Suppose that $f$ is a function of bounded variation on [0, 1], and let $V(x)$ be the total variation function for $f$, i.e., for any $x \in [0, 1]$, $V(x)$ is the total variation of $f$ on the interval $[0, x]$. Prove that, if $V$ is absolutely continuous on [0, 1], then so is $f$.
My attempt:
First, I note that $\sum_{k = 1}^{n} |V(b_{k}) - V(a_{k})| = \sum_{k = 1}^{n} |TV(f_{[a_{k}, b_{k}]})|$. Then I note that $|f(b_{k}) - f(a_{k})|$ would be the variation for the crudest possible partition of $(a_{k}, b_{k})$, hence $|f(b_{k}) - f(a_{k})| \leq TV(f_{[a_{k}, b_{k}]})$ for each $k$. Thus if we choose the $\delta$ responding to the $\epsilon$ challenge for the absolute continuity of function $V$, we have that if $\sum_{k = 1}^{n} [b_{k} - a_{k}] < \delta$ for a collection of open intervals $\{(a_{k}, b_{k})\}_{k = 1}^{n}$ in [0, 1], then:
$\sum_{k = 1}^{n} |f(b_{k}) - f(a_{k})| \leq \sum_{k = 1}^{n} TV(f_{[a_{k}, b_{k}]}) = \sum_{k = 1}^{n} |V(b_{k}) - V(a_{k})| < \epsilon$