How can we show that $\sigma(\mathcal G)\times\sigma(\mathcal G)\subset \sigma(\mathcal G \times \mathcal G)$? As a part of a larger proof, I'm trying to show that if $\mathcal G \subset \mathcal P(\Omega)$ then
$$\sigma(\mathcal G)\times\sigma(\mathcal G)\subset \sigma(\mathcal G \times \mathcal G).$$
Since $\sigma(\mathcal G)\times\sigma(\mathcal G)$ is not necessarily a $\sigma$-algebra, it seems that we can't use the usual techniques for showing an inclusion for $\sigma$-algebras. We obviously don't have $\sigma(\mathcal G)\times\sigma(\mathcal G)\subset \mathcal G \times \mathcal G$ and I don't see how could we construct $\sigma(\mathcal G)\times\sigma(\mathcal G)$ by using the allowed operations (e.g. countable union, complement) on $\mathcal G \times \mathcal G$.
Any hints are hugely appreciated.
 A: In the remainder of the post, I introduced horizontal lines at some points. If you don't want a complete solution, but merely a hint, stop reading after each of these lines :)

First, observe that your claim is not true in general. To see this,
let $\mathcal{G}=\left\{ \left\{ 0\right\} \right\} $ for $\Omega=\left\{ 0,1\right\} $.
Then $\mathcal{G}\times\mathcal{G}=\left\{ \left\{ \left(0,0\right)\right\} \right\} $,
and hence
$$
\sigma\left(\mathcal{G}\times\mathcal{G}\right)=\left\{ \emptyset,\Omega^{2},\left\{ \left(0,0\right)\right\} ,\Omega^{2}\setminus\left\{ \left(0,0\right)\right\} \right\} ,
$$
but
$$
\left\{ \left(0,0\right),\left(0,1\right)\right\} =\left\{ 0\right\} \times\left\{ 0,1\right\} \in\sigma\left(\mathcal{G}\right)\times\sigma\left(\mathcal{G}\right).
$$

One solution for fixing this problem is to assume that there is a
sequence $\left(G_{n}\right)_{n\in\mathbb{N}}$ of elements $G_{n}\in\mathcal{G}$
such that $\Omega=\bigcup_{n}G_{n}$.

If we assume this, let us define
- for $A\subset\Omega$ arbitrary - the sets
\begin{eqnarray*}
\mathcal{M}_{A} & := & \left\{ B\in\sigma\left(\mathcal{G}\right)\mid A\times B\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)\right\} ,\\
\mathcal{M}^{A} & := & \left\{ B\in\sigma\left(\mathcal{G}\right)\mid B\times A\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)\right\} .
\end{eqnarray*}

As a first step, let us assume that $A\times\Omega\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)$
(or $\Omega\times A\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)$)
and show that $\mathcal{M}_{A}$ (or $\mathcal{M}^{A}$) is a $\sigma$-algebra.
By this assumption, $\Omega\in\mathcal{M}_{A}$. Furthermore, if $B\in\mathcal{M}_{A}$,
then
$$
\sigma\left(\mathcal{G}\times\mathcal{G}\right)\ni\left(A\times\Omega\right)\cap\left(A\times B\right)^{c}=\left(A\times\Omega\right)\cap\left[\left(A^{c}\times\Omega\right)\cup\left(A\times B^{c}\right)\right]=A\times B^{c},
$$
so $B^{c}\in\mathcal{M}_{A}$. Finally, for a sequence of sets $\left(B_{n}\right)_{n\in\mathbb{N}}$
in $\mathcal{M}_{A}$, we have
$$
A\times\left(\bigcup_{n}B_{n}\right)=\bigcup\left[A\times B_{n}\right]\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)
$$
and hence $\bigcup_{n}B_{n}\in\mathcal{M}_{A}$. An analogous proof
holds for $\mathcal{M}^{A}$.

Now, let $A\in\mathcal{G}\subset\sigma\left(\mathcal{G}\right)$ be
arbitrary. Then $A\times G_{n}\in\mathcal{G}\times\mathcal{G}\subset\sigma\left(\mathcal{G}\times\mathcal{G}\right)$
for all $n\in\mathbb{N}$ and hence $A\times\Omega=\bigcup\left(A\times G_{n}\right)\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)$.
As shown above, this implies that $\mathcal{M}_{A}$ is a $\sigma$-algebra.
But for $G\in\mathcal{G}$, we trivially have $A\times G\in\mathcal{G}\times\mathcal{G}\subset\sigma\left(\mathcal{G}\times\mathcal{G}\right)$
and hence $\mathcal{G}\subset\mathcal{M}_{A}$, which yields $\sigma\left(\mathcal{G}\right)\subset\mathcal{M}_{A}$.

Now, let $A\in\sigma\left(\mathcal{G}\right)$ be arbitrary. By the
property just established, we have $A\in\mathcal{M}_{B}$ for all
$B\in\mathcal{G}$ and hence $B\times A\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)$,
which yields $B\in\mathcal{M}^{A}$ for all $B\in\mathcal{G}$. But
$\mathcal{M}^{A}$ is a $\sigma$-algebra, whence $\sigma\left(\mathcal{G}\right)\subset\mathcal{M}^{A}$,
which finally yields $B\times A\in\sigma\left(\mathcal{G}\times\mathcal{G}\right)$
for all $B\in\sigma\left(\mathcal{G}\right)$. As $A\in\sigma\left(\mathcal{G}\right)$
was arbitrary, we are done. $\square$
