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