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We know that a countably additive translation invariant measure with $\mu([0,1]) = 1$ cannot be defined on the power set of $\mathbb R$. This is because $[0,1]$ can be partitioned into countably many congruent sets, with the help of the axiom of choice.

But I was wondering whether a finitely additive measure with these properties would be possible? I know it wouldn't be possible for dimension $n>2$ because of the Banach-Tarski paradox, but I am curious about $n=1$. If such a measure can be constructed on $\mathcal P(\mathbb R)$, would that be unique?

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    $\begingroup$ This is called Banach measure. $\endgroup$ Jun 14 '10 at 11:15
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Banach-Tarski poses a problem for existence of measures that are invariant under all rigid motions, not just translation. The existence of finitely additive translation-invariant measures that agree with Lebesgue measure on Lebesgue-measurable sets is a consequence of the Hahn-Banach theorem. This is exercise 21 in chapter 10 of Royden's Real Analysis. The extensions given by Hahn-Banach don't seem to have any uniqueness properties, so I doubt this measure is unique.

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    $\begingroup$ A very detailed description is given in Chapter 10 of Wagon's book: The Banach-Tarski paradox. $\endgroup$ Jun 13 '10 at 5:24
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In n=1 it exists, but far from unique. This is in Hewitt and Stromberg. I believe they show there are 2^c different extension.

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The question should be stated in the following context: Let a subgroup $G$ of all isometries of $\mathbb{R}^n$ act on the space $\mathbb{R}^n$. When there is a finitely additive $G$-invariant measure $\mu$ defined on all subsets of $\mathbb{R}^n$ and normalizing the unit cube?

It appears that such a measure exists in $\mathbb{R}^n$ if the group $G$ is amenable (this works also in higher dimensions in spite of the Banach-Tarski paradox). This follows from a general theorem of Mycielski on the measures in Boolean algebras (as mentioned above there is also an alternative proof via the Hahn-Banach theorem). The measures $\mu$ constructed in this way are not unique. However, there is a modification of a theorem proved by Tarski which says that there is a set function $f$ (called Tarski's absolute measure) defined on a subclass $\mathcal{A}$ of all bounded subsets of $\mathbb{R}$ which attains the values common for all the measures constructed by Mycielski (that is the class $\mathcal{A}$ is defined as $\mathcal{A} = \{X \in \mathbb{R}^n: \mu_1(X) = \mu_2(X)$, where $\mu_1$ and $\mu_2$ are any two measures obtained via the Mycielski's theorem $\}$.

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