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I'm studying Probability theory, but I can't fully understand what are Borel sets. In my understanding, an example would be if we have a line segment [0, 1], then a Borel set on this interval is a set of all intervals in [0, 1]. Am I wrong? I just need more examples.

Also I want to understand what is Borel $\sigma$-algebra.

fragapanagos
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DaZzz
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6 Answers6

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To try and motivate the technical answers, I'm ploughing through this stuff myself, so, people, do correct me:

Imagine Arnold Schwarzenegger's height was recorded to infinite precision. Would you prefer to try and guess Arnie's exact height, or some interval containing it?

But what if there was a website for this game, which provided some pre-defined intervals? That could be quite annoying, if say, the bands offered were $[0,1m)$ and $[1m,\infty)$. I suspect most of us could improve on those.

Wouldn't it be better to be able to choose an arbitrary interval? That's what the Borel $\sigma$-algebra offers: a choice of all the possible intervals you might need or want.

It would make for a seriously (infinitely) long drop down menu, but it's conceptually equivalent: all the members are predefined. But you still get the convenience of choosing an arbitrary interval.

The Borel sets just function as the building blocks for the menu that is the Borel $\sigma$-algebra.

Cookie
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conjectures
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First let me clear one misconception.

The set of all subintervals is not a Borel set, but rather a collection of Borel sets. Every subinterval is a Borel set on its own accord.

To understand the Borel sets and their connection with probability one first needs to bear in mind two things:

  1. Probability is $\sigma$-additive, namely if $\{X_i\mid i\in\mathbb N\}$ is a list of mutually exclusive events then $P(\bigcup X_i)=\sum P(X_i)$.

    Therefore the collection of all events which we can measure their probability must have the property that it is closed under countable unions; trivially we require closure under complements (i.e. negation) and thus by DeMorgan we have also closure under countable intersections.

    If so, the set of all events which we can measure the probability of them happening is a $\sigma$-algebra.

  2. We wish to extend the idea that the probability that $x\in (a,b)$, where $(a,b)$ is a subinterval of $[0,1]$ is exactly $b-a$. Namely the length of the interval is the probability that we choose a point from it.

Combine these two results and we have that the Borel sets of $[0,1]$ is a collection which is a $\sigma$-algebra, and it contains all the subintervals of $[0,1]$. Since we do not want to add more than we need, then Borel sets are defined to be the smallest $\sigma$-algebra which contains all the subintervals.

Asaf Karagila
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    Don't fully understand what **smallest** means. All subintervals of [0, 2] is also $\sigma$-algebra, but not smallest, right? – DaZzz Oct 24 '12 at 18:43
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    @DaZzz: Is the complement of $(\frac12,\frac34)$ a subinterval of $[0,1]$, or is it the union of *two* subintervals? Is the complement of $(\frac12,1)$ in the set $[0,2]$ a subinterval of $[0,1]$? – Asaf Karagila Oct 24 '12 at 19:07
  • @DaZzz: Also, if I said that the collection of all subintervals of $[0,1]$ is **not** a $\sigma$-algebra, why would the collection of subintervals of $[0,2]$ be a $\sigma$-algebra? – Asaf Karagila Oct 24 '12 at 19:22
  • Complement of $(\frac{1}{2}, \frac{3}{4})$ is the union of two subintervals. Complement of $(\frac{1}{2}, 1)$ is not subinterval of $[0, 1]$. – DaZzz Oct 24 '12 at 19:23
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    @DaZzz: Yes. It is the union of *two* distinct intervals. So the collection of subintervals is not closed under complements, nor under unions. – Asaf Karagila Oct 24 '12 at 19:26
  • @AsafKaragila You probably meant to say "..if ... is a set of pairwise _disjoint or mutually exclusive_ events..." instead of "".... pairwise independent events..." – Dilip Sarwate Oct 24 '12 at 19:27
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    @DilipSarwate: Yes, I didn't remember the exact term and counted on someone correcting me! Thanks! :-) – Asaf Karagila Oct 24 '12 at 19:28
  • Is $[\frac{1}{2}, \frac{1}{2}]$ considered a subinterval of $[0, 1]$, so that the singleton set $\{\frac{1}{2}\}$ is a Borel subset (similarly for any element in $[0,1])$? – ET-phone-homology Sep 14 '16 at 21:27
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    @Esmath: Correct. But if you talk about open intervals, this won't work. Luckily, every singleton is the countable intersection of open intervals. – Asaf Karagila Sep 14 '16 at 21:46
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    Alternatively if you are working with the closed unit interval like you are you can take the union of $(0,a)$ and $(a, 1)$, then take the complement of that to get $\{0, a, 1\}$ and then finally intersect this with $(0,1)$ to get the singleton $\{a\}$ for any value between the two end points. – Colm Bhandal Jun 18 '20 at 14:53
  • @AsafKaragila What is the difference between a field of sets and a $\sigma$ algebra? – user_9 Aug 26 '20 at 13:03
  • @user_9: Field of sets is not countably closed in general. – Asaf Karagila Aug 26 '20 at 14:01
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Here are some very simple examples.

