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Integrability, basic measure theory: seeking help with confusing result

  1. Nov 6, 2011 #1
    The canonical example of a function that is not Riemann integrable is the function f: [0,1] to R, such that f(x)=1 if x is rational and f(x)=0 if x is irrational ( i know some texts put this the other way around, but bear with me because i can reference at least one text that does not). Hence, for any partition P of [0,1], the upper sum is 1 and the lower sum is 0, so that f is not Riemann integrable.

    However, Lebesgue's theorem states that if f is a bounded function on a bounded set A, and we extend f to all of R by letting f(x)=0 for all x not in A, then f is Riemann integrable iff the set of points at which the extended f is discontinuous form a set of measure zero (Marsden, "Elementary Classical Analysis", pg 261). Thus, if f is the function defined above, the extension of f is zero everywhere except for the rational numbers r, such that 0 <= r <= 1.

    Now it is quite common knowledge that the set of rationals between 0 and 1 is a countable set and thus has measure zero. And since the extension of f is zero everywhere except at these rationals, we should be able to apply Lesbegue's theorem and conclude that f actually IS Riemann integrable on [0,1]. Or you could just use the simple Corollary given in Marsden, that a bounded function with a countable number of discontinuities is Riemann integrable.

    Aren't these results contradictory? I know many texts define f as 1 at irrationals and 0 at rationals, so that this issue does not arise. However the text "Measure Theory and Integration" by Taylor presents it the way I described. And besides, it's still a contradictory result, isn't it?

    What am I missing here?
     
    Last edited: Nov 6, 2011
  2. jcsd
  3. Nov 6, 2011 #2

    mathman

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    It looks like you have two different definitions of Riemann integrability. Usual definition of Riemann integral is that this function is Lebesgue integrable, but not Riemann integrable. The theorem you quote (as far as I remember) says a function is integrable (not Riemann integrable), when that condition is met. Riemann integrals don't use measure theory.

    There is something called a generalized Riemann integral, which (from what I remember) includes Lebesgue integral.
     
    Last edited: Nov 6, 2011
  4. Nov 6, 2011 #3
    I also thought there was some confusion of the type which you describe arising from the use of multiple textbooks. However, if I pretend for a minute that the only text in existence is "Elementary Classical Analysis" by Marsden and Hoffman, and I flip to chapter 8 (Integration), I find that while they hint at existence of the Lebesgue integral, the topic of discussion is certainly Riemann integrability. The theorem I quoted earlier as "Lebesgue's theorem" is found on pg 261 in subsection 8.4, also titled "Lebesgue's theorem". This is certainly somewhat confusing to anybody who knows about the Lebesgue integral, but regardless, the way they state the theorem specifically indicates Riemann integrability if the conditions of Lebesgue's theorem are satisfied. And they most certainly do extensively discuss the concept of measure in relation to the Riemann integral.

    So I'm confused. I'll talk with my professor about it. Thank you for responding though! It's reassuring to hear that the confusion is likely a matter of conflicting definitions.
     
  5. Nov 6, 2011 #4

    pwsnafu

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    This

    If you take the characteristic function of the rationals, the set of discontinuities is all of R, so it does not have a set of measure zero.
     
  6. Nov 6, 2011 #5
    So that holds even when f is identically zero everywhere except for the rationals? Because it seems in that situation, the discontinuities only occur at rational values, which is a countable set.

    I definitely see what you're saying if f is identically zero everywhere except for the irrationals! But let me get this straight: even though f is constantly zero everywhere except on a countable set, it is still discontinuous everywhere? I guess that makes sense due to the density of Q, but that's very counter-intuitive to me.
     
  7. Nov 6, 2011 #6
    So for a mostly constant function that takes on a different value on a countable set, the set of discontinuities is not necessarily the same as the set where the function takes on it's different value? If this is true it's probably the most counter-intuitive concept I've come across!
     
  8. Nov 6, 2011 #7

    pwsnafu

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    Try to prove that the function is continuous at the irrationals.
     
  9. Nov 6, 2011 #8
    Yeah, you're absolutely correct. In fact, I now realize I've learned this before, but apparently the idea is so weird to me that I've put the blinders on this time around.
     
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