Let $(\Omega,\Sigma,\mu)$ be a countably generated probability space. Must $(\Omega,\Sigma,\mu)$ be isomorphic modulo null sets to a standard probability space?

I assume not, so here is a more specific question. Let $\Omega$ be an ultraproduct of finite sets and $\Sigma$ a countably generated sub-$\sigma$-algebra of the Loeb $\sigma$-algebra, and let $\mu$ be the Loeb measure. Is $(\Omega,\Sigma,\mu)$ a standard probability space?

I gather from Jin and Keisler. Maharam spectra of Loeb spaces, providing I understand the language, that if your ultraproducts are taken over a countable set in the usual way then the Loeb space is isomorphic modulo null sets to the product $\{0,1\}^\mathbf{R}$.

  • $\begingroup$ About your first question, there's a measure algebra isomorphism between the completion of $(\Omega,\Sigma,\mu)$ and a standard probability space. Do you mean a pointwise isomorphism up to a negligible set ? $\endgroup$ – Stéphane Laurent Mar 6 '16 at 11:42

For the first question, the answer is yes. There is an isomorphism between measure algebras. Let $B$ be the Boolean algebra of all measurable sets modulo the collection of null sets. Then define a metric $\rho$ on $B$ by letting $\rho(x,y)=\mu((x\wedge y')\vee(y\wedge x'))$.

$\mathbf{Theorem}$:(Caratheodory, see Royden Real Analysis third edition Theorem 15.3.4) Suppose that $(B,\rho)$ is separable and $(B,\mu)$ is a probability space. Then there is an injective measure preserving $\sigma$-complete Boolean algebra homomorphism $\Phi:(B,\rho)\rightarrow(C/I,m)$ where $C$ is the collection of all Borel sets on $[0,1]$ and $I$ is the ideal of all measure zero sets. If $B$ is atomless, then the mapping $\Phi$ can is bijection.

For a proof, if $B$ is generated by a countable subalgebra $A$, then one inductively constructs a homomorphism from $A$ to the Boolean subalgebra of $C$ consisting of all finite unions of open intervals and then one extends this homomorphism from $A$ to $B$.

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  • $\begingroup$ I see. So I guess asking for a pointwise isomorphism up to null sets (as in @StéphaneLaurent's comment) is a much taller order? $\endgroup$ – Sean Eberhard Mar 6 '16 at 14:54
  • $\begingroup$ How baroque do you want your spaces to be? Ergodic theorists like Lebesgue spaces, in which what you want holds automatically; from our point of view everything is a Lebesgue space anyway. To construct counterexamples, you need to be an expert in 1930s style point set topology. $\endgroup$ – Anthony Quas Mar 6 '16 at 18:16
  • $\begingroup$ @AnthonyQuas Haha, thanks. As in my question I'm most interested in subalgebras of Loeb spaces, but the answer here is perfectly satisfactory for what I need, I was just curious really. $\endgroup$ – Sean Eberhard Mar 6 '16 at 20:05

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