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Positive integral implies bounded below?

  1. Feb 5, 2007 #1
    1. The problem statement, all variables and given/known data
    The integral of f on [a,b] exists and is positive.
    Prove there is a subinterval J of [a,b] and a constant c such that f(x) >= c > 0 for all x in J.

    Hint: Consider the lower integral of f on [a,b]


    2. Relevant equations



    3. The attempt at a solution
    I don't see how the hint helps. Obviously both the upper and lower integrals are greater than 0. I tried considering refinement partitions of J extending to [a,b] but that doesn't get me back to getting the actual function bounded.

    Since the lower integral is greater than 0, the lower sum for any partition is greater than 0 as well.... I just don't see how this is helping and how to go from the sum/integral back to the original function. Is there a different approach that could be taken? Any help? Thx :)
     
  2. jcsd
  3. Feb 5, 2007 #2

    D H

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    What about the Dirac delta distribution?

    If f is indeed a function, then what can you say about the integral if no such subinterval J, positive constant c exists?
     
  4. Feb 6, 2007 #3
    Have no idea what Dirac Delta Distribution is... haven't learnt that.
     
  5. Feb 22, 2007 #4
    Try proof by contradiction (Lower integral converges to THE integral as norm of partition goes to zero)
     
  6. Feb 22, 2007 #5

    AKG

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    Consider the characeristic function of the irrationals in [a,b] (f(x) = 1 if x is an irrational number in [a,b], and f(x) = 0 if x is a rational number in [a,b]). The (Lebesgue) integral of f on [a,b] exists and is b-a which is positive, but it takes on the value 0 infinitely many times in any subinterval J.

    That was just for your information. You're probably dealing with the Riemann integral. Suppose there is no subinterval with that property. Then on every subinterval [c,d] for a < c < d < b, what's the minimum value f takes on? What is the lower Riemann sum for that partition? Then what is the lower integral, i.e. the supremum of these lower sums? This should contradict the sentence: "The integral of f on [a,b] exists and is positive."
     
  7. Feb 22, 2007 #6
    Well, the function is supposed to be positive, so the characteristic function in AKG's response does not really apply (ya it can be modified). I hope my response was plain enough and will get you started.

    gammamcc
     
  8. Feb 22, 2007 #7

    AKG

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    Nowhere does it say that the function has to be positive.
     
  9. Feb 22, 2007 #8

    D H

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    You can't assume that. (1) The intent of the problem is to prove that the function must be positive over some subinterval. (2) It isn't true. All you know is that the [itex]\int_a^b f(x)dx > 0[/itex]. For example, [itex]\int_{-\pi}^{\pi/2}\cos x dx[/itex] is positive (1) but the integrand is negative on [itex][-\pi,0)[/itex].

    Note that the proposition is false as both AKG and I have given counterexamples. Some other constraints (e.g., continuity) must apply to make to proposition true.
     
    Last edited: Feb 22, 2007
  10. Feb 22, 2007 #9
    OK. It's just that one is supposed to conclude it's positive ..on an interval. OK??? So your are over analyzing it as you throw Lebesgue theory at someone you know is asking a classical analysis question.
     
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