Let \(S\) be any non-empty set. If you have a function \(f\colon \mathbb{R} \to S\) and you tell me its values on an interval \((-\infty,t)\), can I predict which value it will have at the time \(t\)? If the function is continuous, then of course I can; and in general not. Now interestingly, if you ask not me but Hardin and Taylor, then they can predict \(f(t)\) for almost all \(t\) knowing only the values of \(f\) before the time \(t\)!

Let me make their result explicit. Denote by \(S^\mathbb{R}_\bullet\) the set of all functions \(f\colon (-\infty,t_f) \to S\); note that the domain \((-\infty,t_f)\) depends on the function \(f\). We call any mapping \(P\colon S^\mathbb{R}_\bullet \to S\) an \(S\)-predictor: if we throw any function \(f\colon (-\infty,t_f) \to S\) into \(P\), then the result \(P(f)\) is the attempt to predict which value \(f\) will take at the time \(t_f \in \mathbb{R}\).

We can now ask for which functions \(F\colon \mathbb{R} \to S\) and which times \(t \in \mathbb{R}\) the equality \[P(F|_{(-\infty,t)}) = F(t)\] hold true. The result of Hardin and Taylor is that for every set \(S\) there exists an \(S\)-predictor for which this equality holds true for all functions \(F\colon \mathbb{R} \to S\) for almost all times \(t \in \mathbb{R}\) (where the measure-zero set of `bad’ predictions depends on the function \(F\)).

Reference: Hardin-Taylor, A Peculiar Connection between the Axiom of Choice and Predicting the Future, The American Mathematical Monthly vol. 115 (2008).