# Is a regular Borel measure on a locally compact space necessarily $\sigma$-finite?

I am trying to compile a proof of the uniqueness of Haar measure. Usually this is done by multiple-integral mumbo-jumbo, abusing left and right invariance of two potential measures and invoking Fubini's Theorem.

However I have only been able to find a proof of Fubini's Theorem for $\sigma$-finite measure spaces. A locally compact group is only $\sigma$-finite when it has a countable number of components. So in general this version of the proof cannot be applied. In Halmos' Measure Theory he says that a Borel Measure (Which might assume the space is locally compact) is necessarily $\sigma$-finite. But this is included as an excercise, and I have no idea how to prove it. Is it true at all?

So I am looking either for a proof of that, which would let me use the usual Fubini Theorem in $\sigma$-finite spaces -- or a separate proof of Fubini for LC Borel measures, when there is no assumption of $\sigma$-finiteness.

• A locally compact, Hausdorff, second countable space is $\sigma$-compact. Perhaps those assumptions are implicit in Halmos' book. Nov 25, 2013 at 13:59
• "In Halmos' Measure Theory he says that a Borel Measure (Which might assume the space is locally compact) is necessarily $\sigma$-finite"; regular Borel measure, probably. The counting measure on your favourite uncountable locally compact space is not $\sigma$-finite. Nov 25, 2013 at 13:59
Halmos's definition of Borel set differs from the one now in common use. While the common definition is the smallest $\sigma$-algebra generated by the open sets (equivalently by the closed sets), his is the smallest $\sigma$-ring (i.e. not including complements) generated by the compact sets (page 219).
Halmos's definition of $\sigma$-finiteness (page 31) for measures is also not equivalent to the $\sigma$-algebraic definition. A measure $\mu$ on a $\sigma$-ring $\mathcal{R}$ is $\sigma$-finite if for all $S \in \mathcal{R}$, $S$ is the union of a sequence of sets of finite measure. This does not imply that the whole space be $\sigma$-finite in the usual sense, as the whole space might not be an element of $\mathcal{R}$.
Under this definition, counting measure on the Borel sets of an uncountable discrete is $\sigma$-finite (by Halmos's definition, the Borel sets are exactly the countable subsets), and is hence not a counterexample.