Invariant contents and paradoxical sets #
Let a group G act on X, and let m be a finitely additive set function on all subsets of X,
m : MeasureTheory.AddContent M (Set.univ : Set (Set X)), which is invariant under the action:
m (g • s) = m s. Cutting a set into finitely many pieces and moving each piece by an element of
G preserves the total content, so equidecomposable sets have the same content
(TauCeti.Equidecomposable.addContent_eq).
Consequently a paradoxical set S, which splits into two disjoint sets each equidecomposable with
S, satisfies m S + m S = m S. For an ℝ≥0∞-valued content this forces m S to be zero or
infinite. In particular a group action admitting an invariant finitely additive probability on all
subsets of X is not paradoxical on all of X. This is the easy half of Tarski's
characterization of amenability by the absence of paradoxical decompositions.
Main results #
TauCeti.Equidecomposable.addContent_eq: an invariant content takes equal values on equidecomposable sets.TauCeti.Paradoxical.addContent_add_self: an invariant content of a paradoxical setSsatisfiesm S + m S = m S.TauCeti.Paradoxical.addContent_eq_zero: a finite invariantℝ≥0∞-valued content vanishes on paradoxical sets.TauCeti.not_paradoxical_univ_of_addContent: ifXcarries an invariant finitely additive probability on all subsets, thenXis not paradoxical.
References #
- A. Garrido, An introduction to amenable groups, §1.
A content on all subsets which is invariant under the action of G takes equal values on sets
which are equidecomposable using G.
An invariant content of a paradoxical set S satisfies m S + m S = m S: the set splits into
two disjoint sets, each with the same content as S.
A finite invariant ℝ≥0∞-valued content on all subsets vanishes on every paradoxical set.
If X carries a finitely additive probability on all subsets which is invariant under the
action of G, then X is not paradoxical for this action.