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Monotone Convergence Theorem

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Let f be a non-negative measurable function. Prove that

    [itex] \lim _{n \rightarrow \infty} \int (f \wedge n) \rightarrow \int f.[/itex]


    3. The attempt at a solution

    I feel like I'm supposed to use the monotone convergence theorem.

    I don't know if I'm on the right track but I created a sequence of functions so that

    [itex]h_1(x) \leq h_2(x) \cdots [/itex] where

    [itex]h_1(x) = \min(f_1(x), n)[/itex]

    [itex]h_2(x) = \min(f_2(x),n) [/itex]

    [itex]\vdots [/itex]

    [itex]h_n(x) = \min(f_n(x),n) [/itex]

    So the [itex]h(x) = \lim_{n\rightarrow\infty} h_n(x) = \lim_{n\rightarrow \infty} \min(f_n,n) = \lim_{n \rightarrow \infty}\min(f,n) [/itex]
     
  2. jcsd
  3. Apr 29, 2014 #2

    Dick

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    I'm guessing ##f \wedge n## means min(f,n)? Then why not just define ##f_n=\min(f,n)##? What do YOU mean by your ##f_n##??
     
  4. Apr 29, 2014 #3
    yes [itex]f \wedge n = \min(f, n) [/itex]

    so define

    [itex]f_1(x) = \min(f(x),1)[/itex]

    [itex]f_2(x) = \min(f(x),2)[/itex]

    [itex]\vdots [/itex]

    [itex]f_n(x) = \min(f(x),n)[/itex]

    In short we still have that [itex] f_1(x) \leq f_2(x) \leq \cdots \leq f_n(x) [/itex] for all x.

    Well I want [itex] \lim_{n \rightarrow \infty}f_n \rightarrow f[/itex] for all x because if I have this situation I can use the monotone convergence theorem.

    I'm just not sure if that's the case.
     
    Last edited: Apr 29, 2014
  5. Apr 29, 2014 #4

    Dick

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    I think it's all fine, except ##f_1(x) = \min(f_1(x),1)## doesn't do a good job of defining ##f_1##. ##f_1(x) = \min(f(x),1)## is much better. Are you asking why ##f_n \rightarrow f##?
     
    Last edited: Apr 30, 2014
  6. Apr 29, 2014 #5
    Sorry I just made some edits, I saw my mistake.
     
  7. Apr 30, 2014 #6
    I think I'm covered, because I'll have to consider the case when f is finite and f is infinite correct?
     
  8. Apr 30, 2014 #7

    Dick

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    Maybe. f is nonnegative. So either the integral exists or it's '+infinity'. I think in either case the sequence convergence is correct. If you allow things like '+infinity'. Think about a function like f(x)=1/x^2. Define f(0) however you want.
     
    Last edited: Apr 30, 2014
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