Let $(X,\cal{A},\mu)$ be a standard (Lebesgue-Rokhlin) space with complete probabilistic measure (for example, $[0,1]$). Let $\cal{B}\supset \cal{A}$ be a wider then $\cal{A}$ $\sigma$-algebra on $X$, and let $\nu$ be an extension of $\mu$ onto $\cal{B}$ so that $(X,\cal{B},\nu)$ is a standard probability space again. Then $\cal{A}=\cal{B}$. We hope that we may prove it, but this should be definitely known if true, so let me ask either for the reference or for one-sentence proof.
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asked Apr 29, 2012 at 22:43
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$\begingroup$ This should follow from the fact that a standard space is perfect. I expect that there is a reference in Fremlin, measure theory for this (I'll have a look...). $\endgroup$– George LowtherCommented Apr 30, 2012 at 0:10
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$\begingroup$ I don't see it explicitly stated. But, it follows from the fact that that complete perfect measures on a Borel space are Radon, and the domain of a Radon measure is uniquely determined by its values on Borel sets. $\endgroup$– George LowtherCommented Apr 30, 2012 at 1:41
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2$\begingroup$ Your statement is Theorem 2b on the EOM page on standard probability spaces (encyclopediaofmath.org/index.php/…). $\endgroup$– George LowtherCommented Apr 30, 2012 at 1:52
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1$\begingroup$ For Standard Borel spaces and, more generally, analytic spaces, this has been shown by Blackwell in On a Class of Probability Spaces, Corollary 1. The argument should extend to completions of standard Borel spaces. $\endgroup$– Michael GreineckerCommented Apr 30, 2012 at 6:32
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