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!

Show that f is not integrable

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Let f(x) = 0, if x is rational; 2 + x, if x is irrational, for all x in [0,1]. Show that f is not integrable


    2. Relevant equations
    density of rationals/irrationals, equation for L(f,P) and U(f,P), Darboux Integrability criterion


    3. The attempt at a solution
    L(f,P) = 0 because on any subinterval formed from two consecutive points in the partition of [0,1], there exists a rational x.

    For U(f,P), I was thinking that I could just pick any irrational x so that sup{ f(x): x is an element of the subinterval from two consecutive points in the partition of [0,1] } = 2 + x. Then note that since [tex]2+x \geq 2[/tex], [tex]U(f,P) \geq 2(b-a) > 0[/tex] so that

    [tex]sup {L(f,P)} \neq inf {U(f,P)}[/tex]

    I left a few steps in demonstrating U(f,P) > 0 but is this a good approach?
     
  2. jcsd
  3. Jan 10, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sort of. 2+x>=2. x<=1. Can you show any U(f,P)>=2 and any L(f,P)<=1?
     
  4. Jan 10, 2009 #3
    Hmm I was really lax about the bounds. U(f,P) >= 2(b-a) = 2. L(f,P) = 0, no matter what, I think, so L(f,P) is certainly <= 1.

    As for U(f,P), could I just say that 2+x >= 2 for any x in [0,1] so that U(f,P) >= 2*SUMMATION{bunch of points that cancel out to give b -a}?
     
  5. Jan 10, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure U(f,P)>=2, but why do you think L(f,P)=0? Take the partition P of the intervals [0,1/2] and [1/2,1]. Then L(f,P)=1/4, I think.
     
  6. Jan 10, 2009 #5
    Sorry, here's my reasoning. The definition of L(f,P) is the summation of a bunch of products, each of which is of the form m*(d-c), where d and c are consecutive points in the partition and c < d. The quantity m is equal to inf{f(x) : c <= x <= d}. But assuming the density of the rationals, the quantity m will always be 0. Therefore, the summation of these products will yield 0 as well.
     
  7. Jan 10, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Oooops. Right. I misread the problem. Somehow I read it as f(x)=x for x rational instead of f(x)=0. So, yes, L(f,P)=0, U(f,P)>=2.
     
  8. Jan 10, 2009 #7
    Haha ok. Actually when you suggested x <= 1 I thought to myself did I put x instead of 0? Thanks for your help though. Especially in noting that it's probably a good idea to use the best bounds possible (I sometimes forget that b and a are given endpoints and leave b-a as b-a).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?