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Topic: Indefinite integral implies Definite integral?
Replies: 5   Last Post: Aug 7, 2011 3:48 PM

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W^3

Posts: 29
Registered: 4/19/11
Re: Indefinite integral implies Definite integral?
Posted: Aug 6, 2011 4:01 PM
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In article
<63e52cad-33ea-4e71-8774-a0348859283a@l11g2000prh.googlegroups.com>,
Kenshin <ksandkk@gmail.com> wrote:

> Let f : [a,b] -> R be differentiable everywhere and f ' be the
> derivative of f.
>
> I've heard that
>
> 1) f ' may not be Riemann integrable.
>
> 2) f ' may not be even Lebesgue integrable.
>
> Can you give an example of this?


1) Correct. For one thing, f' need not be bounded on [a,b] (hence f'
is not RI on [a,b]). Example: f(x) = x^2*sin(1/x^2) on [0,1]. Even if
f' is bounded on [a,b], it need not be RI there. Recall that a bounded
function on [a,b] is RI iff the set of its discontinuities has measure
0. But one can construct a differentiable f on [a,b] with f' bounded,
but with f' discontinuous on a set of large measure.

2) f' will be Lebesgue measurable, certainly. But we can have int_a^b
|f'(x)| dx = oo. In fact f(x) = x^2*sin(1/x^2) has this property. More
exotic counterexamples can be constructed here.



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