This is an attempt at an infinitary version of my previous post on finite strategies

Let \(S\) be some set with a sigma-algebra on it such that points are measurable. Let \(\mathcal{M}\) be be the set of measurable subsets. A strategy is a function s from \(\mathcal{M}\) to the set of probability measures on \((S, \mathcal{M})\) such that \(s(A)(A) = 1\).

A strategy s is subset consistent if whenever \(U, V, W\) are measureable sets with \(U \subseteq V \subseteq W\) we have \(s(W)(U) = S(W)(V) S(V)(U)\).

It’s reasonably easy to see this reduces to the previous version if \(S\) is finite and the sigma-algebra is \(P(S)\). While random variables are replaced by measures, there’s not really much difference between them.

Now fix s as a subset consistent strategy.

We can convert this to the version we had before to get the relations \(\prec\) and \(\sim\) on \(S\). The finiteness of \(S\) was not used to derive any of the properties of these.

Further when we pass to the quotient we get a totally ordered set \(T\) and a partition \(\{L_t : t \in T\}\) such that if \(x \in L_s, y \in L_t\) then \(x \prec y\) iff \(s < t\). Everything we did after that point was reliant on the finiteness of \(S\), so we'll have to try harder here. Theorem: \(T\) is the reverse of a well-ordered relation. i.e. every set has a maximal element. Proof: Note that we need only show that every countable set has a maximal element, as given a set with no maximal element we can construct a strictly increasing sequence in it \(x_1 < x_2 < \ldots\), which would be a countable subset with no maximal element. Suppose now we have some countable subset \(A \subseteq T\). Pick \(U \subseteq S\) such that \(U\) chooses an element from each of \(\{L_a : a \in A\}\). Then \(s(U)(\{x\}) \neq 0\) for some \(x \in U\), as otherwise \(s(U)(U) = 0\) (because \(U\) is countable). Then it can't be the case that \(x \prec y\) for any \(y \in U\). Let \(x \in L_t\). Then \(t\) must be a maximal element of \(A\). QED Theorem: \(L_t\) is countable. Proof: First note that for any countable \(A \subseteq L_t\) it must be the case that for all \(x \in L_t\), \(s(A)(\{x\}) > 0\), for if not then they all must be \(0\) (because otherwise any zero ones would be \(\prec\) the non-zero ones), and thus we would have \(s(A)(A) = 0\) because of countable additivity.

Now suppose \(L_t\) is uncountable. We can find at least \(\aleph_1\) distinct elements in it. Let \((x_a : a < \omega_1)\) be a sequence of length \(\omega_1\) consisting of distinct elements in \(L_t\).
Let \(X_a = \{x_b : b \leq a\}\) and consider the sequence \(p_a = s(X_a)(\{x_0\})\).
Claim: If \(c < d\) then \(p_c > p_d\).

Proof: By subset compatibility, \(p_d = s(X_d)(X_c) s(X_c)(\{x_0\}) \leq (1 – s(X_d)(\{x_d\})) p_c < p_c\)
So we have constructed a strictly decreasing sequence of real numbers of length \(\omega_1\). But this is impossible, for standard set theoretic reasons.
QED
This gives us a theorem very akin to our finitary version:
Theorem: Let \(s\) be a subset consistent strategy on S. There is a reverse well ordered set \(T\), a partition \(\{L_t : t \in T\}\) and weight function \(w : S : \to (0, \infty)\) such that for countable \(U \subseteq S\), \(s(U)\) is determined as follows:
Let \(L_t\) be the largest \(t\) such that \(U' L_t \cap U \neq \emptyset\). Then \(s(U)(V) = \frac{\sum\limits_{x \in U' \cap V} w(x)}{\sum\limits_{x in U'} w(x)}\).
Proof:
We've essentially done all the work for this. Define \(w(x) = s(L_t)(\{x\})\) where \(x \in L_t\). We need the countability of \(U\) only to get that \(s(U)(V) = \sum\limits_{x \in V} s(U)(\{x\})\), after which everything proceeds as in the finite case.
QED
So the behaviour for countable sets is extremely analagous to the finite case. How do things work for uncountable sets? I currently have no idea at all. I don't currently have any examples where the behaviour for uncountable sets is different, but I'd be moderately surprised if there weren't some. If anything, I'd expect it to be possible to build an example out of any probability measure.

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