Chapter 11: Regulated, rd-Continuous, and Pre-Differentiable Functions

Theorem \(\PageIndex{1}\)

Assume \(f: \mathbb{T} \rightarrow \mathbb{R}\).
1. If \(f\) is continuous, then \(f\) is rd-continuous.
2. If \(f\) is rd-continuous, then \(f\) is regulated.
3. The jump operator \(\sigma\) is rd-continuous.
4. If \(f\) is regulated or rd-continuous, then so is \(f^\sigma\).
5. Assume \(f\) is continuous. If \(g: \mathbb{T} \rightarrow \mathbb{R}\) is regulated or rd-continuous, then \(f \circ g\) has that property.

Theorem \(\PageIndex{2}\)

Every regulated function on a compact interval is bounded.
Proof. Assume that \(f:[a, b] \rightarrow \mathbb{R},[a, b] \subset \mathbb{T}\), is unbounded. Then, for each \(n \in \mathbb{N}\), there exists \(t_n \in[a, b]\) such that \(\left|f\left(t_n\right)\right|>n\). Because \(\left\{t_n\right\}_{n \in \mathbb{N}} \subset[a, b]\), there exists a subsequence \(\left\{t_{n_k}\right\}_{k \in \mathbb{N}} \subset\left\{t_n\right\}_{n \in \mathbb{N}}\) such that

\[\lim _{k \rightarrow \infty} t_{n_k}=t_0\notag \]

Since \(\mathbb{T}\) is closed, we have that \(t_0 \in \mathbb{T}\). Also, \(t_0\) is a right-dense point or a left-dense point. Using that \(f\) is regulated, we get

\[\left|\lim _{k \rightarrow \infty} f\left(t_{n_k}\right)\right| \neq \infty,\notag \]

which is a contradiction.

Theorem \(\PageIndex{3}\); Mean Value Theorem

If \(f, g: \mathbb{T} \rightarrow \mathbb{R}\) are pre-differentiable with \(D\), then

\[\left|f^{\Delta}(t)\right| \leq\left|g^{\Delta}(t)\right| \quad \text { for all } \quad t \in D\notag \]

implies

\[|f(s)-f(r)| \leq g(s)-g(r) \quad \text { for all } \quad r, s \in \mathbb{T}, \quad r \leq s . \]

Proof. Let \(r, s \in \mathbb{T}\) with \(r \leq s\). Assume also

\[[r, s) \backslash D=\left\{t_n: n \in \mathbb{N}\right\}\notag \]

We take \(\varepsilon>0\). We consider the statements

\[S(t):|f(t)-f(r)| \leq g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right)\notag \]

for \(t \in[r, s]\). We will prove, using the induction principle, that \(S(t)\) is true for all \(t \in[r, s]\).

1. Since \(0 \leq \varepsilon \sum_{t_n<t} 2^{-n}\), the statement \(S(r)\) is true.
2. Let \(t \in[r, s]\) be right-scattered and assume that \(S(t)\) is true. Then we have

\[\begin{aligned}
|f(\sigma(t))-f(r)| & =\left|f(t)+\mu(t) f^{\Delta}(t)-f(r)\right| \\
& \leq \mu(t)\left|f^{\Delta}(t)\right|+|f(t)-f(r)| \\
& \leq \mu(t)\left|f^{\Delta}(t)\right|+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
& \leq \mu(t) g^{\Delta}(t)+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
& \leq g(\sigma(t))-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
& <g(\sigma(t))-g(r)+\varepsilon\left(\sigma(t)-r+\sum_{t_n<\sigma(t)} 2^{-n}\right)
\end{aligned}\notag \]

since \(t<\sigma(t)\). Hence, \(S(\sigma(t))\) is true.

3. Let \(t \in[r, s)\) be right-dense.

a. Let \(t \in D\). Then \(f\) and \(g\) are differentiable at \(t\). Hence, there exists a neighbourhood \(U\) of \(t\) such that

\[\left|f(t)-f(\tau)-f^{\Delta}(t)(t-\tau)\right| \leq \dfrac{\varepsilon}{2}|t-\tau|\notag \]

and

\[\left|g(t)-g(\tau)-g^{\Delta}(t)(t-\tau)\right| \leq \dfrac{\varepsilon}{2}|t-\tau|\notag \]

for all \(\tau \in U\). Thus,

\[|f(t)-f(\tau)| \leq\left(\left|f^{\Delta}(t)\right|+\dfrac{\varepsilon}{2}\right)|t-\tau|\notag \]

