1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    Where,

    [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

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    Office,

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Real analysis:Showing a function is integrable
Loading...