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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 :) –  Mariano Suárez-Alvarez Oct 16 '11 at 18:31
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(And you should probably explain what exactly you mean by "integrable"...) –  Mariano Suárez-Alvarez Oct 16 '11 at 18:33
    
Riemann integrable –  August Oct 16 '11 at 18:46
    
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! –  Mariano Suárez-Alvarez Oct 16 '11 at 18:47
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@Jose27: that's not Riemann integrable, it only exists as an improper integral. –  Robert Israel Oct 16 '11 at 19:59
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Yes (for Riemann integral on a closed interval $[a,b]$, not for improper Riemann integrals). This is clear from the Lebesgue characterization of Riemann integrability, but you can also prove it using the fact that for any real interval $(c,d)$, $$\max_{x \in (c,d)} |f(x)| - \min_{x \in (c,d)} |f(x)| \le \max_{x \in (c,d)} f(x) - \min_{x \in (c,d)} f(x)$$

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@Jose27 Wow thanks Robert Israel, Uhm I think that the equality of Jose27, it´s not correct –  August Oct 16 '11 at 20:02
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