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Math Help - [SOLVED] seemingly simple Riemann integral

  1. #1
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    [SOLVED] seemingly simple Riemann integral

    Define f(x)=x if x is rational and f(x)=0 if x is irrational. Compute \overline{\int_0^1}f\;dx and \underline{\int_0^1}f\;dx.
    Okay, so clearly \underline{\int_0^1}f\;dx=0 (since m_i=0), so we can concentrate on \overline{\int_0^1}f\;dx.

    The problem is, I am having trouble interpreting my textbook on how to calculate this. According to it,

    \overline{\int_0^1}f\;dx=\inf\{U(P,f) : P a partition of [a,b]\}.

    Okay, that's all very well. But how in the world do we compute that? Thus far, I have not been able to find a way.

    Any help would be much appreciated. Thanks!
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    On any subinterval in the partition \mathcal{P} there a rational number, r, and an irrational, \gamma.
    So f(r)=r and f(\gamma)=0 giving the upper and lower values. (There are no other values.)
    Does that help you see the answer?
    Last edited by Plato; April 19th 2009 at 07:40 AM.
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  3. #3
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    Quote Originally Posted by Plato View Post
    On any subinterval in the partition \mathcal{P} there a rational number, r, and an irrational, \gamma.
    So f(r)=1 and f(\gamma)=0 giving the upper and lower values. (There are no other values.)
    Does that help you see the answer?
    Thanks, but I do not see how that is correct. f(r)=r, not 1 (unless of course r=1).

    So for any subinterval, M_i(f)=x_i. And that means

    U(P,f)=\sum_{i=1}^{n}x_i(x_i-x_{i-1}).

    I don't see how I can choose P to get this equal to zero.
    Last edited by hatsoff; April 19th 2009 at 07:40 AM.
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  4. #4
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    Quote Originally Posted by hatsoff View Post
    Thanks, but I do not see how that is correct. f(r)=r, not 1 (unless of course r=1).

    So for any subinterval, M_i(f)=x_i. And that means

    U(P,f)=\sum_{i=1}^{n}x_i(x_i-x_{i-1}).

    I don't see how I can choose P to get this equal to zero.
    You are correct. I misread the definition of the function.
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    What about this...

    Since \underline{\int_0^1}f\;dx=0, then if f\in R(x) we will know \overline{\int_0^1}f\;dx=0.

    Choose \epsilon>0 and partition P with n\geq1/\epsilon intervals such that x_i=i/n. Then 1/n\leq\epsilon and

    U(P,f)-L(P,f)=\sum_{i=1}^{n}x_i(x_i-x_{i-1})-0=\sum_{i=1}^{n}\frac{i}{n}(\frac{1}{n})=\frac{1}{  n}\leq\epsilon.

    This shows (I think) that f is integrable, which means that

    \overline{\int_0^1}f\;dx=0.

    Do I have any errors in logic here?
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  6. #6
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    Quote Originally Posted by hatsoff View Post
    What about this...
    U(P,f)-L(P,f)=\sum_{i=1}^{n}x_i(x_i-x_{i-1})-0=\sum_{i=1}^{n}\frac{i}{n}(\frac{1}{n})=\frac{1}{  n}\leq\epsilon.
    How did you get \sum_{i=1}^{n}\frac{i}{n}(\frac{1}{n})=\frac{1}{n}?
    It seems to be \frac{1}{n^2}\frac{n(n+1)}{2}
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  7. #7
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    Quote Originally Posted by Plato View Post
    How did you get \sum_{i=1}^{n}\frac{i}{n}(\frac{1}{n})=\frac{1}{n}?
    It seems to be \frac{1}{n^2}\frac{n(n+1)}{2}
    Oops. I'm clumsy with those summations, as you can see.

    But how about this: Could we say that since g:[0,1]\to R defined by g(x)=x (no matter whether or not x is rational) is integrable, then we know

    \overline{\int_0^1}g\;dx=\overline{\int_0^1}f\;dx=  \frac{1}{2}.

    And so then f is not integrable.

    Does that look fine?
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    Quote Originally Posted by hatsoff View Post
    Okay, so clearly \underline{\int_0^1}f\;dx=0 (since m_i=0), so we can concentrate on \overline{\int_0^1}f\;dx.

    The problem is, I am having trouble interpreting my textbook on how to calculate this. According to it,

    \overline{\int_0^1}f\;dx=\inf\{U(P,f) : P a partition of [a,b]\}.

    Okay, that's all very well. But how in the world do we compute that? Thus far, I have not been able to find a way.

    Any help would be much appreciated. Thanks!
    I thought \overline{\int_0^1}f\;dx=\sup\{U(P,f)
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    Quote Originally Posted by manjohn12 View Post
    I thought \overline{\int_0^1}f\;dx=\sup\{U(P,f)
    Quite right, thanks. I do not believe that affects my solution, however.

    So, what say you all? Is my g(x)=x solution viable? If so, I'll mark this thread as solved.
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    Quote Originally Posted by hatsoff View Post
    Quite right, thanks. I do not believe that affects my solution, however. So, what say you all? Is my g(x)=x solution viable? If so, I'll mark this thread as solved.
    Frankly I am puzzled by your having been given this problem.
    I will elaborate, there is a standard theorem which is not given is most basic developments of the Riemann Integral: if the set of discontinuities is not countable then the function is not integrable.

    What is the set of discontinuities for this function?
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    The set at which f is discontinuous is countable no? the set is just the rationals in [0,1].
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    Quote Originally Posted by KZA459 View Post
    The set at which f is discontinuous is countable no? the set is just the rationals in [0,1].
    Given that \frac{{\sqrt 2 }}{2} = \gamma  \in [0,1] then \gamma is a irrational number.
    Any rational number, q that is ‘close to' \gamma, then q is approximately equal to \gamma.
    Now consider f(\gamma)=0~\& f(r)=r\approx \gamma. Is continuity possible at x=\gamma?
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  13. #13
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    right my bad
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