# What does it mean for a sigma algebra to be generated by something?

I have a definition:

Given $$C \supset 2^{\Omega}$$, the $$\sigma$$- algebra generated by $$C$$ written, $$\sigma(C)$$ is the "smallest" $$\sigma$$- algebra containing $$C$$

I understand what this means but I just don't understand what "generated by $$C$$" means.

Similarly, I am given an example of the Borel $$\sigma$$- algebra as $$\sigma(T)$$ where $$T =$${open sets of $$\mathbb{R}$$}

So a Borel $$\sigma$$- algebra is equal to a $$\sigma$$- algebra generated by all the open sets of $$\mathbb{R}$$

Can someone please explain what the word "generated by" means?

I know a $$\sigma$$- algebra is a collection of subsets of the power set $$2^{\Omega}$$ where $$\Omega$$ is any set. So does this mean that if a $$\sigma$$- algebra is generated by something else, that something else is just the set $$\Omega$$?

Sorry for such a basic question, just looking for some clarification.

• I think of "The sigma-algebra generated by $C$" as "everything reachable from $C$ through unions, intersections, and compliments" – Joe Feb 25 at 18:52
• "everything reachable from $C$" meaning that whatever is in the sigma algebra is whatever is reachable from $C$ through unions, intersections and complements? – user477465 Feb 25 at 18:58
• I wouldn't recommend thinking about it like that. Since infinite unions may occur, such a "construction" of the generated sets is impossible in general. Also, this idea may lead you to seriously underestimate the complexity of a $\sigma$-algebra that may be generated by a very simple generator. – Mars Plastic Feb 25 at 19:01
• @MarsPlastic that's true, but I still think it's an "easy" way to think about the rough idea. Obviously there's the caveat that you can't formally construct the $\sigma(C)$ as "all finite sequences of elements of $C, \cap, \cup,$ and compliments" – Joe Feb 25 at 19:10
• @Joe You are right in that this may help to get a basic idea, but one should be very careful. This interpretation is more appropriate for an algebra. – Mars Plastic Feb 25 at 19:14

You can give meaning to the "smallest" $$\sigma$$-algebra containing $$C\subset2^\Omega$$ by first noting that any intersection of $$\sigma$$-algebras on $$\Omega$$ is again a $$\sigma$$-algebra on $$\Omega$$ and then setting

$$\sigma(C):=\bigcap_{\mathcal A \in \mathbb A}\mathcal A, \quad \text{where \mathbb A:=\{\mathcal A\subset 2^\Omega : \text{\mathcal A is a \sigma-algebra on \Omega and C\subset\mathcal A}\}.}$$

• interesting, i did have some confusion on what the difference was between $C$ and $A$. is $C \subset A$? is $C$ also a $\sigma$- algebra? – user477465 Feb 25 at 19:21
• I edited my answer slightly which should take care of your first question. Concerning the second one: $C$ does not have to be a $\sigma$-algebra, as the whole point is to find the smallest $\sigma$-algebra containing $C$. If $C$ was one already, we'd obviously have $\sigma(C)=C$. – Mars Plastic Feb 25 at 19:27
• thanks. so if $A$ is a $\sigma$- algebra, what would it mean for a subset of A to not be a $\sigma$- algebra? If $\Omega$ is $\{1,2,4\}$ for example and $A =\{\{1\},\{2\}\}$, then can $C$ be $\{\{1\}\}$ it can't be $\{\{1\},\{2\},\{4\}\}$ right? In what instance would $C$ not be a subset of the power set? is it because my example was a finite set? – user477465 Feb 25 at 19:38
• I'm not sure if I understand your question. In your case, $A=\{\{1\},\{2\}\}$ is not a $\sigma$-algebra. And $C$ has to be a subset of $2^\Omega$ for the initial question to even make sense. – Mars Plastic Feb 26 at 11:30

An intersection of $$\sigma$$-algebras is a $$\sigma$$-algebra, similar to the fact that an intersection of subgroups is a subgroup. This fact gives us that there is a unique minimal $$\sigma$$-algebra containing $$C$$ arising from a collection $$C$$, such that it generates the $$\sigma$$-algebra in that sense.

Which is again similar to what it means for a subgroup to be generated by a set, for a topology to be generated by a collection of sets and so on.

At least that's my take on it.