Flows, coalescence and noise. A correction

Abstract

Counterexample to Remark 1.7 in [1]. Let φ be a random variable in F (i.e. φ is a random measurable mapping on a compact metric spaceM) of law Q such thatM×Ω 3 (x, ω) 7→ φ(x, ω) ∈ M is measurable. Suppose that Q is regular and let J be a regular presentation of Q. Let X be a random variable in M independent of φ. Out of φ and X, define ψ ∈ F by ψ(x) = φ(x) is x 6= X and ψ(x) = X is x = X. Then M × Ω 3 (x, ω) 7→ ψ(x, ω) ∈ M is measurable. Suppose also that the law of X has no atoms, then (reminding the definition of F) ψ and X are independent and the law of ψ is Q. Note that ψ(X) = X and (except for very special cases) we won’t have that a.s. J (ψ)(X) = ψ(X) = X.