The below problem appeared on the UW-Madison Analysis qualifying exam in January 2020. The proof that $f$ is absolutely continuous (AC) on the whole interval is still not posted, although the requested counterexample has been given.
I don't seem to find version of this problem in the site, but I am sure this is pretty standard type of question.
The problem goes likes this: Let $f$ be of bounded variation on $[0,1]$ and AC on $[\varepsilon,1]$ for all $\varepsilon >0$. $f$ is also continuous at $0$. Prove $f$ is absolutely continuous on whole interval $[0,1]$.
Moreover I am looking for a counterexample in the case when the bounded variation of $f $ is dropped.
Here is what I think,
Using continuity, I can find $\delta>0$ for given $\epsilon >0$ which bounds the sum in the definition of AC upto $\delta$. Then in the interval $[\delta,1]$, I can use given hypothesis. But I am not totally comfortable writing this rigorously.
For the counter example I can use $f(0)=0$ and $f(x)= x\sin (1/x)$ for $x$ not equal to $0$.
I would love to see the rigorous proof and rigorous proof of counterexample. Thank you in advance.
Best Answer
This is for the counterexample part. If $f$ is absolutely continuous on $[0,1]$, it is of bounded variation. So being of bounded variation is a necessary condition for the conclusion. Any function which is not of bounded variation but satisfies the other hypotheses will provide a counterexample.
As you indicated, the function $f(0)=0$ and $f(x)=x\sin(1/x)$ for $x\ne 0$ is not of bounded variation on $[0,1]$. This can be seen by evaluating $f$ at the points where $\sin$ is $1$ or $-1$, namely $x={1\over\pi(2k+1/2)}$ and $x={1\over\pi(2k+3/2)}$. It follows that the total variation of $f$ on $[0,1]$ is at least equal to a constant times the sum of a harmonic series, which diverges. So $f$ is not of bounded variation.
EDIT: The function $f$ above is also continuous at $0$, and is absolutely continuous on each interval $[\varepsilon,1]$ for $\varepsilon>0$. This last follows from the fact that on each such interval $f$ has a bounded derivative. (Apply the mean value theorem to each subinterval in the definition of absolute continuity.)