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



W^3
Posts:
29
Registered:
4/19/11


Re: Indefinite integral implies Definite integral?
Posted:
Aug 6, 2011 4:01 PM


In article <63e52cad33ea4e718774a0348859283a@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.



