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$f(x) = \begin{cases} \frac{1}{x^2} & \text{for x rational} \\ -\frac{1}{x^2} & \text{for x irrational} \\ \end{cases}$

Is it integrable from $1$ to $\infty$?

A book says it is, citing the reason that $|f(x)|$ is integrable and any absolutely integrable function is integrable.

But I am having doubt because I feel that for any partition P Upper Darboux Sum and Lower Darboux Sum are negative of each other, and they are non-zero. So, they will not converge to the same limit.

Any help in this regard is appreciated.

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    $\begingroup$ You're right about the upper and lower sums. $\endgroup$
    – Arthur
    Jul 29 '19 at 10:16
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It is not true that Riemann integrability of $|f|$ implies that of $f$. Perhaps the book is talking about Lebesgue integral in which case this implications holds for all measurable functions $f$. This function is not Riemann integrable.

See https://en.wikipedia.org/wiki/Riemann_integral

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    $\begingroup$ Even for Lebesgue integrals, that statement is false. Consider the indicator function on a non-meassurable set. $\endgroup$
    – Arthur
    Jul 29 '19 at 10:17
  • $\begingroup$ @Arthur Your function is non-negative. $\endgroup$ Jul 29 '19 at 10:36
  • $\begingroup$ Yeah, you're right. Easily fixable, though. Multiply by 2, add $1$. If you want to integrate over some unbounded interval, you can also multiply by $\frac1{1+x^2}$ or something too. $\endgroup$
    – Arthur
    Jul 29 '19 at 10:37
  • $\begingroup$ Multiply by $2$ and add $1$? $\endgroup$ Jul 29 '19 at 11:43
  • $\begingroup$ Subtract, of course. Apparently I'm not completely present today. Sorry about that. $\endgroup$
    – Arthur
    Jul 29 '19 at 11:47

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