Is completion of measures equivalent to completion of sigma algebras as metric spaces with respect to measures?

Question

An alternative way to get the Lebesgue $\sigma $-algebra $\mathcal{L} $ from the Borel algebra $B$ is to set $E\sim J$ iff $d(E,J):=\lambda(E\mathbin\Delta J)=0$ for $E,J\in B$. Then the completion of $(X,d)$ as a metric space, where $X=B/{\sim}$, is equivalent to completion of $B$.

My question is: Are there known generalizations? Given a measure space $(X,\Sigma ,\mu)$, what are the conditions on $X$ and $\Sigma$ so that the completion of the measure space is equivalent to the completion of $M := \Sigma/{\sim}$ as a metric space, with $\sim $ as above? Do we have to use the symmetric difference to define $d$ on $M$?

Answer 1

In Theorem 1.4 of this paper or of its preprint version, it is shown that, for any measure $m$ on any algebra $\mathcal A$ of subsets of a set $X$, a subset $E$ of $X$ is locally approximable by sets in $\mathcal A$ in the sense of pseudo-metrics $d_A$ if and only if $E$ is Carathéodory-measurable, where $d_A(E,F):=m^*(A\cap(E+F))$, $A$ is a member of $\mathcal A$ with $m(A)<\infty$, $m^*$ is the outer measure corresponding to $m$, and $E+F$ is the symmetric difference between subsets $E$ and $F$ of $X$.

In particular, if $m(X)<\infty$, then one can use just the approximation relative to one pseudo-metric $d_X$ instead of the local approximation relative to the family of pseudo-metrics $d_A$.

(We have to say "pseudo-metrics", because $d_A(E,F)=0$ does not in general imply $E=F$.)

It is also shown in that theorem that the completion $\sigma$-algebra is always contained in the Carathéodory $\sigma$-algebra, and the completion $\sigma$-algebra coincides with the Carathéodory $\sigma$-algebra if $m$ is $\sigma$-finite (and this $\sigma$-finiteness condition cannot be dropped).

Answer 2

Isn't this just Carathéodory's extension theorem?