# Prove/disprove if $f$ is continuous on $[0,1]$, and absolutely continuous on $(a,1], a\in (0,1)$, $f$ is absolutely continuous on $[0,1]$.

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.

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

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.
• @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] Aug 16, 2015 at 3:16
• @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. Aug 16, 2015 at 3:27