Integral of $|f|$ outside a compact set

Question

Let $G$ be a locally compact group. Given $f\in L^1(G)$ and $\epsilon>0$, how to show that there is a compact set $K\subset G$ such that $\int_{G\setminus K}|f|<\epsilon$?

Answer 1

I know this is an old question, but I want to provide an answer that does not require the extra assumption of second countability for the group $G$.

I assume that $L^1(G)$ means integrable functions with respect to the Haar measure. A function $f\in L^1(G)$ vanishes outside a $\sigma$-compact subset (see for instance p49 in Folland's book "A course in abstract harmonic analysis" - second edition). In other words, $f$ is supported inside a $\sigma$-compact set. Hence, there is an increasing sequence of compact sets $\{K_n\}_{n=1}^\infty$ such that $$S:=\{x \in G: f(x) \ne 0\}\subseteq \bigcup_{n=1}^\infty K_n.$$ We have $\chi_{(G\setminus K_n)\cap S}\to 0$ pointwise and thus by the dominated convergence theorem $$\int_{G\setminus K_n} |f| = \int_{G} |f| \chi_{(G\setminus K_n)\cap S} \stackrel{n \to \infty}\longrightarrow 0.$$ so for $n$ large enough the integral $$\int_{G\setminus K_n} |f|$$ becomes as small as we like.