Consider a measurable space $(X,\mathcal{A})$. Let $\mathcal{M}$ denote the family of all countably additive measures $\mu\colon \mathcal{A}\to [0,+\infty]$. This family can be made into a partially ordered set by setting $\mu\leq \nu$ iff $\mu(V)\leq \nu(V)$ for every $V\in \mathcal{A}$. It can be shown, see for instance https://arxiv.org/pdf/2104.06753v1.pdf, that $\mathcal{M}$ forms a complete lattice, which means that every $M\subseteq \mathcal{M}$, possibly uncountable, has the lowest upper bound $\bigvee\limits_{\mu\in M}\mu$, i.e., the lowest measure $\mu_0\in \mathcal{M}$ with the property that $\mu\leq \mu_0$ for each $\mu\in M$, and the greatest lower bound, which is defined similarly.
My questions are as follows: given $M\subseteq \mathcal{M}$, can I found a countable subfamily $M_0\subseteq M$ such that $\bigvee\limits_{\mu\in M}\mu=\bigvee\limits_{\mu\in M_0}\mu$; if not, does this property hold under some additional assumptions on $\mathcal{A}$, for instance, if $\mathcal{A}$ is countably generated, i.e., there exists a countable family $\mathcal{E}\subseteq \mathcal{A}$ such that $\mathcal{A}=\sigma(\mathcal{E})$, where the latter denotes the smallest sigma-algebra containing $\mathcal{E}$.
Let me provide some of my thoughts on this and some additional information.
First, I give an explicit definition of the supremum of an arbitrary family $M\subseteq \mathcal{M}$. For every $V\in \mathcal{A}$, define
\begin{equation*}
\mu_0(V)=\sup \sum\limits_{n=1}^{+\infty} \mu_n(V_n),
\end{equation*}
where the supremum is taken over all pairs $(\{\mu_n\}_{n\in \mathbb{N}}, \{V_n\}_{n\in \mathbb{N}})$, where $\{\mu_n\}_{n\in \mathbb{N}}\subseteq M$ and $\{V_n\}_{n\in \mathbb{N}}$ is a partition of $V$ into $\mathcal{A}$-measurable sets. It turns that thus defined $\mu_0$ lies in $\mathcal{M}$ and is the desired supremum of $M$. From this definition, it is obvious that for every fixed (!) set $V\in \mathcal{A}$ we can find a countable family $M_0(V)\subseteq M$ such that
\begin{equation*}
\mu_0(V)=\bigg(\bigvee\limits_{\mu\in M_0(V)} \mu\bigg)(V_0).
\end{equation*}
This observation gave me a hope that the property from my question may hold for countably generated sigma-algebras.
The second thing I want to mention is that the desired property holds when restricted to those $M\subseteq \mathcal{M}$ all measures from which are absolutely continuous with respect to some fixed measure. Such a property is discussed in https://mathoverflow.net/questions/316651/uncountable-infimum-of-measurable-functions, at least for subset of Euclidean space equipped with the standard sigma-algebra and measure. As I understand, the same holds in general, so let me give some general details. Let $\lambda\in \mathcal{M}$, let me also assume that $\mathcal{A}$ is complete with respect to $\lambda$, let $F$ be a family of $\lambda$-measurable functions $X\to [0,+\infty]$. The lowest upper bound of $F$ can be defined as the $\lambda$-a.e. lowest $\lambda$-measurable function $f_0\colon X\to [0,+\infty]$ such that $f(x)\leq f_0(x)$ for $\lambda$-a.e. $x\in X$. Then, as I believe, it can be shown by the same arguments as in the mentioned discussion that there exists a countable family $F_0\subseteq F$ such that
\begin{equation*}
f_0(x)=\sup\limits_{f\in F_0}f(x)
\end{equation*}
for $\lambda$-a.e. $x\in X$. If this indeed holds, then the desired property also holds in this concrete case.
I will be grateful for any help.
Best Answer
This is not true in general. For instance, if $X$ is uncountable and $\mathcal{A}$ contains all the singletons (say, $X=\mathbb{R}$ and $\mathcal{A}$ is the Borel sets) then the supremum of the delta-measures $\delta_x$ at each $x\in X$ is counting measure but there is no countable subfamily with the same supremum.
It is true if the supremum is a $\sigma$-finite measure. To prove this, suppose $(\mu_i)_{i\in I}$ is a family of measures whose supremum $\mu$ is $\sigma$-finite. Partitioning $X$ into countably many $\mu$-finite sets and restricting our measures to each of those sets, we may assume that $\mu$ is actually finite. Fix a well-order $<$ on $I$ with no greatest element and let $\mu^i=\sup_{j<i}\mu_j$. Then $(\mu^i)$ is an increasing sequence of measures with supremum $\mu$. Let $S$ be the set of $i$ such that $\mu^{i+1}\neq\mu^i$. Then $\mu=\sup_{i\in S}\mu_i$ (to prove this, prove by induction on $i$ that $\mu^i=\sup_{j<i,j\in S}\mu_j$ for all $i$). Also, $S$ must be countable, since $\mu^{i+1}(X)>\mu^i(X)$ for all $i\in S$ and the values $\mu^{i+1}(X)$ are an increasing sequence of real numbers.