A Measure Theoretic Paradox from a continuous colouring rule

preprint2022arXivOpen access

Abstract

Given a probability space (X,B,m)(X, {\cal B}, m), measure preserving transformations g1,…,gkg_1, \dots , g_k of XX, and a colour set CC, a colouring rule is a way to colour the space with CC such that the colours allowed for apoint xx are determined by that point's location and the colours of the finitely g1(x),…,gk(x)g_1 (x), \dots , g_k(x) with gi(x)≠xg_i(x) \not= x for all ii and almost all xx. We represent a colouring rule as a correspondence FF defined on X×CkX\times C^k with values in CC. A function f:X→Cf: X\rightarrow C satisfies the rule at xx if f(x)∈F(x,f(g1x),…,f(gkx))f(x) \in F( x, f(g_1 x), \dots , f(g_k x)). A colouring rule is paradoxical if it can be satisfied in some way almost everywhere with respect to mm, but not in {\bf any} way that is measurable with respect to a finitely additive measure that extends the probability measure mm defined on B{\cal B} and for which the finitely many transformations g1,…,gkg_1, \dots , g_k remain measure preserving. Can a colouring rule be paradoxical if both XX and the colour set CC are convex and compact sets and the colouring rule says if c:X→Cc: X\rightarrow C is the colouring function then the colour c(x)c(x) must lie (mm a.e.) in F(x,c(g1(x)),…,c(gk(x)))F(x, c(g_1(x) ), \dots , c(g_k(x))) for a non-empty upper-semi-continuous convex-valued correspondence FF defined on X×CkX\times C^k? The answer is yes, and we present such an example. We show that this result is robust, including that any colouring that approximates the correspondence by εε for small enough positive εε also cannot be measurable in the same finitely additive way. Because non-empty upper-semi-continuous convex-valued correspondences on Euclidean space can be approximated by continuous functions, there are paradoxical colouring rules that are defined by continuous functions.

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