# What is a countably generated $\sigma$-algebra? Can't find a definition online [closed]

What is a countably generated $\sigma$-algebra? Is Borel $\sigma$-algebra countably generated?

• A countably generated $\sigma$-algebra is a $\sigma$-algebra generated by a countable set. As for the Borel $\sigma$-algebra on $\mathbb R$, a hint is to consider the collection $\{[a,\infty) \mid a\in \mathbb Q\}$. Oct 3 '17 at 21:28
• It is just what a dictionary nor of mathematics but of English can define if you looking for "countably generated" Oct 3 '17 at 21:35

A countably generated $\sigma$-algebra is a $\sigma$-algebra $\mathcal M$ of subsets of a set $X$ such that there exists a countable family $\mathcal E=\{E_n:n\in\mathbb N\}$ such that $\mathcal M=\sigma(\mathcal E)$, where $\sigma(\mathcal E)$ denotes the $\sigma$-algebra generated by $\mathcal E$.

The Borel $\sigma$-algebra won't always be countably generated. The Borel $\sigma$-algebra will be countably generated if the topological space is second countable. For example, the Borel $\sigma$-algebra of a separable metric space is countably generated.

• I'm new to measure theory. So can we start from the case of real number with standard topology? Is Borel $\sigma$-algebra countably generated? Oct 3 '17 at 21:32
• @DanielLi Yes, it says so in the answer: the reals are separable.. Oct 3 '17 at 21:33
• Yes, the open intervals with rational center and rational radii form a countable base for the topology. Oct 3 '17 at 21:33
• By "generating", we allow only countably infinite many operations? Right? Here in order to generate $(\sqrt{2}, \sqrt{3})$, we need to only need to intersects countably infinite amount of set to produce each decimal. Is this the right way to understand it? like in case where countably infinite many operations from any countable set won't be enough to reproduce any element from the $\sigma$-algebra, it means that $\sigma$-algebra is not countably generated Oct 3 '17 at 21:43
• Yes, whenever your talking about generating open sets in $\mathbb R$ (or any other second countable topological space for that matter). Any open set is a countable union of elements of the base. When talking about how sets in a $\sigma$-algebra are generated by a countable generating set, things are slightly more complicated. Oct 3 '17 at 22:02

Any family of subsets $\mathcal{C}$ of some set $X$ generates a unique $\sigma$-algebra $\langle \mathcal{C} \rangle$ on $X$, namely the intersection of all $\sigma$-algebras that contain $\mathcal{C}$ as a subset. (The powerset of $X$ is one of such, and the intersection of $\sigma$-algebras is again a $\sigma$-algebra).

A given $\sigma$-algebra $\mathcal{A}$ is countably generated iff there exists a countable family $\mathcal{C}$ such that $\langle \mathcal{C}\rangle = \mathcal{A}$.

For a second countable space $X$ we have that the Borel algebra is generated by the countable base on $X$ that must exist. So yes for all separable metrisable spaces, including all Euclidean spaces.

• $\langle C \rangle$ here allows up to countably infinite many operations right? I'm asking because in my group theory class, $\langle C \rangle$ only allows finite many operations. Oct 3 '17 at 21:47
• It is not straightforward to write the $\sigma$-algebra generated by some set of subsets $C$ as the set of things that can be obtained from $C$ by some operations, like in group theory. It is more subtle, you can look for "Borel Hierarchy" to learn more about this.
– Plop
Oct 3 '17 at 22:03
• @DanielLi brackets for generating a minimal structure are pretty common. In measure theory $\sigma(\mathcal{C})$ is also used Oct 4 '17 at 5:03