# Is every sigma-algebra generated by some random variable?

Let $\mathcal{A}$ be a $\sigma$-algebra over $\Omega$. Is there a function $f:\Omega\rightarrow\mathbb{R}$ such that $\mathcal{A}=f^{-1}(\mathfrak{B(\mathbb{R})})$? ($\mathfrak{B(\mathbb{R})}$ being the Borel field on the real line)

Not necessarily. The Borel $$\sigma$$-algebra is generated by a countable class of measurable sets, namely $$\mathcal D:=\{(a,b),a,b\in\Bbb Q\}$$. By the transfer property, $$\mathcal A=f^{-1}(\mathcal B(\Bbb R))=f^{-1}(\sigma(\mathcal D))=\sigma(f^{—1}(\mathcal D)),$$ so $$\mathcal A$$ is generated by a countable class.
But not every $$\mathcal A$$ is generated by a countable class, consider for example $$(\Omega,\mathcal A)=([0,1],2^{[0,1]})$$.
• Thanks. Could you indicate an example of a $\sigma$-algebra that cannot be generated by a countable sub-family? Commented Dec 30, 2012 at 12:06
• @Evan: take the $\sigma$-algebra of all subsets of an uncountable set. More generally, a $\sigma$-algebra generated by a set of bounded cardinality has bounded cardinality. Commented Dec 30, 2012 at 12:07
• A countably generated sigma-algebra has cardinal at most $\mathfrak c$. So, for example, the sigma-algebra of Lebesgue-measurable sets is not countably generated, since it has cardinal $2^{\mathfrak c}$. Commented Dec 30, 2012 at 14:05
• @EvanAad: The Lebesgue measurable sets contain all the subsets of the Cantor set. Therefore it is at least $2^\frak c$; on the other hand it's a subset of $\mathcal P(\mathbb R)$, so its cardinality cannot extend $2^c$. It follows that the equality ensues. Commented Dec 31, 2012 at 6:43