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Is this statmenet true

  1. Dec 4, 2008 #1
    I can't make up my mind if this statement is true or not: is it true that if we assume g is integrable and g[tex]\geq[/tex]0 on [a,b], then if g(x)[tex]\geq[/tex]0 for an infinite number of points x is in [a,b] then [tex]\int[/tex] g >0.

    I can't figure out if its true or false, i thought that i had a counter example:
    if g>0 at a single point i.e. g(a)=1 and g=0 otherwise, then g is integrable and non negative, and the set of discontinuities must be finite.
    I don't know if this is exactly right. Any help please?
  2. jcsd
  3. Dec 4, 2008 #2
    What do you mean with g(x)>0 for an infinite number of points x in [a,b]? Are u saying here, that we are assuming that for some points in [a,b] g(x) might not be greater than 0 or?
  4. Dec 4, 2008 #3
    I'm assuming you mean if on [a, b], g is integrable and nonnegative, and g(x) > 0 for an infinite number of points x in [a, b], then [tex]\int_{[a, b]} g > 0[/tex].

    Your example only has g(x) > 0 for a finite number of points, so it is not a counterexample.

    However, here is a counterexample: On [0, 1], let g(x) = 1 if x = 1/n for some positive integer n, and 0 otherwise. Then g(x) = 1 at infinitely many points, but it is integrable and the integral is zero. Reason: each point in S = {1/n | n is a positive integer} is an isolated point of S.
    Last edited: Dec 4, 2008
  5. Dec 5, 2008 #4


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    No, it is not true. For example, if f(x)= 1 for x rational and f(x)= -1 for x irrational, then f(x) is "positive for an infinite number of points" but its (Lebesque) integral, from x=0 to x=1, is -1 since f(x)= -1 except on a set of measure 0.
  6. Dec 5, 2008 #5
    Two issues I have with that argument: First, it doesn't directly address the problem, since there is the assumption that g(x) ≥ 0 everywhere. Second, he didn't specify which kind of integrability we're talking about, so I just assumed we're talking about Riemann-integrable functions, and yours is not one of them.
  7. Dec 6, 2008 #6
    The statement is true if you make it [tex]\int^b_ag(x) \ge 0[/tex]

    To see this evaluate the integral with Riemann sums and we make the distance between each point of subdivision equal. So let [tex] h = \frac{b-a}{n} [/tex]. Then we have [tex] \int^b_ag(x) = \lim_{n \rightarrow \infty} \sum\limits_{i=1}^n h(g(x_i)) [/tex]

    The sum is greater than or equal to zero because each term in the sum is nonnegative.
  8. Dec 7, 2008 #7
    Yes, but that's trivial.
  9. Dec 7, 2008 #8
    It was similar to what the OP was asking. There is no harm done in adding what I added.
  10. Dec 7, 2008 #9
    You can easily make HallsOfIvy's function nonnegative, just make it 0 at every irrational number. Come on, you could make that small modification.
  11. Dec 7, 2008 #10
    You could, and the (Lebesgue) integral would be zero. However, that is not Riemann integrable (which is what I was assuming).
  12. Dec 7, 2008 #11
    Well, working in this situation with the Riemann integral is rather delicate, because even the integrability of such a function is difficult to establish. Sorry if you meant Riemann, I just always think Lebesgue.
  13. Dec 7, 2008 #12
    I think what I was really trying to say was that the statement is false even for Riemann-integrable functions, nothing more.
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