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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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    @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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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 correct2011-10-16