The $I$ in the smallest sigma algebra generated by collection of subsets of $\Omega$

In defining a Borel sigma algebra (and if I understand it right) you can depart from the idea that an arbitrary collection of subsets $\mathcal C$ of the sample space $\Omega$, where $\mathcal C$ will end up being intervals of the real line, generates the smallest sigma algebra $\sigma(\mathcal C)$ containing all elements of $\mathcal C$, which happens to also be unique (?).

To find this smallest sigma algebra, I have come across the following notation:

Let $\{\mathcal F_\color{red}{i},i\in \color{red}{I}\}$ be all the collection of all the sigma algebras containing $\mathcal C$. Then $\sigma(\mathcal C)=\bigcap\limits_{i\in I}\mathcal F_i$.

So I think I get the idea, but I don't know what $I$ represents.

SOURCE: here

• The $I$ is some index set, although I'm actually wondering whether it is a set, and not a class. But you can think of it as an index set. Commented May 6, 2016 at 0:05
• Thanks. I went over this a few times, content in the implicit $I=\text{index}$ idea, but it is so well defined, that there has to be a bit more to it. I'm sure it'll be anti-climatic, but still... Commented May 6, 2016 at 0:09
• @Mathematician42 It's a set, not a class. (It's equivalent to a subset of the power set of the power set of $\Omega$.) Commented May 6, 2016 at 0:19
• @AntoniParellada Nope, nothing more to it. Commented May 6, 2016 at 0:19
• @DavidC.Ullrich: True that, very easy argument as well. Thank you. Commented May 6, 2016 at 0:21

1 Answer

This is a common notation: Suppose you have a set $\mathscr{C}$, all whose elements are also sets. The you might want to consider the intersection of all elements of $\mathscr{C}$. We might denote this as $\bigcap\mathscr{C}$.

Example 1: $\mathscr{C}=\left\{\left\{1,2,3\right\},\left\{1,4\right\}\right\}$. Then $\bigcap\mathscr{C}=\left\{1,2,3\right\}\cap\left\{1,4\right\}=\left\{1\right\}$.

Example 2: $\mathscr{C}=\left\{\left\{k\in\mathbb{N}:k\leq n\text{ or }k\text{ is not divisible by }n\right\}:n\in\mathbb{N}, n\geq 2\right\}$. Then $\bigcap\mathscr{C}=\bigcap_{n=2}^\infty\left\{k\in\mathbb{N}:k\leq n\text{ or }k\text{ is not divisible by }n\right\}$ is the collection of prime numbers

However, the notation $\bigcap\mathscr{C}$ is not universal, and when dealing with intersections of large collections of sets, some people prefer to use a notation such as $\bigcap_{n=1}^\infty$ or $\bigcap_{n\in\mathbb{N}}$.

So, in general, when we write $\mathscr{C}=\left\{C_i:i\in I\right\}$, we are simply saying that $I$ is some set and the map $I\to\mathscr{C}$, $i\mapsto C_i$ is a bijection. We can always do that in a trivial way: take $I=\mathscr{C}$ and $C_i=i$.

Then we use the notation $\bigcap_{i\in I} C_i$ instead of $\bigcap\mathscr{C}$.

In your particular problem, we consider the class $\mathscr{F}$ of all $\sigma$-algebras containing $\mathcal{C}$ (do not confuse this $\mathcal{C}$ with the $\mathscr{C}$ above). We then define $\sigma(\mathcal{C})=\bigcap\mathscr{F}$. Just as above we are simply writing $\mathscr{F}=\left\{\mathcal{F}_i:i\in I\right\}$, and using the notation $\sigma(\mathcal{C})=\bigcap_{i\in I}\mathcal{F}_i$ instead of $\bigcap\mathscr{F}$.

I do not know if we can calculate the cardinality of $\mathscr{F}$ (which is the same as the cardinality of $I$) in general, or at least it is not obvious how to do so, so we cannot really replace $I$ by any known set.

• Beautiful, Luiz! Just one minor follow-up question. It seems like in this particular instance we have possibly infinitely uncountable sets in $\mathcal C$ with equally huge numbers of intersections, so would that imply, $\bigcap_{n\in\mathbb{R}}$ in the alternative notation? I know it couldn't make sense... Commented May 6, 2016 at 0:20
• @AntoniParellada I will add some more comments on your particular problem. Commented May 6, 2016 at 0:24
• Silly question before I approve: what is the difference between $\mathcal C$ and $\mathscr{C}$? And in the particular case in the OP, which one should I have used? Commented May 6, 2016 at 0:44
• @AntoniParellada $\mathcal{C}$ is the collection of sets you begin with, so $\mathcal{C}\subseteq\mathcal{P}(\Omega)$ (power set of $\Omega$), whereas $\mathscr{C}$ and $\mathscr{F}$ are the collections which we are intersecting. In this case, $\mathscr{F}$ is a collection of $\sigma$-algebras, so $\mathscr{F}\subseteq\mathcal{P}(\mathcal{P}(\Omega))$ (that is, $\mathcal{C}$ is a set of sets, but $\mathscr{F}$ is a set of sets of sets. This is not the case of $\mathscr{C}$ in the examples, but again, they were used just to introduce the definition.) Please do ask anything that was not clear. Commented May 6, 2016 at 1:33
• A bit of a tangent... using $LaTeX$ as a crutch... \mathscr would be used for the "king" collection of sets, whichever the context happens to be, while \mathcal is reserved for less important collections of sets, again within whichever context we happen to need notation for? Commented May 6, 2016 at 2:02