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)
Is every sigma-algebra generated by some random variable?
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measure-theory
1 Answers
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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.
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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2Thanks. Could you indicate an example of a $\sigma$-algebra that cannot be generated by a countable sub-family? – 2012-12-30
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3@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. – 2012-12-30
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0@QiaochuYuan: You mean to say that every countably generated sigma algebra is countable? – 2012-12-30
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5A 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 $2^{\mathfrak c}$. – 2012-12-30
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0@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. – 2012-12-30
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0@GEdgar: How can you tell the cardinality of the $\sigma$-algebra of Lebesgue-measurable sets? – 2012-12-31
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3@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. – 2012-12-31
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0@AsafKaragila: Nice. Thanks. – 2012-12-31
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0I'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"](http://www.amazon.com/Real-Analysis-Techniques-Applications-Mathematics/dp/0471317160/ref=sr_1_1?ie=UTF8&qid=1356933965&sr=8-1&keywords=folland+real+analysis), Note 1.2, pp. 40-41. – 2012-12-31
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0Equality **ensues** ... OK. – 2012-12-31