I at first didn’t understand your argument for claim (2), so I wrote an alternate proof that’s a bit more obvious/careful. I now see why it works, but I’ll give my version below for anyone interested. In any case, what you really mean is the probability of deciding a sentence outside of Φby having it announced by nature; there may be a high probability of sentences being decided indirectly via sentences in Φ.
Instead of choosing Φ as you describe, pick Φ so that the probability μ(Φ) of sampling something in Φ is greater than 1−μ(ψ)⋅ε/2. Then, the probability of sampling something in Φ−{ψ} is at least 1−μ(ψ)⋅(1+ε/2). Hence, no matter what sentences have been decided already, the probability that repeatedly sampling from μ selects ψ before it selects any sentence outside of Φ is at least
Furthermore, this argument makes it clear that the probability distribution we converge to depends only on the set of sentences which the environment will eventually assert, not on their ordering!
Oh, I didn’t notice that aspect of things. That’s pretty cool.
I at first didn’t understand your argument for claim (2), so I wrote an alternate proof that’s a bit more obvious/careful. I now see why it works, but I’ll give my version below for anyone interested. In any case, what you really mean is the probability of deciding a sentence outside of Φ by having it announced by nature; there may be a high probability of sentences being decided indirectly via sentences in Φ.
Instead of choosing Φ as you describe, pick Φ so that the probability μ(Φ) of sampling something in Φ is greater than 1−μ(ψ)⋅ε/2. Then, the probability of sampling something in Φ−{ψ} is at least 1−μ(ψ)⋅(1+ε/2). Hence, no matter what sentences have been decided already, the probability that repeatedly sampling from μ selects ψ before it selects any sentence outside of Φ is at least
∞∑k=0(1−μ(ψ)⋅(1+ε/2))k⋅μ(ψ)=μ(ψ)μ(ψ)⋅(1+ε/2)>1−ε/2
as desired.
Oh, I didn’t notice that aspect of things. That’s pretty cool.