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


  • 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.

  • 2
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    – 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


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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