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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.

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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

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