[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 
>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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