What is the advantage of Borel sigma algebras in defining probability spaces? I'm trying to get the central concepts correct, so I'm going to express them without embellishment.


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*A Borel $\sigma$ algebra is defined as a sigma algebra generated by a topological space
    $(M,\mathcal{T})$. If $M$ is $\mathbb{R}$, it will typically
    be the "standard topology" defined as the set of all open intervals. 

*Because a $\sigma$ algebra is closed under intersection of infinitely-countably many sub-sets, we can prove that $\displaystyle\bigcap_{n\geq1}(a -\frac{1}{n},\,b)= [a,b)$. This step would not be possible within the topological space where only finite intersections are allowed. So we "recover" closed and half-closed intervals, thanks to the Borel $\sigma$-algebra.

*Now, the half-open intervals are in the sigma algebra (and so are closed intervals), and can be assigned a measure.


So we went through quite a bit of maneuvering (assuming all this is correct) for what purpose? In particular, if we focus on probability spaces, the probability of a close interval is the same as an open interval, since the probability of any point in the real line is zero.
Here is a tentative, unsophisticated, and crass way of looking at it (I'm posting it here to get further feedback):
We can't assign a probability measure to $[0,1]$ because of Cantor's different types of infinites, and $2^Ω$ of $[0,1]$ being far too big, etc. But we want a smooth domain, so that we can get a nice $pdf$ to integrate over... Solution: start with the standard topology over the real line, and then "fill" the gaps by turning it into a sigma algebra.
Reference: On minute:sec 57:39 of this video the speaker ends up the segment with the words: "That very often is very useful." This prompted my question. Where is it needed in probability?
 A: Defining a measure is defining a rule to measure things. The first thing you do in order to do that is to decide what sets you want to measure: picking a sigma-algebra is precisely the act of deciding what sets you will measure.
Now picking the Borel sigma-algebra is exactly saying that you will measure open sets.
There are no acrobatics involved...
A: That's a very good question. The answer is not so simple, but I would say that there are two main reasons why we like a lot Borel measures on the line (or preferably their Lebesgue counterpart):


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*they are regular, which is a wonderful property to establish many properties of the measures, such as when dealing with the relation with topological properties;

*they allow defining conditional measures, which is an important concept in further developments of probability theory, such as stochastic processes.
It should be emphasized that both properties (somehow respectively of semilocal and local nature) fail in general for arbitrary Borel measures on arbitrary metrizable spaces.
There other also less simple reasons why we like measures of the type that you mention. These include stisfying a Lebesgue differentiation theorem (not only for the Lebesgue measure, but for the completion of any Borel measure on the line) and in fact being all that is needed to consider Borel measures on $\mathbb R^n$ (which turn out to be superfluous). But this would be a long story.
