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Suppose $G$ is a locally compact abelian group, with Haar Measure $\mu$, then is $\mu(E)=\mu(E^{-1})$ for all subsets $E$ of $G$?

I have seen that this is true for all Borel subsets of $G$, but I am in need of a proof and also an example showing why it doesn't hold for other subsets.

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A measure comes with a $\sigma$-algebra of measurable sets. For the case of a locally compact group, this $\sigma$-algebra is often taken to be the Borel $\sigma$-algebra. In this level of generality, the only other reasonable $\sigma$-algebra I can see is its completion. Is that what you mean? – Pete L. Clark Aug 11 '10 at 20:23
up vote 3 down vote accepted

Not necessarily. This condition is equivalent to each left-invariant Haar measure being right-invariant. A group having this property is called unimodular. Since the concept has a name, then there must be groups where it fails. See the Wikipedia article on Haar measure for an example.

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Do you mean to say that it doesn't hold true for Borel subsets – anonymous Aug 11 '10 at 19:57
But the OP mentioned that the group is locally compact abelian. So it is certainly unimodular and the answer is yes, at least for Borel subsets. – Pete L. Clark Aug 11 '10 at 20:22
@Pete L. Clark: Yes absolutely, its true for Borel subsets. – anonymous Aug 12 '10 at 4:04

Of course, a measure is in general only defined on some $\sigma$-algebra of sets, and for Haar measure that's the Borel sets. As Pete mentioned you can complete the measure if you like, but that still won't yield all sets. So for an arbitrary set $E$, $\mu(E)$ may not even be defined.

You can see this is necessary with an example you already know: let $G = S^1$ thought of as $[0,1]$ mod endpoints, and let $E$ be the usual example of a non-measurable set (the Vitali set). There's no value for $\mu(E)$ that will let $\mu$ be translation invariant, finite, and nontrivial, which are required for a Haar measure, and so there's no "Haar measure" defined for all subsets. To be able to have a Haar measure, we need to define it only on "reasonable", i.e. Borel, sets.

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