Let $(X,\mathcal{N},\mu)$ be a complete probability space, and assume that the $\sigma$-algebra $\mathcal{N}$ has no atoms, that is, for every set $A \in \mathcal{N}$ with $\mu(A) > 0$ there is a set $B \in \mathcal{N}$ with $B \subset A$ and $0 < \mu(B) < \mu(A)$.
If $(X,\mathcal{N},\mu)$ is separable, that is, if one can find a countable family of elements $\{A_n\} \subset \mathcal{N}$ such that for every $B \in \mathcal{N}$ there exists $i$ with $\mu(A_i \Delta B) = 0$, is $(X,\mathcal{N},\mu)$ almost isomorphic to a standard probability space?
I am aware of the following corollary in Fremlin's Volume 3 of Measure Theory:
344K. A measure space $(X,\mathcal{N},\mu)$ is isomorphic to the Lebesgue measure on $[0,1]$ iff it is an atomless countably separated compact (or perfect) complete probability space.
Here isomorphism is understood to be measurable and inverse-measure-preserving bijection with measurable and inverse-measure-preserving inverse. I am not aware of a analogous condition for almost isomorphism. Indeed, I see no reason for $(X,\mathcal{N},\mu)$ to be countably separated in general, though one might try to pass to a quotient set of $X$ which makes the topology generated by the $\{A_n\}$ to be Hausdorff. Then, one still has to deal with compactness, and again a candidate for a compact family of measurable sets would be the $\{A_n\}$, and finiteness + completeness of the measure could somehow imply compactness, but I don't see it either.
For context, I'm reading A. Carderi's paper Ultraproducts, weak equivalence and sofic entropy, where in Lemma 2.16 he finds a separable $\sigma$-subalgebra $\mathcal{M}$ of $\mathcal{N}$ with $(X,\mathcal{N},\mu)$ being a complete atomless probability space and to my understanding claims that $(X,\mathcal{M},\mu)$ is standard.