TITLE: Absolutely continuous functions and the fundamental theorem of calculus
QUESTION [5 upvotes]: Wikipedia says that the function
$$\begin{equation}
 f(x) = 
 \begin{cases}
 0  & \mbox{if } x = 0 \\
        x\sin(1/x) & \mbox{if } x \neq 0
 \end{cases}
\end{equation}
$$
is not absolutely continuous on any finite interval containing the origin. Its derivative when $x\neq 0$ is easy to compute:
$$f'(x) = \sin(1/x)-\frac{\cos(1/x)}{x}$$
Wikipedia says thats if $f$ has a derivative $f'$ almost everywhere, the derivative is Lebesgue integrable and $f(x) = f(0) +\int_0^x f'(t)dt$ then $f$ is absolutely continuous.
Since the example I gave above is not absolutely continuous and has $f'$ defined almost everywhere does that mean the derivative is not Lebesgue integrable? Presumable there is a way to see that near the origin the positive area and the negative area of $f'$ are both infinite? So that $f$ can't be breaking the theorem that states the equivalence of the fundamental theorem of calculus and absolutely continuity.

REPLY [2 votes]: You are correct. $f'$ is not integrable on any interval containing the origin. To see this, since $f'$ is an odd function, it suffices to show that $f'$ is not integrable on $[0,a]$ for every $a>0$. 
Note that for any $0<x<y$, 
$$\int_x^y|f'(t)|dt\ge|\int_x^yf'(t)dt|=|f(y)-f(x)|.$$
Therefore, if $n\in\mathbb{N}$ and $a\ge\frac{1}{n\pi}$, then 
$$\int_0^a|f'(t)|dt\ge \sum_{k=n}^\infty\int_{\frac{1}{(k+\frac{1}{2})\pi}}^{\frac{1}{k\pi}}|f'(t)|dt\ge\sum_{k=n}^\infty\big|f(\frac{1}{k\pi})-f(\frac{1}{(k+\frac{1}{2})\pi})\big|=\sum_{k=n}^\infty\frac{1}{(k+\frac{1}{2})\pi}=+\infty,$$
i.e. $f'$ is not integrable on $[0,a]$. $\quad\square$

Remark: $\sum_{k=n}^\infty\big|f(\frac{1}{k\pi})-f(\frac{1}{(k+\frac{1}{2})\pi})\big|=+\infty$ implies that $f$ is not of bounded variation on $[0,a]$, i.e. the argument above shows that if $f$ is absolutely continuous, then it must be of bounded variation.