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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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This is called Banach measure. –  Victor Protsak Jun 14 '10 at 11:15

3 Answers 3

up vote 4 down vote accepted

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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A very detailed description is given in Chapter 10 of Wagon's book: The Banach-Tarski paradox. –  Péter Komjáth Jun 13 '10 at 5:24

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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You need some additional restrictions to make this problem interesting. The standard counting measure is translation invariant and actually countably additive. This is of course close to cheating. What one cannot have is a countably additive measure on the real line that takes intervals to their length.

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But there is the condition $\mu([0,1]) = 1$ in my post. –  AgCl Jun 14 '10 at 7:30

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