supremum, for all , one can choose such that which implies

Since , there exists a subsequence such that as . Since is regulated, belongs to and the proof is complete.

The composition of regulated functions may not be a regulated function as shown by the next example proposed by Dieudonné as an exercise. See, for instance, [58, Problem 2, p. 140].

Example 1.7: Consider, for instance, functions given by , for , and , that is, is the sign function. Both and belong to . However, the composition does not.

The next result, borrowed from [209, Theorem 10.11], gives us an interesting property of left‐continuous regulated functions. Such result will be used in Chapters 8 and 11. We state it here without any proof.

Proposition 1.8: Let . If for every , there exists such that for every , we have , then

We end this first section by introducing some notation for certain spaces of regulated functions defined on unbounded intervals of the real line . Given , we denote by the space of regulated functions from to . In order to obtain a Banach space, we can intersect the space with the space of bounded functions from to , in which case, we write and equip such space with the supremum norm,

where . Alternatively, we can consider a subspace of formed by all functions such that

The next result shows that is a Banach space with respect to a special norm. This result, whose proof follows ideas similar to those of [124, 220], will be largely used in. Chapters 5 and 8

Proposition 1.9: The space , equipped with the norm

is a Banach space.

Proof. Let be the linear mapping defined by

for all and .

Claim. is an isometric isomorphism. Indeed, is an isometry because

for all . Moreover, if , then defined by

is such that and , since

Therefore, is onto and the Claim is proved.

Once is an isometric isomorphism and is a Banach space, we conclude that is also a Banach space.