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I don't seem to find version of this problem in the site, but I am sure this is pretty standard type of question.

$f$ be of bounded variation on $[0,1]$, and $f$ is absolutely continuous (AC) on $[\varepsilon,1]$ for all $\varepsilon >0$ and $f$ is continuous at $0$. Now the goal is to 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.

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up vote 4 down vote accepted

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.)

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You have to be careful when you say any function of bounded variation. You also want the function be absolutely continuous on $[\varepsilon,1]$. – JSchlather Dec 25 '12 at 3:57
Yes, I meant not of bounded variation. The condition on $[\varepsilon,1]$ is very relevant to the discussion at hand. We're looking for a counterexample to a particular theorem, we've only dropped the assumption that $f$ is of bounded variation. In particular we still want $f$ to be continuous on $[0,1]$ and absolutely continuous on $[\varepsilon,1]$, but absolutely continuity to fail on $[0,1]$. The function you give, indeed satisfies all of these. But any function not of bounded variation does not. – JSchlather Dec 25 '12 at 4:25
What @JacobSchlather is saying is absolutely correct. We do want the function which is continuous on $[0,1]$ and absolutely continuous on $[\epsilon,1]$. – Deepak Dec 25 '12 at 5:17
I really want to see how one can prove the function mentioned above is absolutely continuous on $[\epsilon,1]$ – Deepak Dec 25 '12 at 5:19
Thanks @JacobSchlather, I have edited the answer to make that clear. – PatrickR Dec 25 '12 at 22:46

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