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Let $(\Omega, \mathcal{F}, P)$ be a probability space. The completion of $\mathcal{F}$ w.r.t. $P$ is the smallest sigma-algebra that contains $\mathcal{F}$ and all subsets of $P$-nullsets in $\mathcal{F}$. The universal completion is defined as the intersection of all universal completions of $\mathcal{F}$ w.r.t all probability measures on $(\Omega, \mathcal{F})$. What is a (as minimal as possible) working example of a sigma-algebra that differs from its universal completion.

Equi
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    Perhaps the $\sigma$-algebra of the Borel sets? – Berci Jun 30 '18 at 19:23
  • What set is in $\mathcal{F}$ but not in the universal completion? – Equi Jun 30 '18 at 19:28
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    As far as I know, the non-Borel sets in the completion (with respect to Lebesgue measure) cannot be explicitly described. We know they exist because the cardinality of all Borel sets is $\mathfrak c$, whereas the cardinality of null sets is $2^{\mathfrak c}$, as the Cantor set has cardinality $\mathfrak c$ and all of its subsets are null. – Mike Earnest Jun 30 '18 at 19:41
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    But what about the universal completion? I have seen a proof of the existence of Lebesgue but non Borel sets here on SE, but I don't see how it helps in the case of universal completions. – Equi Jun 30 '18 at 20:13

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Example: Let $\Omega$ be the space of continuous functions mapping $[0,1]$ to $\Bbb R$. When endowed with the uniform norm, $\Omega$ is a Banach space. The Borel $\sigma$-algebra on $\Omega$ coincides with the $\sigma$-algebra generated by the projection maps $\Omega\ni f\to f(t)\in\Bbb R$, $t\in [0,1]$. It is known ["The set of continuous nowhere differentiable functions," Pacific J. Math. 83 (1979) 199–205] that the subset $M\subset\Omega$ comprising the nowhere differentiable elements of $\Omega$ is not a Borel set but is co-analytic (that is, the complement of an analytic subset of $\Omega$). Thus, because analytic sets are universally measurable, $M$ is universally measurable.

John Dawkins
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