# Cartesian product of two $\sigma$-algebras is generally not a $\sigma$-algebra?

When reading the book Measures, Integrals and Martingales written by R.L. Schilling, I saw a statement as below: $$\mathcal{A} \times \mathcal{B}:= \{A \times B: A\in \mathcal{A}, B\in \mathcal{B}\}$$ where $\mathcal{A}$ and $\mathcal{B}$ are $\sigma$-algebras, is not $\sigma$-algebra in general.

However, I cannot construct a counterexample. Could anyone offer help here?

Kind regards

• Hint: it is not closed under unions or complements. Any nontrivial $\sigma$-algebras will suffice for a counterexample. Commented Apr 5, 2011 at 1:21

Consider the union of two rectangles in $$\mathbb{R}^2$$, such as $$[0,1]\times[0,1]\cup[2,3]\times[2,3]$$. Is this a product? What about the complement of a square, $$\mathbb{R}^2\backslash[0,1]^2$$? Is this a product?
Simple example: $A$ and $B$ are both $\mathbb{R}$ and $\mathcal{A}$ and $\mathcal B$ are both Borel $\sigma$-algebra. Then sets like $(-\infty,a)\times(-\infty,b)$ are in $\mathcal A\times \mathcal B$, but not their complements.
• $$\begin{gathered} ( A \times B ) \backslash ( ( - \infty ,a) \times ( b,\infty ) ) = ( A\backslash ( - \infty ,a ) \times B ) \cup ( A \times B\backslash ( b,\infty ) ) \\ = ( [a,\infty ) \times B ) \cup ( A \times ( - \infty ,b] ) \\ \end{gathered}$$ $( [a,\infty ) \times B ) \in \mathcal{A} \times \mathcal{B}$ and $( A \times ( - \infty ,b] ) \in \mathcal{A} \times \mathcal{B}$ but their union is not in $\mathcal{A} \times \mathcal{B}$ Commented Oct 3, 2018 at 12:54
Take the Borel Algebra $B$ in $R$ - generated by the open intervals of the form $(a,b)$. Then, every set of the type $\{a\}$ belongs to $B$, where $a\in R$. Then $(a,a)\in B\times B, \forall a\in R$. But for $a\neq b$ we have that the set $\{(a,a), (b,b)\}$ does not belong to $B\times B$.
In general, $(X\times Y)\cup (X'\times Y')$ is not of the form $X''\times Y''$.