# What is the relationship between Haar Measure and the Birkhoff-Kakutani Theorem?

Let $G$ be a topological group. We know that if $G$ is locally compact, then there exists a left-invariant measure on $G$ which assigns finite measure to compact sets (a.k.a. the Haar measure).

Analogously, the Birkhoff-Kakutani theorem says that if $G$ has a countable base at the identity (and therefore everywhere), then there exists a left-invariant metric on $G$ which generates the same topology.

My question is: if $G$ both has a countable base and is locally compact, do we know anything about how this metric relates to the Haar measure? For example, are there conditions under which the closed unit ball (with respect to the metric) is compact (and therefore assigned finite volume)? Or conversely, are there conditions under which we know that a set which has finite measure also has finite diameter?

• Interesting question, though I think the example questions have silly negative answers as stated. The unit ball for the metric constructed in this exposition of Birkhoff-Kakutani is the whole space, so compact iff the space is compact. (Even without looking at a particular construction, if a topology is induced by a left-invariant metric $d(x,y)$ then the same topology is induced by the left-invariant metric $\min(\tfrac 1 2,d(x,y))$ whose unit ball is the whole space.) This also means all sets have finite diameter. – Dap Oct 7 '17 at 18:16

Struble's theorem: on a locally compact group with countable basis, there always exists a left-invariant proper compatible metric.

Compatible means: defining the topology. Proper means: whose closed balls are compact (and thus of finite Haar measure). See Theorem 2.B.4 in my book with Pierre de la Harpe (arXiv link).

(As mentioned by other people, when $$G$$ is not compact, other compatible metric can fail to be proper, e.g., can be bounded.)

Since your group is locally compact (and Hausdorff!), the given metric $d$ on $G$ always has the property that for some $r>0$ the closed ball $\bar{B}(e,r)\subset G$ is compact. This does not imply that the same holds for $r=1$. For instance, take an infinite discrete group equipped with discrete metric. If you really like $r=1$, you can just rescale $d$ by replacing it with $r^{-1}d$. As for your last question about bounded diameter for finite measure sets: This is a very unnatural requirement. Just think of the Lebesgue measure on ${\mathbb R}^2$. As you know from calculus, there are many unbounded subsets of finite area in the plane. Of course, you can replace the Euclidean metric with an equivalent invariant bounded metric so that every subset has bounded diameter, but why would you want to do this?

There are other ways in which metric and measure can be related as in "metric measure spaces"; the doubling property will be sometimes true (e.g. for Lie groups), but sometimes it will fail, e.g. for some infinite-dimensional compact groups (yes, there are such things!).

Struble's theorem, as quoted by YCor, was basically what I was looking for. But I want to mention a related result that I came across recently.

Theorem. (Bandt, via ) Let $G$ be a topological group with left-invariant, compatible metric $d$. Then, any two Borel sets which are $d$-isometric have the same Haar measure.

 The joys of Haar measure, Diestel & Spalsbury, AMS, Ch. 8.