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Math Help - i just want to see the proof..

  1. #1
    MHF Contributor kalagota's Avatar
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    i just want to see the proof..

    i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..

    Let A = \{ x\in \Re | x > 0 \}

    let h(x) = \left\{ {\begin{array}{cc} 0 &, x \in Q' \\ \, \\ \dfrac{1}{n} &, x = \dfrac{m}{n}\\ \end{array}} \right.

    (why cant i remove the right brace?)
    where x is always nonnegative and m,n are natural numbers relatively prime..

    show that this function is Riemann integrable on any interval \left[ a,b \right] \subseteq A

    (This function is called Thomae's Function)..
    Last edited by kalagota; December 20th 2007 at 05:01 AM. Reason: forgot (x).. i should have specified that it should be Riemann Integrability..
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kalagota View Post
    i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..

    Let A = \{ x\in \Re | x > 0 \}

    let h = \left\{ {\begin{array}{cc} 0 &, x \in Q' \\ \, \\ \dfrac{1}{n} &, x = \dfrac{m}{n}\\ \end{array}} \right.

    (why cant i remove the right brace?)
    where x is always nonnegative and m,n are natural numbers relatively prime..

    show that this function is integrable on any interval \left[ a,b \right] \subseteq A

    (This function is called Thomae's Function)..
    The function is 0 a.e.

    RonL
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The function is 0 a.e.

    RonL
    huh? i don't get your response.. i think you were saying that the integral is 0, which i also believe to be true..
    but, what i want to see is the proof of its Riemann integrability..
    Last edited by kalagota; December 20th 2007 at 05:12 AM.
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  4. #4
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    Opalg's Avatar
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    Given ε>0, choose n with 1/n < ε/(ba). In the interval [a,b], there are only finitely many rational numbers with denominator ≤n. Let's say there are N of them. Surround each of these N points with an interval of length ε/N, and use the endpoints of these intervals to form a dissection of the interval [a,b]. It's not hard to check that the upper Riemann sum of the function h(x) for this dissection is less than 2ε (the function has an upper bound 1 in each of the N intervals, whose combined length is ε; and an upper bound 1/n in all the remaining intervals of the dissection, whose combined length is less than ba). Since the lower Riemann sum for any dissection is zero, you can deduce from this that h is Riemann integrable over [a,b] (with integral zero).
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  5. #5
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    Quote Originally Posted by Opalg View Post
    Given ε>0, choose n with 1/n < ε/(b–a). In the interval [a,b], there are only finitely many rational numbers with denominator ≤n. Let's say there are N of them. Surround each of these N points with an interval of length ε/N, and use the endpoints of these intervals to form a dissection of the interval [a,b]. It's not hard to check that the upper Riemann sum of the function h(x) for this dissection is less than 2ε (the function has an upper bound 1 in each of the N intervals, whose combined length is ε; and an upper bound 1/n in all the remaining intervals of the dissection, whose combined length is less than b–a). Since the lower Riemann sum for any dissection is zero, you can deduce from this that h is Riemann integrable over [a,b] (with integral zero).
    What you say is a special case of a more general case.

    A set S has Lebesque measure zero in \mathbb{R} if for any \epsilon > 0 there exists finitely many compact intervals I_k such that thier union is a cover on S and the total length of the intervals is < \epsilon.
    (Again, I am not sure if "measure zero" is the correct term).

    Functions that are discontinous on a Lebesque measure zero set are always integrable.

    i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..
    Integrability is not an elementary question to ask.
    Last edited by ThePerfectHacker; December 20th 2007 at 10:27 AM.
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