Chapter 11: Absolute Continuous Functions

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Let $a, b\in \mathbb{T}$, $a<b$. For a given function $f: \mathbb{T}\to \mathbb{R}$, consider an auxiliary function $\overline{f}: [a, b]\to \mathbb{R}$ defined by \begin{equation} \label{80.6} \overline{f}(t)=\left\{ \begin{array}{l} f(t)\quad \text{if}\quad t\in \mathbb{T}\\ \\ f(t_i)+\frac{f(\sigma(t_i))-f(t_i)}{\mu(t_i)}(t-t_i)\quad \text{if}\quad t\in (t_i, \sigma(t_i)),\quad \text{for}\quad \text{some}\quad i\in I, \end{array} \right. \end{equation} with $I\subset \mathbb{N}$ and $\{t_i\}_{i\in I}$ given by

\[ R=\{t\in \mathbb{T}: t<\sigma(t)\}=\{t_i\}_{i\in I}. \notag\]

Theorem

Let $f: \mathbb{T}\to\mathbb{R}$ and $\overline{f}: [a, b]\to \mathbb{R}$ is the extension of $f$, defined by

\[ \overline{f}(t)=\left\{ \begin{array}{l} f(t)\quad \text{if}\quad t\in \mathbb{T}\\ \\ f(t_i)+\frac{f(\sigma(t_i))-f(t_i)}{\mu(t_i)}(t-t_i)\quad \text{if}\quad t\in (t_i, \sigma(t_i)),\quad \text{for}\quad \text{some}\quad i\in I, \end{array} \right.\notag\]

Then the following statements are equivalent

1. $f$ maps every $\Delta$-null subset of $\mathbb{T}$ into a null set.

2. $\overline{f}$ maps every null subset of $[a, b]$ into a null set.

Proof

1. Let $E\subset [a, b]$ be a null set. We will prove that $\overline{f}(E)$ is a null set.

(a) Let $b\notin E$. Then

\[ E= \left(\bigcup_{i\in I} (E\bigcap [t_i, \sigma(t_i))\right)\bigcup \left(E\bigcap (\mathbb{T}\backslash \mathbb{R} )\right) \notag\]

and \begin{equation} \label{80.8} \begin{array}{lll} \mu(\overline{f}(E))&\leq& \sum_{i\in I} \mu\left(\overline{f}(E\bigcap [t_i, \sigma(t_i)])\right)\\ &&+ \mu(\overline{f}(E\bigcap (\mathbb{T}\backslash R))). \end{array} \end{equation} Fix $i\in I$. By the definition of the function $\overline{f}$, it follows that $\overline{f}\Bigl|_{[t_i, \sigma(t_i)]}$ is absolutely continuous. Hence,

\[ \mu((E\bigcap [t_i, \sigma(t_i)))=0. \notag\]

Now, applying the Banach-Zarecki Theorem, we get \begin{equation} \label{80.9} \mu(\overline{f}(E\bigcap [t_i, \sigma(t_i))))=0. \end{equation} Moreover, since the set $E\bigcap (\mathbb{T}\backslash R)$ does not have any right-scattered points and $b\notin E\bigcap (\mathbb{T}\backslash R)$, we get

\[ \begin{aligned} \mu_{\Delta}(E\bigcap (\mathbb{T}\backslash R))=& \mu(E \bigcap (\mathbb{T}\backslash R))\\ =& 0. \end{aligned} \notag\]

Hence and $\overline{f}\Bigl|_{\mathbb{T}}=f$ and our assumptions, we obtain

\[ \begin{aligned} \mu(\overline{f}(E\bigcap (\mathbb{T}\backslash R)))=& \mu(f(E\bigcap (\mathbb{T}\backslash R))\\ =& 0. \end{aligned} \notag\]

From the last relation and from \eqref{80.8}, \eqref{80.9}, we get

\[ \mu(\overline{f}(E))=0. \notag\]

