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Math Help - measure theory

  1. #1
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    measure theory

    f and g are non negative measurable functions.

    (X, \Gamma , \mu ) is the measure space
    for A \in \Gamma
    \mu_f (A)  = \int_A f d\mu
    Prove \int_A g   d\mu_f  =   \int_A (g\cdot f) d\mu
    so i have by monotone conv thm \int \lim f_n = \lim \int f_n
    and there exists a increasing seq of simple functions s.t. \lim s_n  \rightarrow  g
    and \lim t_m \rightarrow f
    now \int_A g d\mu_f = \lim I_A(s_n)
    if each s_n takes on finite number of values \{c_i\}_{i=1}^k k dependent on n and each t_m takes on finite number of values \{d_j\}_{j=1}^l l dependent on m
    and I_A(s_n) = \sum_{i=1}^k  c_i \cdot \mu_f (s^{-1}(c_i)) and let  (s^{-1}(c_i)) = E_i
    = \sum_{i=1}^k  c_i \cdot \int_{E_i} f d\mu
    = \sum_{i=1}^k  c_i \cdot \lim I_{E_i}(t_m)
    this following part is where i am kind of unsure if i did it correctly
    \int_A (g\cdot f) d\mu = \lim_{\substack{m\rightarrow \infty\\n\rightarrow \infty}} \sum_{i,j}^{k,l} c_i \cdot  d_j \cdot \mu (E_i \cap F_j)
    let  (t^{-1}(d_j)) = F_j
    so if i is fixed and j runs from 1 to l, then it becomes
    \lim_{\substack{m\rightarrow \infty\\n\rightarrow \infty}} \sum_{i}^{k} c_i \cdot I_{E_i}(t_m)
    the limiting operations dont necessarily commute i.e lettin n to go infinity and then m to infinity is not the same as m going to infinity and then n going to infinity as the sum need not converge absolutely. any hints or help would be appreciated
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  2. #2
    Super Member girdav's Avatar
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    Re: measure theory

    First of all we will show the result if g is a simple function, namely \displaystyle g(x) =\sum_{k=0}^Na_k\mathbf 1_{A_k}(x) where A_k\in\Gamma are disjoint measurable sets and a_k\geq 0.
    Now, in the general case, take \left\{g_n\right\}_n an increasing sequence of simple functions which converges pointwise to g. We have for all n that \int_A g_nd\mu_f =\int_A g_nfd\mu, and since f is non negative the sequence \left\{fg_n\right\} is also increasing. You can conclude by the monotone convergence theorem.
    (it's what you tough, but here we didn't need to approach f by simple functions).
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  3. #3
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    Re: measure theory

    so if g is a simple function
    g = \sum_{i=1}^k c_i \cdot \1_{E_i}(x)  and E_i's partition A

    then \int_A g d\mu_f = \sum_{i=1}^k c_i \cdot d\mu({E_i})

    = \sum_{i=1}^k c_i \cdot \int_{E_i} f d\mu

    now \int_A (g\cdot f) d\mu now the only way i know how to define that is using the increasing seq of monotone functions, s_n \le f

    and \int_A (g\cdot f) d\mu = \lim_{n \rightarrow\infty}\sum_{i,j}^{k,l_{(n)}} c_i \cdot d_j \mu (E_i \cap F_j)

    now b/c all terms are positive if the series converges then it is absolutely convergent so changing the order of summation does not affect the limit so i can be fixed and sum j from 1 to \infty

    so \lim_{n\rightarrow \infty}\sum_{i,j}^{k,l_{(n)}} c_i \cdot d_j \mu (E_i \cap F_j) = \sum_{i=1}^k c_i \cdot \lim_{n \rightarrow \infty} I_{E_i}(s_n)
    = \sum_{i=1}^k c_i \cdot \int_{E_i} f d\mu

    so wen g is simple we \int_A gf d\mu = int_A g d\mu_f
    so now take a increasing seq , g_n converging to g
    then \int_A g_n f d\mu = \int_A g_n d\mu_f for all n
    since limits are unique in R and monotone conv them we have
    \int_A \lim g_n f d\mu = \lim \int_A g_n f d\mu = \lim \int_A g_n d\mu_f = \int_A  \lim g_n d\mu_f

    did i follow your properly there?
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  4. #4
    Super Member girdav's Avatar
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    Re: measure theory

    I didn't look all the details, but that's the idea.
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