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##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon##

  1. Jun 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Suppose that ##f_{n} \rightarrow f## uniformly on [a,b] and that each ##f_{n}## is integrable on [a,b]. Show that given ##\epsilon > 0##, there exists a partition ##P## and a natural number ##N## such that ##\left|L(f_{n}, P) - L(f,P)\right| < \epsilon##.

    2. Relevant equations


    3. The attempt at a solution
    I let P be a partition. And
    $$m_{k} = inf\{f(x) : x \in [x_{k-1},x_{k}\}$$
    $$m_{k}^{'} = inf\{f_{n}(x): x \in [x_{k-1}, x_{k}]\}$$
    I am thinking maybe I can say ##|m_{k} - m_{k}^{'}| \leq \frac{\epsilon}{b - a}##, but i do not really know how to justify it. formally anyway


     
  2. jcsd
  3. Jun 8, 2015 #2

    Mark44

    Staff: Mentor

    Can you refresh our memory as to what L in L(f, P) represents? Is it Lebesgue integral?
     
  4. Jun 8, 2015 #3
    oh my bad. L(f,p) is the lower sum of the Reimann integral

    $$L(f,P) = \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1})$$
     
  5. Jun 8, 2015 #4

    Zondrina

    User Avatar
    Homework Helper

    Just a thought, use the triangle inequality, and say the sum is ##< \frac{\varepsilon}{2} + \frac{\varepsilon}{2}## using the information given.
     
    Last edited: Jun 8, 2015
  6. Jun 8, 2015 #5
    what exactly is less than ##\epsilon/2##? I was thinking something like...

    $$|L(f_{n},P) - L(f,p)| = |\sum_{k = 1}^{n} m_{k}^{'}(x_{k} - x_{k - 1}) - \sum_{k = 1}^{n} m_{k} (x_{k} - x_{k - 1}|$$

    we then have
    $$\sum_{k = 1}^{n} m_{k}^{'} - m_{k} (x_{k} - x_{k - 1})$$

    now if i can bound ##|m_{k}^{'} - m_{k}## by epsilon some how then i think that would work?
     
  7. Jun 8, 2015 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    If you can guarantee that ##|f_n(x) - f(x)|<\varepsilon## for all ##x##, does this imply some bound on ##|m_k - m_k'|##?
     
  8. Jun 8, 2015 #7
    well since ##m_{k}## and ##m_{k}^{'}## are the infs they are definitely bounded by ##\epsilon## if ##|f_n(x) - f(x)|<\varepsilon## for all ##x##; which is true because we have uniform convergence
     
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