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

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.

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.