Completeness of the space of sets with distance defined by the measure of symmetric difference Let $m$ be the measure defined on the set semiring $\mathfrak{S}_m$ and $m'$ its extension to the minimal ring $\mathfrak{R}(\mathfrak{S}_m)$. I read that $m'(A\triangle B)$ can be used as a distance in $\mathfrak{R}(\mathfrak{S}_m)$.
I would rather say that it can be used as a distance in the quotient $\mathfrak{R}(\mathfrak{S}_m)/\sim$ where we identify $A\sim A'$ when $m'(A\triangle A')=0$ [EDIT Mar 7'15: which is an equivalence relation, I would say, thanking future answerers for confirming or refuting this]: am I right?
I also read, in Kolmogorov-Fomin's Elements of theory of functions and functional analysis, that, with such a metric, the space $\mathfrak{M}$ of Lebesgue measurable sets, where $A$ and $A'$ are identified when $\mu(A\triangle A')=0$,  is complete, but I cannot see why, if $\{A_n\}\subset\mathfrak{M}$ is a sequence such that$$\forall\varepsilon>0\quad\exists N\in\mathbb{N}^+:\forall n,m\geq N\quad\mu(A_n\triangle A_m)<\varepsilon$$then a measurable $A$ must exists such that $\forall\varepsilon>0\quad\exists N\in\mathbb{N}^+:\forall n\geq N\quad\mu(A_n\triangle A)<\varepsilon$.
EDIT: Care Bear, whom I thank very much, has provided a proof for the case that $\bigcup_n A_n\in\mathfrak{M}$. According to Kolmogorov-Fomin's, $\bigcup_{n=1}^\infty A_n$ is Lebesgue-measurable only under the restrictive condition that $\exists K\geq 0:\forall N\in\mathbb{N}^+\quad\mu(\bigcup_{n=1}^N A_n)\leq K$ (problem 5 (b) here). Can anybody explain whether the lemma is valid in any case?
I $\infty$-ly thank you all for any answer!!!
 A: Quotient
Yes, you are right: in order for the distance to be a metric, and not just a "pseudometric", we need to take the quotient by the equivalence relation that you stated.
Completeness
Let's write  $d(A,B)=\mu(A\triangle B)$. To prove completeness of the space, it suffices to show that every sequence $A_n$ with $$d(A_n,A_{n+1})\le 2^{-n}\tag1$$ has a limit. This is a general metric space fact: from any Cauchy sequence one can extract a subsequence with property (1), and therefore obtain the convergence of the whole Cauchy sequence.
Let $B_n=\bigcup_{m\ge n }A_m$ and $C_n=\bigcap_{m\ge n}A_m$; both these sequences are nested. From (1), summing a geometric series,  we obtain $d(A_n,B_n)\le 2^{1-n}$ and $d(A_n,C_n)\le 2^{1-n}$. Hence $\mu(B_n\setminus C_n)\le 2^{2-n}$. Let $E=\bigcap B_n$. Since $C_n\subset A_n\subset B_n$ and $C_n\subset E\subset B_n$, it follows that $$
\mu(A_n\triangle E) \le \mu(B_n\setminus C_n)\to 0 
$$
which proves $\lim A_n=E$ when $\mathfrak{M}$ is a $\sigma$-algebra.
