How do i prove that $\mathfrak{M}\oplus\Sigma$ is the sigma algebra generated by products of elements of generating sets? Let $(X,\mathfrak{M}),(Y,\Sigma)$ be measurable spaces.
Let $\mathscr{A},\mathscr{B}$ be sets such that $\sigma(\mathscr{A})=\mathfrak{M}$ and $\sigma(\mathscr{B})=\Sigma$.
How do i prove that $\mathfrak{M}\oplus\Sigma$ is the sigma-algebra generated by $\{A\times B : A\in \mathscr{A}, B\in\mathscr{B}\}$?
Below is what i tried:
Define $G=\{A\times B : A\in \mathscr{A}, B\in\mathscr{B}\}$.
Define $R=\{E\subset X : E\times Y \in \sigma(G)\}$.
I showed that $R$ is a sigma-algebra, but i cannot prove whether $\mathscr{A}\subset R$.

Define $H := \{ A \times B : A \in \mathfrak M, B \in \Sigma\}$. The goal is to show that $\sigma(H) = \sigma(G)$. The inclusion $\sigma(G) \subseteq \sigma(H)$ follows from $G \subseteq H$. For the other inclusion,
it suffices to prove that $H \subseteq \sigma(G)$. Note that for all $B \in \mathscr B$ the family $R_B := \{ E \subseteq X : E \times B \in \sigma(G)\}$ is a $\sigma$-algebra on $X$ which contains $\mathscr A$, so that $\mathfrak M = \sigma(\mathscr A) \subseteq R_B$, i.e., $A \times B \in \sigma(G)$ for all $A \in \mathfrak M$ and $B \in \mathscr B$.
Now we let $A \in \mathfrak M$ be arbitrary and consider $R_A := \{ F \subseteq Y : A \times F \in \sigma(G) \}$. This is a $\sigma$-algebra on $Y$ by the same reasoning as before. Moreover, by the above paragraph, it contains $\mathscr B$. Together this implies $\Sigma = \sigma(\mathscr B) \subseteq R_A$. Since $A \in \mathfrak M$ was arbitrary, we find $H \subseteq \sigma(G)$ as desired.
