Let $f$ be of bounded variation on $[a,b]$ then $f'$ exist a.e., $f'$ is Lebesgue integrable and $\int_a^b f'dm = f(b)-f(a)$.
Looking for a counterexample to this as:
I know if $f$ is monotone increasing and continuous the $\int_a^b |f'|dm = f(b)-f(a)$.
If $f$ is simply BV but continuity not assumed then $\int_a^b f'dm \le f(b)-f(a)$.
I was trying to think of a counter example to the original statement.
Best Answer
Take the Cantor-Lebesgue function $F$ on $[0,1]$
AKA the Devil's Staircase.
Note that $F'=0$ almost everywhere and $F(1)=1$ and $F(0)=0$ and also $F$ has bounded variation.
Look at this for reference:
https://en.wikipedia.org/wiki/Cantor_function