I am trying to solve the following question:

Let $(\mathbb{R}^k, \tau)$ be a topological space. Consider the classes of sets $\mathcal{O}_1 = \left\lbrace (a_1, b_1) \times \dots \times (a_k, b_k) \ s.t. \ - \infty \leq a_i < b_i \leq \infty \ \forall \ i \in \{1, \dots, k \} \right\rbrace$ and $\mathcal{O}_2 = \left\lbrace (- \infty, x_1) \times \dots \times (- \infty, x_k) \ s.t. \ x_i \in \mathbb{R} \ \forall \ i \in \{1, \dots, k \} \right\rbrace$. If $\mathcal{C}$ is a class of subsets of $ \mathbb{R}^k$ let $\sigma (\mathcal{C})$ be the $\sigma$-algebra generated by $\mathcal{C}$, i.e. $$\sigma(\mathcal{C}) = \bigcap_{\mathcal{F} \in I(\mathcal{C})} \mathcal{F} \quad \mathrm{where} \ I(\mathcal{C}) = \{\mathcal{F : \mathcal{F} \ \mathrm{is \ a} \ \sigma \mathrm{-algebra \ and } \ \mathcal{F} \supseteq \mathcal{C}} \}$$

Then $\sigma(\mathcal{O}_1) = \sigma(\mathcal{O}_2) = \sigma(\tau)$.

I have already shown that $\sigma(\mathcal{O}_1) = \sigma(\tau)$ and that $\sigma(\mathcal{O}_2) \subseteq \sigma(\tau)$. So the only thing left to show is that $\sigma(\mathcal{O}_2) \supseteq \sigma(\tau)$.

The hint from the question tell me that I should use the fact that any interval $(a,b)$ in $\mathbb{R}$ can be written as $$(a, b) = \bigcup_{n = 1}^\infty [(- \infty, b) \backslash (- \infty, a + n^{-1})]$$

  • $\begingroup$ Is $\tau$ usual metric topology? $\endgroup$ – Lev Bahn Nov 12 '18 at 1:25
  • $\begingroup$ Yes, sorry I missed that. $\endgroup$ – F.Vitiello Nov 12 '18 at 10:09

Since the title is Borel $\sigma-$ algebra, I am pretty sure $\tau$ is usual metric topology.

Then $\sigma(\tau)$ is uaual Boral $\sigma$ algebra.

I will show that $\sigma(\tau)\subseteq \sigma(\mathcal{O}_2)$ and it suffices to prove that $\sigma(\mathcal{O}_1)\subseteq \sigma(\mathcal{O}_2)$ since you already prove that $\sigma(\mathcal{O}_1)=\sigma(\tau)$.

So our goal is to prove $\mathcal{O}_1\subseteq \sigma(\mathcal{O}_2)$.

Let $(a_1,b_1)\times \cdots \times (a_k,b_k)\in \mathcal{O}_1$ be given.

And note that $$(a_1,b_1)\times \cdots \times (a_k,b_k)=\bigcup_{n=1}^{\infty} [(-\infty,b_1)\setminus \left(-\infty,a_1+\frac{1}{n}\right)]\times \cdots \times \bigcup_{n=1}^{\infty} [(-\infty,b_k)\setminus \left(-\infty,a_k+\frac{1}{n}\right)] . $$

Clearly, right hand side is an element of $\sigma(\mathcal{O}_2)$ since it is closed under set subtraction and countable union and each $(-\infty,b_i)$ and $(-\infty,a_i+n^{-1})$ are in $\mathcal{O}_2$.

Therefore, $\mathcal{O}_1\subseteq \sigma (\mathcal{O}_2)$.

Since $\sigma(\mathcal{O}_2)$ is a sigma algebra containing $\mathcal{O}_1$ and $\sigma(\mathcal{O}_1)$ is the smallest sigma algebra containing $\mathcal{O}_1$, we have $\sigma(\tau)=\sigma(\mathcal{O}_1)\subseteq \sigma(\mathcal{O}_2)$.

We are done.

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