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Nonintegrable bounded function

  1. Feb 10, 2005 #1
    could someone give me an example of a function that is bounded but is nonintegrable?


    i need to know what a nonintegrable function bounded on [a,b] is as said in my preperation file for a test? urgent help needed
     
    Last edited: Feb 10, 2005
  2. jcsd
  3. Feb 10, 2005 #2

    dextercioby

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    How about
    [tex] f:(0,1)\rightarrow R [/tex],

    [tex] f(x)=0,x\in ((0,1) \cap R-Q) [/tex]

    [tex] f(x)=1,x\in ((0,1) \cap Q) [/tex]

    Daniel.
     
  4. Feb 10, 2005 #3
    yup thnx, i forgot we did an example of dirichilet function in class, this function is an example of alot of things used in calculus, lol,
     
  5. Feb 10, 2005 #4

    dextercioby

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    Always remember that continuity is a necessary condition for integrability...

    Daniel.
     
  6. Feb 11, 2005 #5

    AKG

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    Continuity is a sufficient condition for integrability, not a necessary one. A function that is discontinuous on a set of points of measure 0 is integrable, and vice versa (i.e. this gives a necessary and sufficient condition). Clearly, a continuous function is discontinuous on an empty set which of course has measure 0, so it is integrable. The example you gave is discontinuous on (0, 1), a set that doesn't have measure 0, which is why f is not integrable. Of course, this also depends on how you define integration and integrability.
     
    Last edited: Feb 11, 2005
  7. Feb 11, 2005 #6

    dextercioby

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    Are u talking about Lebesgue,or Riemann integrability...?

    Daniel.
     
  8. Feb 11, 2005 #7

    HallsofIvy

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    dextercioby: The function f(x)= 0 if x< 0; 1 if 0< x< 1; 0 if x> 1 is (Riemann) integrable over any interval but is not continuous at 0 and 1.

    The function: f(x)= 0 if x is rational; 1 if x is irrational is (Lebesque) integrable over any interval but is not continuous anywhere.
     
  9. Feb 11, 2005 #8

    Hurkyl

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    AKG's right, I'm pretty sure. IIRC, A bounded function is Riemann integrable over a compact set iff it's discontinuous on a set of measure zero.
     
  10. Feb 11, 2005 #9

    dextercioby

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    Got it.Thank you.

    Daniel.
     
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