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Does continuity prove integrability?

  1. Jan 15, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I am now asked to prove that f: [0,1]->[0,1] defined thus
    f(0)=0 and f(x)=1/10n for every 1/2n+1<x<1/2n for natural n,
    is integrable.


    2. Relevant equations



    3. The attempt at a solution
    Would it suffice to show that f is continuous? I.e. that lim x->0 f(x) = f(0) = 0, since as x->0 n->infinity?
     
  2. jcsd
  3. Jan 15, 2013 #2

    SammyS

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    You have all strict inequalities.

    What is f(x) for x = 1/2n ?

    How is this function continuous?
     
  4. Jan 15, 2013 #3
    For x=1/2^n f(x)=10^(ln2/lnx), if I am not mistaken.
    For integrability f has to be continuous, doesn't it?
    Alternatively I could demonstrate that f(x) is bound and monotone, couldn't I?
     
  5. Jan 15, 2013 #4

    SammyS

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    Then I take it that [itex]\displaystyle f(x)=\frac{1}{10^n}\ \text{ for }\ x\in\left(\frac{1}{2^{n+1}}\,,\ \frac{1}{2^n}\right]\ .[/itex]

    This function is discontinuous at x = 2-n, for all natural numbers, n .

    However, all bounded piecewise continuous functions are Riemann integrable on a bounded interval.

    What happens near x=0 ?
     
  6. Jan 15, 2013 #5
    F(x) would converge to zero by squeeze theorem, would it not?
    But if 10^(lnx/ln2)<f(x)<10^(1+lnx/ln2) then f(x) is bounded and monotone in [0,1], so why could I not simply use that to show that f is integrable in this interval?
     
  7. Jan 15, 2013 #6

    SammyS

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    Well, yes, f(x) is integrable.

    You can integrate it from 0 to 1 by setting up an infinite sum (infinite series) .
     
  8. Jan 15, 2013 #7
    I know that f is integrable. My question still stands, I believe - how shall I demonstrate that rigorously? Will it be by showing that it is bounded and monotone?
     
  9. Jan 16, 2013 #8

    SammyS

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    That seems like the thing to do.
     
  10. Jan 16, 2013 #9
    Is there a better way to go about demonstrating that? I am not sure which infinite series might serve me ideally. Could it be one whose general term is 1/2^n?
     
  11. Jan 16, 2013 #10

    SammyS

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    Do you know how to set up a Riemann sum to represent an integral?
     
  12. Jan 16, 2013 #11
    Apparently not well. I'd appreciate some guidance.
     
  13. Jan 16, 2013 #12
    Actually, since the interval is 1, dividing it to n segments would yield 1/n. Wouldn't it? Am I in the right direction?
     
  14. Jan 16, 2013 #13

    HallsofIvy

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    "Yield 1/n" for what?

    You said a couple of times "For integrability f has to be continuous, doesn't it?"

    No, it doesn't. If you are going to prove "f is integrable" then you had better review what properties a function must have to be integrable. What theorems do you know about when a function is integrable?
     
  15. Jan 16, 2013 #14

    SammyS

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    The title of your thread is: Does continuity prove integrability?
    The answer to this question is, Yes!, but continuity is not necessary for integrability.

    At any rate, the function you have here is not continuous, so the question asked in the title doesn't apply.​

    If you were to partition the interval [0, 1] into n equal length partitions, then, yes, the length of each partition would be 1/n.
    However, it is not necessary to use equal length partitions, and for this function it would make much sense to do so.

    Do you know what a graph of f(x) looks like?

    f(x):
    (n=0)
    f(x) = 1, for x ∈ (1/2, 1]​
    (n=1)
    f(x) = 1/10, for x ∈ (1/4, 1/2]​
    (n=2)
    f(x) = 1/100, for x ∈ (1/8, 1/4]​
    (n=3)
    f(x) = 1/1000, for x ∈ (1/16, 1/8]​
    (n=4)
    f(x) = 1/104, for x ∈ (1/32, 1/16]​
    ...​
     
  16. Jan 16, 2013 #15
    Alright, but even if I divided the area under 1/10^n into rectangles of height 1/10^n and length 1/2^n+1 - 1/2^n, I would get an infinite geometric series which would still not converge to the actual area! Should I then integrate that expression in order to obtain the actual area between 0 and 1?
     
  17. Jan 16, 2013 #16

    SammyS

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    What is that series? Why wouldn't it converge?

    Actually it does converge to the area between 0 and 1 .
     
  18. Jan 16, 2013 #17
    But then the sum will be a1/1-q, where q is the ratio. Which will be equal to 10/19, which seems a bit too much for such a small area (as presented by Wolfram).
     
  19. Jan 16, 2013 #18
    Moreover, what about the right angle triangles between the curve and each of the rectangles?
     
  20. Jan 16, 2013 #19

    SammyS

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    That is the answer.

    The area under the function in the interval 1/2 < x ≤ 1 is 1/2 .

    10/19 is just 1/38 greater than 1/2 .
     
  21. Jan 16, 2013 #20
    You claim that 10/19 is the answer, and yet I still don't quite understand how that could possibly be! What about the triangles between each rectangle and the curve itself? Were they taken under consideration?
     
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