  1. The set of all rational numbers in $ [0,1] $ is a Borel subset of $ [0,1] $.

    More generally, any countable subset of $ [0,1] $ is a Borel subset of $ [0,1] $.

  2. The set of all irrational numbers in $ [0,1] $ is a Borel subset of $ [0,1] $.

    More generally, the complement of any Borel subset of $ [0,1] $ is a Borel subset of $ [0,1] $.

Haskell Curry
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Another strange Borel set: the set of all numbers in [0,1] whose decimal expansion does not contain 7.

One more: The set of all real numbers in [0,1] whose decimal expansion contains 2 or 5. (Cantor type set)

Last but not least: The set of all real numbers in [0,1] whose decimal expansion contains only finitely many 6.

seriously divergent
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Yes, you're wrong. There's a lot more to Borel sets than intervals.
Basically any subset of $[0,1]$ that you are likely to think of is a Borel set.
The Borel $\sigma$-algebra of $[0,1]$ is the set of all Borel subsets of $[0,1]$.

Robert Israel
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Borel sets are those obtained from intervals by means of the operations allowed in a $\sigma$-algebra. So we may construct them in a (transfinite) "sequence" of steps:

  1. Start with finite unions of closed-open intervals. These sets are completely elementary, and they form an algebra.
  2. Adjoin countable unions and intersections of elementary sets. What you get already includes open sets and closed sets, intersections of an open set and a closed set, and so on. Thus you obtain an algebra, that is still not a $\sigma$-algebra.
  3. Again, adjoin countable unions and intersections to 2. Observe that you get a strictly larger class, since a countable intersection of countable unions of intervals is not necessarily included in 2. Explicit examples of sets in 3 but not in 2 include $F_\sigma$ sets, like, say, the set of rational numbers.
  4. And do the same again.

$\dots$ And again and again.

And even after a sequence of steps we are not yet finished. Take, say, a countable union of a set constructed at step 1, a set constructed at step 2, and so on. This union may very well not have been constructed at any step yet. By axioms of $\sigma$-algebra, you should include it as well - if you want, as step $\infty$ (or, technically, the first infinite ordinal, if you know what that means).

And then continue in the same way until you reach the first uncountable ordinal. And only then will you finally obtain the generated $\sigma$-algebra.

Alexander Shamov
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  • To understand how it works I decided to take all this steps by myself. I took a real segment [0, 1] and divided it on a few intervals, but after step 2 I didn't get new elements in my set. So I don't understand why we need step 3 and all other steps. What am I doing wrong? – DaZzz Oct 25 '12 at 19:08
  • @DaZzz: Well, it is not so easy to actually show that this does not terminate early, but there is an easy explanation why it should not. Namely, the algebra obtained at (say, finite) step $2k$ consists of, say, sets like $\bigcup_{i_1} \bigcap_{i_2} \bigcup_{i_3} \dots \bigcap_{i_{2k}} I_{i_1 i_2 \dots i_{2k}}$ (or vice versa, starting from $\bigcap$). If you take a countable intersection of those, and try to reduce the number of $\bigcap$'s and $\bigcup$'s via distributivity, you will run into a problem: infinite distributivity involves an uncountable union or intersection. – Alexander Shamov Oct 26 '12 at 05:19
  • @DaZzz: Is this obstacle clear to you? – Alexander Shamov Oct 26 '12 at 05:21