Talk:Haar measure

The article says:

It turns out that there is, up to a multiplicative constant, only one translation invariant measure on X which is finite on all compact sets.

This can't be quite right. On R, for example, Borel measure and Lebesgue measure both have this property. Perhaps it's correct if "measure" is replaced with "complete measure". --Zundark, 2002 Mar 6

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You're mistaken. "X" was defined as the sigma algebra generated by the compact sets. Lebesgue measure has a more extensive domain than that; Borel measure does not.