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.
Answer
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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