for all \(\tau \in U\) and

\[g(\tau)-g(t)+g^{\Delta}(t)(t-\tau) \geq-\dfrac{\varepsilon}{2}|t-\tau|\notag \]

for all \(\tau \in U\), i.e.,

\[g(\tau)-g(t)-g^{\Delta}(t)(\tau-t) \geq-\dfrac{\varepsilon}{2}|t-\tau|\notag \]

for all \(\tau \in U\). Hence, for all \(\tau \in U \cap(t, \infty)\), we have

\[\begin{array}{l}
|f(\tau)-f(r)|=|f(\tau)-f(t)+f(t)-f(r)| \\
\leq|f(\tau)-f(t)|+|f(t)-f(r)| \\
\leq\left(\left|f^{\Delta}(t)\right|+\dfrac{\varepsilon}{2}\right)|t-\tau|+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
\leq\left(g^{\Delta}(t)+\dfrac{\varepsilon}{2}\right)|t-\tau|+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
=g^{\Delta}(t)(\tau-t)+\dfrac{\varepsilon}{2}(\tau-t)+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
\leq g(\tau)-G(T)+\dfrac{\varepsilon}{2}|t-\tau|+\dfrac{\varepsilon}{2}(\tau-t)+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
=g(\tau)-g(r)+\varepsilon(\tau-t)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
=g(\tau)-g(r)+\varepsilon\left(\tau-r+\sum_{t_n<t} 2^{-n}\right)
\end{array}\notag \]

Therefore, \(S(\tau)\) is true for all \(\tau \in U \cap(t, \infty)\).

b. Let \(t \notin D\). Then \(t=t_m\) for some \(m \in \mathbb{N}\). Since \(f\) and \(g\) are pre-differentiable, they both are continuous. Therefore, there exists a neighbourhood \(U\) of \(t\) such that

\[|f(\tau)-f(t)| \leq \dfrac{\varepsilon}{2} 2^{-m} \quad \text { for all } \quad \tau \in U\notag \]

and

\[|g(\tau)-g(t)| \leq \dfrac{\varepsilon}{2} 2^{-m} \quad \text { for all } \quad \tau \in U\notag \]

Therefore,

\[g(\tau)-g(t) \geq-\dfrac{\varepsilon}{2} 2^{-m} \quad \text { for all } \quad \tau \in U\notag \]

Consequently,

\[\begin{aligned}
|f(\tau)-f(r)| & =|f(\tau)-f(t)+f(t)-f(r)| \\
& \leq|f(\tau)-f(t)|+|f(t)-f(r)| \\
& \leq \dfrac{\varepsilon}{2} 2^{-m}+g(t)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right)
\end{aligned}\notag \]

\[\begin{array}{l}
\leq \dfrac{\varepsilon}{2} 2^{-m}+g(\tau)+\dfrac{\varepsilon}{2} 2^{-m}-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
=\varepsilon 2^{-m}+g(\tau)-g(r)+\varepsilon\left(t-r+\sum_{t_n<t} 2^{-n}\right) \\
\leq \varepsilon 2^{-m}+g(\tau)-g(r)+\varepsilon\left(\tau-r+\sum_{t_n<\tau} 2^{-n}\right),
\end{array}\notag \]

so \(S(\tau)\) is true for all \(\tau \in U \cap(t, \infty)\).

4. Let \(t\) be left-dense and assume that \(S(\tau)\) is true for \(\tau \in[r, t)\). Then

\[\begin{aligned}
\lim _{\tau \rightarrow t-}|f(\tau)-f(r)| & \leq \lim _{\tau \rightarrow t-}\left\{g(\tau)-g(r)+\varepsilon\left(\tau-r+\sum_{t_n<\tau} 2^{-n}\right)\right\} \\
& \leq \lim _{\tau \rightarrow t-}\left\{g(\tau)-g(r)+\varepsilon\left(\tau-r+\sum_{t_n<t} 2^{-n}\right)\right\}
\end{aligned}\notag \]

implies that \(S(t)\) is true since \(f\) and \(g\) are continuous at \(t\).

Thus, using the induction principle, if follows that \(S(t)\) is true for all \(t \in[r, s]\). Consequently, (10.1) holds for all \(r \leq s, r, s \in \mathbb{T}\).

