# $f$ integrable implies $\left| f \right|$ in distinct intervals

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
(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
@Jose27: that's not Riemann integrable, it only exists as an improper integral. –  Robert Israel Oct 16 '11 at 19:59
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 $(c,d)$, $$\max_{x \in (c,d)} |f(x)| - \min_{x \in (c,d)} |f(x)| \le \max_{x \in (c,d)} f(x) - \min_{x \in (c,d)} f(x)$$