I understand that Borel measurable sets are subsets of a Borel sigma algebra, which is generated by countable unions and intersections, as well as complements, of all open intervals on the real line, plus the empty set. For instance $\mathscr B(0,1]$ would be the borel sigma algebra on the interval $(0,1]$. I have also encountered the definition of a Borel sigma-algebra on a topological space $(M,\mathcal O)$ as the sigma-algebra generated by the open sets of $M$. So that $\mathscr B(0,1]$ would be generated by all the open sets in the standard topology on $(0,1]$.

However, the concept "Borel set" has also come up in finite spaces, as in this example by @whuber discussing a Polya urn problem, and making reference to the event of a red ball coming up. I can see that there is mention to the real line in the post: $\mathcal{R}_i \subset \mathbb{R}$ signifying the event that we draw a red ball at step $i$. So I can imagine the event $0_1,0_2,0_3,\cdots,1_i$ to signify that a ball was drawn on round $i$, but this is still a far cry from the "open intervals" on the real line.

How do you start considering as "open" events like drawing a red ball from an urn?

Is the use of the "Borel" label warranted because we are really referring not so much to the event, but rather the random variable, mapping the event to the real line?


The key concept is "open set". How can something so trivial as drawing balls from an urn constitute open sets? There is no soft-ball around the elementary outcome or the logical sigma algebra of $2^\Omega$.

Two clues: Firstly, there are "topological balls" around these discrete space sets, such that every singleton is an open set. And now these open sets just need to fulfill the requisites of a topological space. Done! No need for random variables!

  • $\begingroup$ It seems your the link in the "pre-answer" answers the question about how you "start considering as 'open' events like drawing a red ball from an urn?", if by open events you mean open sets. Is it something else you are after? I also don't understand your last conclusion that random variables aren't needed. Would it be possible to state it and the question more clearly? $\endgroup$ – ekvall May 10 '16 at 13:10
  • $\begingroup$ Of course! I didn't mean any offense. I just thought there is a question here that I may answer, but I don't quite understand it. As I understand your question presently, you want to know how open sets make sense in discrete spaces. Your link shows how to define a metric on such a space that renders all subsets open, and thus measurable in the Borel sigma algebra which is the sigma algebra generated by the open sets. $\endgroup$ – ekvall May 10 '16 at 13:20
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    $\begingroup$ Go it, I'll see if I can answer that. $\endgroup$ – ekvall May 10 '16 at 13:26

If $\Omega$ is countable, then we may without loss of generality label the outcomes by the integers and set $\Omega = \{1, 2, \dots\}$. This follows from the definition of countability.

That is, even if we are interested in an experiment where we pick balls from an urn, we can label the outcomes in the sample space by the integers. For example, maybe we let "$1$" denote the outcome that all balls are red, "$2$" the outcome that the first is blue and the rest are red, and so on in some coherent manner.

It suffices, thus, to consider the case where $\Omega$ is the natural numbers, or some subset thereof if we want to also deal with finite spaces. The metric on $\Omega$ is taken to be $d(x, y) = I(x \neq y)$, taking the value 1 if $x \neq y$ and 0 otherwise.

Now you may check$^*$ that all points in $\Omega$ are open sets, and that all unions of open sets are open sets. But that means that every subset of $\Omega$ is a Borel set. Remember, the Borel sets are those in the Borel $\sigma-$algebra, $\mathcal B = \sigma(\mathcal O)$, where $\mathcal O$ are the open subsets of $\Omega$.

Since all subsets are measurable, one usually does not bother with the Borel $\sigma-$algebra on discrete spaces, but instead directly declares all subsets of $\Omega$ to be measurable.

$^*$ Let's prove this. In a metric space, a set $A$ is open if for every $x\in A$ there exists an $\epsilon >0$ such that all points in the $\epsilon-$ball around $x$ are also in $A$.

In our example, take $A = \{x\}$ for an arbitrary $x \in \Omega$ and fix an $\epsilon < 1$, say $\epsilon = 1/2$. Then, $x$ is the only point in the open $1/2-$ball around $x$ (recall, the metric is 1 or 0), and $x\in A$ by definition so we conclude $A$ is open. That is, any point is an open set.

  • $\begingroup$ Can you include how you check that "all points in $\Omega$ are open sets"? $\endgroup$ – Antoni Parellada May 10 '16 at 13:45
  • $\begingroup$ Added @AntoniParellada. Let me know if you want further clarification. $\endgroup$ – ekvall May 10 '16 at 13:55
  • $\begingroup$ Yes, reworded slightly, hope this makes it clear. Of course I don't mind. $\endgroup$ – ekvall May 10 '16 at 14:12

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