Let X be a Banach space with norm || - ||_X. A function f : [0, T] → X is said to be a regulated function if one (and hence both) of the following two equivalent conditions holds true:^[1]

• for every t in the interval [0, T], both the left and right limits f(t−) and f(t+) exist in X (apart from, obviously, f(0−) and f(T+));
• there exists a sequence of step functions φ_n : [0, T] → X converging uniformly to f (i.e. with respect to the supremum norm || - ||_∞).

It requires a little work to show that these two conditions are equivalent. However, it is relatively easy to see that the second condition may be re-stated in the following equivalent ways:

• for every δ > 0, there is some step function φ_δ : [0, T] → X such that

${\displaystyle \|f-\varphi _{\delta }\|_{\infty }=\sup _{t\in [0,T]}\|f(t)-\varphi _{\delta }(t)\|_{X}<\delta ;}$ 

• f lies in the closure of the space Step([0, T]; X) of all step functions from [0, T] into X (taking closure with respect to the supremum norm in the space B([0, T]; X) of all bounded functions from [0, T] into X).

Let Reg([0, T]; X) denote the set of all regulated functions f : [0, T] → X.

• Sums and scalar multiples of regulated functions are again regulated functions. In other words, Reg([0, T]; X) is a vector space over the same field K as the space X; typically, K will be the real or complex numbers. If X is equipped with an operation of multiplication, then products of regulated functions are again regulated functions. In other words, if X is a K-algebra, then so is Reg([0, T]; X).
• The supremum norm is a norm on Reg([0, T]; X), and Reg([0, T]; X) is a topological vector space with respect to the topology induced by the supremum norm.
• As noted above, Reg([0, T]; X) is the closure in B([0, T]; X) of Step([0, T]; X) with respect to the supremum norm.
• If X is a Banach space, then Reg([0, T]; X) is also a Banach space with respect to the supremum norm.
• Reg([0, T]; R) forms an infinite-dimensional real Banach algebra: finite linear combinations and products of regulated functions are again regulated functions.
• Since a continuous function defined on a compact space (such as [0, T]) is automatically uniformly continuous, every continuous function f : [0, T] → X is also regulated. In fact, with respect to the supremum norm, the space C⁰([0, T]; X) of continuous functions is a closed linear subspace of Reg([0, T]; X).
• If X is a Banach space, then the space BV([0, T]; X) of functions of bounded variation forms a dense linear subspace of Reg([0, T]; X):

${\displaystyle \mathrm {Reg} ([0,T];X)={\overline {\mathrm {BV} ([0,T];X)}}{\mbox{ w.r.t. }}\|\cdot \|_{\infty }.}$ 

• If X is a Banach space, then a function f : [0, T] → X is regulated if and only if it is of bounded φ-variation for some φ:

${\displaystyle \mathrm {Reg} ([0,T];X)=\bigcup _{\varphi }\mathrm {BV} _{\varphi }([0,T];X).}$ 

• If X is a separable Hilbert space, then Reg([0, T]; X) satisfies a compactness theorem known as the Fraňková–Helly selection theorem.
• The set of discontinuities of a regulated function of bounded variation BV is countable for such functions have only jump-type of discontinuities. To see this it is sufficient to note that given ${\displaystyle \epsilon >0}$ , the set of points at which the right and left limits differ by more than ${\displaystyle \epsilon }$  is finite. In particular, the discontinuity set has measure zero, from which it follows that a regulated function has a well-defined Riemann integral.
• Remark: By the Baire Category theorem the set of points of discontinuity of such function ${\displaystyle F_{\sigma }}$  is either meager or else has nonempty interior. This is not always equivalent with countability.^[2]

• The integral, as defined on step functions in the obvious way, extends naturally to Reg([0, T]; X) by defining the integral of a regulated function to be the limit of the integrals of any sequence of step functions converging uniformly to it. This extension is well-defined and satisfies all of the usual properties of an integral. In particular, the regulated integral
  • is a bounded linear function from Reg([0, T]; X) to X; hence, in the case X = R, the integral is an element of the space that is dual to Reg([0, T]; R);
  • agrees with the Riemann integral.

• Aumann, Georg (1954), Reelle Funktionen, Die Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen mit besonderer Berücksichtigung der Anwendungsgebiete, Bd LXVIII (in German), Berlin: Springer-Verlag, pp. viii+416 MR0061652
• Dieudonné, Jean (1969), Foundations of Modern Analysis, Academic Press, pp. xviii+387 MR0349288
• Fraňková, Dana (1991), "Regulated functions", Math. Bohem., 116 (1): 20–59, ISSN 0862-7959 MR1100424
• Gordon, Russell A. (1994), The Integrals of Lebesgue, Denjoy, Perron, and Henstock, Graduate Studies in Mathematics, 4, Providence, RI: American Mathematical Society, pp. xii+395, ISBN 0-8218-3805-9 MR1288751
• Lang, Serge (1985), Differential Manifolds (Second ed.), New York: Springer-Verlag, pp. ix+230, ISBN 0-387-96113-5 MR772023

• "How to show that a set of discontinuous points of an increasing function is at most countable". Stack Exchange. November 23, 2011.
• "Bounded variation functions have jump-type discontinuities". Stack Exchange. November 28, 2013.
• "How discontinuous can a derivative be?". Stack Exchange. February 22, 2012.