It seems from your presentation like first agent can only hold one of E1 or E2, etc., where E1, E2, etc. are disjoint subsets of the sample space. But the same result holds if the agent can hold the state of knowledge that is a union of such sets, so the first agent could believe E1+E2, etc. There seems to be no reason to not make this generalization. IIRC, that's how Aumann's paper was describing it.
IIRC, that's how Aumann's paper was describing it.
No, not as I read it.
My C & E_i correspond to blocks in the partition that Aumann assigns to agent 1. Aumann has agent 1 evaluating the posterior probability of A in world ω by computing p(A | P₁(ω)), where P₁(ω) is the element of 1's partition containing ω. So the agent will never condition on a union P₁(ω₁) ∪ P₁(ω₂) of distinct blocks. (Here I'm following Aumann's notation as closely as possible.) Correspondingly, my agent never conditions on a disjunction [(C & E₁) ∨ (C & E₂)].
But I ...
I've written up a 2-page explanation and proof of Aumann's agreement theorem. Here is a direct link to the pdf via Dropbox.
The proof in Aumann's original paper is already very short and accessible. (Wei Dai gave an exposition closely following Aumann's in this post.) My intention here was to make the proof even more accessible by putting it in elementary Bayesian terms, stripping out the talk of meets and joins in partition posets. (Just to be clear, the proof is just a reformulation of Aumann's and not in any way original.)
I will appreciate any suggestions for improvements.
Update: I've added an abstract and made one of the conditions in the formal description of "common knowledge" explicit in the informal description.
Update: Here is a direct link to the pdf via Dropbox (ht to Vladimir Nesov).
Update: In this comment, I explain why the definition of "common knowledge" in the write-up is the same as Aumann's.
Update 2020-05-23: I fixed the Dropbox link and removed the Scribd link.