# Relation between total variation and absolute continuity

Happy Thanksgiving to you all.

I have made an attempt to a homework problem, that I'll need someone look over for me. The question is a follows.

If $f$ is of bounded variation on $[a,b]$ then $f'(x)$ exists a.e. and$$\int_a^b |f'|~dx\leq T_a^b(f),$$ where $T(f)$ is the total variation.

This is my attempt.

Since $f$ is of bounded variation, we can write $f(x)=g(x)-h(x)$ where $g$ and $h$ are monotone increasing functions on $[a,b]$. Thus $f'(x)=g'(x)-h'(x)$ a.e. Furthermore, $|f(x)|\leq |g'(x)|+|h'(x)|=g'(x)+h'(x)$. So \begin{align*} \int_a^b|f'(x)|~dx & \leq \int_a^b g'(x) + \int_a^b h'(x)\\ & \leq g(b)-g(a)+h(b)-h(a)\\ & = T_a^b(f). \end{align*} I'm also required to find additional conditions on $f$ needed for equality to hold.
Is it enough to say that equality holds if $f$ is absolutely continuous, or must I justify it?

The last inequality is not correct for a general $h$ and $g$. Try using the positive variation function and negative variation function instead. Notice these are increasing functions and so differentiable a.e... bla bla bla. In the end you will conclude with $P(f,[a,b])+N(f,[a,b])=T(f,[a,b])$.
Also a sufficient condition for equality is that $f$ is $C^{1}$