Results 1 to 3 of 3

Math Help - Today's integration proof #1

  1. #1
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13

    Today's integration proof #1

    1. Let f:[a,b]\to\mathbb R be a monotone function on the compact interval [a,b]. Prove that f is integrable on [a,b].
    2. Let f:[a,b]\to\mathbb R be a continuous function on the compact interval [a,b]. Prove that f is integrable on [a,b].

    ------

    Please, be adviced, don't post URL where the proof is, if you have a proof, post it, otherwise, don't post.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Three things

    1. I am only currently finishing up Riemann-Stieltjes integration...I am still a novice. I would not even be posting if this were not for "fun"

    2.I assume that a knowledge of the defintion of the Riemann integral is knownw

    3. I assume you mean integrable in respect to x











    Quote Originally Posted by Krizalid View Post
    1. Let f:[a,b]\to\mathbb R be a monotone function on the compact interval [a,b]. Prove that f is integrable on [a,b].
    Let me prove a stronger result.

    Suppose that \left|\sum_i \left\{M_i-m_i\right\}\right|\leqslant M then f\in\mathfrak{R}

    Proof: We must merely prove that given \varepsilon>0 we may find a partion P of [a,b] such that U\left(P,f\right)-L\left(P,f\right)<\varepsilon.

    Now for each partition P=\left\{a=x_0,x_1,\cdots,x_n=b\right\} of [a,b] choose \Delta x_i=\frac{b-a}{n} (note this can be done because x is continuous).

    So

    \begin{aligned}U-L&=\sum_{i=1}^{n}\left[M_i-m_i\right]\Delta x_i\\<br />
&=\frac{b-a}{n}\sum_{i=1}^{n}\left[M_i-m_i\right]\\<br />
&\leqslant M\frac{b-a}{n}<\varepsilon\quad\blacksquare\end{aligned}

    The last step is true for sufficiently large n.

    Now I will show that if f is monotone then \sum\left[M_i-m_i\right] is bounded:

    Without loss of generality assume that f is montonically increasing. Then we can see that

    M_i=\sup_{[x_{i-1},x_i]}f=f(x_i) and

    m_i=\inf_{[x_{i-1},x_i]}f=f\left(x_{i-1}\right)

    So

    \begin{aligned}\sum_{i=1}^{n}\left[M_i-m_i\right]&=\sum_{i=1}^{n}\left[f(x_i)-f\left(x_{i-1}\right)\right]\\<br />
&=\left(f(x_1)-f(x_0)\right)+\left(f(x_2)-f(x_1)\right)+\cdots+\left(f(x_n)-f(x_n-1)\right)\\<br />
&=f(x_n)-f(x_0)\\<br />
&=f(b)-f(a)\end{aligned}


    And because Riemann integration is only applicable to bounded functions we can see that if f is montonically increasing that \sum_{i}\left[M_i-m_i\right] is bounded. The proof for montonically decreasing functions is almost identical.

    Let f:[a,b]\to\mathbb R be a continuous function on the compact interval [a,b]. Prove that f is integrable on [a,b].
    Similarly we must just show that for any \varepsilon>0 there exists a corresponding partition P of [a,b] such that U\left(P,f\right)-L\left(P,f\right)<\varepsilon.

    So for every \varepsilon>0 choose \eta>0 such that \eta[b-a]<\varepsilon. Now since continuity and uniform continuity are equivalent on compact spaces we may choose a \delta>0 such that |x-y|<\delta\implies|f(x)-f(y)|<\eta {\color{blue}(*)}. Now choose a partition P of [a,b] such that \max\left\{\Delta x_i\right\}<\delta, so then

    \begin{aligned}U-L&=\sum_{i=1}^{n}\left[M_i-m_i\right]\Delta x_i\\<br />
&\leqslant \eta\sum_{i=1}^{n}\Delta x_i{\color{red}(*)}\\<br />
&\overbrace{=}^{\text{telescopes}}\eta[b-a]<\varepsilon\quad\blacksquare\end{aligned}

    \color{red}(*) is justified by \color{blue}(*)
    Last edited by Mathstud28; January 6th 2009 at 10:28 PM. Reason: Typo and clarification
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Krizalid View Post
    1. Let f:[a,b]\to\mathbb R be a monotone function on the compact interval [a,b]. Prove that f is integrable on [a,b].
    2. Let f:[a,b]\to\mathbb R be a continuous function on the compact interval [a,b]. Prove that f is integrable on [a,b].
    These are well-known and nice results from analysis.

    1)Without lose of generality say f is non-decreasing. Notice that f(a) \leq f(x) \leq f(b) for x\in [a,b] therefore |f| \leq M for some M>0 hence the function is bounded. Now for a partition P = \{ x_0,x_1,...,x_n\} with x_j < x_{j+1} where 0\leq j\leq n-1 and x_0=a,x_n=b define \Delta_j = x_j - x_{j-1} and M_j = \sup \{ f(x) : x\in [x_{j-1},x_j]\} and m_j = \inf \{ f(x) : x\in [x_{j-1},x_j] \}. Let \Delta = \max \{ \Delta_j : 1\leq j \leq n\}. Let U_P be the upper sum and L_P be the lower sum. This means U_P - L_P  = \sum_{j=1}^n (M_j - m_j) \Delta_j . Now M_j = f(x_j) and m_j = f(x_{j-1}). Therefore, U_P - L_P = \sum_{j=1}^n (f(x_j) - f(x_{j-1})  ) \Delta_j. Therefore, U_P - L_P \leq \sum_{j=1}^n (f(x_j) - f(x_{j-1})) \Delta  = \Delta (f(x_n) - f(x_0)) = \Delta(f(b)-f(a)). This means for \epsilon > 0 pick \Delta < \frac{\epsilon}{f(b)-f(a)} (assuming that f(b)\not = f(a), but if f(b) = f(a) then the function is constant!) Therefore, U_P - L_P < \epsilon which forces f be to integrable by Cauchy's criterion.

    2)Here we use the fact that f is uniformly continous. Thus, for any \epsilon > 0 there is \delta > 0 so that if |x-y| < \delta \text{ and }x,y\in [a,b]\implies |f(x)-f(y)|<\epsilon. Let P be a partition \Delta < \delta. Now M_j = \max \{ f(x) : x\in [x_{j-1},x_j] \} and m_j = \min \{ f(x) : x\in [x_{j-1},x_j]\} by extreme-value theorem. Consequently, M_j - m_j < \epsilon since \Delta_j \leq \Delta < \delta. Thus, U_P - L_P = \sum_{j=1}^n (M_j - m_j) \Delta_j  \leq \sum_{j=1}^n \epsilon \Delta_j = \epsilon (b-a). And since this is true for any \epsilon > 0 it means we can make U_P - L_P arbitrary small and it is integrable by Cauchy's criterion.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I need this today! Please
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 6th 2010, 07:59 AM
  2. Question from a test i had today about group proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 3rd 2010, 05:37 PM
  3. Today's integration proof #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 7th 2009, 04:02 PM
  4. LinAlg - please help! Due today!
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 23rd 2008, 12:20 AM
  5. Need Help, Final TODAY
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 27th 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum