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Problem statement: Suppose $f$ is a real-valued, continuous function on $[0,1]$, and $f$ is absolutely continuous on $(a,1]$ for every $a \in (0,1)$. Is $f$ necessarily absolutely continuous on $[0,1]$? If $f$ is also of bounded variation on $[0,1]$, is $f$ absolutely continuous on $[0,1]$? If not, give counterexamples.

My attempt at a solution: I think that I have proved that $x \cdot sin(1/x)$ works for the first part, however, I'm having a hard time proving the second part definitively one way or another. It seems like there should be a counterexample, but I can't think of one, and I'm hoping that someone can either help me out with a counterexample, or nudge me in the right direction to prove it.

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  • $\begingroup$ I'm not sure, but I think a Cantor function will work for the second part. $\endgroup$
    – user84413
    Aug 15, 2015 at 0:19
  • $\begingroup$ @user84413: I'm not sure what kind of Cantor function you have in mind, but whatever it is would contradict my answer. $\endgroup$
    – user21820
    Aug 15, 2015 at 3:45
  • $\begingroup$ @user21820 I was completely ignoring one of the main hypotheses, that f was absolutely continuous on $(a,1]$ for all $a\in(0,1)$. $\endgroup$
    – user84413
    Aug 15, 2015 at 19:57
  • $\begingroup$ @user84413: You are right. I realize my proof requires that hypothesis. I've fixed it. Thanks! $\endgroup$
    – user21820
    Aug 16, 2015 at 3:46

1 Answer 1

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The second part is true. I originally had an incorrect proof because as user84413 noted in a comment the Cantor function does not satisfy the given absolute continuity hypothesis and would be a counter-example.

Using the decomposition theorem $f = g + h$ for some $g,h$ such that $g$ is absolutely continuous and $h$ has derivative $0$ almost everywhere. But $h = f - g$ is absolutely continuous on $[a,1]$ for any $a \in (0,1)$ and hence $h$ is constant on $[a,1]$. Thus $h$ is constant on $(0,1]$ and hence $f = g$ is absolutely continuous.

And contrary to the other answer, the first part is false as you have indeed found a correct counter-example. More generally, you could simply take any function with bounded derivative on any closed interval that does not include zero but unbounded derivative near zero. But of course these are not the only possible counter-examples. (You may also want to look at Wikipedia to see an overview of inclusions of classes of common kinds of functions.)

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  • $\begingroup$ Awesome, thanks for the link! I don't know the decomposition theorem you're, using, however, is there a name for it? $\endgroup$
    – poppy3345
    Aug 15, 2015 at 20:56
  • $\begingroup$ @gesa: I'm not sure whether there is a name. Proof outline: (0) Real integral is absolutely continuous (ac) by Vitali covering theorem. (1) Monotonic function is differentiable almost everywhere (ae). (2) Function $f$ of bounded variation (bv) on $[a,b]$ is $f(a) + P(f;[a,x]) - N(f;[a,x])$ where $P,N$ are positive and negative variation respectively. (3) $P(f;[a,x]) , N(f;[a,x])$ are monotonic and hence differentiable ae, and so $f$ is differentiable ae ... [continued] $\endgroup$
    – user21820
    Aug 16, 2015 at 3:16
  • $\begingroup$ @gesa: ... (1) For monotonic function I forgot to say that the integral of its derivative is at most the difference between its endpoints. (4) $\int_{[a,x]} |f'|$ $\le \int_{[a,x]} P(f;[a,x])' + \int_{[a,x]} N(f;[a,x])'$ $\le P(f;[a,x]) + N(f;[a,x])$ $= V(f;[a,x])$. (5) Thus $f'$ is Lebesgue integrable so let $g(x) = \int_{[a,x]} f'$. Then $f' = g'$ ae by Lebesgue differentiation theorem and $g$ is ac (6) Let $h = f - g$. Then $h' = f' - g' = 0$ ae. (*) I'm not going to sketch the proof of the Lebesgue differentiation theorem but note that it also uses the Vitali covering theorem. $\endgroup$
    – user21820
    Aug 16, 2015 at 3:27
  • $\begingroup$ @gesa: Sorry I made a mistake in my proof as we need absolute continuity and not just continuity to conclude that zero derivative implies constant. The proof of this uses the Vitali covering theorem. If you need a sketch of that tell me. $\endgroup$
    – user21820
    Aug 16, 2015 at 3:48
  • $\begingroup$ Ah, ok, got it... I don't need a sketch of the Vitali covering, I have a proof of that written down somewhere. Thank you for your help! $\endgroup$
    – poppy3345
    Aug 16, 2015 at 4:05

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