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When you start reading measure theory, the first motivation presented is

we want to find some nice sets we can measure.

The solution is $\sigma$-algebras.

Then we define a random variable as $X: \Omega \rightarrow \mathbb{R}$. Both $\Omega$ and $\mathbb{R}$ are equipped with $\sigma$-algebras.

However, the thing we want to measure is $X \in A$, for $A \in \mathbb{B}$, i.e. we want $P(X^{-1}(A))$. But $X^{-1}(A) \in \mathcal{F}$, the $\sigma$-algebra in $\Omega$. So what we want to measure is the sets in $\Omega$.

So, the only purpose of $\mathbb{B}$ seems to be to ensure that when we take $X^{-1}$ on a Borel set, we get something that lies in $\mathcal{F}$ (since $X$ is measurable).

But ... that can be accomplished by any other $\sigma$-algebra as well. Let "The Bublu Sets" be any other $\sigma$-algebra on $X$ which contains enough sets of interest. Then let us define a random variable as a map from $\Omega$to $\mathbb{R}$ which is measurable with respect to The Bublu Sets. Then we can ask questions like $P(X \in A)$ where $A$ is a Bublu Set. This theory seems identical to the one with the Borel sets?

So, what is so special about the Borel sets? Why have I heard this name "Borel" so much in probability and measure theory, when all it seems to do is accomplish something that any other $\sigma$-algebra would also do, such as The Bublu Sets.

Cinad
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    This is answered in this MSE-question. What are Bublu sets? – Dietrich Burde Oct 12 '17 at 19:58
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    The set of Borel sets is exactly the $\sigma$-algebra generated by open sets, in particular, open intervals. If we care about measuring the length/measure of an interval, then we automatically get that we can talk about the measure of Borel sets. And no matter what $\sigma$-algebra we choose that contains the open intervals, the Borel sets will also be in there (by virtue of being generated by the open intervals). – Hayden Oct 12 '17 at 20:00
  • No, that answer is flawed. It states that we want to find a set which we can find the "size" of. But no, that's not what we want. We don't want the size of Borel sets, we want the size of $X^{-1}(A)$, which is a set in $\mathcal{F}$, not a Borel set! – Cinad Oct 12 '17 at 20:00
  • The Borel $\sigma$-algebra is generated by the open sets... meaning that you will be able to use topological properties of $\mathbb{R}$... – Julien Oct 12 '17 at 20:00
  • @Julien In other words, one can hardly imagine another way of picking "enough sets of interest" in $\Bbb R$ than at least the open sets. – Hagen von Eitzen Oct 12 '17 at 20:03

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Because we want to connect Topology with Measure.

Most spaces we deal with are (in some natural way) topological (or metric) spaces where we can talk about open, closed, compact,etc sets. If these spaces are also equipped with any arbitrary measure, by default, there is not necessarily any link between the topology and measure structure of the space. If we want to have some connection and for instance be able to say that "continuous functions (a notion depending on the topology) are measurable (a measure-theoretic notion)" we are lead to require that the open sets be measurable. And once all open sets are in a sigma-algebra then all borel sets are by the closedness of measurability under set theoretic operations.

Radon measure, so pivotal in the theory, are by definition Borel measures.

Behnam Esmayli
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