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Math Help - Riemann integral and null sets

  1. #1
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    Riemann integral and null sets

    Let $f$ be a nonnegative Riemann integrable function on [a,b].

    Prove that $int_{a}^{b}f(x) dx = 0$

    if and only if

    $E = {x \in [a,b] | f(x) > 0}$ is a null set.
    ___________________

    <= If E is a null set I want to be able to define a partition
    that allows me to control the contribution to the integral on those
    intervals which contain points in E by arguing that these intervals have
    arbitrarily small length and f(x) is bounded. Then I can control the
    contribution to the integral on those intervals which do not contain
    points in E by observing that the function is equal to 0 at these points.
    But given that E is a null set I get a cover of E with arbitrarily small
    length. But this cover is the countable union of intervals. I need to refine
    this to a finite subcover if I hope to construct a partition as described
    above. E needn't be compact as far as I can tell thus a finite subcover is
    not guaranteed.

    => No idea.
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  2. #2
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    Nevermind, was unfamiliar with the 'null set' terminology.
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  3. #3
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    Quote Originally Posted by iknowone View Post
    Let $f$ be a nonnegative Riemann integrable function on [a,b].

    Prove that $int_{a}^{b}f(x) dx = 0$

    if and only if

    $E = {x \in [a,b] | f(x) > 0}$ is a null set.
    ___________________

    <= If E is a null set I want to be able to define a partition
    that allows me to control the contribution to the integral on those
    intervals which contain points in E by arguing that these intervals have
    arbitrarily small length and f(x) is bounded. Then I can control the
    contribution to the integral on those intervals which do not contain
    points in E by observing that the function is equal to 0 at these points.
    But given that E is a null set I get a cover of E with arbitrarily small
    length. But this cover is the countable union of intervals. I need to refine
    this to a finite subcover if I hope to construct a partition as described
    above. E needn't be compact as far as I can tell thus a finite subcover is
    not guaranteed.

    => No idea.
    Correct me if I'm wrong but I don't think this is true

    Set E=\mathbb{Q} \cap [a,b]

    Since the rational numbers are countable this mean we can we can index them with the natural numbers and consider the denumeration

    q_1,q_2,q_3,....
    and let \epsilon > 0 then

    E \subset Q but

    Q \subset \bigcup_{n=1}^{\infty}(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+  1}})

    This last set have measure zero i.e it is a null set and contains E.
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  4. #4
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    Quote Originally Posted by iknowone View Post
    Let $f$ be a nonnegative Riemann integrable function on [a,b].

    Prove that $int_{a}^{b}f(x) dx = 0$

    if and only if

    $E = {x \in [a,b] | f(x) > 0}$ is a null set.
    ___________________

    <= If E is a null set I want to be able to define a partition
    that allows me to control the contribution to the integral on those
    intervals which contain points in E by arguing that these intervals have
    arbitrarily small length and f(x) is bounded. Then I can control the
    contribution to the integral on those intervals which do not contain
    points in E by observing that the function is equal to 0 at these points.
    But given that E is a null set I get a cover of E with arbitrarily small
    length. But this cover is the countable union of intervals. I need to refine
    this to a finite subcover if I hope to construct a partition as described
    above. E needn't be compact as far as I can tell thus a finite subcover is
    not guaranteed.

    => No idea.
    \Rightarrow ) If \mu (E) >0 then 0< \int_{E} f \leq \int_{a}^{b} f which is a contradiction.

    \Leftarrow ) If E is a null set then \int_{a}^{b} f=\int_{a}^{b} 0=0 (because the integral doesn't see null sets)

    One question though: Have you seen the Lebesgue integral? because otherwise my 'proof' might not be so adequate.
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  5. #5
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    No, I want to show this result using only the definition of the Riemann integral as the inf of the set upper sums over all partitions of [a,b].
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