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I was reading the answer posted here, and just can't understand one thing. Quote:

the collection of all subintervals of $[0,1]$ is not a $\sigma$-algebra

however, earlier it has been stated that:

the Borel sets of $[0,1]$ is a collection which is a $\sigma$-algebra

The first quote is false, because the complement of, say, $[0.5, 0.7]$ is not a subinterval of $[0,1]$, so the collection of all subintervals of $[0,1]$ is not a sigma algebra - correct? (*)

Is the second quoted statement true because we're not restricting ourselves to subintervals of $[0,1]$, but allow sets that are 'outside' $[0,1]$, so for instance the complement of $[0.5, 0.7]$ can be one of the Borel sets?

What it intuitively tells me is that Borel sets of $[0,1]$ contain a lot more than just subintervals of $[0,1]$.

Edit: I'm actually not sure how this

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 $[0,1]$.

follows from the two points mentioned earlier the answer (1) sigma additivity of probability, 2) probability of choosing a point from interval is the length of that interval).

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user5539357
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1 Answers1

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The complement of an interval is an interval, or a union of two intervals. It follows that the set of finite unions of intervals is closed under complementation (and, obviously, finite unions).

But it's not a $\sigma$-algebra, however, because it is not closed under countable unions. So the first quote is true, and it's true even if we look at finite unions of intervals.

The collection of Borel sets is a $\sigma$-algebra basically by definition, since a Borel set is constructed from intervals using countable operations. In fact, the collection of Borel sets is exactly the smallest $\sigma$-algebra containing the set of open intervals.

Andrew Dudzik
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  • So the collection of Borel sets on [0,1] is made of all intervals from [0,1] plus non-intervals necessary to make it a sigma algebra? – user5539357 Nov 27 '15 at 19:55
  • @user5539357 You can think of it that way, though you might have trouble turning that into a rigorous definition. I would either say that it is a) the $\sigma$-algebra consisting of countable unions of intervals, or b) the intersection of all $\sigma$-algebras containing the intervals. – Andrew Dudzik Nov 27 '15 at 19:56
  • Thanks. I've edited the question, as there's one more thing I'd like to clarify. – user5539357 Nov 27 '15 at 20:30
  • @user5539357 I don't know why you're asking such specific questions about an answer in another thread. I think that Asaf's point is that the Borel sets are the smallest $\sigma$-algebra which allow for a notion of probability, but his point is informal and so nobody but him can really answer this question. – Andrew Dudzik Nov 27 '15 at 20:45
  • Last question - borel sigma algebra is the smallest sigma algebra containing all open sets (if it contains open sets, then it must also contain all their complements - all closed sets). But does it also mean it's the smallest sigma algebra containing all intervals? If it's true, I don't really see why. – user5539357 Nov 27 '15 at 22:06
  • @user5539357 Any open set in $\mathbb{R}$ is a countable union of open intervals, so any $\sigma$-algebra containing open intervals must contain all opens. – Andrew Dudzik Nov 27 '15 at 22:57
  • Actually the set ${[0,0.3] \cup [0.5,1]}$ is not an interval, but a set. https://en.wikipedia.org/wiki/Interval_(mathematics) Complement of an interval usually is not an interval. – user5539357 Dec 23 '15 at 16:39
  • @user5539357 You're quite right about that, this was an oversight. What I've written should now be correct. – Andrew Dudzik Dec 23 '15 at 18:52