absolutely continuous on $[0,1]$ versus absolutely continuous on $[\epsilon ,1]$ for every $\epsilon >0$

Lemma.

Define $f\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{R}$ by

Then $f$ is absolutely continuous (http://planetmath.org/AbsolutelyContinuousFunction2) on $\mathrm{[}\epsilon \mathrm{,}\mathrm{1}\mathrm{]}$ for every $\epsilon \mathrm{>}\mathrm{0}$ but is not absolutely continuous on $\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{]}$.

Proof.

Note that $f$ is continuous on $[0,1]$ and differentiable on $(0,1]$ with $f^{\prime }(x)=\sin \left({\displaystyle \frac{1}{x}}\right)-{\displaystyle \frac{1}{x}}\cos \left({\displaystyle \frac{1}{x}}\right)$.

Let $\epsilon >0$. Then for all $x\in [\epsilon ,1]$:

Since $f$ is continuous on $[\epsilon ,1]$ and differentiable on $(\epsilon ,1)$, the mean value theorem (http://planetmath.org/MeanValueTheorem) can be applied to $f$. Thus, for every $x_1,x_2\in (\epsilon ,1)$ with $x_1\ne x_2$, $\left|{\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}}\right|\le 1+{\displaystyle \frac{1}{\epsilon }}$. This yields $|f(x_2)-f(x_1)|\le \left(1+{\displaystyle \frac{1}{\epsilon }}\right)|x_2-x_1|$, which also holds when $x_1=x_2$. Thus, $f$ is Lipschitz on $(\epsilon ,1)$. It follows that $f$ is absolutely continuous on $[\epsilon ,1]$.

On the other hand, it can be verified that $f$ is not of bounded variation on $[0,1]$ and thus cannot be absolutely continuous on $[0,1]$.∎