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

• Nov 29th 2009, 08:39 AM
Diego
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

$\displaystyle \int_a^b\! f(x) \, dx = g(b) - g(a)$
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so I was wondering if there is a function such that it is not Riemann integrable ( $\displaystyle \exists \epsilon, U(f,P) - L(f,P) \geq \epsilon$, or equivalent formulations), and f = g' for some function g.
• Nov 29th 2009, 10:41 AM
Laurent
Quote:

Originally Posted by Diego
so I was wondering if there is a function such that it is not Riemann integrable ( $\displaystyle \exists \epsilon, U(f,P) - L(f,P) \geq \epsilon$, or equivalent formulations), and f = g' for some function g.

Yes there is. The function $\displaystyle f:x\mapsto |x|^{3/2}\sin\frac{1}{x}$ with $\displaystyle f(0)=0$ is derivable on $\displaystyle \mathbb{R}$ (with $\displaystyle f'(0)=0$), but its derivative is unbounded on any neighbourhood of 0. Therefore, $\displaystyle f'$ is not Riemann integrable on $\displaystyle [-1,1]$ for instance.

(but it is Lebesgue integrable; on the other hand, $\displaystyle x\mapsto\frac{x}{\log|x|}\cos\frac{1}{x}$ satisfies the same properties like $\displaystyle f$ and is not even Lebesgue integrable)
• Nov 30th 2009, 07:11 AM
Diego
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)?
• Nov 30th 2009, 08:48 AM
Plato
Quote:

Originally Posted by Diego
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 $\displaystyle f(x)=F^{~'}(x)$ then $\displaystyle C\int_a^b {f(x)dx} = F(b) - F(a)$