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.
Best Answer
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:
$$\varphi((b_{d})_{d\in D})\leq\bigwedge_{d\in D}\bigvee_{e\geq d}b_{e}$$ for each tuple $(b_{d})_{d\in D}\in B^{D}$.
$$\bigvee_{d\in D}\bigwedge_{e\geq d}b_{e}\leq\varphi((b_{d})_{d\in D})$$ for each tuple $(b_{d})_{d\in D}\in B^{D}$.
$$\varphi(x)=0$$ whenever $x\in I$, and $\varphi((b)_{d\in D})=b$.
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.