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Homework Help: Prove not integrable. Is this correct?

  1. Jul 4, 2010 #1
    Let f:[0,1] be defined as f(x)= 0 for x rational, f(x)=x for x irrational

    Show f is not integrable

    m=inf(f(x) on [Xi-1, Xi])
    M=sup(f(x) on [Xi-1, Xi])

    Okay so my argument goes like this:
    I need to show that the Upper integral of f does not equal the lower integral of f

    Because rationals and irrationals are dense in R for any interval [a,b] of f,


    therefore, for the partition X0=0, X1=1 the Lower Darboux sum = 0 and the Upper Darboux sum = x

    therefore the sup{Lower Darboux sum}=0
    and the inf{Upper Darboux sum} is not equal to 0, (it is the smallest irrational number greater than 0 technically right? although there is no smallest irrational greater than 0, right?)

    therefore the Upper integral of f does not equal the lower integral of f

    and therefore f is not integrable.

    I think this is right, but I just want to make sure I didn't miss anything.

  2. jcsd
  3. Jul 4, 2010 #2


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    Staff Emeritus
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    Gold Member

    For your upper sum you said that M on an interval has to be greater than the smallest irrational inside the interval. When you break up [0,1] into tiny pieces you're essentially integrating the function x, so you expect to get as an upper sum 1/2

    All you have to prove is that the upper sum can't give you 0; saying it has to be larger than the smallest irrational number greater than 0 is not good enough (since as yo usaid there is no such number, so that has no meaning)
  4. Jul 4, 2010 #3
    Well, what I was trying to say was that the upper sum cannot equal 0 because for any interval [a,b] M will not be 0 because there is always an irrational greater than 0 no matter how small the partition [Xo, X1] gets. Because M will never be 0 the inf of the Upper Darboux Sums will never be 0 and thus the upper integral will never equal the lower integral, even though I haven't defined what the upper actually is...

    I wasn't trying to say anything the smallest irrational number greater than 0, just that it couldn't be equal to 0 because there would always be an irrational greater than 0 making M on that interval greater than 0.
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