Darboux's theorem (analysis)

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Synonym:
intermediate value property of the derivative
Type of Math Object:
Theorem
Major Section:
Reference

Mathematics Subject Classification

26A06 One-variable calculus

is the converse true?

is the convese of darboux's theorem true? i mean, if f(x) does not have any jump discontinuity, then is it certain that there exists a function F(x) such that F'(x) = f(x) ?

Re: is the converse true?

But if f(x) does not have any jump on its domain, f(x) is integrable.

Re: is the converse true?

Or I missing something?

Re: is the converse true?

The original question asked is quite specialized.
We might think to try to construct the F
such that F' = f by

F(x) = \int_a^x f(t) dt + F(a)

This works as long as $f$ is integrable; apparently
there might be some $f$ that are not, for example:

f(x) = d/dx [ x^2 \sin(x^{-2}) ] =

This is a standard textbook example; f is not in Lebesgue
integrable on [0,1] because of the (non-jump) discontinuity
at 0. Of course one can construct F by taking limits
and integrating from (-epsilon, 1) or something like that.

But other than this simple counterexample,
I don't know the complete answer though.

Re: is the converse true?

Thanks stevecheng, I misunderstood your question. Sorry.
perucho

Re: is the converse true?

Hi Gorkem,
According to which stevecheng explained to me, it seems to be that Darboux's theorem deals with a kind of "weak" continuity (really I didn't know about that) that is not the conventional concept of continuity. So f(x) not necessarily is integrable and your question is not as trivial as I thought.
Sincerely,
perucho