Idea for proof of "convexity" for generic probabilities [efr-9LPI]

Consider the two ways of sampling one of three values:

1. Sample t \in [0,1] uniformly, decide t < \lambda , \lambda < t < \lambda + \rho - \lambda \rho , \lambda + \rho - \lambda \rho < t
2. Sample t_1, t_2 uniformly, decide t_1 < \lambda (if yes return a), else decide t_2 < \rho and return b or c

The latter corresponds to the formal convex combination a +_\lambda (b +_\rho c), whereas the former is an "unbiased" representation of the same thing. It seems clear that if we can prove their equivalence, we can prove the equivalence with (a + b) + c (whatever coefficients) also.

Now for dyadic rationals q,r, let f^{q,r} : 2^\mathbb {N} \to 3_\bot be the function parameterizing the unbiased choice, and let g^{q,r}: (2^\mathbb {N})^2 \to 3_\bot parameterize the "biased" choice. Note both of these are actually uniformly continuous and can be shown by finitary means to give the same distribution. Moreover we can construct a permutation \sigma ^{q,r} : 2^\mathbb {N} \to (2^\mathbb {N})^2 which witnesses this identity of probability.

Finally note that for generic reals, in both cases, we can write f^{\lambda ,rho}(s) = \sup _{q<\lambda , r<\rho }f^{q,r}(s) (choosing the ordering a > b > c on our choice set), and similarly for g. (This is because in both cases increasing \lambda or \rho can only move your choice c \to b or b \to a, not the other direction). Moreover (this is the nonobvious part) we can hopefully prove that there is a limiting permutation \sigma _{\lambda ,\rho }, which witnesses the identity between these limiting distributions (in the usual way, because we are with probability one in the fragment determined by some specific q,r, hence the two maps are equal because equal at that point).

We may be able to arrange it so we can just take the supremum (or maybe infimum) of the maps (this will only land in \Omega ^\mathbb {N}, but we can try to show that with probability one we actually land in 2^\mathbb {N})