Let $\Sigma = \Sigma _2 \times \Sigma _2 $
Given that $x $-and $y $ sections are $\Sigma _1 $ - and $\Sigma _2 $ measurable respectively,
and that for $F \in \Sigma $, $\int ( \int I _F \, d \mu _1 ) \, d \mu _2=\int ( \int I _F \, d \mu _2 ) \, d \mu _1$
Then show $\lambda ( F ) = \int ( \int I _F \, d \mu _1 ) \, d \mu _2 $ defines a measure on $\Sigma $.
My approach is to try to show finite additivity and then use monotone convergence theorem.
First suppose $\bigcup _{i=1}^n A_i = \bigcup_{i=1}^n (B_i \times C_i )$, where $A _n \cap A _m $ if $m \neq n $.
Then $I_{\bigcup A _n }= \sum I_{A_n}$
$$\int \left( \int I_{\cup A_i} \, d \mu _1\right) \, d \mu_2 = \int \left(\sum \mu _1 (B_i) I_{C_i} \right) \, d \mu_2=\sum \mu_1 ( B_i ) \mu_2 ( C_i ) ,$$ (where $1 \le i \le n $).
(correct?)
And now countable additivity follows since
$$\lim _{n \to \infty } \int \left(\sum \mu_1 (B_i ) I_{C_i } \right) \, d \mu_2=\int \lim_{n \to \infty } \left(\sum \mu_1 (B_i) I_{C_i } \right) \, d \mu_2$$
is this also correct?
Thanks in adevance!