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Math Help - Integrability

  1. #1
    Senior Member I-Think's Avatar
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    Integrability

    Using these 2 facts for integrable functions f, g
    1. \int_a^b (f+g)=\int_a^b f+\int_a^b g<br />

    2. \int_a^b cf=c\int_a^b f for a constant c

    Prove that any integrable function f can be changed at a finite number of points in the interval [a,b] without changing its integrability or its integral.

    Start of Proof
    I know that I only have to prove it for changing 1 point, and an induction argument will take care of the rest.
    But how do I go about doing this?

    I started with the definition of integrability
    sup{ }L(f,P)=inf{ }U(f,Q), where P and Q are partitions of [a,b]

    and so dealing with only the sup and a partition a<x_1<b)" alt="Pa<x_1<b)" /> and a<x_2<b)" alt="Qa<x_2<b)" />
    I need to prove
    sup{ }f[a,x_1](x_1-a)+sup{ }f[x_1,b](b-x_1)=sup{ }f[a,x_2](x_2-a)+sup{ }f[x_2,b](b-x_2)<br />
    How do I use the given facts to solve this problem. Or do I need a new approach
    Thanks in advance.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    Using these 2 facts for integrable functions f, g
    1. \int_a^b (f+g)=\int_a^b f+\int_a^b g<br />

    2. \int_a^b cf=c\int_a^b f for a constant c

    Prove that any integrable function f can be changed at a finite number of points in the interval [a,b] without changing its integrability or its integral.

    Start of Proof
    I know that I only have to prove it for changing 1 point, and an induction argument will take care of the rest.
    But how do I go about doing this?

    I started with the definition of integrability
    sup{ }L(f,P)=inf{ }U(f,Q), where P and Q are partitions of [a,b]

    and so dealing with only the sup and a partition a<x_1<b)" alt="Pa<x_1<b)" /> and a<x_2<b)" alt="Qa<x_2<b)" />
    I need to prove
    sup{ }f[a,x_1](x_1-a)+sup{ }f[x_1,b](b-x_1)=sup{ }f[a,x_2](x_2-a)+sup{ }f[x_2,b](b-x_2)<br />
    How do I use the given facts to solve this problem. Or do I need a new approach
    Thanks in advance.
    I would assume the intended method was as follows. I'll assume for ease that \alpha is continuous. You can generalize


    Lemma: Let c\in[a,b] and \mathbf{1}_c the indicator function on \{c\}. Then, \mathbf{1}\in\mathcal{R}(\alpha) on [a,b] and \displaystyle \int_a^b\mathbf{1}_c\text{ }d\alpha(x)=0.
    Proof: Note that evidently \displaystyle L(P,\mathbf{1}_c,\alpha)=0 for every partition P of [a,b]. Thus, it suffices to show that for every \varepsilon>0 there exists some partition P of [a,b] such that U\left(P,\mathbf{1}_c,\alpha\right)<\varepsilon. To do this choose \displaystyle x_1,x_2 such that \alpha(x_2)-\alpha(x_1)<\varepsilon and c\in(x_1,x_2). It's clear then that if P=\{a,x_1,x_2,b\} then U\left(P,\mathbf{1}_c,\alpha\right)<\varepsilon from where integrability follows. The integral then follows by recalling that \displaystyle \int_a^b\mathbf{1}_c\;d\alpha(x)=\sup_{P\in\wp}L(P  ,\mathbf{1}_c,\alpha)=0. The conclusion follows. \blacksquare


    From this note then that if f\in\mathcal{R}(\alpha) on [a,b] and \tilde{f} is such that f-\tilde{f}=\left(f(c)-\tilde{f}(c)\right)\mathbf{1}_c where f(c)\ne\tilde{f}(c) then evidently \tilde{f} is continuous since by our lemma we have that \displaystyle \frac{1}{f(c)-\tilde{f}(c)}\left(f-\tilde{f}\right) is integrable and thus so is \left(f(c)-\tilde{f}(c)\right)\cdot\frac{1}{f-\tilde{f}(c)}\left(f-\tilde{f}\right)f=\tilde{f} and


    \displaystyle \int_a^b\tilde{f}(x)\text{ }d\alpha(x)=(f(c)-\tilde{f}(c))\int_a^b\mathbf{1}_c\text{ }d\alpha(x)+\int_a^bf(x)\text{ }d\alpha(x)=\int_a^bf(x)\text{ }d\alpha(x)
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