0
$\begingroup$

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 $\{0,1\}^\mathbf{R}$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $B$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.