I am having a hard time understanding the proof found in Federer's "Geometric measure theory" that every left $\chi$ covariant measure over a locally compact Hausdorff groupe $G$ is a Radon measure (2.7.14).
Perhaps it is better to recall what Federer calls a Radon measure on $G$. It is a Borel regular outer measure $\alpha \colon 2^G \to [0, \infty]$ that is finite on compact sets, and outer and inner regular in the following sense:
- for every subset $A \subset G$, we have $$\alpha(A) = \inf \{ \alpha(U) : A \subset U \text{ open}\}$$
- for every $\alpha$ measurable set $A \subset G$, we have $$\alpha(A) = \sup \{ \alpha(K) : K \subset A \text{ is compact}\}$$
Not that this definition of Radon measure is not equivalent to others found in the literature.
A left $\chi$ covariant Haar integral is a monotone Daniell integral $\lambda \colon \mathscr{K}(G) \to \mathbb{R}$ (where $\mathscr{K}(G)$ denotes the vector space of continuous functions on $G$ with compact support) such that $$ \lambda(h \mapsto u(gh)) = \chi(g) \lambda(h) \qquad \text{for all } u \in \mathscr{K}(G) \text{ and } g \in G$$ A left $\chi$ covariant measure is the measure associated to a left $\chi$ covariant Daniell integral as in the proof of 2.5.2. Such a measure is not automatically a Borel measure (open sets need not be measurable), hence Federer proves in 2.7.14 that open sets are $\alpha$ measurable.
Then, the Borel regularity of $\alpha$ follows from its so-called $\mathscr{K}(G)$ regularity (see 2.5.3). Outer regularity with respect to open sets follows from 2.5.14, so does the inner regularity with respect to compact sets for $\alpha$ measurable sets of finite measure. However, it does not seem obvious to me that this property holds for open sets of infinite measure. In other words, how to prove that $$ \sup \{\alpha (K) : K \subset U \text{ is compact}\} = \infty \qquad \text{if } U \text{ is open and } \alpha(U) = \infty \text{ ?}$$