# Is Lebesgue measure the only measure that is locally finite, strictly positive and translation-invariant on $\mathbb{R}^n$?

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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No.

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 at 4:33