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.

  • $\begingroup$ I think of "The sigma-algebra generated by $C$" as "everything reachable from $C$ through unions, intersections, and compliments" $\endgroup$ – Joe Feb 25 at 18:52
  • $\begingroup$ "everything reachable from $C$" meaning that whatever is in the sigma algebra is whatever is reachable from $C$ through unions, intersections and complements? $\endgroup$ – user477465 Feb 25 at 18:58
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    $\begingroup$ 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. $\endgroup$ – Mars Plastic Feb 25 at 19:01
  • $\begingroup$ @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" $\endgroup$ – Joe Feb 25 at 19:10
  • $\begingroup$ @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. $\endgroup$ – 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

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

  • $\begingroup$ 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? $\endgroup$ – user477465 Feb 25 at 19:21
  • $\begingroup$ 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 $\sigma(C)=C$. $\endgroup$ – Mars Plastic Feb 25 at 19:27
  • $\begingroup$ 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? $\endgroup$ – user477465 Feb 25 at 19:38
  • $\begingroup$ 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. $\endgroup$ – 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.


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