I am having problem with this question (from Erhan Cinlar's Probability and Stochastic Chapter 1 Section 5 Question 24b):
Let $\mathscr{A}$ be a collection of disjoint intervals of the form $(,]$ whose union is $(s,t]$.
Let $V_f(s,t)=\sup\limits_{\mathscr{A}}\sum\limits_{(u,v]\in\mathscr{A}}|f(v)-f(u)|$.
If $f$ is differentiable and its derivative is bounded by $b$ on $[s,t]$, then $V_f(s,t)\leq(t-s)\cdot b$.
My attempt:
$\forall \mathscr{A}$ By Triangle Inequality, $\sum\limits_{(u,v]\in\mathscr{A}}(f(v)-f(u))\leq\sum\limits_{(u,v]\in\mathscr{A}}|f(v)-f(u)|$
Thus $\forall \mathscr{A}, f(t)-f(s)\leq\sum\limits_{(u,v]\in\mathscr{A}}|f(v)-f(u)|$, since union of intervals in $\mathscr{A}$ is $(s,t]$
Thus $f(t)-f(s)\leq V_f(s,t)$, by definition of supremum
$\implies$ $\frac{f(t)-f(s)}{t-s}\leq\frac{V_f(s,t)}{t-s}$, since $t>s$
$\implies$ $f'(c)\leq \frac{V_f(s,t)}{t-s}$, By Mean Value theorem, $\exists c\in(s,t), f'(c)=\frac{f(t)-f(s)}{t-s}$
But I am unable to use the derivative is bounded since the inequality seems to be in the wrong direction
Can I have a hint please?
$$\sum_{(u,v]\in \mathscr A} |f(u)-f(v)|=\sum_{(u,v]\in \mathscr A}|f'(c_{u,v})||u-v|$$$$\leq \sum_{(u,v]\in \mathscr A}M |u-v|\leq M(t-s).$$
Here, $|f'|\leq M$. Note that the intervals in $\mathscr A$ are disjoint and contained in $(s,t]$. So sum of lengths of the intervals in $\mathscr A$ is less or equals to length of $(s,t]$.