# Uniqueness of the Haar measure.

I am aware that uniqueness of a Haar measure on a locally compact topological group is a well known fact in the sense below:

"If $$v,\mu$$ are Radon measures that are left invariant then there exists a positive constant $$a$$ such that $$v=a\mu$$. (Where a Radon measure is a locally finite Borel measure that satisfies the weak inner regularity and outer regularity conditions)."

For the Lebesgue measure on $$\mathbb{R}^n$$ the above can be strengthened in the sense that the assumptions of regularity are redundant. So, the following holds:

"If $$\mu$$ is a locally finite Borel measure on $$\mathbb{R}^n$$ such that $$\mu(A+x)=\mu(A)$$ for any $$A \in \mathcal{B}\left(\mathbb{R^n}\right)$$ and $$x \in \mathbb{R^n}$$ then $$\mu=a\lambda$$, where $$a\geq0$$ and $$\lambda$$ is the Lebesgue measure."

Is an analogous statement true for an arbitrary locally compact topological group $$G$$ and a Haar measure $$v$$ on $$G$$. In other words, does the following hold?

"If $$G$$ is a locally compact topological group, $$v$$ a Haar measure on $$G$$ and $$\mu$$ is a locally finite measure on $$G$$ that is left invariant, then there exists a positive constant $$a$$, such that $$\mu=av$$."

• What you are effectively asking is "Is every left invariant locally finite measure Radon?" The answer is in general no if the groups under consideration are "too large" (not second countable). For second countable groups, any locally finite measure is Radon. I will see tomorrow if I can find a counterexample. There should be one in the book by Folland. – PhoemueX Nov 6 '18 at 22:09
• @PhoemueX I would very much appreciate a reference for the claim about second countable groups. – exosphere Nov 8 '18 at 18:25
• Sorry for my delay in answering. The claim about second countable groups follows from Theorem 7.8 in Folland's book "Real Analysis". There it is shown that if $X$ is any LCH space in which every open set is $\sigma$-compact (e.g., if $X$ is 2nd countable), then every Borel measure on $X$ that is finite on compact sets is regular and Radon. This is a very nice result which is not as widely known as it should be (I think). – PhoemueX Nov 9 '18 at 17:23
• @PhoemueX. Indeed that is a very interesting result, thank you. – exosphere Nov 9 '18 at 17:43