I want to prove the following:
Proposition$\quad$If $F:\mathbb{R}\to\mathbb{R}$ is absolutely continuous and of finite variation, then $V_F$ is absolutely continuous.
Here is my attempt so far:
Let $\epsilon>0$. By the absolute continuity of $F$, we can choose a $\delta$ such that $\sum_{i=1}^n|F(t_i)-F(s_i)|<\epsilon$ holds whenever $\{(s_i,t_i)\}_{i=1}^n$ is a finite sequence of disjoint open intervals for which $\sum_{i=1}^n(t_i-s_i)<\delta$. Then each finite sequence $\{(u_j,v_j)\}_{j=i}^m$ of disjoint open subintervals of $\bigcup_{i=1}^n(s_i,t_i)$ satisfies $\sum_{j=1}^m(v_j-u_j)<\delta$ and so satisfies $\sum_{j=1}^m|F(v_j)-F(u_j)|<\epsilon$.
From here I want to write:
Since\begin{align*}\sum_{i=1}^nV_F[s_i,t_i] = \sup\left\{\sum_{j=1}^m|F(v_j)-F(u_j)|:\ \text{$\{(u_j,v_j)\}_{j=1}^m$ is a finite sequence of disjoint open subintervals of $\bigcup_{i=1}^n(s_i,t_i)$}\right\},\end{align*}we have that $\sum_{i=1}^nV_F[s_i,t_i]\leq\epsilon$, and thus$$\sum_{i=1}^n|V_F(t_i)-V_F(s_i)| = \sum_{i=1}^nV_F[s_i,t_i] \leq \epsilon.$$
But I am not sure if the "supremum step" is correct. (It intuitively makes sense to me, but I am not sure how prove it formally.) Could someone please help me out? Thank you very much in advance!
I realized that perhaps the following would make the question more clear:
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes the interval $[a,b]$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} a \leq t_0 < t_1 < \dots < t_n \leq b.\end{align*}Then $V_F[a,b]$, the variation of $F$ over $[a,b]$} is defined by\begin{align*} V_F[a,b] = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{i=0}^n\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) on $[a,b]$ if $V_F[a,b]$ is finite.
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes the interval $(-\infty,b]$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} -\infty < t_0 < t_1 < \dots < t_n \leq b.\end{align*}Then $V_F(-\infty,b]$, the variation of $F$ over $(-\infty,b]$, is defined by\begin{align*} V_F(-\infty,b] = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{i=0}^{n}\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) on $(-\infty,b]$ if $V_F(-\infty,b]$ is finite.
Definition$\quad$ Suppose that $F$ is a real-valued function whose domain includes $\mathbb{R}$. Let $\mathscr{S}$ be the collection of finite sequences $\{t_i\}_{i=0}^n$ such that\begin{align*} -\infty < t_0 < t_1 < \dots < t_n < +\infty.\end{align*}Then $V_F(-\infty,+\infty)$, the variation of $F$ over $\mathbb{R}$, is defined by\begin{align*} V_F(-\infty,+\infty) = \sup\left\{\sum_{i=1}^n|F(t_i)-F(t_{i-1})|:\{t_i\}_{t=0}^n\in\mathscr{S}\right\}.\end{align*}The function $F$ is of finite variation (or of bounded variation) if $V_F(-\infty,+\infty)$ is finite. If $F:\mathbb{R}\to\mathbb{R}$ is of finite variation, then the variation of $F$ is the function $V_F:\mathbb{R}\to\mathbb{R}$ defined by $V_F(x)=V_F(-\infty,x]$.
In addition, we have:
Proposition$\quad$Suppose that $F:\mathbb{R}\to\mathbb{R}$ is of finite variation. If $-\infty<a<b<+\infty$, then\begin{align} V_F(-\infty,b] = V_F(-\infty,a]+V_F[a,b].\end{align}