This is a problem form Tao's note here: https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/comment-page-3/#comment-679829.
Exercise $80$: Show that any function of bounded variation has an (almost everywhere defined) derivative that is absolutely integrable.
Attempt: Let $F: \mathbb{R} \rightarrow \mathbb{R}$ be of bounded variation. By the BV differentiation theorem, $F'$ exist a.e. By Jordan decomposition, we can write $F = g - h$, where $g, h: \mathbb{R} \rightarrow \mathbb{R}$ are monotone non-decreasing functions, so $F' = g' - h'$. In particular, on any compact interval $[a, b]$ we have $\int_{[a,b]} |F'| = \int_{[a,b]} |g' - h'| \leq \int_{[a,b]} |g'| + \int_{[a,b]} |h'| = \int_{[a,b]} g' + \int_{[a,b]} h'$, since $g'$ and $h'$ are both unsigned. By Proposition $79$ (refer to the note in the link above), $\int_{[a,b]} g' \leq g(b) - g(a)$ and $\int_{[a,b]} f' \leq f(b) - f(a)$. Hence for any compact $[a, b]$ we have:
$\displaystyle \int_{[a,b]} |F'| \leq g(b) - g(a) + f(b) - f(a) \leq M_1 + M_2$,
assuming $g$ is bounded above by some $M_1 > 0$ and $h$ by some $M_2 > 0$. Then $\int_\mathbb{R} |F'| = \lim_{N \to \infty} \int_{[-N, N]}| F'| \leq \lim_{N \to \infty} (M_1 + M_2) = M_1 + M_2 < +\infty$. In other words, $F'$ is absolutely integrable.
Is this a valid argument for the proof?