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In an infinite interval this is not true. But in a finite interval is this true? Or at least in a closed interval?

$\textbf{EDITED:} $ Ok, suppose that $$ f:\left[ {a,b} \right] \to \mathbb{R} $$ is Riemann integrable, is it true that the function $$ \left| f \right|:\left[ {a,b} \right] \to \mathbb{R} $$ is Riemann integrable? Where $$ \left| f \right| $$ denotes the function $$ \left| {f\left( x \right)} \right| $$ this is my first question, the other is with other kinds of finite length intervals, like open intervals, or semi-opens intervals.

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    What you wrote is missing something... As it stands, it is not even grammatically correct :)2011-10-16
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    (And you should probably explain what exactly you mean by "integrable"...)2011-10-16
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    Riemann integrable2011-10-16
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    August, please add the information to the question itself. You *must* tell us what you want to know about $|f|$, as your title is surely missing the key part of the question!2011-10-16
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    How about $\int_1^{\infty} \frac{\sin(x)}{x}dx=\int_0^1\frac{\sin(\frac{1}{y})}{y}dy$?2011-10-16
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    @Jose27: that's not Riemann integrable, it only exists as an improper integral.2011-10-16
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    Can you also edit the title so that it makes sense?2011-10-16
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    @Robert Yes, it's not bounded, and so can't be Riemann integrable, but isn't the integral over infinite intervals also a kind of improper Riemann integral (this is what led me to believe that the question was open to improper examples).2011-10-16

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