Bounded variation and differentiability implies absolute continuity Problem: Let $f$ be a real-valued function everywhere differentiable function on $[0,1]$. If $f$ is of bounded variation on $[0,1]$, then it is absolutely continuous on $[0,1]$.

This follows from the Banach–Zaretsky theorem, the mean value theorem, and the observation that Lipschitz functions take sets of measure zero to measure zero.

This follows from the Banach–Zaretsky theorem, which I will state it here for convenience: 


  Let $F$ be a function of bounded variation on $[a,b]$, then the
  following are equivalent:
  
  (i) F is absolutely continuous on $[a,b]$.
  
  (ii) $m(A) = 0 \implies m(F(A)) = 0$ for all measurable $A \subset [a,b]$.


The proof of this statement along with the posed problem can be found here.

Question: The original problem appeared in a qualifying exam a few months ago, and I am very curious if there is an alternative approach here for an examinee who might not have been aware of Banach–Zaretsky theorem. Any suggestion?
If f is differentiable everywhere on $(0,1)$ we can say $$f(1)-f(0) =\int^1_0 f'(x)dx \ \ (*)$$ If we knew that $f$ is a monotone function, (*) would immediately imply that $f$ is AC. However we know that $f$ is of bounded variation, thus can be written as the difference of two monotonically increasing functions on $[0,1]$ : $$f(x)=[f(x)+{TV}(f_{[0,x]})] - TV(f_{[0,x]})$$ Where $TV(f_{[0,x]})$ is the total variation function, which is differentiable everywhere on $[0,1]$, since it is monotone on $[0,1]$ and by Lebesgue's theorem is differentiable.
If f is differentiable everywhere on $(0,1)$ we can say $$f(1)-f(0) =\int^1_0 f'(x)dx \ \ (*)$$ If we knew that $f$ is a monotone function, (*) would immediately imply that $f$ is AC. However we know that $f$ is of bounded variation, thus can be written as the difference of two monotonically increasing functions on $[0,1]$ : $$f(x)=[f(x)+{TV}(f_{[0,x]})] - TV(f_{[0,x]})$$ Where $TV(f_{[0,x]})$ is the total variation function, which is differentiable everywhere on $[0,1]$, since it is monotone on $[0,1]$ and by Lebesgue's theorem is differentiable.