Sectioning sets in product measures I'm reading through Capinski and Kopp's Measure, Integral and Probability and stumbled across Theorem 6.4 about sectioning sets in product measures:

If $A$ is in the product $\sigma$-field $\mathcal{F} = \mathcal{F}_1 \times \mathcal{F}_2$, then for each $\omega_2$, $A_{\omega_2} \triangleq
\left\{ \omega_1 \in \Omega_1 \ \middle|\ (\omega_1, \omega_2) \in A\right\}
\in \mathcal{F}_1$ (similarly for $\omega_1$, $A_{\omega_1} \in \mathcal{F}_2$).

Intuitively, this makes sense, however the proof is a bit puzzling: the authors define the set
$$
\mathcal{G} = \left\{ A \in \mathcal{F} \ \middle|\ A_{\omega_2} \in \mathcal{F}_1, \forall \omega_2 \right\}
$$
and proceed to show that $\mathcal{G} = \mathcal{F}$ by showing that $\mathcal{G}$ is a $\sigma$-field containing the "rectangles" of $\mathcal{F}_1 \times \mathcal{F}_2$. However, the first step reads:

Take $A = A_1 \times A_2, A_1 \in \mathcal{F}_1$. In that case:
  $$
A_{\omega_2} = \begin{cases}
  A_1, & \text{ if } \omega_2 \in A_2 \\
  \emptyset, & \text{ otherwise}
\end{cases}
$$
  is in $\mathcal{F}_1$ (since $A_1, \emptyset \in \mathcal{F}_1$) so the rectangles are in $\mathcal{G}$ (i.e. $A \in \mathcal{G}$).

Question: is that step not sufficient for concluding that $A_{\omega_2} \in \mathcal{F}_1$?
Why do we need to go through the process of proving that
$\mathcal{G}$ is a $\sigma$-field and then concluding $\mathcal{G} = \mathcal{F}$ to say that?
 A: The step you highlighted only proves that $A_{\omega_2}$ is in $\mathcal F_1$ in the special case where $A$ is a rectangle (i.e. when $A$ can be written as a product, $A_1 \times A_2$, where $A_1 \in \mathcal F_1$ and $A_2 \in \mathcal F_2$). But many measurable sets $A$ in $\mathcal F$ are not rectangles!
[For example, consider endowing $\mathbb R^2 = \mathbb R \times \mathbb R$ with the product measure induced by the Lebesgue measure on each of the two $\mathbb R$'s. Then then disk $\{(x,y) \in \mathbb R^2 : \sqrt{x^2 + y^2 } < 1 \}$ is measurable w.r.t. the product measure, but it is not a rectangle.]
So to go from the easy special case of rectangles to the general case of all measurable sets, the authors structure their argument as follows:
(i) $A_{\omega_2} \in \mathcal F_1$ in the special case where $A$ is a "rectangle" (i.e. a set of the form $A_1 \times A_2$ where $A_1 \in \mathcal F_1$ and $A_2 \in \mathcal F_2$).
(ii) The collection of sets $A\in\mathcal F$ such that $A_{\omega_2} \in \mathcal F_1$ is a sigma-algebra, i.e. it contains the empty set, and it is closed under taking complements and countable unions.
(iii) Since $\mathcal F$ is, by definition, the smallest sigma-algebra containing all rectangles, it must be the case that the collection of sets $A \in\mathcal F$ such that $A_{\omega_2} \in \mathcal F_1$ is the whole of $\mathcal F$.
