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Thread: measure theory

  1. #1
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    measure theory

    f and g are non negative measurable functions.

    (X,$\displaystyle \Gamma , \mu$ ) is the measure space
    for A $\displaystyle \in \Gamma$
    $\displaystyle \mu_f (A) = \int_A f d\mu $
    Prove $\displaystyle \int_A g d\mu_f = \int_A (g\cdot f) d\mu$
    so i have by monotone conv thm $\displaystyle \int \lim f_n$ = $\displaystyle \lim \int f_n$
    and there exists a increasing seq of simple functions s.t. $\displaystyle \lim s_n \rightarrow g$
    and $\displaystyle \lim t_m \rightarrow f$
    now $\displaystyle \int_A g d\mu_f$ = $\displaystyle \lim I_A(s_n)$
    if each $\displaystyle s_n$ takes on finite number of values $\displaystyle \{c_i\}_{i=1}^k$ k dependent on n and each $\displaystyle t_m$ takes on finite number of values $\displaystyle \{d_j\}_{j=1}^l$ l dependent on m
    and $\displaystyle I_A(s_n) = \sum_{i=1}^k c_i \cdot \mu_f (s^{-1}(c_i)) $ and let $\displaystyle (s^{-1}(c_i)) = E_i$
    =$\displaystyle \sum_{i=1}^k c_i \cdot \int_{E_i} f d\mu$
    =$\displaystyle \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
    $\displaystyle \int_A (g\cdot f) d\mu$ = $\displaystyle \lim_{\substack{m\rightarrow \infty\\n\rightarrow \infty}} \sum_{i,j}^{k,l} c_i \cdot d_j \cdot \mu (E_i \cap F_j)$
    let $\displaystyle (t^{-1}(d_j)) = F_j$
    so if i is fixed and j runs from 1 to l, then it becomes
    $\displaystyle \lim_{\substack{m\rightarrow \infty\\n\rightarrow \infty}} \sum_{i}^{k} c_i \cdot I_{E_i}(t_m)$
    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 $\displaystyle g$ is a simple function, namely $\displaystyle \displaystyle g(x) =\sum_{k=0}^Na_k\mathbf 1_{A_k}(x)$ where $\displaystyle A_k\in\Gamma$ are disjoint measurable sets and $\displaystyle a_k\geq 0$.
    Now, in the general case, take $\displaystyle \left\{g_n\right\}_n$ an increasing sequence of simple functions which converges pointwise to $\displaystyle g$. We have for all $\displaystyle n$ that $\displaystyle \int_A g_nd\mu_f =\int_A g_nfd\mu$, and since $\displaystyle f$ is non negative the sequence $\displaystyle \left\{fg_n\right\}$ is also increasing. You can conclude by the monotone convergence theorem.
    (it's what you tough, but here we didn't need to approach $\displaystyle f$ by simple functions).
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  3. #3
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    Re: measure theory

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

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

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

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

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

    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 $\displaystyle \infty$

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

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

    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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