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I'm in trouble with this exercise tough I thought it was not that difficult at the begininng, now I definitely changed my mind. So I hope someone could help me to solve it. Give the function:
$f:[0,1] \rightarrow \mathbb R$ defined as
$$ f(x)= \begin{cases} x^2 \cos\left(\frac{1}{x^2}\right), \qquad x\neq 0\\ 0 , \qquad\qquad\qquad \text{elsewhere} \end{cases} $$
I have to show that this function is derivable, but that $f'$ is not integrable and deduce that $f$ is not absolutely continuous.
I started proving that the functions is continuous on $[0,1]$. Then I calculate the derivative: $$ f'(x)= \begin{cases} 2x \cos\left(\frac{1}{x^2}\right)+\frac{2}{x}\sin\left(\frac{1}{x^2}\right) \qquad x\neq 0\\ 0 \qquad\qquad\qquad\qquad\qquad\;\;\, \qquad x=0 \end{cases} $$ so it exist in every point and I can guess that the $f$ is derivable but also that $f'$ is not continuous since the limit as $x\rightarrow 0$ doesn't exist and $f$ isn't $C^1$. Is it right?
Now to prove that is not integrable. I can't understand how to do it, I mean $f'$ is not continuous but has a finite number of points of non-continuity so I'd say that the integral exist but could take infinite value.
Now if I calculate the integral and I got that it has a infinite value.
How can I desume the non absolute continuity of $f$? And also is my trial correct up to now? Especially the compute of the integral.

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Integral does not exist at discontinuity points. –  Seyhmus Güngören Nov 26 '12 at 12:57
    
Ok but there's just one point, so I can say that the integral exist almost everywhere? I think that if a function is continuous except from a finite number of point is possible to calculate is integral but maybe it is infinite,and that's the case.I think. –  Laura Nov 26 '12 at 13:01
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