2
$\begingroup$

Let C denote the Lie group (ℝ/ℤ, +), and let ℤn denote the subgroup of C generated by [1/n].

Let (X,n) be an ordered pair where X ⊂ C is an arbitrary subset, and n ∊ ℤ+, such that that the set of images {g+X | g ∊ ℤn} of X by the action of the group ℤn forms a partition of C.

Call such an X an "1/n equal part" of C.

We now try to create a finitely additive measure 𝜇 "from the top down" on C by assigning to each 1/n equal part X of C the value 𝜇(X) = 1/n.

Assume that, for all positive integers n, the assignment 𝜇(X) = 1/n is made for all (X,n) such that X is a 1/n equal part of C.

Questions:

  1. Does 𝜇 extend to a finitely additive measure on an algebra of sets containing all such X ?

  2. What about a 𝜎-additive measure on a 𝜎-algebra of sets containing all such X ?

  3. And if this is possible, can such a 𝜇 extend ordinary Lebesgue measure?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

There is a finitely additive, translation invariant measure on the entire power set of $\mathbf C$ (see, for example, Amenability, the ping-pong lemma, and the Banach–Tarski paradox). Clearly, it satisfies your desired condition (1). Since rational-length, half-open ‘intervals’ in $\mathbf C$ are finite, disjoint unions of $1/n$-equal parts that are intervals for various $n$, and generate the Borel $\sigma$-algebra, and since Lebesgue (I would rather say Haar) measure assigns measure $1/n$ to a $1/n$-equal part that is an interval, we have that any countably additive extension as in (2) extends Haar measure. I suspect that a Vitali-type argument shows that no extension as in (2) exists, but I do not currently see it.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .