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Prove that this function is Riemann integrable

  1. Jan 14, 2006 #1
    Let f:[a,b]->R be bounded. Further, let it be continuous on [a,b] except at points a1, a2, ...,an,... such that a1>a2>a3>...>an>...> a where an converges to a. Prove that f is Riemann integrable on [a,b].

    It suffices to prove that f is integrable on [a,a1) (I've worked out that part). And that's what I'm having trouble with.
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  3. Jan 15, 2006 #2


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    Are you dealing strictly with Riemann integration, or is this proof done in the context of Lebesgue integration?

    One might start out by noting that:

    [tex]\int_{a}^{b} f(x)dx = \int_{a_1}^{b} f(x)dx + \lim_{n\rightarrow\infty}\sum_{k=2}^{n} \int_{a_{k-1}}^{a_{k}} f(x)dx [/tex]
  4. Jan 15, 2006 #3
    Riemann. And I had already figured out that part. What I'm having trouble with is that I must show the terms [tex]\int_{a_n}^{a_{n+1}}f \rightarrow 0[/tex] as [tex]n\rightarrow \infty[/tex]. Since the series is bounded, showing that the terms converge to 0 will imply the convergence of the series.
  5. Jan 15, 2006 #4


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    What is the definition of convergence of a sequence, say a_n ?
  6. Jan 16, 2006 #5
    If I got by antiderivatives, that is F(a_n+1)-F(a_n), there's no way to garantee the continuity of F (where F' = f). However since F is bounded by M, the integral is bounded below and above by +/-M(a_n+1 - a_n) which squeezes the integral to zero by convergence of a_n.
  7. Jan 16, 2006 #6


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    So you have an infinite sequence of points an[/b] that converge to a such that f is discontinuous at those points? You need more than that: you will also need that there is only a finite step discontinuity at those points.
  8. Jan 17, 2006 #7
    I found a flaw in my argument. If a series is bounded, and the terms converge to zero, it DOES NOT imply the convergence of the series. I'll have to try another approach.
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