Let $G$ be an (abelian) locally compact Hausdorff group. Consider the following fragment from Folland's text "A course in abstract harmonic analysis" (second edition, p102).
Why is the boxed line true? I guess formally, the statement we should prove is the following:
If $\mu \in M(G)$, there exists a sequence $\{\mu_n\}_{n=1}^\infty \subseteq M(G)$ such that $\|\mu_n -\mu\|\to 0$ and such that $\mu_n$ has support on a compact set (i.e. there exists a compact subset $K_n\subseteq G$ such that $\mu_n(A)=\mu_n(A\cap K_n)$ for all Borel subsets $A\subseteq G$.
Is this statement true? If so, how can we prove this? I have no idea how to construct such a sequence of measures (especially if $G$ is not assumed to be $\sigma$-compact, but even that simpler case remains elusive to me).
Let $\mu $ be in $M(G)$, so that $\mu $ is a Radon measure with finite total variation. This said we see that $|\mu |$ is a finite Radon measure and hence regular, so there are compact sets $K_n$ such that $|\mu|(G\setminus K_n)\to 0$. You may then define $\mu_n$ by $$\mu_n(E):=\mu(E\cap K_n),$$ in which case $$ \|\mu-\mu_n\|= |\mu-\mu_n|(G) = $$$$= |\mu-\mu_n|(K_n) + |\mu-\mu_n|(G\setminus K_n) =$$$$= 0+|\mu|(G\setminus K_n)\to 0. $$