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Riemann integration

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the function specified below is Riemann integrable and that its integral is equal to zero.


    2. Relevant equations
    f(x)=1 for x=1/n (n is a natural number) and 0 elsewhere on the interval [0,1].


    3. The attempt at a solution
    I have divided the partition into two subintervals, the first with tags different from x=1/n and the second with tags at x=1/n. But, given an epsilon>0, I am not sure how to choose my delta (the norm of the partition) such that the points where the function is not zero doesn't make a contribution.

    Or, is my approach all wrong?

    Thanks!
     
  2. jcsd
  3. Feb 18, 2008 #2

    quasar987

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    Maybe not all wrong, but I would say overly complicated. :)

    Consider any partition of [0,1]. Note that every subinterval from your partition contains a point not of the form 1/n.
     
  4. Feb 18, 2008 #3
    Yeah, it is sometimes like that if you study independently. :P

    So, considering any partition of [0,1]. I should then tag the points different from 1/n, then making all the contributions zero. Right?
     
  5. Feb 18, 2008 #4

    quasar987

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    Wait, a sec, have you seen the result that if a function is discontinuous at a countable number of points then it is integrable?
     
  6. Feb 18, 2008 #5
    No, i have not. But I will definetly look for it now.

    Thanks!
     
  7. Feb 18, 2008 #6

    quasar987

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    Well, if your book hasn't covered this yet, try to do without it.

    You want to show that the lower and upper integrals are 0. Prove that the lower riemann sums s(f;P) are 0 for any partition P of [0,1]. This will of course imply that the lower integral is 0.

    For the upper integral, you want to show that the inf over every P partition of [0,1] of the upper riemann sums S(f,P) is 0. Show that for every epsilon>0, you can always find a partition P' such that S(f;P')<epsilon.
     
  8. Feb 18, 2008 #7
    What you are describing now feels much better, the squeeze theorem. :)

    But I don't understand at all, how to deal with the upper integral...? When finding the inf over every partition. I would like to do it in the same way as i treat the lower integral.

    :S
     
    Last edited: Feb 18, 2008
  9. Feb 18, 2008 #8

    quasar987

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    And I don't understand your question. :P
     
  10. Feb 18, 2008 #9
    Ok. :)

    I don't understand this!
     
  11. Feb 18, 2008 #10
    Now, I understand. (I hope so anyway)

    When you wrote 'inf' I thought you meant infimum... so I thought that I was really lost since I have never heard of infimum in the context as Riemann integrals. But you must have meant int as in integral, right?

    And, yes! I am an analysis-rookie. ;)
     
  12. Feb 18, 2008 #11

    quasar987

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    I sure meant infimum.

    How do you define the upper integral? For me, the upper integral is defined as

    [tex]\inf_{P\in \mathcal{P}[0,1]}S(f;P)[/tex]

    where [tex]\mathcal{P}[0,1][/tex] is the set of all partitions of [0,1] and

    [tex]S(f;P)=\sum_{x_i\in P}\max_{x\in[x_i-1,x_i]}f(x)[/tex]
     
  13. Feb 18, 2008 #12
    Sorry!

    I don't define the 'upper integral' at all. For me the 'Riemann integral' is defined as a limit of the Riemann sums as the norm tend to zero. That is why I am talking about the partitions and their tags. I can't find any section with upper Riemann sums either. It is only the Riemann sum.
     
  14. Feb 18, 2008 #13

    quasar987

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    I see!

    Well in that case it's even simpler! Consider epsilon>0, then for any delta>0, we have that any Riemann sum associated with a partition whose norm is lesser than delta is 0 because in every subinterval of [0,1], there is a point not of the form 1/n!
     
  15. Feb 18, 2008 #14

    lurflurf

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    I would procede thusly
    clearly the lower integral is 0
    take any partition and choose taggeg point where f=0
    consider the upper integral
    suppose the norm is h where 1>h>0
    The idea is we want to make a large sum
    f1h1+f2h2+f3h3+...+fNhN
    f=0,1 so we choose f=1 whenever possible
    to take maximum advantage consiger our taggged partition
    0=x0=<x*1=<x1=<x*2=<...=<x*N-1=<xN-1=<x*N=<xN=1
    we would like n for our tagged partition to include points where f=1 when possible
    so we begin
    1>1-h>...>h+1/2>1/2>-h+1/2>...>h+1/3>1/3>-h+1/3>...
    however at some point intervals chosen in this way begin to intersect and an adjustment is needed
    we need to know when
    1/n-1/(n+1)<2h
    elementary algebra tells us this happens when
    n>N=floor(-1+sqrt(1+1/(2h)))/2
    thus
    the upper integral
    UI<1/(N+1)+2hN
    we want something in h alone
    I leave that to you
     
  16. Feb 19, 2008 #15
    Finally, I understand! Thank you so much! :)
     
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