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Math Help - Is there a function f = g' but not Riemann Integrable

  1. #1
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    Is there a function f = g' but not Riemann Integrable

    The second fundamental theorem of calculus states that:

    If f is integrable in [a, b] and f = g' for some function g, then

    \int_a^b\! f(x) \, dx = g(b) - g(a)
    -----------------------

    so I was wondering if there is a function such that it is not Riemann integrable ( \exists \epsilon, U(f,P) - L(f,P) \geq \epsilon, or equivalent formulations), and f = g' for some function g.
    Last edited by Diego; November 29th 2009 at 09:30 AM.
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  2. #2
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    Quote Originally Posted by Diego View Post
    so I was wondering if there is a function such that it is not Riemann integrable ( \exists \epsilon, U(f,P) - L(f,P) \geq \epsilon, or equivalent formulations), and f = g' for some function g.
    Yes there is. The function f:x\mapsto |x|^{3/2}\sin\frac{1}{x} with f(0)=0 is derivable on \mathbb{R} (with f'(0)=0), but its derivative is unbounded on any neighbourhood of 0. Therefore, f' is not Riemann integrable on [-1,1] for instance.

    (but it is Lebesgue integrable; on the other hand, x\mapsto\frac{x}{\log|x|}\cos\frac{1}{x} satisfies the same properties like f and is not even Lebesgue integrable)
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    Henstock-Kurzweil integral case

    By any chance do you know Laurent, or any one else if that is also the case with the Henstock-Kurzweil integral (i.e. there's a function such that it is derivable, but it's derivative is not henstock-kurzweil integrable)?
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    Quote Originally Posted by Diego View Post
    By any chance do you know Laurent, or any one else if that is also the case with the Henstock-Kurzweil integral (i.e. there's a function such that it is derivable, but it's derivative is not henstock-kurzweil integrable)?
    With the Henstock-Kurzweil (also known as the complete integral) a derivative is 'integrable'.
    That if f(x)=F^{~'}(x) then C\int_a^b {f(x)dx}  = F(b) - F(a)
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