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What are all the translation invariant measures on $\mathbb{R}$?

Except Lebesgue measure on $\mathbb R$ I didn't find any translation invariant measure. So I put this question?

I know that if $\mu$ is a measure then $c \times \mu$ is again a measure where $c>0$.

t.b.
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users31526
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    Up to a multiplicative constant, Lebesgue measure is the only translation-invariant measure on the Borel sets that puts positive, finite measure on the unit interval. I don't have a reference at hand, though. – Michael Greinecker May 19 '12 at 17:00
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    This is a special case of Haar's theorem: http://en.wikipedia.org/wiki/Haar_measure. I suppose it might've been known for $\mathbb{R}$ earlier. – Kevin Ventullo May 19 '12 at 17:14
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    @Kevin: yes, this was known to Lebesgue already. He also asked explicitly whether it was possible to extend Lebesgue measure to the entire power set of $\mathbb{R}$ and whether such an extension was unique. This became known as Le problème de la mesure and influenced Banach's early work. The Banach-Tarski paradox is the most famous outgrowth of these investigations. – t.b. May 19 '12 at 19:00

4 Answers4

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Here is a way to argue out. I will let you fill in the details.

  1. If we let $\mu([0,1))=C$, then $\mu([0,1/n)) = C/n$, where $n \in \mathbb{Z}^+$. This follows from additivity and translation invariance.
  2. Now prove that if $(b-a) \in \mathbb{Q}^+$, then $\mu([a,b)) = C(b-a)$ using translation invariance and what you obtained from the previous result.
  3. Now use the monotonicity of the measure to get lower continuity of the measure for all intervals $[a,b)$.

Hence, $\mu([a,b)) = \mu([0,1]) \times(b-a)$.

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    If we want to generalize this result to every Borel set where $m$ is the Lebesgue measure and $\mu$ is the measure of Borel invariant by translations? Should I also start with the interval $[0,1]$? – Ilovemath Aug 23 '18 at 12:17
  • I don't see any way to get the equaltity $ \mu([a,b))=C(b-a)$ just using the result from first and translation invariance to get "b-a" outside as a facor. – hallo007 Nov 09 '23 at 02:26
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Let $\lambda$ be a translation-invariant measure on the Borel sets that puts positive and finite measure on the right-open unit interval $[0,1)$ then $\lambda$ is a positive multiple of Lebesgue measure. Here is an outline of the proof: Every Borel measure is determined by its behavior on finite intervals. By translation invariance, you know that a right-open interval of length $1/2^n$ has measure $1/2^n \lambda[0,1)$, since $2^n$ such pieces form a disjoint cover over $[0,1)$ and every such piece can be translated into every other other such piece. Now you can approximate every interval by such pieces to pin down the measure of each interval.

Michael Greinecker
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$\mathbb{R}$ is a locally compact group with respect to addition and the translation invariant measures are the Haar measures on this group. A general theorem by Von Neumann states that such a measure is unique up to a multiplicative constant.

Hui Yu
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Lebesgue measure and its multiples are not the only measures that are invariant under translations. Take, for example, the counting measure over the integers or over the rationals. Sure, they are infinite, but your question doesn't mention finiteness.

John B
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    The counting measure is not translation invariant on $\mathbb{R}$. – Alex R. Jul 01 '15 at 23:04
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    @AlexR. Why isn't the counting measure translation invariant? It is surely not regular and also does not put finite mass on the unit cube, but translation invariance is not a problem. Other Hausdorff measures with fractional exponents between zero and one should also fit the bill. – Dirk Jun 09 '17 at 11:31
  • @Dirk maybe he was referring to the counting measure over the integers or over the rationals. A measure that is translation invariant (and includes singletons among its measurable sets) should assign the same measure to every singleton. – Anguepa Jul 06 '17 at 16:09
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    The Hausdorff measures are also translation-invariant, and in some sense "in between" the counting and Lebesgue measures. I would like to know the answer to the original question (and not just in the particular case that the approved answer assumes). – Gro-Tsen Jan 23 '19 at 18:55
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    PS: The paper https://doi.org/10.1017/S0305004100040378 suggests that the full answer to the original question is probably quite complicated. – Gro-Tsen Jan 23 '19 at 18:58