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Prove the function is Riemann-integrable

  1. May 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##f:[-1,5]→ℝ## such that ##f(x)= \left\{
    \begin{array}{l}
    3, -1≤x<0 \\
    0 , 0≤x≤2 \\
    -2, 2<x≤5
    \end{array}
    \right. ##. Prove f is Riemann integrable.

    3. The attempt at a solution

    Obviously f is bounded.
    Let ##ε>0## arbitrarily.
    Let partition of interval [-1,5] be P={-1,0,2,5}.

    Then ##U(P,f)=3*(0-(-1))+0*(2-0)+(-2)*(5-2)=-3##
    and ##L(P,f)=3*(0-(-1))+0*(2-0)+(-2)*(5-2)=-3##

    Therefore ##U(P,f)-L(P,f)=0<ε##
    Here comes the question: Is the partition I chose valid and correct?
     
    Last edited: May 7, 2016
  2. jcsd
  3. May 7, 2016 #2

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    Without knowing what a particular teacher might want, your partition looks very valid and appropriate. I think it must be the answer that was expected.
     
  4. May 7, 2016 #3
    I am asking because my teacher and one of my classmates said my partition is not valid and I don't get why. May be it has something to do with how the function is defined and/or infimum and supremum of the given function? I am little confused. Or maybe they are just wrong? I chose this particular partition because it seemed to be straightforward and simple. They had included epsilons in their partitions.
     
    Last edited: May 7, 2016
  5. May 7, 2016 #4

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    They must want a partition with equal sized steps of Δx. That is how the Riemann integral is usually defined. You can break up your partition into smaller, Δx=1, steps and keep essentially the same logic. Just sum more terms for the smaller steps.
     
  6. May 7, 2016 #5

    micromass

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    What is the definition of Riemann integrable?
     
  7. May 7, 2016 #6

    LCKurtz

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    Stop right there. The sup on [0,2] is not 0 and the sup on [2,5] is not -2. There are similar errors for your lower sum.
     
  8. May 8, 2016 #7
    Let ##f:[a,b]→ℝ## be bounded function and ##P=##{##x_0, x_1, x_2, ..., x_n##} partition of interval ##[a,b]## so ##a=x_1<x_2<...<x_n=b##. Let ##\Delta x_i=x_i -x_{i-1}##, where ##i=1,2,3,...,n## and let's denote ##M_i=sup## {##f(x)|x∈[x_{i-1},x_i]##}. Then upper sum is ##U(P,f)=\sum_{n=1}^{n} M_i \Delta x_i##
     
  9. May 8, 2016 #8
    What is it then?
     
  10. May 8, 2016 #9

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    The sup on [0,2] is 0. The sup on (2,5] is -2, but on [2,5] it is 0. For the formal definition of Riemann integral, you probably should surround the points of discontinuity by arbitrarily small intervals and show that the difference between L(P,f) and U(P,f) can be made arbitrarily small. Since the summation for the Riemann integral is countable, it is unfortunate that the endpoints of the intervals are included in the definition of the Mi. Any countable set of points can be covered by countable many arbitrarily small intervals and will not effect the integral.
     
  11. May 8, 2016 #10
    I don't understand the problem statement. The integral of a step function defined on a segment is a fundamental definition rather than a property. It seems to me that you are asked to prove a definition, which seems to be nonsense to me.
     
  12. May 8, 2016 #11

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    The problem is to prove that the function is Riemann integrable. The property of being Riemann integrable is defined (see http://math.feld.cvut.cz/mt/txtd/1/txe3da1a.htm ) and it is necessary to prove that the given function has that property. Once that property is proven, the Riemann integral is defined.
     
  13. May 8, 2016 #12
    But why would one want to prove integrability of a step function defined on a segment ? Such functions are integrable by definition. It is like if someone said 'Let ABCD a rectangle. Show that angle ABC equals 90 degrees'. I don't get the subtlety.
     
  14. May 8, 2016 #13

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    Apparently they are not integrable by definition because that is not how the Riemann integral has been defined in that class. The link I gave above tells what it takes to prove that a Riemann integral exists in terms the OP is using . That is a common definition of the Riemann integral. It is necessary to prove at least once that a step function satisfies the conditions. As far as the subtlety goes, this example is a good exercise on how to do such proofs. It's nice to practice on easy examples because there may be much harder examples later.
     
  15. May 8, 2016 #14

    micromass

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    Why is everybody guessing? Shouldn't we wait for the OP to first tell us what his definition of Riemann integrable is?
     
  16. May 8, 2016 #15

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    It's a standard definition; he is using the standard terminology; the teacher's objection is absolutely correct.
     
  17. May 8, 2016 #16

    micromass

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    It's not. I know at least 4 different definitions of Riemann integrability. All are equivalent of course. I just want to hear which one he adopts.
     
  18. May 8, 2016 #17

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    In my opinion, posts 1 and 7 leave no room for doubt. And the inclusion of both endpoints in the Mi of post 7 causes the problem that the original approach in post 1 refers to.
     
  19. May 8, 2016 #18

    micromass

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    Maybe not. But since this thread is about proving Riemann integrability, I don't see it as a bad thing that the OP actually states what this means.
     
  20. May 8, 2016 #19

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    Agree. You make a good point.
     
  21. May 8, 2016 #20

    LCKurtz

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    Since we are speculating here, I'm guessing the OP may be taking a course from little Rudin where, if my memory from long ago serves me, whatever particular definition he gave, he proves a theorem that ##f## is integrable on ##[a,b]## if you can show that given ##\epsilon > 0## you can find a partition ##P## of the interval such that$$
    U(P,f) - L(P,f) < \epsilon$$It's pretty clear that is what the OP is trying to show. Under ideal circumstances the OP would have stated the definition of integrability and given the theorem in the home work template.
     
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