[FOM] Query concerning measure.
joeshipman@aol.com
joeshipman at aol.com
Sun Feb 19 09:13:18 EST 2006
Taylor asks:
>Is it consistent with ZFC that there is a finitely additive set
function
>on ALL subsets of [0,1) that agrees with length for intervals,
>and is translation (i.e. rotation) invariant?
Not only is it consistent, it is provable! In fact, for any n, there is
a FINITELY additive measure on all subsets of [0,1]^n, or on R^n if you
allow +infinity as a value, that extends Lebesgue measure and is
TRANSLATION invariant. This follows from the Hahn-Banach theorem (which
requires the Axiom of Choice).
In dimensions higher than 2, you can't have it be ROTATION invariant
because of the Banach-Tarski paradox. In dimensions 1 and 2, you can
have it be both translation and rotation invariant.
The key difference is that the symmetry group is nonabelian in
dimensions 3 and higher. (Technically, it is not only not abelian, but
not "amenable", a property equivalent to "not having a paradoxical
decomposition".) This allows you to embed, for example, the free
product of Z/2 and Z/3 into the rotation group; this subgroup is
generated by two rotations, and has a Banach-Tarski-like property that
it properly contains two of its own cosets, which allows you to
construct the Hausdorff paradoxical sphere decomposition by taking
equivalence classes of points on the sphere based on whether they are
related by a rotation in one of the cosets, choosing one point from
each equivalence class, and fiddling around to deal with countably many
exceptional points.
The Hahn-Banach argument doesn't work in R^3 when you allow rotations
because you no longer have a correspondence between vectors and
transformations on the vector space -- as long as you are only allowing
translations you can identify the points in R^n with the translations,
so the space is acting on itself and you can apply the theorem.
-- JS
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