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Is f Integrable?

  1. Nov 23, 2008 #1
    The function f is defined on [0,1] so that f(1/n)=n^(-1/2) for n=1,2,3,... and f(x)=0 if x is not a reciprocal of a positive integer. Is f integrable on [0,1]? If so, prove it and compute the integral. If not then give an argument for why not.

    See, I read this question over a hundred times, and the thing is... f(x) is always going to be a reciprocal of a positive integer... so the second statement is saying that any number inbetween the reciprocal of a positive integer, meaning all the irrational numbers inbetween 0 and 1 are equal to zero, no? Therefore this function is not continuous? Therefore this function cannot be integrable right? On top of that, if the function is 1 over a squareroot, then there will be two values for every reciprocal of a positive integer. (keeping in mind that a squareroot gives a positive and a negative answer right?

    Am I wrong?

    Jonnah Song
  2. jcsd
  3. Nov 23, 2008 #2


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    The function is integrable, Riemann or Lebesgue. Lebesgue is easy, since f=0 except on a set of measure 0. To show Riemann integrable, divide [0,1] into m subintervals, most of them will not contain a point where x=1/n. The upper Darboux sum can be seen to be < [sum(1,m) 1/n^1/2]/m ~ 1/m^1/2 -> 0. The lower Darboux sum is 0.
  4. Nov 23, 2008 #3
    sorry wrong statement, he said... "first notice that the function f is not continuous only at
    countably many points."

    sorry about that.
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