Restriction of product $\sigma$-algebra vs. product $\sigma$-algebra of restrictions Let $(X,\mathscr M)$ be a measurable space and $A,B\in\mathscr M$ be two non-empty measurable subsets. Consider the following constructions:
\begin{align*}
\mathscr{M}_A\equiv&\,\{E\in\mathscr M\,|\,E\subseteq A\},\\
\mathscr{M}_B\equiv&\,\{E\in\mathscr M\,|\,E\subseteq B\},\\
(\mathscr{M}\otimes\mathscr M)_{A\times B}\equiv&\,\{F\in\mathscr M\otimes\mathscr M\,|\, F\subseteq A\times B\},
\end{align*}
where $\mathscr M\otimes\mathscr M$ is the product $\sigma$-algebra on $X\times X$. It is clear that $\mathscr M_A$, $\mathscr M_B$, and $(\mathscr M\otimes\mathscr M)_{A\times B}$ are $\sigma$-algebras on $A$, $B$, and $A\times B$, respectively. I am wondering if
\begin{align*}
\mathscr{M}_A\otimes\mathscr M_B=(\mathscr{M}\otimes\mathscr M)_{A\times B}.
\end{align*}
NB: $\mathscr{M}_A\otimes\mathscr M_B$ is the product $\sigma$-algebra corresponding to the measurable spaces $(A,\mathscr M_A)$ and $(B,\mathscr M_B)$.
I think I can see $\subseteq$ easily, but whether $\mathscr{M}_A\otimes\mathscr M_B\supseteq(\mathscr{M}\otimes\mathscr M)_{A\times B}$ is true got the better of me.
Any help is appreciated.
 A: Let's prove that $(\mathscr{M}\otimes\mathscr M)_{A\times B}\subseteq\mathscr{M}_A\otimes\mathscr M_B$.
Let $j:A\times B\to X\times X$ be the canonical inclusion given by $j(x)=x$ and $\mathscr{R}\subseteq 2^{X\times X}$ be the set of measurable rectangles. The measurable rectangles in $A\times B$ are exactly $\mathscr{S}=\{j^{-1}(R):R\in\mathscr{R}\}$. Since $\mathscr{M}\otimes\mathscr{M}=\sigma(\mathscr{R})$ and $\mathscr{M}_A\otimes\mathscr{M}_B=\sigma(\mathscr{S})$, it suffices to prove the following lemma:
Lemma: Let $S$ and $T$ be sets, $f:S\to T$ be a function and $\mathscr{T}$ be a $\sigma$-algebra generated by the family $\mathscr{G}$. Then $$\sigma\Big(\{f^{-1}(G):G\in\mathscr{G}\}\Big)=\{f^{-1}(B):B\in\mathscr{T}\}.$$
Proof: Call the $\sigma$-algebra on the left $\mathscr{S}$. The family $\mathscr{F}=\{B\subseteq T:f^{-1}(B)\in\mathscr{S}\}$ is a $\sigma$-algebra on $T$ containing $\mathscr{G}$. Hence, $\sigma(\mathscr{G})=\mathscr{T}\subseteq\mathscr{F}$ and
$$\{f^{-1}(B):B\in\mathscr{T}\}\subseteq\{f^{-1}(B):B\in\mathscr{F}\}\subseteq\mathscr{S}=\sigma\Big(\{f^{-1}(G):G\in\mathscr{G}\}\Big).$$
Since $\mathscr{G}\subseteq\mathscr{T}$, we also have $$\sigma\Big(\{f^{-1}(G):G\in\mathscr{G}\}\Big)\subseteq\sigma\Big(\{f^{-1}(B):B\in\mathscr{T}\}\Big)=\{f^{-1}(B):B\in\mathscr{T}\},$$
so $$\mathscr{S}=\{f^{-1}(B):B\in\mathscr{T}\}.$$
