Why is the Borel-$\sigma$-field in $n$ dimensions the $\sigma$-field generated by the products of half open intervals? 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 $\sigma\{(a, b] : a \leq b\}$.
 A: 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 $\mathcal{B}_1^2 \supseteq \mathcal{B}^2$.
A: Call the first $\sigma$-algebra $\mathcal{A}$ and the second $\mathcal{C}$. It is clear that $\mathcal{C}\subset \mathcal{A}$. To show the other inclusion, fix a half open interval $(a, b]$ and consider the set $\{A : A\times (a, b]\in \mathcal{C}\}$. Clearly this set contains the half open intervals and since $\mathcal{C}$ is a $\sigma$-algebra it is also easily seen that this set is a $\sigma-$algebra, so it contains the Borel sets. The conclusion we draw from this is that for any Borel set $A$ and any half open interval, $(c, d]$, we have $A \times (c, d] \in \mathcal{C}$.
Now fix a Borel set $B$ and consider the set $\{C : B\times C\in \mathcal{C}\}$. By what we showed this set contains the half open intervals and again it is clear that this is a $\sigma-$algebra, so it contains the Borel sets. We conclude that for Borel sets, $A$ and $B$, we have $A\times B\in \mathcal{C}$. These products generate $\mathcal{A}$ so we must have $\mathcal{A}\subset \mathcal{C}$ also.
A: When $T$ is a family of subsets of $X$ we  define $\sigma (T,X)$ as the smallest $\sigma$-algebra $S$ on $X$ such that $T\subset S.$
When $T$ is a topology on $X$ we define the set $B(T)$ of Borel sets on $X$  as $\sigma (T,X).$ With the usual abuse of notation this is often denoted $B(X)$. And abbreviated as $B.$
$(1).$ Prove that if $U,V$ are families of subsets of $X$ and if $H=\sigma (U,X)=\sigma (V,X)$ then $$\sigma (\{u\times u':u,u'\in U\},X^2)=\sigma (\{v\times v':v,v'\in V\},X^2)=\sigma (\{h\times h':h,h'\in H\},X^2).$$
$ (2). $Let $T$ be the usual topology on $\Bbb R.$ Let $C=\{(a,b): a,b\in \Bbb R\}$ and  $D=\{(a,b]:a,b\in \Bbb R\}.$ 
Every $d\in D$ is $\cap E$ for some countable $E\subset C,$  and  $C\subset T,$ so $$\sigma (D,\Bbb R)\subset \sigma (C,\Bbb R)\subset \sigma (T,\Bbb R)=B(T).$$ Every $t\in T$ is $\cup F$ for some countable $F\subset C,$ and every $c\in C$ is $\cup G$ for some countable $G\subset D,$ so we have $$B(T)=\sigma (T,\Bbb R)\subset \sigma (C,\Bbb R)\subset \sigma (D,\Bbb R).$$
$(3).$  Now let $T'$ be the usual topology on $\Bbb R^2.$  Let $C'=\{c_1\times c_2: c_1,c_2\in C\}$ and  $\quad D'=\{d_1\times d_2: d_1,d_2\in D\}.$
$\quad (3a). $ Every $d'\in D'$ is equal to $\cap  E'$ for some countable $E'\subset C',$ and $C'\subset T',$ so we have  $$\sigma (D',\Bbb R^2)\subset \sigma (C',\Bbb R^2)\subset \sigma (T',\Bbb R^2)=B(T').$$  Every $t'\in T'$ is $\cup F'$ for some countable $F'\subset C', $  and $C'\subset T'$ , so $$B(T')=\sigma (T',\Bbb R^2)\subset \sigma (C',\Bbb R^2)\subset \sigma (T',\Bbb R^2).$$ $\quad (3b). $  By $(1)$  with $U=C$ and $V=B(T)$ we have $$\sigma (C',\Bbb R^2)=\sigma (\{A\times A':A,A'\in B(T)\},\Bbb R^2). $$
