On the complete boolean algebra wikipedia page, I found the following statement :
The algebra of all measurable subsets of a $\sigma$-finite measure space, modulo null sets, is a complete Boolean algebra. [...]
I am interested in this statement and literature about it, find below an attempt proof.
Here is a statement of the problem :
Let $(X,\mathcal X, \mu)$ be a finite measure space, then for any $A\in\mathcal X$, the equivalence class of $A$ modulo the null sets is \begin{align*} [A]=\{ B\in\mathcal X : \mu(A\,\triangle\, B)=0 \} \end{align*}
the algebra is defined as $[A]\land[B]=[A\cap B]$, $[A]\lor[B]=[A\cup B]$ and $\lnot[A]=[X\setminus A]$. the partial order is defined as $[A]\leq [B]$ whenever $[A]\lor [B]=[B]$, or equivalently $\mu(A\setminus B)=0$.
I am also interested in the following question about continuity :
If $\{[B_\alpha]\}_{\alpha\in A}$ is a net in the quotient set of $\mathcal X$ with null sets, such that $[B_\alpha]\to [B]$, then $\mu(B_\alpha)\to \mu(B)$.
Let $\mathcal A\subseteq \mathcal X$ and $\mathcal B=\{ B\in\mathcal X : \forall A\in\mathcal A,~ \mu(A\setminus B)=0 \}$ be the set of upper bound on $\mathcal A$. In view of using Zorn's lemma, let $\mathcal C$ be a chain in $\mathcal B$, i.e. for any $A,B\in \mathcal C$, either $\mu(A\setminus B)=0$ or $\mu(B\setminus A)=0$. Let $x=\inf \{ \mu(A) : A\in\mathcal C \}$, then for any $n\in\mathbb N$, there is $A_n\in \mathcal C$ such that $\mu(A_n) < x+\frac1n$ and we can suppose WLOG that $\{ A_n \}_{n\mathbb N}$ is a chain. Let $A=\bigcap_{n\in\mathbb N} A_n$, then $\mu(A\setminus A_n)=0$ for any $n$ and for all $B\in\mathcal A$, $\mu(B\setminus A)=\lim_{n\to\infty} \mu(B\setminus A_n)=0$ by countable additivity, therefore $A\in\mathcal B$ and $A$ is a lower bound on $\mathcal C$, by Zorn's Lemma, $\mathcal B$ has a minimal element $B$. Let $C$ be another minimal element of $\mathcal B$, then $B\cap C$ is also a minimal element with $\mu((B\cap C)\setminus B)=\mu(\emptyset)=0$, therefore by minimality of $B$, $\mu(B\,\triangle\, (B\cap C))=0$ but $B\,\triangle\,(B\cap C)=B\setminus C$ so that $[B]\leq [C]$ so that $B$ is actually the unique minimal element of $B$ and is therefore the supremum of $\mathcal A$.
Does that sound correct ?