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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.

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    $\begingroup$ I have posted an answer to your question. Please, let me know if you have any question regarding my answer. If my answer provides relevant / helpful information regarding your question, please, upvote it. If my answer actualy answers your question, accept it too, please. To upvote, click the triangle pointing upward above the number (of votes) in front of the question. To accept the answer, click on the check mark beside the answer to toggle it from greyed out to filled in. $\endgroup$
    – Ramiro
    Mar 17, 2021 at 11:47
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    $\begingroup$ Thanks very much. It is very clear. Don't worry. I've just upvoted. I'm just rewriting. You were very kind. Thanks. $\endgroup$
    – Silvinha
    Mar 17, 2021 at 11:50

1 Answer 1

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

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