1. Not finding help here? Sign up for a free 30min 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!

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


    User Avatar
    Science Advisor
    Gold Member

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

Have something to add?