Let $(\Omega,\Sigma,\mu)$ be a countably generated probability space. Must $(\Omega,\Sigma,\mu)$ be isomorphic modulo null sets to a [standard probability space](https://en.wikipedia.org/wiki/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](http://www.jstor.org/stable/2586555?seq=1#page_scan_tab_contents "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 $\{0,1\}^\mathbf{R}$.



Countably generated $\sigma$-algebra
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 $B$. 


