Proof of equivalent definitions of sigma algebra of Borel sets in $R^n$

Question

I'm trying to prove that the n-dimensional Borel $\sigma$-algebra

$$B(R) = \sigma(\{\times_{j=1}^n (a_j,b_j) : \forall a_j < b_j \text{ and } a_j,b_j \in R \})$$

is equivalent to the following definition

$$B(R) = \sigma(\{\times_{j=1}^n (-\infty, a_j] : \forall a_j \in R \})$$

I've tried doing the analogous proof for the one-dimensional case, where we get the complement of $(-\infty,a_j]$ and intersect it with some $(-\infty,b_j]$ to obtain $(a_j,b_j] \in B(R)$ and then build a countable union of sets of that form that are equal to the open interval $(a_j,b_j)$.

I've figured out that the complement of $\times_{j=1}^n (-\infty, a_j]$ is the union

$$(a_1,\infty)\times R \times R \ldots \cup R \times (a_2,\infty) \times \ldots \cup \ldots$$

and its intersection with an arbitrary $\times_{j=1}^n (-\infty, b_j]$ interval yields

$$(a_1,\infty)\times (-\infty,b_2] \times (-\infty,b_3] \ldots \cup (-\infty,b_1] \times (a_2,\infty) \times \ldots \cup \ldots$$

but I don't know how to proceed. This doesn't seem to be the correct way to prove this, right?

Answer 1

In two dimensions,

$$ (a_1,b_1] \times (a_2,b_2] = \big((-\infty,b_1] \times (-\infty,b_2]\big) \cap \big((-\infty, a_1] \times (-\infty, b_2]\big)^c \cap \big((-\infty, b_1] \times (-\infty, a_2]\big)^c. $$

See if you can generalize this to higher dimensions.

Answer 2

If $\mathcal{F}_i$ on $X_i$ is generated by $\mathcal{E_i}$ (i.e $\mathcal{F}_i=\sigma(\mathcal{E}_i)$) and $X_i\in\mathcal{E}_i$ (or each $\mathcal{E}_i$ contains a sequence $\{E_{i,j}\}$, $E_{i,j}\uparrow X_i$), then $\otimes_{i=1}^n\mathcal{F}_i$ is generated by $\mathcal{E}=\{\times_{i=1}^n E_i:E_i\in \mathcal{E}_i\}$.

In your case $\mathcal{B}_{\mathbb{R}^n}=\otimes_{i=1}^n\mathcal{B}_{\mathbb{R}}$ and $\mathcal{B}_{\mathbb{R}}$ is generated by both $\{(a,b):a,b\in\mathbb{R};a<b\}$ and $\{(-\infty, a]:a\in \mathbb{R}\}$.