Explanation of Aumann's “agreeing to disagree” in modern notation

I'm attempting to understand Aumann's classic 1976 paper Agreeing to Disagree, which claims, under certain assumptions, that if two Bayesian agents share knowledge of each others' posteriors then they must always eventually agree on the same posterior.

Unfortunately, this paper has a tendency to make up its own notation as it goes along rather than using standard probability theory notation. This bad habit seems to have infected all the papers that reference it as well, with the result that I'm finding it really hard to work out what's going on.

My question is whether an explanation exists in more modern notation, i.e. where the result is described in terms of random variables and the relationships between them, rather than constantly reinventing the notion of a random variable using set theory notation.

As an example of what I mean, the original paper contains the sentence

Let $A$ be an event, and let $\mathbf{q}_i(.)$ denote the posterior probability $p(A \mid \mathscr{P}_i)$ of $A$ given $i$’s information: i.e., if $\omega \in \Omega$, then $\mathbf{q}_i(\omega) =p(A \cap P_i(\omega))/p(P_i(\omega)).$

Well firstly, if it's a conditional probability then why wouldn't you just always write it as $p(A\mid \mathscr{P}_i)$ rather than defining a new notation for no apparent reason, and secondly, if it's equal to $p(A\mid \mathscr{P}_i)$ then why is $\mathbf{q}_i$ a function of $\omega$ and not a function of $A$? The paper is full of this sort of thing, and while I can slowly work through it and convert these set theory level definitions into a more familiar probability theory notation, I'm hoping that someone else has already done this and presented a more readable explanation.

Shortly after posting the question, I came across exactly what I was looking for, in a blog post by Tyrell McAllister. The post contains a pdf of the explanation, with a link to a comment where he explains why his version corresponds to Aumann's. Since link-only answers are discouraged on SE, I post here my own even shorter version, further simplifying McAllister's summary. (But possibly over-simplifying it.)

Introduction: Suppose two agents initially share the same prior about some random variable $X$. They then go off and receive some different information about the world. That is, agent $\mathbf{A}$ learns the value of a random variable $A$ (that is in general correlated with $X$, according to the shared prior), and agent $\mathbf{B}$ learns the value of a different random variable $B$ (that is in general correlated with $X$ and $A$, according to the shared prior). I'll write $a$ for the value of $A$ learned by $\mathbf{A}$, and $b$ for the value of $B$ learned by $\mathbf{B}$.

Definition: the two agents' posteriors, $p(X=x\mid A=a)$ and $p(X=x\mid B=b)$, are common knowledge if there exists some information $c$ such that

1. $p(c\mid A=a)=1$ and $p(c\mid B=b) = 1.$ (i.e. both agents know $c$ with certainty, after learning their information.)

2. $p(X=x\mid c,\,A=a') = p(X=x\mid A=a)$ for all $a'$ in the domain of $A$. That is, knowing $c$ completely determines $\mathbf{A}$'s posterior for $x$, independently of any other information that $A$ might have learned.

3. $p(X=x\mid c,\,B=b') = p(X=x\mid B=b)$ for all $b'$ in the domain of $B$. (As above, but for $\mathbf{B}$.)

McAllister doesn't really spell this out, but I think the idea here is that $c$ is supposed to be the information that $\mathbf{A}$'s posterior for $X$ has its particular value, and that $\mathbf{B}$'s posterior has its particular value. Conditions 2 and 3 are what's required for $c$ to have this meaning: if $\mathbf{A}$ knows some information that asserts that its posterior has a particular value, then for consistency its posterior must indeed have that particular value, not matter what else $\mathbf{A}$ might know.

Claim: if the two agents' posteriors are common knowledge, then their posteriors are the same.

Proof: We have, as an identity in probability theory, that $$p(X=x\mid c) = \sum_{a'} p(X=x\mid c, A=a')p(A=a'\mid c),$$ where the sum is over the domain of $A$. But since every $p(X=x\mid c, A=a')=p(X=x\mid A=a)$ we have that $\mathbf{A}'s$ posterior $p(X=x\mid A=a)=p(X=x\mid c)$. By the same reasining, $p(X=x\mid B=b)$ is also equal to $p(X=x\mid c)$, completing the proof.

In this notation the result seems trivial given the definition, but I'm still struggling to see clearly how this definition of "common knowledge" corresponds to the intuitive description that Aumann gives, which is that not only do $\mathbf{A}$ and $\mathbf{B}$ know each others' posteriors, but they also know that each other know, know that they know that they know, etc. For this reason I won't accept my own answer and I'll hope someone else can offer a better explanation.