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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)

BCLC
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Evan Aad
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1 Answers1

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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 $(\Omega,\mathcal A)=([0,1],2^{[0,1]})$.

Maximilian Janisch
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Davide Giraudo
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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
  • @Evan: no, I mean to say that a $\sigma$-algebra generated by a set of cardinality at most $S$ itself has cardinality at most $f(S)$ for some function of $S$, not necessarily the identity. – Qiaochu Yuan Dec 30 '12 at 21:18
  • @GEdgar: How can you tell the cardinality of the $\sigma$-algebra of Lebesgue-measurable sets? – Evan Aad Dec 31 '12 at 05:59
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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 06:43
  • @AsafKaragila: Nice. Thanks. – Evan Aad Dec 31 '12 at 07:05
  • I'd like to add (to complement Davide's answer as well as Qiaochu's and DEdgar's comments) that the cardinality of a countably geberated $\sigma$-algebra is derived in Folland's "Real Analysis", Note 1.2, pp. 40-41. – Evan Aad Dec 31 '12 at 07:09
  • Equality ensues ... OK. – GEdgar Dec 31 '12 at 14:51