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:

  1. 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 $A$?
  2. 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?
  3. 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 $P$?
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    $\begingroup$ There is a countably generated sigma-algebra on $\omega_1$ which admits no diffused probability measure. You can have a copy of this measure inside any set so that the answer to Q2 is no. The statement of Q3 is strange since you can always extend any measure to any disjoint family of sets so the answer is no. A more interesting formulation of such problems would require that the base measure be Radon like Lebesgue measure on $2^{\omega}$ - For example, see section 8 here: www1.essex.ac.uk/maths/people/fremlin/rvmc.pdf $\endgroup$ – Ashutosh May 13 '18 at 17:45
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    $\begingroup$ @Ashutosh I would encourage you to post your comment as an answer, with a fuller explanation. $\endgroup$ – Joel David Hamkins May 14 '18 at 2:18
  • $\begingroup$ Every infinite sigma algebra $\mathcal{A}$ contains an infinite disjoint family so the answer to Q1 is trivially yes. $\endgroup$ – Ashutosh May 15 '18 at 0:12
  • $\begingroup$ @Ashutosh I want to include the generators of the algebra, which are not necessarily disjoint. $\endgroup$ – Zoorado May 15 '18 at 1:03
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    $\begingroup$ @Ashutosh While you can have a copy of a $\sigma$-algebra on $\omega_1$ not admitting a diffuse probability measure inside every uncountable set, the copy of $\omega_1$ need not have positive measure. Can you explain how to finish the argument? $\endgroup$ – Michael Greinecker May 16 '18 at 5:19

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.

  • $\begingroup$ Thanks for the answer. Unfortunately, the argument in the first paragraph went over my head somehow. I am struggling with the following two points: 1. How can to choose a countable set that generates the graph of the well ordering of $\omega_1$, and 2. How to end the argument assuming the graph is measurable. Is there a reference for this so I can read in more detail? $\endgroup$ – Zoorado May 22 '18 at 1:25
  • $\begingroup$ I added more details and a reference to Rao's result. $\endgroup$ – Ashutosh May 22 '18 at 9:53
  • $\begingroup$ Ah, it makes perfect sense to me now! Thanks again! $\endgroup$ – Zoorado May 22 '18 at 17:36

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