My professor defines $\mathcal{B}^2 \equiv \sigma\{A \times E : A,E \in \mathcal{B} \}$ where $\mathcal{B}$ is the Borel sets on the real line.
He then says an equivalent definition is that $\mathcal{B}^2 \equiv \sigma\{(a_1, b_2] \times (a_2, b_2] : a_1 \leq b_1, a_2 \leq b_2\}$.
I am not seeing how these two are equivalent. I understand that the Borel sets on the real line are defined as $\sigma\{(a, b] : a \leq b\}$.
Best Answer
Let $\mathcal{B}^2 = \sigma\{A \times E : A,E \in \mathcal{B}\}$, as you defined. Let $\mathcal{B}_1^2$ be the $\sigma$-algebra generated by products of half-open intervals. We want to show that $\mathcal{B}^2 = \mathcal{B}_1^2$.
Clearly, $\mathcal{B}_1^2 \subseteq \mathcal{B}^2$. Now, to show the reverse inclusion, let $A,E \in \mathcal{B}$. Note that
$$ A \times E = (A \times \mathbb{R}) \cap (\mathbb{R} \times E). $$
Clearly, $A \times \mathbb{R}, \mathbb{R} \times E$ are in $\mathcal{B}_1^2$ so that $A \times E \in \mathcal{B}_1^2$ as well. This means that the generator of $\mathcal{B}^2$ is contained in $\mathcal{B}_1^2$; this implies that $\mathcal{B}_1^2 \supseteq \mathcal{B}^2$.