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Proving a Function is Rieman Integrable

  1. Jan 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that
    f(x) = 1 x (element) E = {1/n : n (element) N}
    0 x (element) [0,1]\E
    is integrable on [0, 1] by using the definition of integrability

    2. Relevant equations

    Definition of Integrability: Let f be a bounded function. for each epsilon greater than 0 there exists a partition P such that. U(f,P)- L(f,P)< epsilon




    3. The attempt at a solution

    Fix [tex]\epsilon[/tex] >0

    I think that for any partition P, L(f,P) is gunna be zero. Since mk will always be 0 for all k.

    so all we need to do is find a partition P such that U(f,P) < [tex]\epsilon[/tex]

    i could be completely wrong up to this step, and even if im not, i don't know how to go about choosing P. HELP MEH PLZ!!!
     
  2. jcsd
  3. Jan 21, 2008 #2
    A partition of [0,1] may have the form {0,1/k,t_1,...,t_s = 1}. On [0,1/k] sup(f)=1. On [1/k,1] there are now only a finite number of elements for which f = 1, namely, x = 1,1/2,1/3,...,1/k. If you choose k large enough so that 1/k < [itex] \epsilon/2 [/itex] and you have a finite number of elements remaining for which f = 1, you need to make sure that the subintervals surrounding 1,...1/(k-1) can also have a total summed length < [itex] \epsilon/2 [/itex]. I leave the rest to you.
     
  4. Jan 21, 2008 #3
    Ok im still having trouble with this.

    from what i can tell the picture looks like this: http://i31.photobucket.com/albums/c373/SNOOTCHIEBOOCHEE/Graph.jpg

    Sorry for the crude drawing.

    Anyways. so 1/k <epsilon/2

    that means that the whole area from [0,1/k] must be < epsilon/2. I understand this and can write it out using sigma notation...

    Now choosing the the intervals around the other parts are proving to be difficult for me. I know they have to sum up to be less than Epsilon/2.

    So what i am thinking is that if we make the with of the partitions epsilon/k, that the maximum value of the area of the graph will also be epsilon/2. Is this correct?
     
  5. Jan 21, 2008 #4
    Close. I think you want to make sure that the subintervals have length less than [itex] \epsilon/2k [/itex].
     
  6. Jan 21, 2008 #5
    ya thats what i meant. Thank you sir.
     
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