# Prove that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)=\sigma(S_2)$.

QUESTION: Given the following collection of subsets of $$\mathbb{R}^n$$,

$$S_1=\{F\subset\mathbb{R}^n; F\; \text{is closed}\;\}$$ and $$S_2=\{(a_1, b_1)\times\cdots \times (a_n, b_n)\in \mathbb{R}^n; \; \text{where}\; (a_i, b_i)\subset \mathbb{R}, \; i=1, \cdots, n. \}.$$ Prove that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)=\sigma(S_2)$$.

MY ATTEMPTY:

• First let's prove that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)$$.

We have $$\mathcal{B}(\mathbb{R}^n)=\sigma(\mathcal{O})$$ where $$\mathcal{O}$$ is a collections of open subsets in $$\mathbb{R}^n$$. Now, remembering that a subset in $$\mathbb{R}^n$$ is open iff its complement is closed. Considering $$S_1=\mathcal{O}^c$$, since the Borel $$\sigma$$-algebra in $$\mathbb{R}^n$$ is also a $$\sigma$$-algebra then $$\mathcal{B}(\mathbb{R}^n)=\sigma(\mathcal{O})=\sigma(\mathcal{O}^c)=\sigma(S_1).$$

• Now let's prove that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$$. Remembering that open rectangles $$(a_i, b_i)\times\cdots\times (a_j, b_j)\in \mathbb{R}^n$$ provides a generator bases of the topology of $$\mathbb{R}^n$$, such that any open subset can be represented by a countable union of rectangles, hence, writting $$E_i=\displaystyle\bigcup_{i=1}^{\infty}\left[(a_i, b_i)\times \cdots \times (a_j, b_j)\right]_i$$ and $$\displaystyle\prod_{k=1}^{n}(a_k, b_k)=\left[(a_i, b_i)\times \cdots \times (a_j, b_j)\right].$$ Thereby, $$E_i=\displaystyle\bigcup_{i=1}^{\infty}\prod_{k=1}^{n}(a_k, b_k).$$ On the one hand ones has $$S_2\subset S_1^c\implies \sigma(S_2)\subset\sigma (\mathcal{O})=\mathcal{B}(\mathbb{R}^n).$$ On the other hand, if $$E_i\in S_1^c=\mathcal{O}$$ then exists open rectangles $$\prod_{k=1}^{n}(a_k, b_k), \; k=1, 2, \cdots, n$$ such that $$E_i=\displaystyle\bigcup_{i=1}^{\infty}\prod_{k=1}^{n}(a_k, b_k).$$ Thus, $$E_i\in \sigma(S_2)$$, this is, $$S_1^c=\mathcal{O}\subset\sigma(S_2)$$, therefore $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_1^c)\subset\sigma(\sigma(S_2))=\sigma(S_2).$$

MY DOUBT: Would someone scan for mistakes in my proof? I feel that it is not completely right. For example, I'm not sure about the first step if I can conclude that equality. And didn't persuade myself with the proof I've provided in the second equality.

Your proof that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)$$ is fine.
The proof that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$$ is OK, but it can be done in a simpler way.
Proof that $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$$ :
Since $$S_2 \subset \mathcal{O}$$, it follows immediately that $$\sigma(S_2) \subseteq \sigma(\mathcal{O}) = \mathcal{B}(\mathbb{R}^n)$$
Since, any open set can be written as a countable union of rectangles, we have that $$\mathcal{O} \subseteq \sigma(S_2)$$. So, we have $$\mathcal{B}(\mathbb{R}^n)= \sigma( \mathcal{O}) \subseteq \sigma(S_2)$$ So, we have $$\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$$.