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Homework Help: Integral question

  1. Jul 28, 2007 #1

    daniel_i_l

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    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. jcsd
  3. Jul 28, 2007 #2

    Dick

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    {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.
     
  4. Jul 29, 2007 #3

    daniel_i_l

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    When I wrote {0,0.2,0.6,1} I meant {[0,0.2].[0.2,0.6],[0.6,1]}.
     
  5. Jul 29, 2007 #4

    Dick

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    Ok. The previous advice still stands. Put 'small intervals' around each of the points 1/n.
     
  6. Jul 29, 2007 #5

    daniel_i_l

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    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)?
     
  7. Jul 29, 2007 #6

    Dick

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    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.
     
  8. Mar 10, 2011 #7
    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?
     
  9. Mar 10, 2011 #8

    Dick

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    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.
     
  10. Mar 10, 2011 #9
    Well, it's not me who needs a finite partition, it's the definition :smile:

    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.
     
  11. Mar 10, 2011 #10

    Dick

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    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.
     
  12. Mar 10, 2011 #11
    Now you're talking :biggrin: 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.
     
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