Theorem \(\PageIndex{4}\)

Suppose \(f: \mathbb{T} \rightarrow \mathbb{R}\) is pre-differentiable with \(D\). If \(U\) is a compact interval with endpoints \(r, s \in \mathbb{T}\), then

\[|f(s)-f(r)| \leq\left\{\sup _{t \in U^\kappa \cap D}\left|f^{\Delta}(t)\right|\right\}|s-r| .\notag \]

Proof. Without loss of generality, we suppose that \(r \leq s\). We set

\[g(t)=\left\{\sup _{\tau \in U^\kappa \cap D}\left|f^{\Delta}(\tau)\right|\right\}(t-r), \quad t \in \mathbb{T}\notag \]

Then

\[g^{\Delta}(t)=\left\{\sup _{\tau \in U^\kappa \cap D}\left|f^{\Delta}(\tau)\right|\right\} \geq\left|f^{\Delta}(t)\right|\notag \]

for all \(t \in D \cap[r, s]^\kappa\). Thus, using Theorem 0.8, we obtain

\[|f(t)-f(r)| \leq g(t)-g(r) \quad \text { for all } \quad t \in[r, s]\notag \]

whereupon

\[|f(s)-f(r)| \leq g(s)-g(r)=g(s)=\left\{\sup _{t \in U^\kappa \cap D}\left|f^{\Delta}(t)\right|\right\}(s-r)\notag \]

Theorem \(\PageIndex{5}\)

Let \(f\) be pre-differentiable with \(D\). If \(f^{\Delta}(t)=0\) for all \(t \in D\), then \(f\) is a constant function.

Proof. Let \(U\) be a compact interval with endpoints \(r, s \in \mathbb{T}\). From Theorem 0.9, it follows that for all \(r, s \in \mathbb{T}\),

\[|f(s)-f(r)| \leq\left\{\sup _{t \in U^\kappa \cap D}\left|f^{\Delta}(t)\right|\right\}|s-r|=0\notag \]

i.e., \(f(s)=f(r)\). Therefore, \(f\) is a constant function.

Theorem \(\PageIndex{6}\)

Let \(f\) and \(g\) be pre-differentiable with \(D\). If \(f^{\Delta}(t)=g^{\Delta}(t)\) for all \(t \in D\), then

\[g(t)=f(t)+C \quad \text { for all } \quad t \in \mathbb{T}\notag \]

where \(C\) is a constant.

Proof. Let \(h(t)=f(t)-g(t), t \in \mathbb{T}\). Then

\[h^{\Delta}(t)=f^{\Delta}(t)-g^{\Delta}(t)=0 \quad \text { for all } \quad t \in D\notag \]

Theorem \(\PageIndex{7}\)

Suppose \(f_n: \mathbb{T} \rightarrow \mathbb{R}\) is pre-differentiable with \(D\) for each \(n \in \mathbb{N}\). Assume that for each \(t \in \mathbb{T}^\kappa\), there exists a compact interval neighbourhood \(U(t)\) such that the sequence \(\left\{f_n^{\Delta}\right\}_{n \in \mathbb{N}}\) converges uniformly on \(U(t) \cap D\).

(i) If \(\left\{f_n\right\}_{n \in \mathbb{N}}\) converges at some \(t_0 \in U(t)\) for some \(t \in \mathbb{T}^\kappa\), then it converges uniformly on \(U(t)\).
(ii) If \(\left\{f_n\right\}_{n \in \mathbb{N}}\) converges at some \(t_0 \in \mathbb{T}\), then it converges uniformly on \(U(t)\) for all \(t \in \mathbb{T}^\kappa\).
(iii) The limit mapping \(f=\lim _{n \rightarrow \infty} f_n\) is pre-differentiable with \(D\) and

\[f^{\Delta}(t)=\lim _{n \rightarrow \infty} f_n^{\Delta}(t) \quad \text { for all } \quad t \in D\notag \]

Proof.

1. (i) \(]\) Since \(\left\{f_n^{\Delta}\right\}_{n \in \mathbb{N}}\) converges uniformly on \(U(t) \cap D\), there exists \(N \in \mathbb{N}\) such that

\[\sup _{s \in U(t) \cap D}\left|\left(f_m-f_n\right)^{\Delta}(s)\right|\notag \]

is finite for all \(m, n \in \mathbb{N}\). Let \(m, n \geq N\) and \(r \in U(t)\). Then

\[\begin{aligned}
\left|f_n(r)-f_m(r)\right| & =\left|f_n(r)-f_m(r)-\left(f_n\left(t_0\right)-f_m\left(t_0\right)\right)+\left(f_n\left(t_0\right)-f_m\left(t_0\right)\right)\right| \\
& \leq\left|f_n\left(t_0\right)-f_m\left(t_0\right)\right|+\left\{\sup _{s \in U(t) \cap D}\left|\left(f_n-f_m\right)^{\Delta}(s)\right|\right\}\left|r-t_0\right|
\end{aligned}\notag \]

Hence, \(\left\{f_n\right\}_{n \in \mathbb{N}}\) converges uniformly on \(U(t)\), i.e., \(\left\{f_n\right\}_{n \in \mathbb{N}}\) is a locally uniformly convergent sequence.

