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

**Diego** · November 29th 2009, 08:39 AM

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.

**Laurent** · November 29th 2009, 10:41 AM

Originally Posted by Diego

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)

**Diego** · November 30th 2009, 07:11 AM

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)?

**Plato** · November 30th 2009, 08:48 AM

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

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

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