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Homework Help: Real analysis:Showing a function is integrable

  1. Apr 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f:[0,2]-> R [/itex] be defined by[itex] f(x)= 1 [/itex] if [itex]x≠1[/itex], and [itex]f(1)= 0[/itex]. Show that [itex]f[/itex] is integrable on [itex][0,2][/itex] and calculate its integral

    2. Relevant equations
    Lower integral of f

    [itex]L(f)= sup {(P;f): P \in P(I)} [/itex]

    Upper integral of f

    [itex]U(f)= inf {U(P;f): P \in P(I)} [/itex]


    [itex]L(P;f)= \sum m_{k}(x_{k}-x_{k-1})[/itex]

    [itex]U(P;f)= \sum M_{k}(x_{k}-x_{k-1})[/itex]

    And lastly, [itex]U(f)=L(f)[/itex] if the integral exists

    3. The attempt at a solution
    So it seems pretty obvious the integral is equal to 2, but I am not sure how to deal with this function at x=1. I tried splitting the interval up into sections, but it turns out partitions only work if the interval is closed and bounded on R, so I couldn't do any open interval stuff. Any ideas?
  2. jcsd
  3. Apr 7, 2012 #2


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    Given a partition, what can you say about Mk on every sub-interval? What about mk on every sub-interval except for 1?
  4. Apr 7, 2012 #3
    Mk is always 1, and so is mk except for 1 subinterval it will be 0
  5. Apr 8, 2012 #4


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    And the length of every sub-interval goes to 0.
  6. Apr 9, 2012 #5
    The sub-intervals width depends on how large "n" is doesn't it? I don't see what you mean Halls of Ivy.
  7. Apr 9, 2012 #6


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    You're looking at the supremum and infimum over all partitions. Can you sumac partitions where the interval containing 1 had arbitrarily small length?
  8. Apr 10, 2012 #7

    Are you saying that you could consider the interval containing 1 infintesimally small so that mk and Mk are zero on it? So this would be summing zero for the interval? I am not sure
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