(ii) Assume \(\left\{f_n\left(t_0\right)\right\}_{n \in \mathbb{N}}\) converges for some \(t_0 \in \mathbb{T}\). Consider the statement

\[S(t):\left\{f_n(t)\right\}_{n \in \mathbb{N}} \quad \text { converges. }\notag \]

a. Since \(\left\{f_n\left(t_0\right)\right\}\) converges, the statement \(S\left(t_0\right)\) is true.

b. Let \(t\) be right-scattered and assume that \(S(t)\) is true. Then

\[f_n(\sigma(t))=f_n(t)+\mu(t) f_n^{\Delta}(t)\notag \]

converges by the assumption, i.e., \(S(\sigma(t))\) is true.

c. Let \(t\) be right-dense and assume that \(S(t)\) is true. Then, by (i), \(\left\{f_n\right\}_{n \in \mathbb{N}}\) converges on \(U(t)\), and so \(S(r)\) is true for all \(r \in U(t) \cap(t, \infty)\).

d. Let \(t\) be left-dense and assume that \(S(r)\) is true for all \(t_0 \leq r<t\). Since \(U(t) \cap\left[t_0, t\right) \neq \emptyset\), using again part (i), we have that \(\left\{f_n\right\}_{n \in \mathbb{N}}\) converges on \(U(t)\); in particular, \(S(t)\) is true.

Consequently, \(S(t)\) is true for all \(t \in\left[t_0, \infty\right)\). Using the dual version of the induction principle for the negative direction, we have that \(S(t)\) is also true for all \(t \in\left(-\infty, t_0\right]\) (we note that the first part of this has already been shown, the second part follows by \(f_n(\rho(t))=f_n(t)-\mu(\rho(t)) f_n^{\Delta}(\rho(t))\), the third part and the fourth part follow again by (i)).

(iii) Let \(t \in D\). Without loss of generality, we can assume that \(\sigma(t) \in U(t)\). We take \(\varepsilon>0\) arbitrarily. Using (i), there exists \(N \in \mathbb{N}\) such that

\[\left|\left(f_n-f_m\right)(r)-\left(f_n-f_m\right)(\sigma(t))\right| \leq\left\{\sup _{s \in U(t) \cap D}\left|\left(f_n-f_m\right)^{\Delta}(s)\right|\right\}|r-\sigma(t)|\notag \]

for all \(r \in U(t)\) and all \(m, n \geq N\). Since

\[\left\{f_n^{\Delta}\right\}_{n \in \mathbb{N}}\notag \]

converges uniformly on \(U(t) \cap D\), there exists \(N_1 \geq N\) such that

\[\sup _{s \in U(t) \cap D}\left|\left(f_n-f_m\right)^{\Delta}(s)\right| \leq \dfrac{\varepsilon}{3} \quad \text { for all } \quad m, n \geq N_1\notag \]

Hence,

\[\left|\left(f_n-f_m\right)(r)-\left(f_n-f_m\right)(\sigma(t))\right| \leq \dfrac{\varepsilon}{3}|r-\sigma(t)|\notag \]

for all \(r \in U(t)\) and for all \(m, n \geq N_1\). Now, letting \(m \rightarrow \infty\), we obtain

\[\left|\left(f_n-f\right)(r)-\left(f_n-f\right)(\sigma(t))\right| \leq \dfrac{\varepsilon}{3}|r-\sigma(t)|\notag \]

for all \(r \in U(t)\) and all \(n \geq N_1\). Let

\[g=\lim _{n \rightarrow \infty} f_n^{\Delta}\notag \]

