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 $v(E)=0$.

link to definition
definition can also be found in Royden's Real Analysis
 A: 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. 
A: 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.
A: 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?
