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Math Help - Prove integrability

  1. #1
    Senior Member Pinkk's Avatar
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    Prove integrability

    Let f(x) = \sin(\frac{1}{x}) for x\ne 0 and f(0)=0. Show that f is Riemann integrable on [-1,1].

    I have a feeling that I have to use some fact like f is monotone on some intervals but I really don't know. Any help would be appreciated.
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    Quote Originally Posted by Pinkk View Post
    Let f(x) = \sin(\frac{1}{x}) for x\ne 0 and f(0)=0. Show that f is Riemann integrable on [-1,1].

    I have a feeling that I have to use some fact like f is monotone on some intervals but I really don't know. Any help would be appreciated.
    I'ts easy to prove that if a function is discontinous only at a finite number of points (and bounded of course) then it is Riemann integrable (use induction and isolate the point of discontinuity). Or more generally Lebesgue's criterion for integrability which says a function is Riemann integrable iff the set on which said function is discontinous has measure 0.
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  3. #3
    Senior Member Pinkk's Avatar
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    Well, the point of discontinuity is x=0, but could you show how only a finite number of discontinuities and boundedness implies integrability?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let f(x) = \sin(\frac{1}{x}) for x\ne 0 and f(0)=0. Show that f is Riemann integrable on [-1,1].

    I have a feeling that I have to use some fact like f is monotone on some intervals but I really don't know. Any help would be appreciated.
    Hint:

    Spoiler:
    This is an outline for the case when we are integrating over [0,1], I leave it to you to generalize. Let \varepsilon>0 be given. Then, since for the interval [\tfrac{\varepsilon}{2},1] f is continuous we may find a parition P of it such that U\left(P,f\mid_{[\frac{\varepsilon}{2},1]}\right)-L\left(P,f\mid_{[\frac{\varepsilon}{2},1]}\right)<\frac{\varepsilon}{2}. But, clearly \sup_{x\in[0,\frac{\varepsilon}{2}]}f(x)=1,\inf_{x\in[0,\frac{\varepsilon}{2}]}f(x)=0 and so U\left(\left\{0,\frac{\varepsilon}{2}\right\},f\mi  d_{[0,\frac{\varepsilon}{2}]}\right)-L\left(\left\{0,\frac{\varepsilon}{2}\right\},f\mi  d_{[0,\frac{\varepsilon}{2}]}\right)=\frac{\varepsilon}{2}. So, what can we do now?

    Quote Originally Posted by Jose27 View Post
    Or more generally Lebesgue's criterion for integrability which says a function is Riemann integrable iff the set on which said function is discontinous has measure 0.
    Haha, every freshman analysis student hates when people say it because it's so tempting to use but they can't!
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  5. #5
    Senior Member Pinkk's Avatar
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    I don't see how the inf of the function on that interval would be 0 since it can take on negative values on such an interval. But I guess the general idea would be that U(f, [0,1]) = U(f,[0,a]) + U(f,[a,1]) and the same for the lower sums, correct? (I know that is technically an incorrect statement since we have to state that the union of the two partitions is in fact the partition of the larger interval and that a is an element in both partitions, etc, etc).
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    Quote Originally Posted by Pinkk View Post
    Well, the point of discontinuity is x=0, but could you show how only a finite number of discontinuities and boundedness implies integrability?
    It's easy: Given 1>r>0 just divide [a,b] into [a,c-dr], [c-dr,c+dr], [c+dr,b] where d=\min \{ \frac{1}{2(M-m)} , \frac{c-a}{2} ,\frac{c-b}{2} \} and m\leq f(x)\leq M for all x\in [a,b] then in the first and last intervals the function is continous and in the second just use an argument analogous to Drexel's (if the discontinuity is on one of the endpoints the argument is the same). The result follows by induction.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    I don't see how the inf of the function on that interval would be 0 since it can take on negative values on such an interval.
    Damnit, this is why I always lose points on quizzes. Try [0,\tfrac{\varepsilon}{4}] then you should get U-L=\frac{\varepsilon}{4}-\frac{-\varepsilon}{4}=\frac{\varepsilon}{2}



    But I guess the general idea would be that U(f, [0,1]) = U(f,[0,a]) + U(f,[a,1]) and the same for the lower sums, correct?
    Yes, and same for Ls
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