Why $[0,1)$, i.e. "closed/open" in defining Borel sigma algebra?

The Borel sigma algebra over the interval $\Omega=\color{red}{[}0,1\color{blue}{)}$ can be defined as the sigma algebra generated by all open subintervals $\mathcal C_o$, i.e. $\sigma(\mathcal C)$, and is denoted as $\mathscr B\,\color{red}{[}0,1\color{blue}{)}$.

I think I understand a bit the idea of the standard topology and open intervals, i.e. $\color{blue}{(}\color{blue}{)}$, but why do we need to use the closed-open (or open-close) for the sample space, $\Omega$? Why can't we also use open-open brackets, or conversely, what is the advantage of closing one of the ends?

SOURCE: HERE

"Note to self": It is likely that this choice is explained here at the point where the idea is not just to generate a sigma algebra of the same cardinality as $\mathbb R$, which is the aim in defining Borel sets on [0,1) as the algebra (sigma) generated by all open intervals $\mathcal C_{\text{o for open}}$, but to assign a measure to the Borel sets.

It is at that point where an algebra (not sigma) $\mathcal F_o$ is defined as the collection of the null set and all subsets resulting from the finite union of disjoint intervals of the form (a,b] with the idea that any element, such as $(a_1,b_1]\cup(a_2,b_2]\cdots(a_n,b_n])$, where $0\leq a_1<b_1\leq a_2\leq\cdots \leq a_n<b_n$. Has it's complement of the same form, including $\emptyset^c=\Omega$. Applying Caratheodory's extension theorem we get to apply a measure of the form $b-a$ to all these subsets, and any countable unions and intersections of them, including less straightforward sets, such as Cantor's. In fact this sigma algebra generated by $\sigma(\mathcal F_o)$ is the Borel sigma: $\sigma(\mathcal C_o)=\sigma(\mathcal F_o)$.

• I think for most applications you could use $(0,1)$ or $[0,1]$ instead. What context did you encounter this in? Commented May 6, 2016 at 0:56
• Since a sigma algebra is closed under intersections, complements and countable unions, it does not matter if you start with $[a,b)$s or with $[a,b]$s or with $(a,b)$s: you get the same sigma algebra in all cases. Commented May 6, 2016 at 0:57
• Thanks, Mariano. It's just that the lecturer on the video promises that something will become clear later regarding his choice, but it doesn't to me. And he is an MIT CalTech graduate, so I have to extend a lot of credit. Commented May 6, 2016 at 1:01
• @MarianoSuárez-Alvarez The question seems to be about the choice of sample space, not the generating intervals. At least, this is how I'm interpreting it. Commented May 6, 2016 at 1:02
• @AntoniParellada Maybe if you watch more, it will become clearer? I don't think this is a major point. Commented May 6, 2016 at 1:03

You have the technical advantage that if $a_n$ increases to $b$, then $$[a_0,b)=\bigcup [a_n,a_{n+1}).$$ A disjoint union. This is used in verifying some of the basic properties, such as countable additivity.
Yes, I remember now, the above fact is important for showing additivity of measure, true. But for Borel algebras the real point is that subtraction is obtained that is: $[a,b)-[c,d)$ is a again a disjoint union of left closed/right open sets.
If for example one started with open sets then you are forced to have half open sets for example $(2,4)-(1,3)=[3,4)$, and more... But with just the half open its so much neater.
• In the example above the "neatness" would come from $[2,4)-[1,3)=[3,4)$? Commented May 6, 2016 at 3:38
• Or something like $[1,4)-[2,3)=[1,2)\cup [3,4)$ you stay with the half open. Later some sort of completion will take place and you get many more types of sets. Commented May 6, 2016 at 3:45