Monotone Convergence

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  1. TheEd's Avatar
    1 18-04-2012  20:54
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    Monotone Convergence


    Monotone Convergence Theorem:



    Does this follow straight from the theorem?

    Have I got to prove something to show it follows from the theorem? If so, what?
  2. nuodai's Avatar
    2 18-04-2012  22:52
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    Re: Monotone Convergence
    It almost follows from the theorem. But first you have to write \displaystyle \int_{I_n} f = \int_S f_n for some f_n in such a way that (f_n) is an increasing sequence tending to f. (Then you have something that satisfies the hypotheses of the theorem.) Can you do that?

    Then the case for g is similar by considering positive and negative parts.
  3. TheEd's Avatar
    3 18-04-2012  23:38
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    Re: Monotone Convergence
    (Original post by nuodai)
    It almost follows from the theorem. But first you have to write \displaystyle \int_{I_n} f = \int_S f_n for some f_n in such a way that (f_n) is an increasing sequence tending to f. (Then you have something that satisfies the hypotheses of the theorem.) Can you do that?

    Then the case for g is similar by considering positive and negative parts.
    Let f_n = f \chi_{I_n} where \chi is the characteristic function.

    Then since f \geqslant 0, (f_n) is an increasing sequence of functions which converges everywhere to f.

    So the result now follows from the MCT. Right?

    Any function g can be written as g = g^+ - g^- where g^+,g^- \geqslant 0 are the +ve and -ve parts of g.

    Then we know that if g \in L^1 (I_n) then g^+ , g^- \in L^1(I_n).

    \displaystyle \int_{I_n} g = \int_{I_n} (g^+ - g^-) = \int_{I_n} g^+ - \int_{I_n} g^-

    If I let g_n^+ = g^+\chi_{I_n} then (g_n^+) is increasing and converging everywhere to g^+ and let g^-_n = g^- \chi_{I_n} so (g_n^-) is an increasing sequence of functions which converges everywhere to g^-.

    So \displaystyle \int_{I_n} g = \int_S g_n^+ - \int_S g_n^- and the results for g_n^+ and g_n^- follow from the MCT.

    What is the significance of the question saying \int_{I_n} |g| converge?

    Does the result for g now follow from the MCT?
    Last edited by TheEd; 18-04-2012 at 23:56.
  4. nuodai's Avatar
    4 19-04-2012  09:14
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    Re: Monotone Convergence
    (Original post by TheEd)
    Let f_n = f \chi_{I_n} where \chi is the characteristic function.

    Then since f \geqslant 0, (f_n) is an increasing sequence of functions which converges everywhere to f.

    So the result now follows from the MCT. Right?
    That's right.

    (Original post by TheEd)
    Any function g can be written as g = g^+ - g^- where g^+,g^- \geqslant 0 are the +ve and -ve parts of g.

    Then we know that if g \in L^1 (I_n) then g^+ , g^- \in L^1(I_n).

    \displaystyle \int_{I_n} g = \int_{I_n} (g^+ - g^-) = \int_{I_n} g^+ - \int_{I_n} g^-
    You could essentially have stopped here. Since g^+,g^- are nonnegative, you could have just said that the result follows from the first part of the question.

    (Original post by TheEd)
    What is the significance of the question saying \int_{I_n} |g| converge?
    That's just the statement that g\chi_{I_n} is integrable for each n. It ensures that \displaystyle \int_{I_n} g^+ and \displaystyle \int_{I_n} g^- are all finite so that you're not taking infinity away from infinity or anything horrible.
  5. TheEd's Avatar
    5 19-04-2012  23:18
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    Re: Monotone Convergence
    (Original post by nuodai)
    That's right.
    OK. So how do I exactly structure the argument?

    Let f_n = f \chi_{I_n}.

    Then since f \geqslant 0 we have that (f_n) is an increasing sequence of functions which converges everywhere to f.

    Now we are supposing \int_{I_n} f converges.

    So by MCT, \int_{I_n} f = \lim_{n\to\infty} \int_{I_n} f_n

    How does the results that f \in L^1 (S) and \int_S f = \lim_{n\to\infty} \int_{I_n}f now follow from the MCT?
    Last edited by TheEd; 19-04-2012 at 23:21.
  6. nuodai's Avatar
    6 19-04-2012  23:22
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    Re: Monotone Convergence
    (Original post by TheEd)
    OK. So how do I exactly structure the argument?

    Let f_n = f \chi_{I_n}.

    Then since f \geqslant 0 we have that (f_n) is an increasing sequence of functions which converges everywhere to f.

    Now we are supposing \int_{I_n} f converges.

    How does the results that f \in L^1 (S) and \int_S f = \lim_{n\to\infty} \int_{I_n}f now follow from the MCT?
    Let f_n = f \chi_{I_n} where f \ge 0. Then f_n \in L^1(S) since f \in L^1(I_n).

    Then (f_n) is an increasing sequence in L^1(S) tending to f

    So by the monotone convergence theorem

    \displaystyle \int_{I_n} f = \int_S f_n \to \int_S f and f \in L^1(S)
    Last edited by nuodai; 19-04-2012 at 23:24.
  7. TheEd's Avatar
    7 19-04-2012  23:48
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    Re: Monotone Convergence
    (Original post by nuodai)
    Let f_n = f \chi_{I_n} where f \ge 0. Then f_n \in L^1(S) since f \in L^1(I_n).

    Then (f_n) is an increasing sequence in L^1(S) tending to f

    So by the monotone convergence theorem

    \displaystyle \int_{I_n} f = \int_S f_n \to \int_S f and f \in L^1(S)
    OK. So why do we have to assume that the integral of the absolute value of g converges for the 2nd part?

    \displaystyle \int_{I_n} |g| = \int_{I_n} |g^+ - g^-| = \int_S |g_n^+ - g_n^-| = ...

    Can I split it into 2 integrals like I did before with |\cdot | there?
    Last edited by TheEd; 19-04-2012 at 23:50.
  8. nuodai's Avatar
    8 19-04-2012  23:51
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    Re: Monotone Convergence
    (Original post by TheEd)
    OK. So why do we have to assume that the integral of the absolute value of g converges for the 2nd part?

    \displaystyle \int_{I_n} |g| = \int_{I_n} |g^+ - g^-| = ...

    Can I split it into 2 like I did before with |.| there?
    \left| g \right| = g^+ + g^-, and g^+,g^- \ge 0, so if \displaystyle \int_{I_n} \left| g \right| \left(= \int_{I_n} g^+ + \int_{I_n} g^-\right) < \infty then \displaystyle \int_{I_n} g^+ < \infty and \displaystyle \int_{I_n} g^- < \infty, so that the difference \displaystyle \int_{I_n} g^+ - \int_{I_n} g^- is well-defined. This is something you should have come across in the definition of (\mu-)integrability.
    Last edited by nuodai; 19-04-2012 at 23:52.
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