Let $(X,\cal{A},\mu)$ be a standard (LebesgueRokhlin) 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 onesentence proof.
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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 Lowther Apr 30 '12 at 0:10

$\begingroup$ 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 Lowther Apr 30 '12 at 1:41

2$\begingroup$ Your statement is Theorem 2b on the EOM page on standard probability spaces (encyclopediaofmath.org/index.php/…). $\endgroup$ – George Lowther Apr 30 '12 at 1:52

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 Greinecker Apr 30 '12 at 6:32