Why are Haar measures finite on compact sets?

Question

I'm working through the answer by t.b. to another user's question here:

A net version of dominated convergence?

because I am trying to work through a related problem and I think it will be illuminating.

From step two, "Then $KK'$ is compact and thus has finite Haar measure."

why is this true?

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$\bf{\text{My attempt so far}}$:

Let $\mu$ be a left Haar measure on a locally compact group $G$. Then $\mu$ is non-zero, regular, and left $G$-invariant. Let $K\subset G$ be compact.

$\color{red}{(!)}$ If I could construct set $U$ with the properties:

(i) $1\in U$

(ii) $0 < \mu(U) < \infty$

(iii) $U$ is open.

Then I could cover $K$ with $\{xU : x\in K\}$ and choose finitely many $k_{1}, ... , k_{n}\in K$ such that $K\subset \bigcup_{j=1}^{n}xU$ which forces $\mu(K)\leq \sum\limits_{j=1}^{n}\mu(xU) = n\mu(U) < \infty$.

I don't know how to construct such a $U$, though it seems innocent enough to ask for.

Does such a $U$ exist? or is the proof much more difficult?

Answer 1

A Haar measure is finite on compact sets by definition: A Borel measure $\mu$ on a topological group $G$ is a (left) Haar measure if

1. $\mu$ is regular.

2. $\mu$ is left-invariant, i.e. $\mu(xU) = \mu(U)$ for any measurable $U \subset G$.

3. $\mu(U) > 0$ for all $U \subset G$ open.

4. $\mu(K) < \infty$ for all $K \subset H$ compact.

Answer 2

It is in fact possible to deduce a Haar measure is finite on compact sets from the rest of the axioms, if you assume there is some nonempty subset of finite measure.

Let $A\subseteq G$ such that $\mu(A)<\infty$. By regularity there is some open set $U\supseteq A$ such that $\mu(U)\leqslant\mu(A)+\varepsilon<\infty$ (for every $\varepsilon>0$, say $\varepsilon=1$). Since $U$ is open we also have $\mu(U)>0$. Now, pick some $u_0\in U$ and let $V:=u_0^{-1}U$; by left invariance $\mu(V)=\mu(U)$. We conclude that $V$ is open, $1\in V$ and $0<\mu(V)<\infty$.

The rest you've already showed: if $K$ is compact, then $\{kV\}_{k\in K}$ is an open cover of $K$, and hence there are $k_1,...,k_n$ such that $K\subseteq \bigcup_{i=1}^n k_iV$, and therefore $\mu(K)\leqslant n\cdot \mu(V)<\infty$.

It is not clear to me why some people put this in the definition of a Haar measure. Incidentally, it is also not necessary to assume a Haar measure is positive on open sets if you assume it is non-zero: if you had some open set $V\subseteq G$ such that $\mu(V)=0$, then clearly you'd have that $\mu(K)=0$ for every compact $K$, which means by regularity that $\mu(G)=0$.