# Assistance: "Generated Borel $\sigma$-algebra"

I am attempting to work through the problem below and am experiencing some difficulty in solving it, and moreover, in understanding it. I would appreciate any insights.

Problem

Use the fact that $$\mathbb{R}$$ is a countable union of open intervals to show that $$\mathcal{B}_{\mathbb{R}}$$ is generated by the collection of all open intervals.

Confusions

My naive belief is that this statement just follows from the definition of $$\mathcal{B}_{\mathbb{R}}$$ - "Let $$E$$ be a topological space, then the $$\sigma$$-algebra generated by the collection of all open subsets of $$E$$ is called the Borel $$\sigma$$-algebra on $$E$$; it is denoted $$\mathcal{B}_{\mathbb{E}}$$" - as each open interval of $$\mathbb{R}$$ is an open subset of $$\mathbb{R}$$. For this problem should I verify that for any open interval in $$\mathcal{B}_{\mathbb{R}}$$, the complement of that interval is also in $$\mathcal{B}_{\mathbb{R}}$$? As you can see, I am somewhat lost.

Attempt Following Assistance

$$\text{ }$$ Let $$\mathcal{C}$$ be the collection of all open intervals of $$\mathbb{R}$$. By definition, $$\sigma \mathcal{C}$$ is the smallest $$\sigma$$-algebra containing $$\mathcal{C}$$. Let $$\mathcal{D}$$ denote the collection of all open subsets of $$\mathbb{R}$$. We know that $$\mathcal{B}_{\mathbb{R}}$$ is the smallest $$\sigma$$-algebra containing $$\mathcal{D}$$.

$$\text{ }$$ To begin, consider any open interval in $$\mathcal{C}$$. Every open interval of $$\mathbb{R}$$ is also an open subset of $$\mathbb{R}$$, this open interval is contained in $$\mathcal{D}$$, thus $$\mathcal{C} \subseteq \mathcal{D}$$.

$$\text{ }$$ Now consider any open subset in $$\mathcal{D}$$. Since every open subset in $$\mathcal{D}$$ can we written as a countable union of open intervals of $$\mathbb{R}$$ and $$\sigma \mathcal{C}$$ is closed under countable unions of open intervals of $$\mathbb{R}$$, we have that $$\mathcal{D} \subseteq \sigma \mathcal{C}$$.

$$\text{ }$$ Since $$\mathcal{C} \subseteq \mathcal{D} \subseteq \sigma \mathcal{C}$$, we have that $$\sigma \mathcal{C} = \mathcal{B}_{\mathbb{R}}$$, i.e. $$\mathcal{B}_{\mathbb{R}}$$ is generated by all open intervals.

• See the difference? Generated by all open intervals; generated by all open sets. Apr 15, 2021 at 20:13

Let $$\Delta = \{(a,b): a,b \in \mathbb{R}, \; a < b\}$$ What the problem asks essentially is to show that $$\sigma(\Delta) = \mathcal{B}(\mathbb{R})$$. What follows form the definition, as you said, is that $$\sigma(\Delta) \subseteq \mathcal{B}(\mathbb{R})$$. That's because every interval $$(a,b) \in \Delta$$ is an open subset of $$\mathbb{R}$$ so it belongs to $$\mathcal{B}(\mathbb{R})$$. Since $$\sigma(\Delta)$$ is the minimal $$\sigma$$-algebra that contains $$\Delta$$ and $$\mathcal{B}(\mathbb{R})$$ is a $$\sigma$$-algebra that contains $$\Delta$$ from the minimality of $$\sigma(\Delta)$$ we get $$\sigma(\Delta) \subseteq \mathcal{B}(\mathbb{R})$$.

For the other inclusion remember that every open subset of $$\mathbb{R}$$ can be written as a countable union of open intervals. This should give you that the collection of open subsets of $$\mathbb{R}$$ is a subset of $$\sigma(\Delta)$$. Then use a similar minimality argument like the above.

• This makes sense. The idea of showing $\sigma \mathcal{C} \subseteq \mathcal{B}_{\mathbb{R}}$ and the converse to show equality just didn't really cross my mind. I haven't full read through your answer but will use this idea to try to attempt it myself and add it as a comment. I have upvoted the question and will check it a bit later to motivate others to participate. Apr 15, 2021 at 20:29
• @rodeo_flagellum Happy to have helped! If something is still unclear after attempting a proof do let me know. Apr 15, 2021 at 20:35
• I attempted the proof. What are your thoughts on it? Also, thank you again for this thorough response and the time you spent helping me learn, I appreciate it. Also, question: I am a bit confused on the second part --> is there a way to go directly from saying that every open subset is a countable union of open intervals so each element of $\mathcal{D}$ is also is $\sigma \mathcal{C}$ to saying that $\mathcal{B}_{\mathbb{R}} \subseteq \sigma \mathcal{C}$? Apr 15, 2021 at 21:18
• @rodeo_flagellum I don't think you can say it directly without using such a minimality argument. It's quite short though. Let $U$ be the collection of open sets. $\mathcal{B}(\mathbb{R}) = \sigma(U)$ and since $\sigma(U)$ is the minimal $\sigma$-algebra that contains $U$ and $\sigma(\Delta)$ contains $U$ we have the required inclusion. Apr 15, 2021 at 21:24
• That explanation is great. Once again, thank you. Apr 15, 2021 at 21:26

Recall that the $$\sigma$$-algebra generated by a collection $$F$$ of subsets of $$X$$ is, by definition, the intersection of all $$\sigma$$-algebras containing $$F$$, (this intersection is a $$\sigma$$-algebra).

So you would have to show that if $$\Sigma$$ is a $$\sigma$$-algebra of subsets of $$\mathbb{R}$$ containing every open interval, then $$\Sigma$$ contains every open subset of $$\mathbb{R}$$. Indeed, $$\mathcal{B}_\mathbb{R}$$ is defined as the $$\sigma$$-algebra generated by the collection of open subsets of $$\mathbb{R}$$.

One way to prove this is to note that an open subset of $$\mathbb{R}$$ is a union of open intervals with rational endpoints. And since $$\mathbb{Q}$$ is countable this is necessarily a countable union. So you are done since $$\sigma$$-algebras are closed under countable unions.

N.B. I'm not sure what the problem as stated wants you to do, after all $$\mathbb{R}$$ is not just a countable union of open intervals, it is itself an open interval.

• Thank you for your thoughts and comments! Apr 15, 2021 at 21:19