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Math Help - Darboux Sums, Upper/Lower integral

  1. #1
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    Darboux Sums, Upper/Lower integral

    If the function f : [a,b] \rightarrow R is bounded and that P = \{\{x_0\},.....,\{x_n\}\} is a partition of its domain [a,b]. For each index i \geq 1 we have:

    {m_i} \equiv \mbox{inf} \{f(x) | x \in [x_{i-1}, x_i] \}

    and

    M_i \equiv \mbox{ sup} \{f(x) | x \in [x_{i-1}, x_i] \}
    -----------------------------------------------------
    Then for a function f: [0,1] \rightarrow R defined by:

    f(x)=\left\{\begin{array}{cc}0,&\mbox{ if the point x in [0,1] is rational}\\1, & \mbox{ if the point x in [0,1] is irrational}\end{array}\right.

    in this case, we suppose P = \{{x_0},....., {x_n}\} a partition of its domain [0,1]. since the rationals and irrationals are dense in R, it follows that for each index i \geq 1, if m_i & M_i are defined as above, then m_i = 0 and M_i = 1.

    Therefore, the collection of the lower Darboux sums consists of the single number 0,

    \int_{*a}^b f = 0

    and the upper collection of darboux sums consists of the single number 1, therefore:

    \int_a^{*b} f = 1

    note: the * sign at the bottom and top denotes the lower integral of f on [a,b] and the higher integral of f on [a,b] respectively.

    -----------------------------
    So, if we have another function

    f(x)=\left\{\begin{array}{cc}x,&\mbox{ if the point x in [0,1] is rational}\\0, & \mbox{ if the point x in [0,1] is irrational}\end{array}\right.


    then how does

    \int_{*a}^b f = 0 and \int_a^{*b} f \geq \frac{1}{2} in this case???

    Following the example above, isn't the collection of lower darboux sum supposed to be x here, and how is it shown that the upper integral of f is greater than (1/2)?
    Last edited by serious331; May 3rd 2010 at 04:46 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    If the function f : [a,b] \rightarrow R is bounded abd that P = \{\{x_0\},.....,\{x_n\}\} is a partition of its domain [a,b]. For each index i \geq 1 we have:

    m_i \equiv \inf \left\{f(x) | x \in [x_{i-1}, x_i]\right\}

    and

    M_i=\sup \left\{f(x) | x in [x_{i-1}, x_i]\right\}
    -----------------------------------------------------
    Then for a function f: [0,1] \rightarrow R defined by:

    f(x)=\begin{cases} 0 & \mbox{if} \quad x\in\mathbb{Q} \\ 1 & \mbox{if}\quad x\notin\mathbb{Q}\end{cases}

    in this case, we suppose P = \{{x_0},....., {x_n}\} a partition of its domain [0,1]. since the rationals and irrationals are dense in R, it follows that for each index i \geq 1, if m_i & M_i are defined as above, then m_i = 0 and M_i = 1.

    Therefore, the collection of the lower Darboux sums consists of the single number 0,

    \int_{*a^b} f  = 0

    and the upper collection of darboux sums consists of the single number 1, therefore:

    \int_{a^*b} f  = 0

    note: the * sign at the bottom and top denotes the lower integral of f on [a,b] and the higher integral of f on [a,b] respectively.
    I'm not sure what you mean, but the upper sums are always one and the lower zero.

    -----------------------------
    So, if we have another function

    f(x)=\begin{cases}x & \mbox{if} \quad x\in\mathbb{Q} \\ 0 &\mbox{if}\quad x\notin\mathbb{Q}\end{cases}

    then how does
    \int_{*a^b} f  = 0 and \int_{a^*b} f  \geq \frac{1}{2} in this case??? Following the function above, isn't the collection of lower darboux sum supposed to be x here, and how is it shown that the upper integral of f is greater than (1/2)?
    I actually really like this problem, give it some more thought but here's the idea

    Spoiler:


    So, let us denote \mathcal{P}_n=\left\{0,\frac{1}{n},\frac{2}{n},\cd  ots,1\right\}.

    So, we cut [0,1] into subintervals of the form \left[\tfrac{k}{n},\tfrac{k+1}{n}\right]. And so clearly \sup_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k+1}{n}. And so, we get U\left(\mathcal{P}_n,f\right)=\sum_{j=0}^{n}\mathc  al{M}_j\Delta x_j=\sum_{j=0}^{n}\frac{j+1}{n}\cdot\frac{1}{n}=\f  rac{1}{n^2}\sum_{j=0}^{n}j+1=\frac{(n+1)(n+2)}{2n^  2}\geqslant\frac{1}{2}

    So?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    I'm not sure what you mean, but the upper sums are always one and the lower zero.


    I actually really like this problem, give it some more thought but here's the idea

    Spoiler:


    So, let us denote \mathcal{P}_n=\left\{0,\frac{1}{n},\frac{2}{n},\cd  ots,1\right\}.

    So, we cut [0,1] into subintervals of the form \left[\tfrac{k}{n},\tfrac{k+1}{n}\right]. And so clearly \sup_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k+1}{n}. And so, we get U\left(\mathcal{P}_n,f\right)=\sum_{j=0}^{n}\mathc  al{M}_j\Delta x_j=\sum_{j=0}^{n}\frac{j+1}{n}\cdot\frac{1}{n}=\f  rac{1}{n^2}\sum_{j=0}^{n}j+1=\frac{(n+1)(n+2)}{2n^  2}\geqslant\frac{1}{2}

    So?
    then,
     U\left(\mathcal{P}_n,f\right) \geq \frac{1}{2} proves that the upper integral of f on [a,b] is greater than or equal to 1/2.

    and are we using \inf_{x\in[\frac{k}{n},\frac{k+1}{n}]}f(x)=\frac{k}{n} to prove that the lower integral of f on [a,b] = 0?

    Thank you so much for the help!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by serious331 View Post
    then,
     U\left(\mathcal{P}_n,f\right) \geq \frac{1}{2} proves that the upper integral of f on [a,b] is greater than or equal to 1/2.
    Well, it proves that \overline{\int_0^1}f(x)\text{ }dx\geqslant\frac{1}{2}, for a general interval I leave you to generalize.
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