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: $\sigma(\mathcal C_o)=\sigma(\mathcal F_o)$.
Best Answer
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.