[HaouLelouch]

http://www.maths.cam.ac.uk/undergrad.../PaperIA_1.pdf

the last part of question 12.

'Let f:R->R be differentiable, and set g(x)=f'(x) for x in [0,1]. Must the Riemann integral of g from 0 to 1 exist?'

My guess is no since I know that a real function is integrable iff the set of its discontinuous points is at most countable (actual statement more rigorous than this obviously) and I seem to remember seeing somewhere before that a derivative can have uncountably many discontinuities. I wonder what would be a good counterexample would be? or am I missing something from the previous parts of the question which may help with the last part?

thanks.

[nuodai]

Isn't this just part of the fundamental theorem of calculus? Namely, if is differentiable and then .

[HaouLelouch]

FTC requires g to be continuous, in this question the only condition given is f is differentiable where g=f'. f differentiable is not equivalent to g continuous I dont think. eg f=x^2sin(1/x) for x=/=0, f=0 for x=0 is differentiable everywhere yet its derivative is not continuous at x=0

[Glutamic Acid]

Perhaps look for a function with a horribly unbounded, discontinuous derivative?

It doesn't have to be that horrible. You're pretty close with your x^2*sin(1/x) suggestion.

[HaouLelouch]

(Original post by Glutamic Acid)
Perhaps look for a function with a horribly unbounded derivative?

what do you mean by 'horribly unbounded derivative'?
I'm thinking that mean the function's derivative will tend to infinity for some value of x? but surely that's impossible given the function is differentiable for all the reals? (so the derivative is finite at each value of x?)

[Glutamic Acid]

(Original post by HaouLelouch)
what do you mean by 'horribly unbounded derivative'?
I'm thinking that mean the function's derivative will tend to infinity for some value of x? but surely that's impossible given the function is differentiable for all the reals? (so the derivative is finite at each value of x?)

That cannot happen if the derivative is continuous. But if it's discontinuous we could have some function f with f'(0) = 0 (or any finite value), yet does not exist.

[HaouLelouch]

(Original post by Glutamic Acid)
That cannot happen if the derivative is continuous. But if it's discontinuous we could have some function f with f'(0) = 0 (or any finite value), yet does not exist.

does the function you're thinking of have uncountably many discontinuities?

[DFranklin]

Well, the function I'm thinking of doesn't...

[Glutamic Acid]

(Original post by HaouLelouch)
does the function you're thinking of have uncountably many discontinuities?

Nope, just one. So it should be clear what sort of discontinuity it is.

[HaouLelouch]

(Original post by Glutamic Acid)
Nope, just one. So it should be clear what sort of discontinuity it is.

I'm pretty certain a function g with just one discontinuous point is integrable.
Ie take a partition where you put an interval of arbitrarily small width around that point and the rest of the function is integrable by continuity.

Unless you're talking about f where f'=g but in that case a discontinuous point implies it is not differentiable over R.

Nvm I realised I should be looking for an integral that is infinite over [0,1] rather than some not integrable function.

[Glutamic Acid]

(Original post by HaouLelouch)
I'm pretty certain a function g with just one discontinuous point is integrable.
Ie take a partition where you put an interval of arbitrarily small width around that point and the rest of the function is integrable by continuity.

Unless you're talking about f where f'=g but in that case a discontinuous point implies it is not differentiable over R.

Nvm I realised I should be looking for an integral that is infinite over [0,1] rather than some not integrable function.

There do exist differentiable functions with bounded derivative that are not Riemann-integrable, e.g. Volterra's function. But it'd be a little harsh to expect this in an exam.

[HaouLelouch]

For unbounded derivative I'm thinking f=x^1/2 but this is not differentiable at x=0 which is a pain? I don't see how one can construct an unbounded derivative without at least having one point where the function is not differentiable.

[Glutamic Acid]

(Original post by HaouLelouch)
For unbounded derivative I'm thinking f=x^1/2 but this is not differentiable at x=0 which is a pain? I don't see how one can construct an unbounded derivative without at least having one point where the function is not differentiable.

is close: it's differentiable yet the derivative is discontinuous at 0. Unfortunately, it's not unbounded so we can't conclude that it's not Riemann integrable. Perhaps modify it?

[HaouLelouch]

Actually I think it's impossible to find an unbounded integral.

Because if f is differentiable everywhere then g is finite everywhere so if S=supg(x), F=infg(x) x in [0,1] so the integral of g from 0 to 1 must be less than S but bigger than F.

[DFranklin]

(Original post by HaouLelouch)
Actually I think it's impossible to find an unbounded integral.

Because if f is differentiable everywhere then g is finite everywhere so if S=supg(x), F=infg(x) x in [0,1] so the integral of g from 0 to 1 must be less than S but bigger than F.

Suppose g(x) = 1/x (for x non zero), g(0) = 0. g is finite everywhere, but it's not bounded.

[Glutamic Acid]

(Original post by HaouLelouch)
so if S=supg(x) ... x in [0,1]

This could be infinite. Consider f defined by f(x) = 1/x for x in (0,1] yet f(0) = 0. f is defined everywhere in [0,1] yet is not bounded.

[HaouLelouch]

Does x^1.5sin(1/x) work?

[DFranklin]

Ayup.

[MatchDancer]

Timothy Gowers is providing some model answers for IA Analysis I and Numbers and Sets questions http://gowers.wordpress.com/2012/04/...s-questions-i/, which might be helpful.