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?

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    $\begingroup$ I don't think your argument works, you're proving the much stronger statement that $f$ is Lipschitz continuous, but a counterexample to this is given by $f(x)=x^{3/2}\sin(1/x)$ which is everywhere differentiable with integrable derivative (hence AC) but it's not Lipschitz. $\endgroup$ – Jose27 Dec 31 '13 at 21:29
  • $\begingroup$ Another thing: In the Banach-Zaretsky theorem you need $F$ to be a-priori continuous (as in the book), since otherwise $F=1_{[0,1/2]}$ is a counterexample. $\endgroup$ – Jose27 Jan 2 '14 at 8:27

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.

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    $\begingroup$ Doesn't equation $(*)$ only hold if we already know that $f$ is AC? $\endgroup$ – yoknapatawpha Dec 31 '13 at 18:44
  • $\begingroup$ There is another theorem that says: if (*) holds and if $f$ is a monotone function, then $f$ is AC. $\endgroup$ – the8thone Dec 31 '13 at 18:53
  • $\begingroup$ If we have that : Sum of two AC functions is AC, we're done. $\endgroup$ – the8thone Dec 31 '13 at 18:54
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    $\begingroup$ But only $f$ is continuous, not its derivative (which is what you use). $\endgroup$ – Jose27 Dec 31 '13 at 21:59
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    $\begingroup$ If $f$ is Riemann integrable then $F$ is Lipschitz (since $f$ is bounded) and there is no problem. $\endgroup$ – Jose27 Dec 31 '13 at 23:20

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