# Borel $\sigma$-algebra on $R^k$

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 $$(a, b) = \bigcup_{n = 1}^\infty [(- \infty, b) \backslash (- \infty, a + n^{-1})]$$

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