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).
Best Answer
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.