On the failure of extending a probability measure on uncountable $\Omega$ It is a well known fact that if $(\Omega, \mathcal{F}, P)$ is a probability triple and $\{A_i : i < k\}$ is a finite collection subsets of $\Omega$, then there is a $P' \supset P$ and $\mathcal{F'} \supset (\mathcal{F} \cup \{A_i : i < k\})$ such that $(\Omega, \mathcal{F'}, P')$ is a probability triple. In this case, let's say that $P$ can be extended to include $\{A_i : i < k\}$, or that $\{A_i : i < k\}$ can be adjoined to $P$.
I would like to know what is the best, and the worst, we can say when we replace the word "finite" in the statement above with "countable". More precisely, I am asking the following:


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*What is the biggest class of probability triples and countable collections (that has been proven in $ZFC$) for which the statement above holds? For example, is it true that if $A \subset \mathscr{P}(\Omega)$ is countable, and the complete subalgebra of $(\mathscr{P}(\Omega), \subset)$ generated by $A$ includes no dense linear order, then any probability measure can be extended to include $A$?

*Modulo large cardinal assumptions, what is the best case scenario consistent with $ZFC$? For example, is it consistent that we can extend any probability measure to include any countable collection of subsets of the sample space? 

*Modulo large cardinal assumptions, what is the worst case scenario consistent with $ZFC$? For example, is it consistent that for each uncountable cardinal $\kappa$ there is a probability triple $(\kappa, \mathcal{F}, P)$ such that no countable $S \subset \mathscr{P}(\kappa)$ can be adjoined to $P$?

 A: Claim: Let $\mathcal{F}$ consist of all countable and co-countable subsets of $\omega_1$ and $m: \mathcal{F} \to \{0, 1\}$ be defined by $m(X) = 0$ iff $X$ is countable. There is a countable family $\mathcal{A}$ of subsets of $\omega_1$ such that there is no probability measure defined on the sigma-algebra generated by $\mathcal{F} \cup \mathcal{A}$ that extends $m$.
Proof: Recall that the sigma-algebra generated by $\{A \times B: A, B \subseteq \omega_1\}$ is $\mathcal{P}(\omega_1 \times \omega_1)$ (B. V. Rao). Let $W = \{(\alpha, \beta): \alpha \leq \beta < \omega_1\}$. Since the sigma-algebra generated by any family is the union of the sigma-algebras generated by its countable subfamilies, we can choose a countable family $\mathcal{A} \subseteq \mathcal{P}(\omega_1)$ such that $W$ belongs to the sigma-algebra generated by $\{A \times B: A, B \in \mathcal{A}\}$. Let $\mathcal{G}$ be the sigma-algebra generated by $\mathcal{F} \cup \mathcal{A}$ and towards a contradiction, suppose $m':\mathcal{G} \to [0, 1]$ is a probability measure that extends $m$. As $W \in\mathcal{G} \otimes \mathcal{G}$, every horizontal section $W^{\beta} = \{\alpha: (\alpha, \beta) \in W\}$ of $W$ has $m'$-measure zero and every vertical section $W_{\alpha} = \{\beta: (\alpha, \beta) \in W \}$ of $W$ has $m'$-measure one, we get a contradiction by Fubini's theorem.
Answer 1: Every infinite sigma algebra $\mathcal{F}$ contains an infinite disjoint family. It follows that $(\mathbb{R}, \leq)$ embeds into $(\mathcal{F}, \subseteq)$. This makes Question 1 trivial.
Answer 2. By the first paragraph, the answer is no. We can assume that $\omega_1 \subseteq X$. Choose $(X, \mathcal{F}, m)$ such that $\omega_1 \in \mathcal{F}$, $m(\omega_1) = 1$ and $m \upharpoonright \omega_1$ is the countable cocountable measure.
Answer 3. Since any measure can be extended to the sigma-algebra generated by a disjoint family of sets, the answer is no.
So to avoid trivialities, we must put restrictions on the base measure. With this in mind, let me mention the following.
(1) Carlson showed that in the random real model, the Lebesgue measure on $[0, 1]$ can be extended to any countably generated sigma-algebra containing the Borel algebra.
(2) Section 8 in Fremlin's "Real valued measurable cardinals" addresses similar issues when the underlying measure space is Radon.
