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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)

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up vote 12 down vote accepted

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.

It may be not the case, for example when $(\Omega,\mathcal A,\mu)=([0,1],2^{[0,1]},\delta_0)$ (no need to specify a measure).

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Thanks. Could you indicate an example of a $\sigma$-algebra that cannot be generated by a countable sub-family? –  Evan Aad Dec 30 '12 at 12:06
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@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. –  Qiaochu Yuan Dec 30 '12 at 12:07
    
@QiaochuYuan: You mean to say that every countably generated sigma algebra is countable? –  Evan Aad Dec 30 '12 at 13:51
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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 $2^{\mathfrak c}$. –  GEdgar Dec 30 '12 at 14:05
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@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. –  Asaf Karagila Dec 31 '12 at 6:43

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