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.


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.)

  • $\begingroup$ You have to be careful when you say any function of bounded variation. You also want the function be absolutely continuous on $[\varepsilon,1]$. $\endgroup$
    – JSchlather
    Dec 25 '12 at 3:57
  • $\begingroup$ 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. $\endgroup$
    – JSchlather
    Dec 25 '12 at 4:25
  • $\begingroup$ What @JacobSchlather is saying is absolutely correct. We do want the function which is continuous on $[0,1]$ and absolutely continuous on $[\epsilon,1]$. $\endgroup$
    – Deepak
    Dec 25 '12 at 5:17
  • $\begingroup$ I really want to see how one can prove the function mentioned above is absolutely continuous on $[\epsilon,1]$ $\endgroup$
    – Deepak
    Dec 25 '12 at 5:19
  • $\begingroup$ Thanks @JacobSchlather, I have edited the answer to make that clear. $\endgroup$
    – PatrickR
    Dec 25 '12 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.