Problem Statement: If $f$ is of bounded variation on $[a,b]$, show that $$\newcommand{\nc}{\newcommand} \nc{\R}{\mathbb{R}} \nc{\BV}{\mathrm{BV}} \nc{\PP}{\mathcal{P}} \nc{\abs}[1]{\left|#1\right|} \nc{\set}[1]{\left\{ #1 \right\}} \nc{\para}[1]{\left( #1 \right)} \nc{\br}[1]{\left[ #1 \right]} \nc{\ve}{\varepsilon} \nc{\vp}{\varphi} \int_a^b \abs{f'} \le V[a;b] $$ This is the first part of Problem $7.9$ in Measure & Integral: An Introduction to Real Analysis by Richard Wheeden and Antoni Zygmund.
Notes / Definitions:
Here, the authors choose to denote total variation by $V[a;b]$; other common notations include $V[f;a,b]$ (specifying $f$), $\mathrm{TV}[f;a,b]$, $V_a^b(f)$, and so on.
Integration is meant in the Lebesgue sense; we are dealing with functions $f : [a,b] \to \R$.
Given $f : [a,b] \to \R$, we define its variation by $$ V[f;a,b] := \sup_{P \in \PP_{a,b}} \sum_i \abs{ f(x_i) - f(x_{i-1}) } $$ We may associate a variation to a given partition by $$ V(f;P) := \sum_i \abs{ f(x_i) - f(x_{i-1}) } $$ If $V[f;a,b] < \infty$, then $f$ is of bounded variation and we say $f \in \BV[a,b]$.
Some Thoughts:
It is known that $\BV[a,b]$ functions are differentiable a.e. with derivative in $L^1$ (Corollary $7.23$ of the source text), so the well-definedness of the integral is a non-issue.
I've had a couple of ideas that went nowhere, however.
Jordan Decomposition: We use that $\BV[a,b]$ functions may be written as the difference of monotone-increasing functions, and that monotone functions are differentiable a.e. Note that, since $f \in \BV[a,b]$, we may write $f = \vp - \psi$ for $\vp,\psi$ monotone-increasing functions (and hence differentiable a.e.). Note that this means $\vp',\psi' \ge 0$. Then $$\begin{align*} \abs{f'} &= \abs{\vp' - \psi'} \\ &\le \abs{\vp'} + \abs{\psi'} \\ &= \vp' + \psi' \end{align*}$$ giving us $$\begin{align*} \int_a^b \abs{f'} &\le \int_a^b \vp' + \int_a^b \psi' \\ &\le \vp(b) - \vp(a) + \psi(b) - \psi(a) \\ &\le V(\vp;P) + V(\psi;P) \end{align*}$$ for any partition $P = \set{x_i}_{i=0}^n$. As $\vp,\psi$ are increasing, we may drop the absolute values in their variation: $$\begin{align*} V(\vp;P) + V(\psi;P) &= \sum_i \abs{ \vp(x_i) - \vp(x_{i-1}) } + \sum_i \abs{ \psi(x_i) - \psi(x_{i-1}) }\\ &= \sum_i \vp(x_i) - \vp(x_{i-1}) + \sum_i \psi(x_i) - \psi(x_{i-1}) \\ &= \sum_i \vp(x_i) - \vp(x_{i-1}) + \psi(x_i) - \psi(x_{i-1}) \\ &= V(\vp+\psi;P) \end{align*}$$ This does not help us, as $f \ne \vp + \psi$. This approach via decomposition is noted in this MSE post and related ones, but not to a sufficient level of detail to grasp what I need to finish the proof.
Limit Definitions: Consider a sequence of partitions $P_n := \set{x_i^{(n)}}_{i=0}^{k_n}$ of $[a,b]$, ordered in the usual sense. For simplicity, let $$\begin{align*} \Delta_n x_i &:= x_i^{(n)} - x_{i-1}^{(n)} \\ \Delta_n f &:= f \para{ x_i^{(n)} } - f \para{ x_i^{(n)} } \end{align*}$$ Then for each $i$, by the mean value theorem, $\exists \xi_i^{(n)} \in \br{ x_{i-1}^{(n)} , x_{i}^{(n)} }$ such that $$ \frac{\Delta_n f}{\Delta_n x_i} = f' \para{ \xi_i^{(n)} } $$ and hence $$ \abs{\Delta_n f} = \abs{f' \para{ \xi_i^{(n)} } \Delta_n x_i}= \abs{f' \para{ \xi_i^{(n)} }} \Delta_n x_i $$ and thus $$ \sum_i \abs{\Delta_n f} = \sum_i \abs{f' \para{ \xi_i^{(n)} }} \Delta_n x_i $$ Taking the limit $n \to \infty$ then gives $$ V[f;a,b] = \int_a^b \abs{f'} $$ However, this is obviously stronger than necessary and does not clearly use that $f \in \BV[a,b]$. I suspect the core issue lies with the existence of the $\xi_i^{(n)}$; perhaps there is a scenario in which they do not always exist? (I suppose issues on a zero-measure dense set might be enough to break this argument.) Still, there is an appeal in proving this for first principles, so it's hard for me to let go of this approach.
Does anyone have ideas -- be it in terms of alternative approaches, or ways to salvage these?
