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Let $\mu$ be a map from Lebesgue measurable sets to $[0, \infty]$ such that $\mu(\emptyset) = 0$, $\mu$ obeys countable additivity, and $\mu$ is translation invariant (that is $\mu(E + x) = \mu(E)$ for any $x \in \mathbb{R}^{d}$). Then must $\mu$ be a multiple of the standard Lebesgue measure?

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This is not true. $\mu$ might be the counting measure, i.e. $\mu(E)=|E|$ if $E$ is finite and $\mu(E)=\infty$ otherwise.

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  • $\begingroup$ How about if we also assume $\mu$ is a measure on Lebesgue set and we have $\mu[0,1]<\infty$ in addition to translation invariant property. Then, is it true that $\mu$ is a multiple of Lebesgue measure? $\endgroup$ – seriously divergent Feb 6 '14 at 9:38

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