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Lebesgue measure on Euclidean space $\mathbb{R}^n$ is locally finite, strictly positive and translation-invariant.

Is Lebesgue measure the only such measure on $\mathbb{R}^n$? Thanks!

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up vote 7 down vote accepted


The trouble is that strict positivity and local finiteness only concern open and compact sets. Adding translation-invariance will yield a unique measure on the Borel sets (up to a positive multiplicative constant). However, there are many translation-invariant $\sigma$-algebras $\Sigma$ between the Borel sets and the Lebesgue measurable sets, and restriction of Lebesgue measure to $\Sigma$ will yield a measure different from Lebesgue measure. Such a restriction still is locally finite and strictly positive and translation invariant.

Moreover, there are are strict extensions of Lebesgue measure which are translation invariant: see Extension of the Lebesgue measurable sets. Again, these extensions are strictly positive and locally finite.

Thus, the three properties you mention are not enough to ensure uniqueness. Either you need to add regularity and completeness or require the measure to be defined exactly on the Borel sets in order to obtain the uniqueness statement up to a multiplicative constant also mentioned in the other answer.

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It's good know that somebody thought this through. Thanks for clearing that up. – Christopher A. Wong Feb 7 '13 at 4:33

Up to a multiplicative constant, yes.

See here : Haar Measure.

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Note that the definition of Haar measure requires some form of regularity, which should be afforded by the "strictly positive" condition; at least, I think. – Christopher A. Wong Feb 7 '13 at 4:22
@ChristopherA.Wong: Do you mean "strictly positive" implies "some form of regularity"? – Ethan Feb 7 '13 at 4:27
Probably that any point in the measure space is contained in a set of positive/finite measure. – Taylor Martin Feb 7 '13 at 4:28
@Wong: You are right; I didn't check that. Since the measure is locally finite, some open neighborhood around zero has finite measure; hence by translation all the open neighborhoods has finite measure. All closed neighborhoods too have finite measure. This should perhaps be enough for regularity. – Severus Snape Feb 7 '13 at 4:31

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