[Math] continuous but not absolutely continuous function

Question

How to show by definition (without the variation of function) that $f \colon [0,1] \to \mathbb{R}$
$$ f(x) = \begin{cases} 0 \quad \mbox{if} \quad x=0 \\ x \cos \frac{\pi}{x} \quad \mbox{if} \quad x \neq 0 \end{cases} $$
is not absolutely continuous?

Answer 1

Ok, another answer not using the variation... We have to show that for every $\delta > 0$ there exist mutually disjoint open intervals $\Delta_i = (a_i,b_i)$, $i=1,\ldots,n$, such that the sum of their lengths is $<\delta$ but $\sum_1^n|f(a_i) - f(b_i)|\ge 1$. So, let $\delta > 0$ be given. Choose $N\in\mathbb N$ odd (we need this later) such that $\frac{1}{N} < \delta$. Now, let $$ \Delta_i = \left(\frac 1 {N(i+1)},\frac{1}{Ni}\right), \quad i=1,\ldots,n, $$ where $n$ is such that $\frac 1 N\sum_{i=1}^n\frac 1 i\ge 1$. Now, we have that $$ \sum_{i=1}^n\operatorname{Length}(\Delta_i) = \sum_{i=1}^n\left(\frac{1}{Ni} - \frac 1 {N(i+1)}\right) = \frac 1 N\left(1 - \frac 1 {n+1}\right) < \frac 1 N < \delta. $$ But on the other hand, \begin{align*} \sum_{i=1}^n\left|f\left(\frac 1 {Ni}\right) - f\left(\frac 1 {N(i+1)}\right)\right| &= \sum_{i=1}^n\left|\frac {\cos(Ni\pi)}{Ni} - \frac {\cos(N(i+1)\pi)}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left|\frac {(-1)^{Ni}}{Ni} - \frac {(-1)^{N(i+1)}}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left|\frac {1}{Ni} - \frac {(-1)^N}{N(i+1)}\right|\\ &= \sum_{i=1}^n\left(\frac {1}{Ni} + \frac {1}{N(i+1)}\right)\,\ge\,1. \end{align*} So, $f$ cannot be absolutely continuous.