L1 of a second-countable locally compact group is separable

Let $\mathcal{B}$ be an at most countable basis for $G$. The relatively compact open sets form a basis by [F2], so the family of all relatively compact open sets covers $G$. By Lindelofness and AC's implication of Countable Choice in [F6], it has an at most countable subcover; enumerate that nonempty subcover as $(V_n{)}_{n\in \mathbb{N}}$, repeating terms if it is finite.

Put $K_n={\bigcup }_{j\le n}\overline{V_j}$. Each closure is compact; an ambient open cover of $K_n$ has a finite subcover on each of the finitely many closures by [F3], and their finite union covers $K_n$. Thus $K_n$ is compact, and it is closed and Borel because $G$ is Hausdorff. The sequence increases and covers $G$. If $f\in C_c(G)$, compactness of $\mathrm{supp}f$ gives a finite subcover from $(V_n)$; taking the largest index in that subcover (or $n=0$ for empty support) shows $\mathrm{supp}f\subseteq K_n$ for some $n$. Each $K_n$ has finite Haar measure by [F1].

The family $\mathcal{G}=\mathcal{B}\cup \{K_n:n\in \mathbb{N}\}$ is countable by [F4]. Set ${\mathcal{A}}_0$ specifically to the algebra of finite Boolean combinations of $\mathcal{G}$. As in the countable-algebra supplier proof in [F4], enumerate $\mathcal{G}$ and let ${\mathcal{C}}_m$ be the finite algebra generated by its first $m$ terms. Then ${\mathcal{A}}_0={\bigcup }_m{\mathcal{C}}_m$: every finite Boolean combination uses some finite prefix. These algebras increase, so their union is an algebra, and [F4] makes it countable. Its generators are Borel, and finite Boolean operations preserve Borel sets, so every member of ${\mathcal{A}}_0$ is Borel. Every open set is a union of basis members, and because $\mathcal{B}$ is countable this is a countable union; therefore $\sigma ({\mathcal{A}}_0)$ contains every open set. Conversely ${\mathcal{A}}_0$ consists of Borel sets, so minimality in [F8] gives $\sigma ({\mathcal{A}}_0)=\mathcal{B}(G)$. Each $K_n\in {\mathcal{A}}_0$ and has finite measure by step 2.1.

Fix $f\in C_c(G)$ and a target $\eta >0$, and choose $n$ with $\mathrm{supp}f\subseteq K_n$ by step 2.1. If $\mu (K_n)=0$, then $f$ is zero as an $L^1$ class because it vanishes outside the null set $K_n$; the zero function is in the required span since $\mathrm{\varnothing }\in {\mathcal{A}}_0$. Otherwise set $\delta =\eta \mathrm{/}(4\mu (K_n))$. Let ${\mathcal{U}}_{\delta }$ be the family of all $U\in \mathcal{B}$ for which there exists $x\in K_n$ such that $\mathrm{\mid }f(y)-f(x)\mathrm{\mid }<\delta$ for every $y\in U$. Continuity and the basis property show that ${\mathcal{U}}_{\delta }$ covers $K_n$; compactness gives a finite subcover $U_1,\dots ,U_m$. For each $i$, choose a witness $x_i\in K_n$ for its defining property, and set $E_1=K_n\cap U_1$ and $E_i=(K_n\cap U_i)\setminus {\bigcup }_{j<i}{U}_j$ for $i>1$. These sets partition $K_n$, belong to ${\mathcal{A}}_0$, and have finite measure by [F7]. For each nonempty $E_i$, choose $y_i\in E_i$ and $q_i\in \mathbb{Q}(i)$ with $\mathrm{\mid }{q}_i-f(y_i)\mathrm{\mid }<\delta$; these are finitely many choices, justified by [F6] and density in [F5]. Then $g={\sum }_{i:E_i\mathrm{\ne }\mathrm{\varnothing }}{q}_i{\mathbf{1}}_{E_i}$ lies in the required span and in $L^1$, since it is measurable and bounded with support in the finite-measure set $K_n$. For $y\in E_i$, $\mathrm{\mid }f(y)-f(y_i)\mathrm{\mid }<2\delta$, hence $\mathrm{\mid }f(y)-g(y)\mathrm{\mid }<3\delta$; outside $K_n$ both functions vanish. Thus $\mathrm{\mid }f-g\mathrm{\mid }\le 3\delta {\mathbf{1}}_{K_n}$ and [F7] gives $\mathrm{\parallel }f-g{\mathrm{\parallel }}_1\le 3\delta \mu (K_n)=3\eta \mathrm{/}4<\eta$.

Let $\mathcal{E}=\{A\in {\mathcal{A}}_0:\mu (A)<\mathrm{\infty }\}$ and let $D_0$ consist of all finite $\mathbb{Q}(i)$-linear combinations of ${\mathbf{1}}_A$ with $A\in \mathcal{E}$. The family $\mathcal{E}$ is countable as a subset of ${\mathcal{A}}_0$; the alphabet $C=\mathbb{Q}(i)\times \mathcal{E}$ is at most countable by [F4,F5]. Every finite power $C^m$, including the one-point $C^0$, is at most countable by [F4]. AC gives Countable Choice by [F6], so [F4] makes ${\bigcup }_{m\in \mathbb{N}}{C}^m$, the set of finite lists of coefficient/set pairs, at most countable. Its image under the finite-sum map is $D_0$, so $D_0$ is countable. Each generator indicator is integrable because its set has finite measure, and finite linear combinations remain in $L^1$. For $h\in L^1(G)$ and $ϵ>0$, choose $f\in C_c(G)$ with $\mathrm{\parallel }h-f{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ by [F1], then choose $d\in D_0$ with $\mathrm{\parallel }f-d{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ by step 4.1. The triangle inequality gives $\mathrm{\parallel }h-d{\mathrm{\parallel }}_1<ϵ$, so $D_0$ is dense in $L^1(G)$.

The set $D_0\times {\mathbb{N}}_{>0}$ is countable by [F4]. For each $(d,k)$ in it, density of $C_c(G)$ in $L^1(G)$ gives a nonempty set of $c\in C_c(G)$ with $\mathrm{\parallel }c-d{\mathrm{\parallel }}_1<1\mathrm{/}k$. AC's implication of Countable Choice [F6] selects one such $c_{d,k}$ for every pair. The resulting set of functions is countable; for any $h\in L^1(G)$ and $ϵ>0$, choose $d\in D_0$ with $\mathrm{\parallel }h-d{\mathrm{\parallel }}_1<ϵ\mathrm{/}2$ and then $k$ with $1\mathrm{/}k<ϵ\mathrm{/}2$. It follows that $\mathrm{\parallel }h-c_{d,k}{\mathrm{\parallel }}_1<ϵ$, so their image in $L^1(G)$ is a countable dense subset contained in the image of $C_c(G)$.

Step 5.1 proves separability of $L^1(G)$, and [F1] gives its completeness, so it is a separable Banach space. Steps 3.1 and 5.1 give the asserted generating Borel algebra and dense $\mathbb{Q}(i)$-linear span, while step 6.1 gives the countable dense subset from $C_c(G)$.