Take the tour ×  Consider a probability space $(\Omega,
\mathcal{F}, P)$. Why in the definition of a random variable $X$
it is required that $X$ is a mapping from the sample space and
not from the $\sigma$-algebra of events? That is, why $X: \Omega
\rightarrow R$ and not $X: \mathcal{F} \rightarrow R$?
share|improve this question You want the preimage of a subset of
$\mathbb{R}$ to be an element of $\mathcal{F}$ (so you can
assign a measure to it), not a collection of elements of
$\mathcal{F}$. –  Qiaochu Yuan Jun 19 '11 at 23:26 A collection
of elements of $\mathcal{F}$, interpreted as the union of these
elements, is an element of $\mathcal{F}$. –  Leo Jun 19 '11 at
23:32 Yes, but that's completely unnecessary when you can just
take a union of points in $\Omega$ instead. –  Qiaochu Yuan Jun
19 '11 at 23:35 You can, however, think of $X$ as a map from $R$
to $F$,by taking preimages. You then get a morphism of sigma-
algebras. Some probabilistic questions can be phrased only in
terms of this morphism. –  Mark