I am having trouble finding an example of a function f:[0,1] onto Reals st f is not integrable on [0,1], but absolute value of f is integrable on [0,1]. Any suggestions?

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- Apr 26th 2008, 11:37 PMstudent1001[SOLVED] Ex of f not integrable
I am having trouble finding an example of a function f:[0,1] onto Reals st f is not integrable on [0,1], but absolute value of f is integrable on [0,1]. Any suggestions?

- Apr 26th 2008, 11:44 PMMoo
[forget it :D]

- Apr 27th 2008, 12:00 AMJhevon
- Apr 27th 2008, 12:04 AMo_O
$\displaystyle f(x) = \left\{ \begin{array}{cl} 1 & x \in \mathbb{Q} \\ -1 & x \in \text{Irrationals} \end{array}\right. \forall x \in [0,1] $

$\displaystyle \left|f(x)\right| = 1 \quad \forall x \in [0,1]$

I was thinking of that too, Jhevon although you beat me to posting it Haha.

I don't quite understand the implications of your edit though. - Apr 27th 2008, 12:14 AMMoo
- Apr 27th 2008, 12:20 AMJhevon
- Apr 27th 2008, 12:29 AMJhevon
- Apr 27th 2008, 12:30 AMMoo
- Apr 27th 2008, 12:31 AMJhevon
- Apr 27th 2008, 02:17 AMIsomorphism
Cant we just tweak this a little bit and say:

$\displaystyle f(x) = \left \{ \begin{array}{lrr} 1 & \mbox{ if } x \in \mathbb{Q} & \\ & & \mbox{ for } x \in [0,1] \\ -1 & \mbox{ if } x \not \in \mathbb{Q} & \end{array} \right.$

(Tweak)

$\displaystyle \forall x \in \mathbb{R} - [0,1], f(x) = x\sin x$

Isnt this onto?

EDIT: Its wrong... misunderstood the question :( - Apr 27th 2008, 02:44 AMCaptainBlack
You want a function from $\displaystyle [0,1]$ onto $\displaystyle \mathbb{R}.$

Note there is no qualification on integrable. Presumably the OP is interested in something like Riemann integrability rather than Lesbegue integrability. Which is just as well as all the functions presented are Lesbegue integrable (since $\displaystyle \mathbb{Q}$ is countable and hence of measure zero).

RonL - Apr 27th 2008, 06:47 PMThePerfectHacker
This problem just does not make any sense. The way the Riemann integral is defined we use the condition that $\displaystyle f$ is bounded on $\displaystyle [0,1]$ which certainly makes it impossible for it to be onto $\displaystyle \mathbb{R}$. Unless you have a different understanding of integration, maybe you are thinking about improper integration?

- Apr 27th 2008, 08:03 PMCaptainBlack