can anyone give me an example of functions that are Riemann Integrable on $[0, +\infty]$ but its absolute value is not Riemann Integrable on $[0, +\infty]$? Thanks!
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1Possible duplicate of Integrable Functions – Did Apr 12 '17 at 00:08
1 Answers
It is not possible when considering a Riemann integrable function on a bounded interval.
If $f$ is Riemann integrable then there is a partition $P$ for any $\epsilon > 0$ such that
$$U(P,|f|) - L(P,|f|) \leqslant U(P,f) - L(P,f) < \epsilon,$$
which implies that $|f|$ is Riemann integrable.
This result follows from the reverse triangle inequality $|\,|f(x| - |f(y)|\,| \leqslant |f(x) - f(y)|,$ implying for any partition interval $I$,
$$\sup_{x \in I}|f(x)| - \inf_{x \in I}|f(x)| \leqslant \sup_{x \in I}f(x) - \inf_{x \in I}f(x) $$
Edited Question
The term Riemann integrable is not relevant for the unbounded interval $[0, \infty)$. The usual example of $\sin x / x$ now gives $f$ where the improper integral exists, but fails to converge to a finite value for $|f|$.

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sorry, I didnt state the condition that it is from 0 to positive infinity, please see the updated question – Yaoxian Qu Apr 11 '17 at 23:19
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Now you are talking about an improper integral. Riemann integrability is only relevant for bounded intervals. – RRL Apr 11 '17 at 23:27
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