Is the total variation function of an absolute continuous function absolute continuous? Say $f$ is absolute continuous on $[a,b]$. Let $F(x) = TV(f_{[a,x]})$ the total variation function from $a$ to $x$. I think the given statement is true. From fundamental theorem of calculus, we know
$$f(x)-f(a) = \int_a^x f'(t)dt$$
And by the definition of total variation, we know:
$$F(x) = \int_a^x |f'(t)|dt$$
However, I am unable to proceed and check how $F(x)$ is absolute continuous. Since we don't have $f'(t)$ bounded. I wonder if you can give me a hint?
 A: The answer as you may suspect, is yes.
Suppose that $f$ is absolutely continuous on $[a,b]$. Let $\delta_1>0$ be such that for any finite collection of non--overlapping subintervals $[a_j,b_j]$ with $\sum^N_{j=1}(b_j-a_j)<\delta_1$, $\sum^N_{j=1}|f(b_j)-f(a_j)|<1$. 
For any subinterval $I\subset [a,b]$ of length less than $\delta_1$, $V_f(I)\leq1$. Splitting $[a,b]$ in $N=\lfloor(b-a)/\delta_1\rfloor+1$ subintervals  of length $(b-a)/N<\delta_1$ we obtain that $V_f[a,b]\leq N$.
Given $\varepsilon>0$, let  $\delta>0$ be such that for any finite collection of non--overlapping intervals $\{[a_j,b_j], j=1,\ldots, N\}$ with
$\sum_jb_j-a_j<\delta$,   $\sum_j|f(b_j)-f(a_j)|<\varepsilon/2$.
For each interval $[a_j,b_j]$, choose a partition
$\mathcal{P}_j=\{t_{jk}\}\subset[a_j,b_j]$ such that
\begin{align*}
V_f(b_j)-V_f(a_j)-\tfrac{\varepsilon}{2^{j+1}}&< \sum_k|f(t_{j,k})-f(t_{j,k-1})|
\end{align*}
Since $\sum_j\sum_k(t_{j,k}-t_{j,k-1})=\sum_k(b_j-a_j)<\delta$, we obtain that
\begin{align*}
\sum_j|V_f(b_j)-V_f(a_j)|& < \tfrac{\varepsilon}{2}+ \sum_j\sum_k|f(t_{j,k})-f(t_{j,k-1})|<\varepsilon
\end{align*}
Comment: In many treatments of the Lebesgue version of the fundamental theorem of Calculus (Rudin, Cohn, Riesz-Nagy, etc.), a version for absolutely continuous monotone functions is proven first. Then, from the result in the OP, one may prove the following version of the fundamental theorem of Calculus:
If $f:[a,b]\rightarrow\mathbb{C}$ is absolutely continuous, then $f'$ exists $\lambda$--a.s., $f'\in L_1([a,b],\lambda)$, and
$$
f(x)-f(a)=\int^x_a f'(t)\,dt, \quad a\leq x\leq b.
$$
Notice that $f=V_f-(V_f-f)$. Since $f$ is absolutely continuous then so are $V_f$ and $V_f-f$. Since $V_f$ and $V_f-f$ are increasing, we  apply the the fundamental theorem of Calculus for monotone increasing absolutely continuous functions.