(b) Let $b\in E$. Then we consider both sets $E\backslash \{b\}$ and $\{b\}$. Since

\[ \mu(\overline{f}(\{b\}))=0, \notag\]

we obtain

\[ \begin{aligned} \mu(\overline{f}(E))\leq& \mu(\overline{f}(E\backslash \{b\}))+\mu( \overline{f}(\{b\}))\\ =& 0, \end{aligned} \notag\]

i.e.,

\[ \mu(\overline{f}(E))=0. \notag\]

2. Let $E\subset \mathbb{T}$ be a $\Delta$-null set.

(a) Let $b\notin E$. Then, we obtain

\[ \begin{aligned} 0=& \mu_{\Delta}(E)\\ =& \sum_{i\in I_E}(\sigma(t_i)-t_i) +\mu(E)\\ =& \sum_{i\in I_E}(\sigma(t_i)-t_i). \end{aligned} \notag\]

Therefore, the set $E$ does not have any right-scattered points. So,

\[ f(E)=\overline{f}(E) \notag\]

and the assertion follows.

(b) Let $b\in E$. Consider both sets $E\backslash \{b\}$ and $\{b\}$. By the previous case, it follows that $E\backslash \{b\}$ does not have any right-scattered points.Then $\overline{f}(E\backslash \{b\})$ is a null set. Since

\[ \mu(\{b\})=0, \notag\]

then, we have that $\overline{f}(\{b\})$ is a null set. Consequently $\overline{f}(E)$ is a null set. This completes the proof.

Definition

For a given function $f: \mathbb{T}\to \mathbb{R}$ and a partition $P=\{x_0, x_1, \ldots, x_n\}$ of $\mathbb{T}$, define

\[ V(P, f)= \sum_{k=1}^N |f(x_i)-f(x_{i-1})|. \notag\]

We define the total variation of the function $f$ as follows.

\[ \bigvee_a^b (f)= \sup \{V(P, f): P\quad \text{partition}\quad \text{of}\quad \mathbb{T}\}. \notag\]

If $\bigvee_a^b(f)<\infty$, we say that the function $f$ is of bounded variation on the time scale $\mathbb{T}$.

Note that if $f: \mathbb{T}\to \mathbb{R}$ is of bounded variation, then $\overline{f}: \mathbb{T}\to \mathbb{R}$, defined by \eqref{80.6}, is of bounded variation.

Definition

A function $f: \mathbb{T}\to \mathbb{R}$ is said to be absolutely continuous on $\mathbb{T}$ if for every $\epsilon>0$ there exists a $\delta>0$ such that if $\{(a_k, b_k)\bigcap \mathbb{T}\}_{k=1}^n$, with $a_k, b_k\in \mathbb{T}$, is a finite partition disjoint family of subintervals of $\mathbb{T}$ satisfying

\[ \sum_{k=1}^n (b_k-a_k)<\delta, \notag\]

then

\[ \sum_{k=1}^n |f(b_k)-f(a_k)|<\epsilon. \notag\]

Theorem

A function $f: \mathbb{T}\to \mathbb{R}$ is absolutely continuous on $\mathbb{T}$ if and only if the following conditions hold.

1. $f$ is continuous and of bounded variation on $\mathbb{T}$.

2. $f$ maps every $\Delta$-null subset of $\mathbb{T}$ into a null set.

Proof

Let $\overline{f}$ be the extension of the function $f$, defined by \eqref{80.6}.

1. Let $f$ is absolutely continuous on $\mathbb{T}$. Then $f$ is continuous on $\mathbb{T}$. Now we will prove that it is of bounded variation on $\mathbb{T}$. Let $\delta>0$ corresponds to $\epsilon=1$ as in the definition of absolutely continuous function on $\mathbb{T}$. Then there exists a partition $P=\{x_0, x_1, \ldots, x_n\}$ such that for any $k\in \{1, \ldots, n\}$ either

\[ x_k-x_{k-1}\leq \frac{\delta}{2} \notag\]

\[ x_k-x_{k-1}>\frac{\delta}{2}\quad \text{and}\quad \sigma(x_{k-1})=x_k. \notag\]

If $x_k-x_{k-1}\leq \frac{\delta}{2}$, then

\[ \bigvee_{x_{k-1}}^{x_k}(f)\leq 1. \notag\]

