TITLE: Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
QUESTION [6 upvotes]: 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

REPLY [3 votes]: 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.