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Let $G$ be a locally compact group with Haar measure $ \mu $, $K \subset G$ a compact subset and $ F \subset G $ any subset of finite Haar measure $\mu (F) < \infty $.

Is the Haar measure of the product $ \mu(KF) $ finite as well?

I know that the compactness of $K$ implies that $\mu(K)<\infty$, and that the above would be true if F was compact (since then $ KF$ would be compact as well).

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up vote 2 down vote accepted

Take $G:=\Bbb R$ (additive group) with Lebesgue measure, $K:=[0,1]$ and $F:=\Bbb Z$. Then $$KF=\{x+y,x\in [0,1],y\in\Bbb Z\}=\Bbb R$$ which has infinite measure.

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Thank you, you are right and I missed this rather simple example. – the_lar Feb 4 '13 at 9:33
In fact, when I said that $F$ has finite Haar measure, I was thinking about a set that is a fundamental domain for some lattice $\Gamma$ in $G$. Then $F$ is "localized" in some sense, even if it is not compact. So perhaps I can reformulate the question, with the additional assumption that $F$ is indeed a fundamental domain for a lattice $\Gamma$ in $G$ ? – the_lar Feb 4 '13 at 9:36

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