Results 1 to 5 of 5

Thread: Integral proof

  1. #1
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303

    Integral proof

    Q: Prove that a bounded function $\displaystyle f$ is integrable on $\displaystyle [a,b]$ if and only if there exists a sequence of partitions $\displaystyle (P_{n})_{n=1}^{\infty}$ satisfying

    $\displaystyle lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0$.

    Here some of my (incomplete) thoughts:

    $\displaystyle (\Leftarrow)$
    Let $\displaystyle \epsilon>0$. Since there exists a sequence of partitions $\displaystyle (P_{n})_{n=1}^{\infty}$ satisfying $\displaystyle lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0$, there exists an $\displaystyle N$ such that, for any $\displaystyle n\geq\\N$ $\displaystyle U(f,P_{n})-L(f,P_{n})<\epsilon$. Thus, $\displaystyle f$ integrable on $\displaystyle [a,b]$.

    $\displaystyle (\Rightarrow)$
    For this direction I am not sure how to show there exists a sequence of partitions. Since we are assuming $\displaystyle f$ is integrable, I know that there is a partion $\displaystyle P_{\epsilon}$ of $\displaystyle [a,b]$ such that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$ for any positive $\displaystyle \epsilon$.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2827
    Awards
    1
    Using a the definition of integral if $\displaystyle f$ on $\displaystyle [a,b]$ then there is a number $\displaystyle I$ such that for any positive $\displaystyle \epsilon>0 $ there is a partition $\displaystyle P$ of $\displaystyle [a,b]$ then $\displaystyle \left|\mathcal{S} (P;f)-I\right|<\epsilon $.

    Now here is the trick. We can do it twice with $\displaystyle \frac{\epsilon}{2}$
    $\displaystyle \left|\mathcal{U} (P;f)-I\right|<\frac{\epsilon}{2} $ and $\displaystyle \left|\mathcal{L} (P;f)-I\right|<\frac{\epsilon}{2} $.
    Now complete it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303
    Define $\displaystyle U(f)=inf\{U(f,P):P\in\\Q\}$ and $\displaystyle L(f)=sup\{L(f,P):P\in\\Q\}$, where $\displaystyle Q$ is the set of all possible partitions.

    We are given $\displaystyle f$ is integrable on $\displaystyle [a,b]$, so there exists a partition $\displaystyle P_{\epsilon}$ of $\displaystyle [a,b]$ such that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$. Since $\displaystyle f$ is integrable $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})=(U(f,P_{\epsilon})-U(f))+(L(f)-L(f,P_{\epsilon})$.

    Let $\displaystyle I=U(f)=L(f)$. Then, $\displaystyle |U(f,P_{\epsilon})-I|<\frac{\epsilon}{2}$ and $\displaystyle |L(f,P_{\epsilon})-I|<\frac{\epsilon}{2}$.

    So, we have a sequence of paritions $\displaystyle (P_{n})_{n=1}^{\infty}$ such that, there exists an $\displaystyle N>0$ for which any $\displaystyle n\geq\\N$ implies the upper and lower sums converge to the same value.

    I am not sure how to introduce the limiting variable, $\displaystyle n$. I understand what is going on, but I am having trouble formalizing the idea of a sequence of partitions.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2827
    Awards
    1
    You cannot do the above for $\displaystyle \epsilon=\frac{1}{n}?$
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Danneedshelp's Avatar
    Joined
    Apr 2009
    Posts
    303
    Sorry to bring up an old thread, but I have been away for a while and have not yeat figured out the proof.

    I am unsure how to prove the existance of the sequence of partitions.

    $\displaystyle f$ is integrable on $\displaystyle [a,b]$, so for any $\displaystyle \epsilon>0$ we have that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon$. Since $\displaystyle \epsilon>0$ is arbitrary it is certianly true that $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})$ can be made less than $\displaystyle \epsilon=\frac{1}{n}$ for each $\displaystyle n\in\mathbb{N}$. Therefore, there exists a sequenct of paritions $\displaystyle P_{n}=\frac{b-a}{n}$ such that $\displaystyle n>N$, where $\displaystyle N\in\mathbb{N}$, implies $\displaystyle (P_{n})\rightarrow\\P_{\epsilon}$.

    Notice,
    $\displaystyle U(f,P_{\epsilon})-L(f,P_{\epsilon})=U(f,P_{\epsilon})-U(f)+L(f)-L(f,P_{\epsilon})$ since $\displaystyle f$ is integrable. Thus, $\displaystyle U(f,P_{\epsilon})-U(f)$ can be made less than half the prescribled $\displaystyle \epsilon$ as can $\displaystyle L(f)-L(f,P_{\epsilon})$. Since $\displaystyle U(f)=I=L(f)$, we have that the lower and upper sums converge to the same limit. Thus,

    $\displaystyle U(f,P_{n})-L(f,P_{n})=0$ as $\displaystyle n\rightarrow\infty$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integral proof
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jun 17th 2009, 05:30 AM
  2. Integral Proof,
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 10th 2009, 09:21 PM
  3. proof an integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 24th 2009, 03:26 PM
  4. Integral proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 28th 2008, 09:55 AM
  5. Replies: 6
    Last Post: Aug 4th 2007, 09:48 AM

Search Tags


/mathhelpforum @mathhelpforum