Thus, $f\Bigl|_{[x_{k-1}, x_k]\bigcap \mathbb{T}}$ is of bounded variation on $[x_{k-1}, x_k]\bigcap \mathbb{T}$. If $x_k-x_{k-1}>\frac{\delta}{2}$, then

\[ \bigvee_{x_{k-1}}^{x_k}(f)= |f(x_k)-f(x_{k-1})| \notag\]

and $f\Bigl|_{[x_{k-1}, x_k]\bigcap \mathbb{T}}$ is of bounded variation on $[x_{k-1}, x_k]\bigcap \mathbb{T}$. Therefore $\overline{f}\Bigl|_{[x_{k-1}, x_k]}$ is of bounded variation on $[x_{k-1}, x_k]$. Consequently $\overline{f}$ is of bounded variation on $\mathbb{T}$. Let now $E\subset \mathbb{T}$ is a $\Delta$-null set. We will prove that $f(E)$ is a null set. Let $\epsilon>0$ be arbitrarily chosen and $\delta>0$ corresponds to $\epsilon$ as in the definition for absolutely continuous function. Because $\mu_{\Delta}(E)=0$, we have that $b\notin E$ and it has not any right-scattered points. Then we may choose a family $\{[a_k, b_k]\bigcap \mathbb{T}\}_{k=1}^{\infty}$ with $a_k, b_k\in \mathbb{T}$, of pairwise disjoint subintervals of $\mathbb{T}$ such that

\[ E\subset \bigcup_{k=1}^{\infty} ([a_k, b_k]\bigcap \mathbb{T}) \notag\]

and

\[ \sum_{k=1}^{\infty} (b_k-a_k)<\delta. \notag\]

Hence,

\[ \begin{aligned} f(E)\subset& f\left(\bigcup_{k=1}^{\infty} ([a_k, b_k)\bigcap \mathbb{T})\right)\\ =& \bigcup_{k=1}^{\infty} f([a_k, b_k)\bigcap \mathbb{T}) \end{aligned} \notag\]

and

\[ \begin{aligned} \mu(f(E))\subset& \mu\left(f\left(\bigcup_{k=1}^{\infty} ([a_k, b_k)\bigcap \mathbb{T})\right)\right)\\ =& \bigcup_{k=1}^{\infty} \mu\left(f([a_k, b_k)\bigcap \mathbb{T})\right). \end{aligned} \notag\]

Because $f$ is continuous on $[a_k, b_k]\bigcap \mathbb{T}$, it follows that there exist $c_k, d_k \in [a_k, b_k]\bigcap \mathbb{T}$, $c_k<d_k$, such that

\[ \begin{aligned} \mu\left(f([a_k, b_k)\bigcap \mathbb{T})\right)=& \mu\left(f([a_k, b_k]\bigcap \mathbb{T})\right)\\ \leq& |f(d_k)-f(c_k)|. \end{aligned} \notag\]

Since $[c_k, d_k]\subset [a_k, b_k]$ and $f$ is absolutely continuous on $\mathbb{T}$, we have

\[ \sum_{k=1}^n |f(d_k)-f(c_k)|<\epsilon \notag\]

and

\[ \sum_{k=1}^{\infty} |f(d_k)-f(c_k)|\leq \epsilon. \notag\]

From here,

\[ \begin{aligned} \mu(f(E))\leq& \sum_{k=1}^{\infty} \mu\left(f([a_k, b_k)\bigcap \mathbb{T})\right)\\ \leq& \sum_{k=1}^{\infty} |f(d_k)-f(c_k)|\\ \leq& \epsilon. \end{aligned} \notag\]

Because $\epsilon>0$ was arbitrarily chosen, we get that

\[ \mu(f(E))=0. \notag\]

2. It follows that $\overline{f}$ maps every null subset of $[a, b]$ into a null set. Hence and the Banach-Zarecki theorem, it follows that $\overline{f}$ is absolutely continuous on $[a, b]$. By the equality $f=\overline{f}\Bigl|_{\mathbb{T}}$, we conclude that $f$ is absolutely continuous on $\mathbb{T}$. This completes the proof.

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