# Monotone increasing continuous function with $\int_a^b f' = f(b) - f(a)$ which is not absolutely continuous

If $f:[a, b] \to \mathbb{R}$ is continuous and real-valued, f' integrable on [a, b], and $\int_a^b f' = f(b) - f(a)$, must f be absolutely continuous? What if f is monotone increasing?

For the first part, I have as a counterexample the Cantor-Lebesgue function joined with itself reflected across the line x = 1; this has derivative f' = 0 everywhere it exists, and f(0) = f(2) = 0, but it's obviously not absolutely continuous.

My instincts tell me there should be a similarly pathological example which is monotone increasing, but I may be wrong. Is there a way either to find such a counterexample or to use the fact that the function is now monotone increasing to show that $\int_a^x f' = f(x) - f(a)$ for every x in [a, b]?

Progress edit:

It seems any reasonable counterexample is a sum of a singular function that begins and ends at 0, but goes up and down in the middle, and an absolutely continuous function that increases faster than the singular function in the part where the latter is going down, but I'm not convinced it's possible for that to exist. Singular non-AC functions seem to inherently have a property that is something like an infinite slope in a set of measure zero.

On the opposite front, I know that any BV function f is equal to the sum of an AC function g and a singular function h, so it suffices to show h is constant to get an absolutely continuous function. Showing that h must be constant means assuming I have a "bump" where it goes up and then down again, or vise versa, and then showing said bump's existence contradicts one of the properties of f somehow...but I don't see how. I've tried looking at the monotonicity-based inequalities and they seem to point the wrong way to be useful.

1. For a monotone increasing function, the total variation on any interval $[c, d]$ is equal to $f(d) - f(c)$.
2. There should be some known inequality relating the total variation to the integral of the absolute value of the derivative; in my case I have $V[f; a, b] \leq \int_a^b |f'(t)|dt$.
For a monotone increasing function, f' is nonnegative, so these together mean that $f(d) - f(c) \leq \int_c^d f'(t) dt$ for any $[c, d] \subseteq [a, b]$. From there the equality $\int_a^b f'(t)dt = f(b) - f(a)$ is finally useful, if the goal is thought of in terms of showing that $\int_a^x f'(t)dt - (f(x) - f(a)) = 0$.
It suffices to show that $$\int_a^x f'=f(x)-f(a).\tag{*}\label{eqn}$$ It is well known that for an increasing function the inequality $$\int_a^x f'\le f(x)-f(a)$$ holds. If for any $$x\in [a,b]$$ we had a strict inequality in \eqref{eqn}, it would follow that $$\int_a^x f' < f(x)-f(a) \quad\mbox{ and }\quad \int_x^b f'\le f(b)-f(x),$$ and adding these two would yield $$\int_a^b f'< f(b)-f(a)$$. Thus, we must have equality in \eqref{eqn}, whence $$f$$ is absolutely continuous.