Then there exists \(M \geq N_1\) such that

\[\left|f_M^{\Delta}(t)-g(t)\right| \leq \dfrac{\varepsilon}{3}\notag \]

and since \(f_M\) is differentiable at \(t\), there also exists a neighbourhood \(W\) of \(t\) with

\[\left|f_M(\sigma(t))-f_M(r)-f_M^{\Delta}(t)(\sigma(t)-r)\right| \leq \dfrac{\varepsilon}{3}|\sigma(t)-r|\notag \]

for all \(r \in W\). From here, for all \(r \in U(t) \cap W\), we get

\[|f(\sigma(t))-f(r)-g(t)| \sigma(t)-r| | \leq\left|\left(f_M-f\right)(\sigma(t))-\left(f_M-f\right)(r)\right|\notag \]

\[\begin{aligned}
& +\left|f_M^{\Delta}(t)-g(t)\right||\sigma(t)-r|+\left|f_M(\sigma(t))-f_M(r)-f_M^{\Delta}(t)\right| \sigma(t)-r| | \\
\leq & \dfrac{\varepsilon}{3}|\sigma(t)-r|+\dfrac{\varepsilon}{3}|\sigma(t)-r|+\dfrac{\varepsilon}{3}|\sigma(t)-r| \\
= & \varepsilon|\sigma(t)-r|
\end{aligned}\notag \]

Consequently, \(f\) is differentiable at \(t\) with \(f^{\Delta}(t)=g(t)\).

Theorem \(\PageIndex{8}\)

Let \(t_0 \in \mathbb{T}, x_0 \in \mathbb{R}\). Assume \(f: \mathbb{T}^\kappa \rightarrow \mathbb{R}\) is regulated. Then there exists exactly one pre-differentiable function \(f\) with \(D\) satisfying

\[F^{\Delta}(t)=f(t) \quad \text { for all } \quad t \in D, \quad F\left(t_0\right)=x_0 .\notag \]

Proof. Let \(n \in \mathbb{N}\) and consider the statement

\[S(t):\left\{\begin{array}{l}
\text { there exists a pre-differentiable } \quad\left(F_{n t}, D_{n t}\right), \\
F_{n t}:\left[t_0, t\right] \rightarrow \mathbb{R} \quad \text { with } \quad F_{n t}\left(t_0\right)=x_0 \quad \text { and } \\
\left|F_{n t}^{\Delta}(s)-f(s)\right| \leq \dfrac{1}{n} \quad \text { for } \quad s \in D_{n t} .
\end{array}\right.\notag \]

1. Let \(t=t_0, D_{n t_0}=\emptyset\) and \(F_{n t_0}\left(t_0\right)=x_0\). Then the statement \(S\left(t_0\right)\) is true.

2. Let \(t\) be right-scattered and assume that \(S(t)\) is true. Define

\[D_{n \sigma(t)}=D_{n t} \cup\{t\}\notag \]

and \(F_{n \sigma(t)}\) on \(\left[t_0, \sigma(t)\right]\) by

\[F_{n \sigma(t)}(s)=\left\{\begin{array}{ll}
F_{n t}(s) & \text { if } \quad s \in\left[t_0, t\right] \\
F_{n t}(t)+\mu(t) f(t) & \text { if } \quad s=\sigma(t)
\end{array}\right.\notag \]

Then

\[\begin{array}{c}
F_{n \sigma(t)}\left(t_0\right)=F_{n t}\left(t_0\right)=x_0 \\
\left|F_{n \sigma(t)}^{\Delta}(s)-f(s)\right|=\left|F_{n t}^{\Delta}(s)-f(s)\right| \leq \dfrac{1}{n} \quad \text { for } \quad s \in D_{n t}
\end{array}\notag \]

and

\[\begin{aligned}
\left|F_{n \sigma(t)}^{\Delta}(t)-f(t)\right| & =\left|\dfrac{F_{n \sigma(t)}(\sigma(t))-F_{n \sigma(t)}(t)}{\mu(t)}-f(t)\right| \\
& =\left|\dfrac{F_{n t}(t)+\mu(t) f(t)-F_{n \sigma(t)}(t)}{\mu(t)}-f(t)\right| \\
& =\left|\dfrac{F_{n t}(t)+\mu(t) f(t)-F_{n t}(t)}{\mu(t)}-f(t)\right|
\end{aligned}\notag \]

\[\begin{array}{l}
=\left|\dfrac{\mu(t) f(t)}{\mu(t)}-f(t)\right| \\
=0 \\
\leq \dfrac{1}{n}
\end{array}\notag \]

and therefore the statement \(S(\sigma(t))\) is true.

