11.5 Regulated Functions
Proposition 11.5.1. Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map.
Let $G: [a, b] \to F$ and $[c, d] \subset [a, b]$ such that $G$ is continuous at $c$ and $d$, then for any $x \in E$, $x \cdot \one_{[c, d]}\in RS([a, b], G)$, and
Proof. Assume without loss of generality that $a < c \le d < b$. Let $U \in \cn_{H}(0)$, then there exists $V \in \cn_{F}(0)$ such that $xV \subset U$. By continuity of $G$, there exists $\delta > 0$ such that $G((c - \delta, c]) - G(c) \subset V$ and $G([d, d + \delta)) - G(d) \subset V$. In which case, for any tagged partition $(P = \bracsn{x_j}_{0}^{n}, t = \seqf{t_j})$ that contains $\bracs{c - \delta, c, d, d + \delta}$,
Definition 11.5.2 (Step Map). Let $[a, b] \subset \real$, $E$ be a TVS, $f: [a, b] \to E$ be a function, and $P = \bracsn{x_j}_{1}^{n} \in \scp([a, b])$, then $f$ is a step map with respect to $P$ if for each $1 \le j \le n$, $f$ is constant on $(x_{j - 1}, x_{j})$.
Definition 11.5.3 (Regulated Function). Let $[a, b] \subset \real$, $E, F, H$ be TVSs over $K \in \RC$, and $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous linear map.
Let $G: [a, b] \to F$ and $f: [a, b] \to E$ be a step map, then $f$ is regulated with respect to $G$ if $f$ is continuous on all discontinuity points of $G$. Let $\text{Reg}([a, b], G; E)$ be closure of all regulated step maps with respect to the uniform norm, then:
Every regulated step map is in $RS([a, b], G)$.
If $E$ is metrisable, then for any $f \in \text{Reg}([a, b], G; E)$, $f$ is continuous at all but at most countably many points, and $f$ does not share any discontinuity points with $E$.
If $E, F, H$ are locally convex, $G \in BV([a, b]; F)$, and $H$ is complete, then $\text{Reg}([a, b], G; E) \subset RS([a, b], G)$.
If $F$ is normed and $G \in BV([a, b]; F)$, then $\text{Reg}([a, b], G; E) \supset C([a, b]; E)$.
The set $\text{Reg}([a, b], G; E)$ is the space of regulated functions with respect to $G$. If $G = \text{Id}$, then $\text{Reg}([a, b]; E) = \text{Reg}([a, b], G; E)$ denotes the space of regulated functions on $[a, b]$.
Proof. (1): If $f: [a, b] \to E$ is a regulated step map, then there exists $\seqf{[a_j, b_j]}\subset 2^{[a, b]}$ and $\seqf{x_j}$ such that
$f = \sum_{j = 1}^{n} x_{j} \one_{[a_j, b_j]}$.
For each $1 \le j \le n$, $G$ is continuous at $a_{j}$ and $b_{j}$.
Thus $f \in RS([a, b], G)$ by Proposition 11.5.1.
(2): Let $f \in \text{Reg}([a, b], G; E)$, then there exists regulated step maps $\seq{f_n}\subset \text{Reg}([a, b], G; E)$ such that $f_{n} \to f$ uniformly. For each $n \in \natp$, let $D_{n}$ be the set of discontinuity points of $f_{n}$. Let $x \in [a, b] \setminus \bigcup_{n \in \natp}D_{n}$. For any symmetric neighbourhood $U \in \cn_{E}(0)$, there exists $n \in \natp$ such that $(f_{n} - f)([a, b]) \subset U$. Since $f_{n}$ is continuous at $x$, there exists $\eps > 0$ such that $f_{n}((x - \eps, x + \eps)) - f_{n}(x) \in U$. In which case,
Therefore $f$ is continuous at $x$. Since each $D_{n}$ is finite, $\bigcup_{n \in \natp}D_{n}$ is countable. Given that $G$ is continuous on every point in each $D_{n}$, $G$ is also continuous on all the discontinuity points of $f$.
(3): Suppose that $E, F, H$ are locally convex, $G \in BV([a, b]; F)$, and $H$ is complete. Let $f \in \text{Reg}([a, b], G; E)$, then there exists regulated step maps $\net{f}\subset \text{Reg}([a, b], G; E)$ such that $f_{\alpha} \to f$ uniformly. By Proposition 11.4.2, $f \in RS([a, b], G)$. Therefore $\text{Reg}([a, b], G; E) \subset RS([a, b], G)$.
(4): If $F$ is normed and $G \in BV([a, b]; F)$, then $G$ has at most countably many discontinuities by Definition 11.2.3, so the continuity points of $G$ are dense in $[a, b]$.
Let $f \in C([a, b]; E)$, then $f \in UC([a, b]; E)$ by Proposition 5.2.8. Let $U \in \cn_{E}(0)$, then there exists a partition $P = \bracsn{x_j}_{1}^{n} \in \scp([a, b])$ such that for each $1 \le j \le n$ and $x, y \in [x_{j-1}, x_{j}]$, $f(x) - f(y) \in U$. In which case, let
then $\phi$ is a regulated step map with $(\phi - f)([a, b]) \in U$.$\square$
Theorem 11.5.4 (Fundamental Theorem of Calculus for Riemann Integrals). Let $[a, b] \subset \real$ and $E$ be a separated locally convex space, then:
For any Riemann integrable function $f: [a, b] \to E$ and $x \in (a, b)$ such that $f$ is continuous at $x$, the function
\[F: [a, b] \to E \quad y \mapsto \int_{a}^{y} f(t)dt\]is differentiable at $x$ with $DF(x) = f(x)$.
For $F \in C^{1}([a, b]; E)$,
\[F(b) - F(a) = \int_{a}^{b} DF(t)dt\]
Proof. (1): Let $x_{0} \in (a, b)$ such that $f$ is continuous at $x$, and $\delta > 0$ such that $(x_{0} - \delta, x_{0} + \delta) \subset (a, b)$, then for any $h > 0$,
As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.
(2): Let $G(x) = \int_{a}^{x} DF(t)dt + F(a)$, then $G - F$ has derivative $0$. By the Mean Value Theorem, $G - F$ is constant. As $G(a) - F(a) = 0$, $G = F$.$\square$