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Thread: 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 $\displaystyle \[
    f(x)=
    \begin{cases}
    1, &\text{if $x > 0$}\\
    -1, &\text{if $x<0$}\\
    \end{cases}
    \]$, then $\displaystyle f$ is integrable on $\displaystyle [-1,1]$ and $\displaystyle \int_{-1}^{1}f=0$.[Hint: Any lower sum is$\displaystyle \leq 0$ and some lower sum $\displaystyle =0$. Hence, the lower integral $\displaystyle \int_{-1}^{1}f=0$].

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

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

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


    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; Jan 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 $\displaystyle \[
    f(x)=
    \begin{cases}
    1, &\text{if $x > 0$}\\
    -1, &\text{if $x<0$}\\
    \end{cases}
    \]$, then $\displaystyle f$ is integrable on $\displaystyle [-1,1]$ and $\displaystyle \int_{-1}^{1}f=0$.[Hint: Any lower sum is$\displaystyle \leq 0$ and some lower sum $\displaystyle =0$. Hence, the lower integral $\displaystyle \int_{-1}^{1}f=0$].

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

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

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


    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 $\displaystyle f(x)=-1$ on $\displaystyle x\leqslant 0$. You clearly have that $\displaystyle f$ is integrable on $\displaystyle [-1,0]$ since it's constant, and it's integrable on $\displaystyle [0,1]$ 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 $\displaystyle \displaystyle \int_{-1}^0f(x)\text{ }dx=-1$ and $\displaystyle \displaystyle \int_0^1 f(x)\text{ }dx=1$ since by that same post we get that $\displaystyle \displaystyle \int_0^1 f(x)\text{ }dx=\int_0^1 1\text{ }dx$.
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  3. #3
    Member
    Joined
    Oct 2010
    Posts
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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 $\displaystyle s$ we see that for $\displaystyle f=\sum s=\sum_{i=1}^{n}A_{i}(x_{i}-x_{i-1})$, when $\displaystyle x<0, A_{i}<0$ and when $\displaystyle x>0,$ we see that the value for $\displaystyle A_{i}$ is equal to $\displaystyle |A_{i}|$ when $\displaystyle x<0$. Therefore, $\displaystyle \int_{-1}^{1}f=0$.
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