3. Let \(t\) be right-dense and assume that \(S(t)\) is true. Since \(t\) is right-dense and \(f\) is regulated,

\[f\left(t^{+}\right)=\lim _{s \rightarrow t, s>t} f(s) \quad \text { exists. }\notag \]

Hence, there exists a neighbourhood \(U\) of \(t\) with

\[\left|f(s)-f\left(t^{+}\right)\right| \leq \dfrac{1}{n} \quad \text { for all } \quad s \in U \cap(t, \infty) . \]

Let \(r \in U \cap(t, \infty)\). Define

\[D_{n r}=\left(D_{n t} \backslash\{t\}\right) \cup[t, r]^\kappa\notag \]

and \(F_{n r}\) on \(\left[t_0, r\right]\) by

\[F_{n r}(s)=\left\{\begin{array}{ll}
F_{n t}(s) & \text { if } \quad s \in\left[t_0, t\right] \\
F_{n t}(t)+f\left(t^{+}\right)(s-t) & \text { if } \quad s \in(t, r] .
\end{array}\right.\notag \]

Then \(F_{n r}\) is continuous at \(t\) and hence on \(\left[t_0, r\right]\). Also, \(F_{n r}\) is differentiable on \((t, r)^\kappa\) with

\[F_{n r}^{\Delta}(s)=f\left(t^{+}\right) \quad \text { for all } \quad s \in(t, r]^\kappa .\notag \]

Hence, \(F_{n r}\) is pre-differentiable on \(\left[t_0, t\right)\). Since \(t\) is right-dense, we have that \(F_{n r}\) is pre-differentiable with \(D_{n r}\).

From here and from (10.2), we also have

\[\left|F_{n r}^{\Delta}(s)-f(s)\right| \leq \dfrac{1}{n} \quad \text { for all } \quad s \in D_{n r}\notag \]

Therefore, the statement \(S(r)\) is true for all \(r \in U \cap(t, \infty)\).

4. Now, let \(t\) be left-dense and suppose that \(S(r)\) is true for \(r<t\). Since \(f\) is regulated,

\[f\left(t^{-}\right)=\lim _{s \rightarrow t, s<t} f(s) \quad \text { exists. } \]

Hence, there exists a neighbourhood \(U\) of \(t\) with

\[\left|f(s)-f\left(t^{-}\right)\right| \leq \dfrac{1}{n} \quad \text { for all } \quad s \in U \cap(-\infty, t) .\notag \]

Fix some \(r \in U \cap(-\infty, t)\) and define

\[D_{n t}=\left\{\begin{array}{ll}
D_{n r} \cup(r, t) & \text { if } r \text { is right-dense } \\
D_{n r} \cup[r, t) & \text { if } r \text { is right-scattered }
\end{array}\right.\notag \]

and \(F_{n t}\) on \(\left[t_0, t\right]\) by

\[F_{n t}(s)=\left\{\begin{array}{ll}
F_{n r}(s) & \text { if } \quad s \in\left(t_0, r\right] \\
F_{n r}(t)+f\left(t^{-}\right)(s-r) & \text { if } \quad s \in(r, t]
\end{array}\right.\notag \]

We note that \(F_{n t}\) is continuous at \(r\) and hence in \(\left[t_0, t\right]\). Since

\[F_{n t}^{\Delta}(s)=f\left(t^{-}\right) \quad \text { for all } \quad s \in(r, t]\notag \]

\(F_{n t}\) is pre-differentiable with \(D_{n t}\) and

\[\left|F_{n t}^{\Delta}(s)-f(s)\right| \leq \dfrac{1}{n} \quad \text { for all } \quad s \in D_{n t}\notag \]

Hence, the statement \(S(t)\) is true.

By the induction principle, it follows that \(S(t)\) is true for all \(t \geq t_0, t \in \mathbb{T}\). Similarly, we can show that \(S(t)\) is valid for \(t \leq t_0\). Hence, \(F_n\) is pre-differentiable with \(D_n, F_n\left(t_0\right)=x_0\), and

\[\left|F_n^{\Delta}(t)-f(t)\right| \leq \dfrac{1}{n} \quad \text { for all } \quad t \in D_n .\notag \]

Now, let

\[F=\lim _{n \rightarrow \infty} F_n \quad \text { and } \quad D=\bigcap_{n \in \mathbb{N}} D_n .\notag \]

Then \(F\left(t_0\right)=x_0, f\) is pre-differentiable on \(D\), and, using Theorem \(\PageIndex{7}\),

\[F^{\Delta}(t)=\lim _{n \rightarrow \infty} F_n^{\Delta}(t)=f(t) \quad \text { for all } \quad t \in D\notag \]