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R integrable ?

  1. May 5, 2008 #1
    f(x) = x , if x is rational
    = 0 , if x is irrational
    on the interval [0,1]

    i just wanted to check if my reasoning is right.

    take the equipartition of n equal subintervals with choices of t_r's as r/n for each subinterval.

    calculating the integral as limit of this sum (and sending the norm to 0) i got 1/2 as my value.

    now if f were to be R - integrable the value of the integral must be 1/2.
    but each subinterval for any partition would contain an irrational so the lower R sum would be 0, for all partitions of [0,1]
    this yields 0 as the value of the integral.

    the two values contradict.
  2. jcsd
  3. May 5, 2008 #2


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    Staff Emeritus
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    Yes, if that function were integrable, then you would get two contradictory values. Conclusion: that function is not (Riemann) integrable.

    (It is Lebesque integrable: since the rational numbers have measure 0, it Lebesque integral is 0. If the function were f(x)= x if is is irrational, 0 if x is rational, then its Lebesque integral would be 1/2.)
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