Product of two trivial sigma algebra is trivial I am reading this note on product measure and on page 196, it states:
Show that the product of two trivial $\sigma$-algebras (on two different spaces $X$, $Y$) is again trivial.
I can't show it.
My attempt: Let $\mathfrak{B}_{X} = \{\emptyset, X\}$ and $\mathfrak{B}_{Y} = \{\emptyset, Y\}$. The pull-back $\sigma$-algebras are $\pi^*_{X}(\mathfrak{B}_X) = \{\emptyset \times Y, X \times Y \}$ and $\pi^*_{Y}(\mathfrak{B}_Y) = \{X \times Y, X \times \emptyset\}$. Therefore the product $\sigma$-algebra is $\mathfrak{B}_X \times \mathfrak{B}_Y = <\pi^*_{X} (\mathfrak{B}_X) \cup \pi^*_{Y}(\mathfrak{B}_Y)> = \{X \times Y, X \times \emptyset,\emptyset \times Y, \emptyset \times \emptyset\}$. This does not looks like $\{\emptyset \times \emptyset, X \times Y\}$, if my understanding of trivial $\sigma$-algebra is correct.
 A: Your proof is correct, the only incomplete part is (as Paul VanKoughnett mentions in the comments) that you did not use the fact that $X \times \emptyset = \emptyset$ and $\emptyset \times Y = \emptyset$.
Here is a proof of those two facts, using the axiom of extensionality (see also Dudley's Real Analysis and Probability). As a result of that axiom, and the definition of the empty set, a set is equal to the empty set if and only if, for every $p$, $p\not\in \emptyset$.
So assume by means of contradiction that there were a $p \in X \times \emptyset$. Then $p$ would be a point in $X \times Y$ so that there would be a point $(x,y) \in X \times \emptyset$. However, for every $y \in Y$, $y \not\in \emptyset$. Therefore this is impossible, i.e. for every point $p = (x,y) \not\in X \times \emptyset$. So by definition of empty set, since $p \not\in X \times \emptyset$ for all $p$, $X \times \emptyset = \emptyset$.
Analogously, if there were a $q \in \emptyset \times Y$, there would be a $(x, y ) \in \emptyset \times Y$, but for every $x \in X$, $x \not\in\emptyset$, so this is impossible, hence for every $p=(x,y) \not\in \emptyset\times Y$, so $\emptyset \times Y = \emptyset$.
See also this comments and answers to this question.
Anyway, in conclusion, you already showed that the product $\sigma$-algebra is  $$ \{ \emptyset, X \times \emptyset, \emptyset \times Y, X \times Y \} = \{ \emptyset, \emptyset, \emptyset, X \times Y \}, = \{ \emptyset, X \times Y \} \,, $$ the trivial $\sigma$-algebra on $X \times Y$.
