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I know that the Haar measure on a locally compact second-countable group $G$ is unique up to a constant multiple.

That is, any nontrivial left-invariant Radon measure is a constant multiple of the Haar measure.

Equivalently (since second-countable), any nontrivial left-invariant locally finite (finite on compact sets) measure is a constant multiple of the Haar measure.

Is it possible that there is a nontrivial left-invariant Borel measure (not locally finite) that is not a constant multiple of the Haar measure?

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Sure. For example, counting measure is trivially invariant and if the group is non-discrete, it is not a constant multiple (or even absolutely continuous with respect to) of the Haar measure.

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  • $\begingroup$ Take a sequence $A_n \supset A_{n+1}$ of neighborhoods of $g=e$ such that $\bigcap_n A_n = \{e\}$ and $\mu(A_n) \ne 0$, consider the operator $A_n \ast E= \{g \in G, \exists h \in A_n, hg \in E\}$. I think your counting measure is $\nu(E) = \lim_{n \to \infty} \displaystyle\frac{\mu(A_n \ast E)}{\mu(A_n)}$, while the Haar measure satisfies $\mu(E) = \lim_{n \to \infty} \mu(A_n \ast E)$ $\endgroup$ – reuns Nov 30 '16 at 11:29
  • $\begingroup$ Now comes the question : isn't it in some sense an extended definition of "a constant multiple of the Haar measure" ? $\endgroup$ – reuns Nov 30 '16 at 11:50

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