Integral of the derivative of a function of bounded variation

Let $f\colon [a,b] \to \mathbb R$ be of bounded variation. Must it be the case that $|\int_a ^b f' (x) |\leq |TV(f)|$, where $TV(f)$ is the total variation of $f$ over $[a,b]$? If so, how can one prove this?

In the standard proof of the monotone differentiation theorem, it is shown tat this holds for increasing functions: if $f$ is increasing, then $\int_a ^b f'(x) \leq f(b) - f(a) = TV(f)$. I am trying to generalize this to functions of bounded variation.

The answer to the question posed here as to whether, for a function $f$ of bounded variation on an interval, $$\left|\int_a^b f'(x)\,dx\right| \leq \int_a^b |f'(x)|\,dx \leq V(f,[a,b]) $$ is of course yes and can be found in numerous textbooks. I don't think I need to list them. Rather more interesting is the further generalization. If $f$ is not absolutely continuous then you would have strict inequality with the variation. But what explains the difference in the values?

This was nicely done years ago by De La Vallee Poussin. His formula looks like this $$V_f(E) = V_f(E_\infty) + \int_E |f'(x)|\,dx$$ where $V_f$ is a measure that describes the total variation of $f$ on a set $E$ and $E_\infty$ are the points in $E$ at which $f'(x)=\pm\infty$. See Saks Theory of the Integral Chapter 4, Section 9 for a classical write-up of these ideas.

In terms of the title of this question (Integral of the derivative of a function of bounded variation) I thought this might be worth mentioning as otherwise someone randomly reading topics would find little else of interest here.

First, we must note that if a function is of bounded variation on $[a,b]$, then it is bounded variation on $[a,x]$ for all $a \leq x \leq b$. Furthermore, $P_a^x(f)-N_a^x(f)= f(x)-f(a)$, where $P_a^x(f)$ and $N_a^x(f)$ are the positive and negative variations of the function from a to x respectively.
Rearranging the terms we note that $f(x)= [P_x^a(f)+f(a)]- [N_x^a(f)]$ for all $x \in [a,b]$ is the difference of two increasing functions, since we're taking the variation over bigger and bigger intervals as x increases.
We can take the derivative of both sides and get $f'(x)= (P_a^x(f))'-(N_a^x(f))'$ for almost all $x \in [a,b]$, since monotone functions are differentiable almost everywhere. So, $|f'(x)| = |(P_a^x(f))'-(N_a^x(f))'|\leq (P_a^x(f))'+ (N_a^x(f))'=(T_a^x(f))'$, where $T_a^x(f)$ is the total variation of the function from a to x. Overall we have $|f'(x)| \leq (T_a^x(f))'$ almost everywhere.
Now, integrating both sides we get $\int_a^b |f'(x)| dx \leq \int_a^b (T_a^x(f))'dx$. Now since $T_a^x(f)$ is monotone increasing, we get $\int_a^b (T_a^x(f))' dx \leq T_a^b(f)-T_a^a(f) =T_a^b(f)$. This gives us the desired result $\int_a^b|f'(x)|dx\leq T_a^b(f)$.
Side note: We can show for an absolutely continuous function that the two are actually equal.