# Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$

On the interval [0,1]. Define $f(x)=x\sin(1/x)$ for $x\in(0,1]$ and $f(0)=0$. I didn't work out the exact details but I'm pretty sure that then $$\Big |\int_0^xf'(t)dt\Big |=\infty,$$ due to a process similar to something of the form $1-2+3-4+5-...$ , as one approaches zero from above.

However according to the measure-theoretic definition of absolute continuity, there should in fact be some set of measure zero $E\in[0,1]$ such that $$\Big |\int_Ef'd\mu\Big | > 0.$$

I wasn't under the impression that this was even possible.

Edit: Maybe I wasn't clear about what my question is. What I want is a proof (constructive or not) that there exists a set of measure zero $E$ such that $\Big |\int_Ef'd\mu\Big | > 0.$ Or if that's not possible then for someone to explain to me what my misconception is concerning the measure theoretic definition of absolute continuity:

For $v(E)=\int_Efd\mu$.

If $\mu(E)=0$ then $v(E)=0$.

definition can also be found in Royden's Real Analysis

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Try to show $f$ is not of bounded variation by breaking down the interval of $(0,1]$ into $(1/(2n\pi+2\pi), 1/(2n\pi)]$, and each one of them into four quadrants, the total variation by this partition tends to infinity, therefore $f$ is not absolutely continuous. – Shuhao Cao May 30 '13 at 4:04
@Shuhao I know it's not of bounded variation, but I can't see how the measure theoretic definition of not being absolutely continuous can possibly be satisfied. – cuckmaster5000 May 30 '13 at 4:05
@srijan this is essentially what I concluded about the function in my post. But I still don't see how there can be a set of measure zero on which the integral of $f'$ is non-zero. – cuckmaster5000 May 30 '13 at 4:15

Update: Corrected the definition type mistake, but it seems the proof is not measure based, as OP said what he/she needed. So further work is needed.

Given positive number $\epsilon$, for every $\delta>0$, if you pick up points $$a_{k}=\frac{1}{2km\pi},a_{k+1}=\frac{1}{(2k+1)m\pi}$$for example, then you have $$f(a_{k})=\frac{1}{2km\pi},f(a_{k+1})=\frac{-1}{(2k+1)m\pi},|f(a_{k})-f(a_{k+1})|\ge \frac{2}{(2k+1)m\pi}$$Here $m\in \mathbb{N}$ is an odd number large enough such that $$\sum_{k=1}^{\infty}|a_{k}-a_{k+1}|<\delta,\forall k\in \mathbb{N}$$ This is possible because we are essentially taking the partial sums of the alternating series. So if we choose $m$ to be large enough, we can "squeeze" the sum to be less than $\delta$.

Now if you pick up points $\{a_{k}\}_{k\rightarrow \infty}$, then $$\sum_{k=1}^{\infty}|f(a_{k})-f(a_{k+1})|>\epsilon$$since the left hand side essentially diverges.

For your question in the comment, the derivative is only undefined when $x=0$. Otherwise it is a perfectly well-defined function. So it is defined almost-everywhere.

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This answer is actually incorrect, as the condition for absolute continuity isn't $|a_{k}-a_{k+1}|<\delta,\forall k\in \mathbb{N}$. While it is true that there exists an odd number large enough for that to be true, you would need to find a sequence of disjoint intervals s.t. $\sum_{k=1}^{\infty}|a_{k}-a_{k+1}|<\delta$. That, you will find, is somewhat hard to do, and I haven't found a way to actually do it. Rather, other means must be used to prove that $f$ is not absolutely continuous. – Ryker Nov 11 '13 at 7:18
@Ryker: I suggest you double check the definition instead. – Bombyx mori Nov 11 '13 at 17:02
No worries, I have, but it seems you haven't done so. Here's a link to the Wikipedia page (en.wikipedia.org/wiki/Absolute_continuity#Definition), or if you don't like that one, perhaps check out the link to a thread on stackexchange where the same definition is given (math.stackexchange.com/questions/191268/…;. – Ryker Nov 11 '13 at 20:18
@Ryker: I see. I miss read the third line as $\sum^{\infty}_{k=1}|f(a_{k})-f(a_{k+1})|<\delta$, which feels like nonsense. I shall correct the proof. – Bombyx mori Nov 11 '13 at 23:52

Note that we are taking as given the fact that $f$ is not absolutely continuous.

I'm going to answer my own question and in the process pose another question:

If a measure $v$ isn't absolutely continuous with respect to $\mu$, then it doesn't have a representation of the form $$v(E)=\int_Egd\mu.$$

Thus given $$f(x) = \left\{\begin{matrix} x\sin(1/x) &\;\;\;\;\;\;\;\;x\in[-1,1]\backslash\{0\} \\ 0&x=0 \end{matrix}\right.$$

Then the measure $v$ on $[-1,1]$ induced by $f$, that is to say $v(E)=\mu(\;\{x\in[-1,1]:f(x)\in E\}\;)$, isn't absolutely continuous on $[-1,1]$ with respect to the Lebesgue measure, this means it can't be written in the integral form above.

However it still holds true that absolute continuity of a measure $v$ with respect to another measure $\mu$ (in this case $\mu$ being Lebesgue measure), is equivalent to the property that $$\mu(E)=0\Rightarrow v(E)=0.$$

**Provided $v$ and $\mu$ are $\sigma$-finite, which they are.

Thus we are forced to conclude that there exists some $E\in[-1,1]$ such that $\mu(E)=0$, but where the Lebesgue measure of the set which maps to $E$ under $f$ is not zero; which honestly doesn't seem possible. The only way it seems possible is if some weird uncountable set of measure zero (such as the Cantor set), happens to get mapped to a set of positive measure. Can anyone find this set?

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