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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.

user782220
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1 Answers1

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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.

Hu Zhengtang
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