Let $\mathbb{B}$ be an infinite $\sigma$-complete Boolean algebra. By $\mathbb{B}^\omega$ we denote the countable product of $\mathbb{B}$ with the coordinate-wise operations. Let us call a homomorphism $\varphi\colon\mathbb{B}^\omega\to\mathbb{B}$ *generalized limit* if for every sequence $(A\_n)\in\mathbb{B}^\omega$ we have:


$\bigvee\_{n\in\omega}\bigwedge\_{k\ge n} A\_k\le\varphi(A\_n)\le\bigwedge\_{n\in\omega}\bigvee\_{k\ge n}A\_k$.


(I guess it's immediately visible, why $\varphi$ is called *limit*.)


Are you aware of some standard way of constructing in ZFC such generalized limits? I am especially interested in the case of $\mathbb{B}$ being a measure algebra, e.g. $\mathbb{B}=\mathrm{Bor}([0,1])/\mathcal{N}$ (here $\mathcal{N}$ stands for the Lebesgue null ideal), but of course any general study will be useful.


I will appreciate any answer, hint, or reference to a research paper.



Generalized limits in Boolean algebras
If the Boolean algebra $B$ is complete, then I claim that there exists generalized limits. I am going to generalize this answer from limits indexed by $\omega$ to a limit of a net indexed by an arbitrary directed set.


Let $B$ be a complete Boolean algebra. Suppose that $D$ is a directed set. Then
let $I$ be the ideal on $B^{D}$ consisting of all $(b\_{d})\_{d\in D}$ where there exists some $e\in D$ where $b\_{d}=0$ for all $d\geq e.$


Proposition: Suppose that $\varphi:B^{D}\rightarrow B$ is a Boolean algebra homomorphism. Then the following are equivalent:


1. $$\varphi((b\_{d})\_{d\in D})\leq\bigwedge\_{d\in D}\bigvee\_{e\geq d}b\_{e}$$ for each tuple $(b\_{d})\_{d\in D}\in B^{D}$.
2. $$\bigvee\_{d\in D}\bigwedge\_{e\geq d}b\_{e}\leq\varphi((b\_{d})\_{d\in D})$$ for each tuple $(b\_{d})\_{d\in D}\in B^{D}$.
3. $$\varphi(x)=0$$ whenever $x\in I$, and $\varphi((b)\_{d\in D})=b$.
4. There is a Boolean algebra homomorphism $f:B^{D}/I\rightarrow B$ such that $f((b)\_{d\in D}+I)=b$ for all $b\in B$ and where if $\pi\_{I}:B^{D}\rightarrow B^{D}/I$ is the quotient mapping, then
$\varphi=f\pi\_{I}.$


Proof outline: The equivalence between 1 and 2 follows by taking complementation. The equivalence between 3 and 4 follows from the first isomorphism theorem applied to Boolean algebras.


$(1\wedge 2)\rightarrow 3$. If $(b\_{d})\_{d\in D}\in I$, then there is some $d\in D$ where $b\_{e}=0$ for $e\geq d$. Therefore,
$$\varphi((b\_{d})\_{d\in D})\leq\bigvee\_{e\geq d}b\_{e}=0.$$
Furthermore, we have $$\varphi((b)\_{d\in D})\leq\bigwedge\_{d\in D}\bigvee\_{e\geq d}b=b\leq\bigvee\_{d\in D}\bigwedge\_{e\geq d}b\leq\varphi((b)\_{d\in D}),$$ so
$\varphi((b)\_{d\in D})=b.$


$(3\wedge 4)\rightarrow 1.$ Let $\varphi,\pi\_{I},f$ be the mappings stated in the proposition where $\varphi=f\pi\_{I}$.


Let $(b\_{d})\_{d\in D}\in B^{D}$, and suppose that $d\_{0}\in D$. Let $b=\bigvee\_{e\geq d\_{0}}b\_{e}$. Then


$\pi\_{I}(b\_{d})\_{d\in D}\leq\pi\_{I}((b)\_{d\in D})$. Therefore,
$$\varphi(b\_{d})\_{d\in D}=f\pi\_{I}(b\_{d})\_{d\in D}\leq f\pi\_{I}((b)\_{d\in D})
=\varphi((b)\_{d\in D})=b=\bigvee\_{e\geq d\_{0}}b\_{e}.$$


We conclude that
$$\varphi(b\_{d})\_{d\in D}\leq\bigwedge\_{d\_{0}\in D}\bigvee\_{e\geq d\_{0}}b\_{e}.$$


Q.E.D


The mappings $f:B^{D}/I\rightarrow B$ such that $f((b)\_{d\in D}+I)=b$ for all $b\in B$ can be obtained using Sikorski's extension theorem, so just let $\varphi=f\pi\_{I}$ to obtain your generalized limits.


