Question about measure theory. Suppose $( \Omega_1, \mathcal{F}_1), ( \Omega_2, \mathcal{F}_2)  $ are measurable spaces then 


My question is with regard to the definition of the collection $\mathcal{G}$. I don't really understand what is the property of this set. My thought: I think this collection has the property that if $A_{\omega_2}$ is $\mathcal{F}_1$- measurable, then the 'rectangle' $A = A_1 \times A_2$ is $\mathcal{F}$-measurable. Is this correct??
Also, Im not seeing why is it enough to prove theorem 6.3 by showing $\mathcal{G} = \mathcal{F}$ Can someone explain this to me?
by the way, $\mathcal{F} = \mathcal{F}_1 \otimes \mathcal{F}_2$ the smallest sigma field generated by rectangles $A_1 \times A_2$ where $A_i \in \mathcal{F}_i$
 A: Starting with your last question, the theorem asserts that, for every $A$ in $\mathcal F$, some property $Q(A)$ holds. Since $\mathcal G$ is the collection of sets $A$ in $\mathcal F$ such that $Q(A)$ holds, indeed the theorem is equivalent to the fact that $\mathcal G=\mathcal F$. To sum up: this is pure logic, not measure theory.

I don't really understand what is the property of this set.

I suggest to read the proof to see which properties this set has. To begin with, the proof shows that, if $A=A_1\times A_2$ with $A_1$ in $\mathcal F_1$ and $A_2\subseteq\Omega_2$, then $A$ is in $\mathcal G$ (which is rather different from what you wrote).
A: The collection $\mathcal G$ is the set of all measurable sets in the product @\sigma$-algebra such that their "horizontal sections" are measurable.
By definition, $\mathcal G\subset\mathcal F$. If it is a $\sigma$-algebra and it contains the rectangles, then $\mathcal F\subset\mathcal G$.
Finally, if you now take $A\in\mathcal F$, then $A\in\mathcal G$ and so it satisfies the statement of the theorem.
