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Math Help - f is integrable

  1. #1
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    f is integrable

    I am working on the following exercise and could use some help.
    Prove that if \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{if $x > 0$}\\<br />
-1, &\text{if $x<0$}\\<br />
\end{cases}<br />
\], then f is integrable on [-1,1] and \int_{-1}^{1}f=0.[Hint: Any lower sum is \leq 0 and some lower sum =0. Hence, the lower integral \int_{-1}^{1}f=0].

    Here is what I have tried so far...
    Let s be a lower step function such that s(x)\leq f(x) for all x\in [-1,1]. Then, let the lower sum for f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1}). Then, we have that A_{i}<f(x) for any lower step function. Also, x_{i-1}<x<x_{i}. Thus, for any x<0, x_{i}-x_{i-1}>0. So, A_{i}(x_{i}-x_{i-1})<0.
    So for \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{if $x > 0$}\\<br />
-1, &\text{if $x<0$}\\<br />
\end{cases}<br />
\], we see that any lower sum \leq 0 and some lower sum is =0. Thus, sup\{ \sum s|\text{s is a lower step function}\} =0
    =lower integral \int_{-1}^{1}f=0.

    We can do something similar for x>0...For x>0 we have that A_{i}>0 and x_{i}-x_{i-1}>0. Therefore, A_{i}(x_{i}-x_{i-1})>0.

    Then conclude that \int_{-1}^{1}f=0.


    I'm not sure if I have said enough for the proof to be complete or if what I have is even good. Please help me out.
    Last edited by zebra2147; January 30th 2011 at 11:20 AM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zebra2147 View Post
    I am working on the following exercise and could use some help.
    Prove that if \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{if $x > 0$}\\<br />
-1, &\text{if $x<0$}\\<br />
\end{cases}<br />
\], then f is integrable on [-1,1] and \int_{-1}^{1}f=0.[Hint: Any lower sum is \leq 0 and some lower sum =0. Hence, the lower integral \int_{-1}^{1}f=0].

    Here is what I have tried so far...
    Let s be a lower step function such that s(x)\leq f(x) for all x\in [-1,1]. Then, let the lower sum for f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1}). Then, we have that A_{i}<f(x) for any lower step function. Also, x_{i-1}<x<x_{i}. Thus, for any x<0, x_{i}-x_{i-1}>0. So, A_{i}(x_{i}-x_{i-1})<0.
    So for \[<br />
f(x)=<br />
\begin{cases}<br />
1, &\text{if $x > 0$}\\<br />
-1, &\text{if $x<0$}\\<br />
\end{cases}<br />
\], we see that any lower sum \leq 0 and some lower sum is =0. Thus, sup\{ \sum s|\text{s is a lower step function}\} =0
    =lower integral \int_{-1}^{1}f=0.

    We can do something similar for x>0...For x>0 we have that A_{i}>0 and x_{i}-x_{i-1}>0. Therefore, A_{i}(x_{i}-x_{i-1})>0.

    Then conclude that \int_{-1}^{1}f=0.


    I'm not sure if I have said enough for the proof to be complete or if what I have is even good. Please help me out.
    Could you explain your proof a little more. Which step function are you talking about?

    Meanwhile...

    See here? I assume this should really be f(x)=-1 on x\leqslant 0. You clearly have that f is integrable on [-1,0] since it's constant, and it's integrable on [0,1] since it's a one-point perturbation of a continuous function, which in the post i linked you to I proved was integrable. Moreover, it's clear that \displaystyle \int_{-1}^0f(x)\text{ }dx=-1 and \displaystyle \int_0^1 f(x)\text{ }dx=1 since by that same post we get that \displaystyle \int_0^1 f(x)\text{ }dx=\int_0^1 1\text{ }dx.
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  3. #3
    Member
    Joined
    Oct 2010
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    Well this is my first attempt at an integral proof and I learned that the way to prove that the integral exists is to prove that the lower integral is equal to the upper integral. So I guess my proof was attempting to prove that the lower integral =0 and then use a similar proof to show that the upper integral =0.
    It gets a little messy but basically i think I wanted to show that for a lower step function s we see that for f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1}), when x<0, A_{i}<0 and when x>0, we see that the value for A_{i} is equal to |A_{i}| when x<0. Therefore, \int_{-1}^{1}f=0.
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