Theorem: Suppose $f \in AC([a,b])$. Prove that $$ V_a^b(f) \leq ||(f')^+||_{L^1([a,b])} + ||(f')^-||_{L^1([a,b])} $$
Proof (my attempt): If $f \in AC([a,b])$ then calculus formula holds and we can write $$ f(x) = f(a) + \int_a^x f'(t) dt $$ Moreover $f' \in L^1([a,b])$. By definition of abstract integral (here we are thinking to the Lebesgue measure on $\mathbb{R}$) we can derive the following bound $$ \int_a^x f'(t) dt = \int_a^x (f')^+(t) dt - \int_a^x (f')^-(t) dt \leq \Biggl| \int_a^x (f')^+(t) dt - \int_a^x (f')^-(t) dt \Biggr| \leq \int_a^x |(f')^+(t)| dt + \int_a^x |(f')^-(t)| dt $$ In particular, letting $x=b$ we have $$ f(b) - f(a) \leq \int_a^b |(f')^+(t)| dt + \int_a^b |(f')^-(t)| dt$ = ||(f')^+||_{L^1([a,b])} + ||(f')^-||_{L^1([a,b])} \qquad (1) $$ The idea is now to derive a bound for $f(b) - f(a)$ involving the total variation of $f$.
Hence let $V_a^b(f)$ the total variation of $f$, by defintion: $$ V_a^b(f) = \sup_{a=t_0 < t_1 < \ldots < t_n = b}{ \sum_{i=1}^n |f(t_i) - f(t_{i-1})|} $$ This last one implies $f(b) - f(a) \leq V_a^b(f)$ beeing just a particular partition of the interval $[a,b]$. But $V_a^b(f)$, beeing a supremum, is the smallest constant for which the inequality $f(b) - f(a) \leq V_a^b(f)$ is satisfied. We conclude, from (1) that $$ V_a^b(f) \leq ||(f')^+||_{L^1([a,b])} + ||(f')^-||_{L^1([a,b])} $$
Is my proof correct? Or there is something wrong?