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For a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, we say $X: \Omega \rightarrow \mathbb{R}$ is a random variable if $X^{-1}(B) \in \mathcal{F} ~\forall B \in \mathcal{B}$, where $\mathcal{B}$ is the Borel $\sigma$-field. In other words, $X$ is a random variable if $X$ is measurable.

The intuition I've gathered so far is that since we can only talk about measurable subsets of $\Omega$, we want the preimage of any set in the image of X to be measurable. However, my question is, why are we only concerned with preimages of Borel sets, rather than all subsets of $\mathbb{R}$?

(For context, my experience with measure theory is only in the view of probability theory)

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    $\begingroup$ Measurability is always related to two sigma-algebras, one on the source set and the other on the target set. There is no reason to chose the power set as the latter (and actually one rarely does so). $\endgroup$
    – Did
    Commented Oct 31, 2016 at 21:37
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    $\begingroup$ The equivalence of the two definitions seems to imply that the Borel $\sigma$-field is the largest collection of measurable subsets of $\mathbb{R}$, but I feel like that's not the case. $\endgroup$
    – dkv
    Commented Oct 31, 2016 at 21:38
  • $\begingroup$ "The equivalence of the two definitions" What? If "the two definitions" are what I think, then they are not equivalent (and one of them is never used). $\endgroup$
    – Did
    Commented Oct 31, 2016 at 21:43
  • $\begingroup$ $X^{-1}(B) \in \mathcal{F} ~\forall B \in \mathcal{B}$ vs "$X$ is measurable." Do these not mean the same thing? $\endgroup$
    – dkv
    Commented Oct 31, 2016 at 21:47
  • $\begingroup$ You might want to check this $\endgroup$
    – Ran Wang
    Commented Feb 23, 2017 at 20:23

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In fact, it is enough to be concerned about sets $\{X > a\}$ with $a$ real.

For $X$ to be considered a "random variable" we need to be able to tell whether or not it is ${} > a$ for any given real number $a$.

You can try to work with "all subsets of $\mathbb R$", but you won't get very far. Why not use only minimal assumptions? We do not want to use the largest possible collection of sets; that is, we do not want to make it most difficult to be a "random variable". Instead we want to use the smallest possible collection of sets; that is, we want to make it very easy to be a "random variable".

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    $\begingroup$ Right, but that's exactly equivalent to the Borel-set definition above since $(a,\infty)$ is a generating set for $\mathcal{B}$. $\endgroup$
    – dkv
    Commented Oct 31, 2016 at 21:55
  • $\begingroup$ So then I guess my question is, why use Borel sets, rather than some simpler sigma-field on R? Why specifically Borel? What makes those sets special? $\endgroup$
    – dkv
    Commented Oct 31, 2016 at 21:56
  • $\begingroup$ The Borel sets are not special. Comparison with a constant, that is what is special. $\endgroup$
    – GEdgar
    Commented Oct 31, 2016 at 21:57
  • $\begingroup$ Why only a constant? What about {a<X<b} or {a<X<b and c<X<d} or {c<X<d and X=4}? Sure, those can be turned into comparisons to a constant, but I can keep going and in the most general case ask about {X $\in$ S} for any S $\subset \mathbb{R}$? $\endgroup$
    – dkv
    Commented Oct 31, 2016 at 22:11

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