measure theory

**bubble86** · July 4th 2011, 09:13 PM

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

**girdav** · July 5th 2011, 01:28 AM

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).

**bubble86** · July 5th 2011, 08:29 AM

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?

**girdav** · July 5th 2011, 08:37 AM

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

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