Integral question

1. Jul 28, 2007

daniel_i_l

1. The problem statement, all variables and given/known data
f is a function defined in [0,1] by:
If there's some n (natural number) so that x = 1/n then f(x) = 1
If not then f(x) = 0.
Prove or disprove:
1) For every E>0 there's some P (P is a division of [0,1], for example
{0,0.2,0.6,1}) so that S(P) < E (Where S(P) is the upper some of f(x) with the division P)

2)f in integrable in [0,1] and the integral is 0.

2. Relevant equations

3. The attempt at a solution

1) True, because we can always choose A<E and: P={0,A,1/n,...,1/2,1} were 1/n is the smallest number of that type bigger than A. And then
S(P) =A < E

2)True, because if for every E>0 there's someP so that S(P) < E than the upper integral of f in [0,1] is 0. And since the lower integral is always bogger or equal to 0 (in this case) but smaller than the upper integral then the upper and lower integrals are both equal to 0 and so the integral exists and is 0.

Are thise right?
Thanks.

2. Jul 28, 2007

Dick

{0,0.2,0.6,1} is not a division of [0,1]. It's a set of points in [0,1]. How do you mean this to be interpreted as a 'division'? What you want is to choose the width of an interval around 1/n (say r_n) in such a way that the sum of the r_n is less than E. There are many explicit ways to do this. Choose one.

3. Jul 29, 2007

daniel_i_l

When I wrote {0,0.2,0.6,1} I meant {[0,0.2].[0.2,0.6],[0.6,1]}.

4. Jul 29, 2007

Dick

Ok. The previous advice still stands. Put 'small intervals' around each of the points 1/n.

5. Jul 29, 2007

daniel_i_l

Thanks, and after I do what you said to prove that (1) is true, then does that also mean that (2) is true (as I explained in the answer to 2)?

6. Jul 29, 2007

Dick

It sure does. You then have that the upper and lower sum are both zero. So what's your choice for a unit interval division. You can be pretty explicit.

7. Mar 10, 2011

Ryker

Sorry to resurrect an old thread, but I have the same problem as the OP's second one, and haven't yet found a good solution for it yet. But I don't think what you suggested here is correct, since the set containing 1/n is infinite, whereas the partitions need to be finite sets. So you can't really put small intervals around each 1/n. It would be great, if you could, and that was my first thought when approaching the problem, as well, but I don't think you can.

Any other suggestions perhaps?

8. Mar 10, 2011

Dick

Nope. Same suggestion. Try an interval size like 1/2^(n+k) and then let k->infinity to shrink the partition. If you need a finite partition then you just have to be a little fussier. Cover the first N points with intervals of those sizes and then cover the remaining points with one of size 1/N. Now let N->infinity as well.

9. Mar 10, 2011

Ryker

Well, it's not me who needs a finite partition, it's the definition

Could you perhaps elaborate on your suggestion? When you say cover the first N points with intervals of those sizes, what size do you mean? Because I can't fix n (it's infinite), I can only fix k. And if I introduce N, as well, then I now have two variables, that are not dependent on one another. But then I can't take the limit, I can only let them approach infinity one at a time.

10. Mar 10, 2011

Dick

Ok, put k=N. Surround the point 1/n with an interval of size 1/2^(n+N) for n from 1 to N. That's finite, yes? The last point is 1/N. So you can include the rest of the points in an interval of size 1/N. That's one more interval. Still finite. Now take N to infinity.

11. Mar 10, 2011

Ryker

Now you're talking Thanks, I think this should work now. I had a similar idea, but couldn't quite wrap my head around how to put it formally.