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Math Help - [SOLVED] Ex of f not integrable

  1. #1
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    [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?
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  2. #2
    Moo
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    [forget it ]
    Last edited by Moo; April 27th 2008 at 01:01 AM.
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    Quote Originally Posted by student1001 View Post
    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
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    o_O
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    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.
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  5. #5
    Moo
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    Quote Originally Posted by Jhevon View Post
    EDIT: Oh, wait, we said "onto" Reals...hmmm
    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" ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by o_O View Post
    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
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    [forget it ]
    why did you delete your answer?

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

    EDIT: Oh yeah, i remember what you said now, nevermind
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  8. #8
    Moo
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    Quote Originally Posted by Jhevon View Post
    why did you delete your answer?

    also, if you were going to delete it, why not just delete the post altogether
    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.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    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
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    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
    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
    Last edited by Isomorphism; April 27th 2008 at 04:09 AM. Reason: Thank you CB
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  11. #11
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    Quote Originally Posted by Isomorphism View Post
    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
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    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?
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  13. #13
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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