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

  • $\begingroup$ 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. $\endgroup$ – Mathematician 42 May 6 '16 at 0:05
  • $\begingroup$ 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... $\endgroup$ – Antoni Parellada May 6 '16 at 0:09
  • 1
    $\begingroup$ @Mathematician42 It's a set, not a class. (It's equivalent to a subset of the power set of the power set of $\Omega$.) $\endgroup$ – David C. Ullrich May 6 '16 at 0:19
  • $\begingroup$ @AntoniParellada Nope, nothing more to it. $\endgroup$ – David C. Ullrich May 6 '16 at 0:19
  • $\begingroup$ @DavidC.Ullrich: True that, very easy argument as well. Thank you. $\endgroup$ – Mathematician 42 May 6 '16 at 0:21

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.

  • $\begingroup$ 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... $\endgroup$ – Antoni Parellada May 6 '16 at 0:20
  • $\begingroup$ @AntoniParellada I will add some more comments on your particular problem. $\endgroup$ – Luiz Cordeiro May 6 '16 at 0:24
  • $\begingroup$ 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? $\endgroup$ – Antoni Parellada May 6 '16 at 0:44
  • $\begingroup$ @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. $\endgroup$ – Luiz Cordeiro May 6 '16 at 1:33
  • $\begingroup$ 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? $\endgroup$ – Antoni Parellada May 6 '16 at 2:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.