# [SOLVED] Ex of f not integrable

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• Apr 27th 2008, 12:37 AM
student1001
[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 27th 2008, 12:44 AM
Moo
[forget it :D]
• Apr 27th 2008, 01:00 AM
Jhevon
Quote:

Originally Posted by student1001
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?

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

EDIT: Oh, wait, we said "onto" Reals...hmmm (Thinking)
• Apr 27th 2008, 01:04 AM
o_O
$f(x) = \left\{ \begin{array}{cl} 1 & x \in \mathbb{Q} \\ -1 & x \in \text{Irrationals} \end{array}\right. \forall x \in [0,1]$

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

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, 01:14 AM
Moo
Quote:

Originally Posted by Jhevon
EDIT: Oh, wait, we said "onto" Reals...hmmm (Thinking)

What do you mean by onto ?

I've seen that it meant surjective, but here, does it work for "the original set is R" or "the image of the original set is R" ?
• Apr 27th 2008, 01:20 AM
Jhevon
Quote:

Originally Posted by o_O
$f(x) = \left\{ \begin{array}{cl} 1 & x \in \mathbb{Q} \\ -1 & x \in \text{Irrationals} \end{array}\right. \forall x \in [0,1]$

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

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.

Yes, as Moo said. I was thinking, "hmmm, onto means surjective. But my function only hits 1 and -1, even if i don't restrict the interval"

as onto is used here, it refers to the image of the function
• Apr 27th 2008, 01:29 AM
Jhevon
Quote:

Originally Posted by Moo
[forget it :D]

why did you delete your answer?

also, if you were going to delete it, why not just delete the post altogether :D

EDIT: Oh yeah, i remember what you said now, nevermind :D
• Apr 27th 2008, 01:30 AM
Moo
Quote:

Originally Posted by Jhevon
why did you delete your answer?

also, if you were going to delete it, why not just delete the post altogether :D

Because I didn't think about that...
I can't get it again, but I can write it back... I just didn't want to confuse the guy.
• Apr 27th 2008, 01:31 AM
Jhevon
Quote:

Originally Posted by Moo
Because I didn't think about that...
I can't get it again, but I can write it back... I just didn't want to confuse the guy.

No, don't worry about it. I remember what you wrote (Shake)
• Apr 27th 2008, 03:17 AM
Isomorphism
Quote:

Originally Posted by Jhevon
How about $f(x) = \left \{ \begin{array}{lrr} 1 & \mbox{ if } x \in \mathbb{Q} & \\ & & \mbox{ for } x \in [0,1] \\ -1 & \mbox{ if } x \not \in \mathbb{Q} & \end{array} \right.$ ?

EDIT: Oh, wait, we said "onto" Reals...hmmm (Thinking)

Cant we just tweak this a little bit and say:
$f(x) = \left \{ \begin{array}{lrr} 1 & \mbox{ if } x \in \mathbb{Q} & \\ & & \mbox{ for } x \in [0,1] \\ -1 & \mbox{ if } x \not \in \mathbb{Q} & \end{array} \right.$

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

Isnt this onto?

EDIT: Its wrong... misunderstood the question :(
• Apr 27th 2008, 03:44 AM
CaptainBlack
Quote:

Originally Posted by Isomorphism
Cant we just tweak this a little bit and say:
$f(x) = \left \{ \begin{array}{lrr} 1 & \mbox{ if } x \in \mathbb{Q} & \\ & & \mbox{ for } x \in [0,1] \\ -1 & \mbox{ if } x \not \in \mathbb{Q} & \end{array} \right.$

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

Isnt this onto?

You want a function from $[0,1]$ onto $\mathbb{R}.$

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 $\mathbb{Q}$ is countable and hence of measure zero).

RonL
• Apr 27th 2008, 07:47 PM
ThePerfectHacker
This problem just does not make any sense. The way the Riemann integral is defined we use the condition that $f$ is bounded on $[0,1]$ which certainly makes it impossible for it to be onto $\mathbb{R}$. Unless you have a different understanding of integration, maybe you are thinking about improper integration?
• Apr 27th 2008, 09:03 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
This problem just does not make any sense. The way the Riemann integral is defined we use the condition that $f$ is bounded on $[0,1]$ which certainly makes it impossible for it to be onto $\mathbb{R}$. Unless you have a different understanding of integration, maybe you are thinking about improper integration?

On-to in this case would mean have to mean that the integral is improper if we are talking Riemann integrals.

RonL