[proofplan]
We prove measurability of the product map by checking preimages on a generating family for the product $\sigma$-algebra. For each measurable rectangle $A \times B$, the preimage under $h=(f,g)$ is exactly $f^{-1}(A)\cap g^{-1}(B)$, which is measurable because $f$ and $g$ are measurable. We then collect all subsets of $E\times G$ whose preimages under $h$ are $\mathcal{M}$-measurable, show that this collection is a $\sigma$-algebra, and use the definition of $\mathcal{E}\otimes\mathcal{G}$ as the $\sigma$-algebra generated by measurable rectangles.
[/proofplan]
[step:Define the product map and the rectangle generators]
Define the map
\begin{align*}
h: X \to E \times G,\quad h(x)=(f(x),g(x)).
\end{align*}
Let $\mathcal{R}$ denote the family of measurable rectangles in $E\times G$:
\begin{align*}
\mathcal{R}:=\{A\times B : A\in\mathcal{E}, B\in\mathcal{G}\}.
\end{align*}
By definition of the product $\sigma$-algebra,
\begin{align*}
\mathcal{E}\otimes\mathcal{G}=\sigma(\mathcal{R}).
\end{align*}
[/step]
[step:Compute preimages of measurable rectangles]
Let $A\in\mathcal{E}$ and $B\in\mathcal{G}$. For every $x\in X$,
\begin{align*}
x\in h^{-1}(A\times B) \iff h(x)\in A\times B \iff f(x)\in A \text{ and } g(x)\in B.
\end{align*}
Therefore
\begin{align*}
h^{-1}(A\times B)=f^{-1}(A)\cap g^{-1}(B).
\end{align*}
Since $f$ is $\mathcal{M}/\mathcal{E}$-measurable, $f^{-1}(A)\in\mathcal{M}$. Since $g$ is $\mathcal{M}/\mathcal{G}$-measurable, $g^{-1}(B)\in\mathcal{M}$. Because $\mathcal{M}$ is a $\sigma$-algebra, it is closed under finite intersections, so
\begin{align*}
h^{-1}(A\times B)\in\mathcal{M}.
\end{align*}
[guided]
The purpose of this step is to connect measurability of $h$ to the already known measurability of $f$ and $g$. Fix $A\in\mathcal{E}$ and $B\in\mathcal{G}$. The set $A\times B$ is a measurable rectangle in $E\times G$. We compute its preimage under $h$ directly from the definition of $h$.
For $x\in X$, the statement $x\in h^{-1}(A\times B)$ means that $h(x)\in A\times B$. Since $h(x)=(f(x),g(x))$, this is equivalent to saying that the first coordinate belongs to $A$ and the second coordinate belongs to $B$. Thus
\begin{align*}
x\in h^{-1}(A\times B) \iff f(x)\in A \text{ and } g(x)\in B.
\end{align*}
This equivalence is exactly the assertion that
\begin{align*}
h^{-1}(A\times B)=f^{-1}(A)\cap g^{-1}(B).
\end{align*}
Now we use the hypotheses. Since $f: X\to E$ is $\mathcal{M}/\mathcal{E}$-measurable and $A\in\mathcal{E}$, the preimage $f^{-1}(A)$ belongs to $\mathcal{M}$. Since $g: X\to G$ is $\mathcal{M}/\mathcal{G}$-measurable and $B\in\mathcal{G}$, the preimage $g^{-1}(B)$ belongs to $\mathcal{M}$. A $\sigma$-algebra is closed under finite intersections, so
\begin{align*}
f^{-1}(A)\cap g^{-1}(B)\in\mathcal{M}.
\end{align*}
Combining this with the preimage identity gives
\begin{align*}
h^{-1}(A\times B)\in\mathcal{M}.
\end{align*}
Thus every measurable rectangle has measurable preimage under $h$.
[/guided]
[/step]
[step:Show that sets with measurable preimage form a sigma algebra]
Define
\begin{align*}
\mathcal{C}:=\{S\in\mathcal{E}\otimes\mathcal{G}: h^{-1}(S)\in\mathcal{M}\}.
\end{align*}
We show that $\mathcal{C}$ is a $\sigma$-algebra on $E\times G$. Since
\begin{align*}
h^{-1}(E\times G)=X
\end{align*}
and $X\in\mathcal{M}$, we have $E\times G\in\mathcal{C}$.
Let $S\in\mathcal{C}$. Then $h^{-1}(S)\in\mathcal{M}$, and the preimage rule for complements gives
\begin{align*}
h^{-1}((E\times G)\setminus S)=X\setminus h^{-1}(S).
\end{align*}
Since $\mathcal{M}$ is closed under complements, $X\setminus h^{-1}(S)\in\mathcal{M}$, so $(E\times G)\setminus S\in\mathcal{C}$.
Let $(S_n)_{n=1}^{\infty}$ be a sequence in $\mathcal{C}$. Then $h^{-1}(S_n)\in\mathcal{M}$ for every $n\in\mathbb{N}$, and the preimage rule for unions gives
\begin{align*}
h^{-1}\left(\bigcup_{n=1}^{\infty}S_n\right)=\bigcup_{n=1}^{\infty}h^{-1}(S_n).
\end{align*}
Since $\mathcal{M}$ is closed under countable unions, this preimage belongs to $\mathcal{M}$. Hence $\bigcup_{n=1}^{\infty}S_n\in\mathcal{C}$. Therefore $\mathcal{C}$ is a $\sigma$-algebra on $E\times G$.
[/step]
[step:Conclude from the generating rectangles]
The previous rectangle computation shows that every $A\times B\in\mathcal{R}$ belongs to $\mathcal{C}$. Hence
\begin{align*}
\mathcal{R}\subset\mathcal{C}.
\end{align*}
Since $\mathcal{C}$ is a $\sigma$-algebra on $E\times G$, it contains the $\sigma$-algebra generated by $\mathcal{R}$:
\begin{align*}
\sigma(\mathcal{R})\subset\mathcal{C}.
\end{align*}
By definition, $\sigma(\mathcal{R})=\mathcal{E}\otimes\mathcal{G}$. Therefore every $S\in\mathcal{E}\otimes\mathcal{G}$ satisfies $h^{-1}(S)\in\mathcal{M}$. This is exactly the statement that $h=(f,g): X\to E\times G$ is $\mathcal{M}/(\mathcal{E}\otimes\mathcal{G})$-measurable.
[/step]
