This question comes from reading Sheldon Axler's Measure, Integration and Real Analysis, and in particular form 5A.1.
Let $\left(X,\,\mathcal{S}\right)$ and $\left(Y,\,\mathcal{T}\right)$ be measurable spaces. Correspondingly $\mathcal{S}$ is the $\sigma$-algebra on $X$, and $\mathcal{T}$ is the $\sigma$-algebra on $Y$.
Let $\mathcal{S}\otimes \mathcal{T}$ be the smallest $\sigma$-algebra that contains all rectangles $P\times Q$ where $P\in\mathcal{S}$ and $Q\in \mathcal{T}$
I would like to show that the only measurable rectangles in $\mathcal{S}\otimes\mathcal{T}$ are of form $A\times B$ where $A,B\neq \emptyset$, with $A\in \mathcal{S}$ and $B\in \mathcal{T}$ .
I am not quite sure how to tackle this. If $\mathcal{S}\otimes \mathcal{T}$ is the smallest $\sigma$-algebra that contains Cartesian products of members of the $\sigma$-algebras ($\mathcal{S},\,\mathcal{T}$), then all members of $\mathcal{S}\otimes \mathcal{T}$ would have to be of form $Q\times P$, where $Q$ can be built from members of $\mathcal{S}$ using (countable) unions and complimentation, and equivalent for $P$. But then if $A\times B \in \mathcal{S}\otimes \mathcal{T}$ then, for example, $A$ can be built from countable union and compliments of members of $\mathcal{S}$, so $A\in\mathcal{S}$, and equivalent for $B$.
Does this logic make sense?
In my (limited) experience, trying to think explicitly in terms of unions and complements doesn't lead anywhere good.
Instead, use the fact that sections of measurable sets are measurable (Axler 5.6, p. 118). Construct functions $\phi$ and $\psi$ whose preimages are the relevant sections, and the result falls out.
Specifically:
Take $a \in A$ and define the function $\phi_a \colon Y \to X \times Y$ by $y \mapsto (a, y)$. Note that $\phi_a^{-1}(A \times B) = \{ y \in Y \colon \phi_a(y) \in A \times B \} = B = [E]_a \in \mathcal T$.
You may wish to do this one. It's almost a copy-paste of 1., so it might not be that edifying. My suggestion, for whatever it's worth, is to draw a picture.