Is $f(x)=x\sin(1/x)$ with $f(0)=0$ of bounded variation on $[0,1]$? - Problem with abs. continuous

Question

I am having the following trouble:

From Is $f(x)=x\sin(\frac{1}{x})$ with $f(0)=0$ of bounded variation on $[0,1]$?, $x\sin(1/x)$ has not bounded variation in $[0,1]$.
$x\sin(1/x)$ has derivative $-\cos(1/x)/x + \sin(1/x)$ a.e. and $\int_0^1 (-\cos(1/x)/x + \sin(1/x)) dx=\sin(1)$, i.e., $x\sin(1/x)$ is abs. continuous.
Every abs. continuous function is of bounded variation.

What is the error?

Thank you, really much.

score 2 · Accepted Answer · answered Feb 19 '22 at 14:34

2

$$g(x) = -\frac{1}{x}\cos\left(\frac{1}{x}\right) + \sin\left(\frac{1}{x}\right)$$ is not integrable on $[0,1]$. Hence the equality

$$\int_0^1 (-\cos(1/x)/x + \sin(1/x)) dx=\sin(1)$$ is wrong and $g$ is not absolutely continuous.

answered Feb 19 '22 at 14:34

1

Not Lebesgue integrable, that is. – aschepler Feb 19 '22 at 14:39
Thank you really much! Please, how can I prove that this function is not Lebesgue integrable on $[0,1]$? – Quiet_waters Feb 19 '22 at 15:22
The result I've posted is from Wolfram... What can be the error? – Quiet_waters Feb 19 '22 at 15:30
2

Wolfram can make errors... It can make computations without checking if those computations make sense. You have to check that $\int_0^1 \vert g \vert = \infty$. – mathcounterexamples.net Feb 19 '22 at 15:55
Thank you so much, and I've forgotten to put the absolute value – Quiet_waters Feb 19 '22 at 18:55

1 Answers1