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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 $v(E)=0$.

link to definition

definition can also be found in Royden's Real Analysis

Set
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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 04:04
  • @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. – Set May 30 '13 at 04:05
  • Follow this link http://math.stackexchange.com/questions/385267/absolutely-continuous-functions-and-the-fundamental-theorem-of-calculus – Srijan May 30 '13 at 04:07
  • http://math.stackexchange.com/questions/264805/proving-f-is-absolutely-continuous-on-0-1 – Srijan May 30 '13 at 04:08
  • @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. – Set May 30 '13 at 04:15
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    @angry The integral on a measure zero set is zero, where did you see the definition? – Shuhao Cao May 30 '13 at 05:47
  • @Shuhao Cao http://en.wikipedia.org/wiki/Absolute_continuity#Generalizations_2 also in Royden's Real Analysis – Set May 30 '13 at 05:51
  • N. L. Carothers Real Analysis Proposition 20.15 (i): If $f \in AC[a,b]$ then $f \in C[a,b] \cap BV[a,b]$. – kahen May 30 '13 at 07:39

3 Answers3

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I feel like a totally formal approach would add some value to this question.

Note that the function is actually uniformly continuous. That it is not absolutely continuous must have something to do with its sporadic behaviour close to zero. The idea then is to choose intervals $(a_k,b_k)$ close to zero where the end points are close to each other in such a way that $|f(b_k)-f(a_k)|$ produces a big value. A natural choice is where $sin(\frac{1}{x})$ achieves max and min. It turns out we may be a little sloppy here.

Set $a_k = \frac{2}{m(k+2)\pi}$, $b_k=\frac{2}{mk\pi}$, for $k=1,2,3,...$ (for $k$ odd, $b_k$ and $a_k$ take turns being max/min), while letting $m\in \mathbb{N}$ be even and large enough so that

$\sum_{k=1}^{N} b_k-a_k = \frac{2}{m\pi} \sum_{k=1}^{N} \frac{1}{k}-\frac{1}{k+2} < \frac{3}{m\pi} < \delta$.

Then $f(a_k) = a_k sin(\frac{m(k+2)\pi}{2}) = a_k(-1)^{(k-1)/2+1}$ and $f(b_k) = b_k sin(\frac{mk\pi}{2}) = b_k(-1)^{(k-1)/2}$. Now, due to how the sign alternates

\begin{align} \sum_{k=1}^{N} |f(b_k)-f(a_k)| = |b_1+a_1| + |-b_2-a_2| + ... = \sum_{k=1}^{N} b_k+a_k =\\ = \frac{2}{m\pi}\sum_{k=1}^{N} \frac{1}{k}+\frac{1}{k+2} \to \infty \end{align}

as $N \to \infty$ (the harmonic series). The partial sums are nondecreasing and so there can be no $\epsilon>0$ that bounds them. Hence $f$ cannot be absolutely continuous.

plebmatician
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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.

Bombyx mori
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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 07:18
  • @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 (http://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 (http://math.stackexchange.com/questions/191268/absolutely-continuous-functions). – Ryker Nov 11 '13 at 20:18
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    @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
  • @Bombyxmori did you mean to take $a_k = 1 / [2(m + k) \pi + \pi / 2$? Otherwise with the current choice it feels like $f(a_k) = 0$, or I am tired? – P. Camilleri Dec 19 '23 at 14:45
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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?

Set
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