L^1 of an LCA group is a commutative Banach star algebra under convolution

(Reduction to $\sigma$-compact supports.) Let $f\in A$. For each $n\ge 1$ the Borel set $E_n:=\{\mathrm{\mid }f\mathrm{\mid }>1\mathrm{/}n\}$ has $m_G(E_n)\le n\mathrm{\parallel }f{\mathrm{\parallel }}_1<+\mathrm{\infty }$, so by [F1] there is a $\sigma$-compact set $S_n$ with $m_G(E_n\setminus S_n)=0$. Then $S:={\bigcup }_n{S}_n$ is $\sigma$-compact and $m_G(\{\mathrm{\mid }f\mathrm{\mid }>0\}\setminus S)=0$, so $f=0$ $m_G$-a.e. outside $S$ and $f\cdot {\mathbf{1}}_S$ represents the same class with $\mathrm{\parallel }f\cdot {\mathbf{1}}_S{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_1$. For the product-measurability needed below, choose $C_c$ approximants converging in $L^1$ and, after a subsequence, almost everywhere, using [F4, F5]. Define a representative by their pointwise limit where it exists, and zero elsewhere. Its support lies in the countable union of their compact supports. Do this for both $f$ and $g$; we may thus assume $f,g$ have $\sigma$-compact supports $S,T$ and are pointwise limits of $C_c$ functions wherever their limits exist, with zero assigned on the remaining null sets. On any product of two compact subsets of $G$, a continuous scalar kernel is uniformly approximable by finite sums of products of bounded Borel functions of the separate coordinates: take finite sufficiently fine covers in each coordinate and disjointify them. Hence it is product measurable there. Applying this to the continuous kernels $g_n(x-y)$ and taking their pointwise limits proves product measurability for $g(x-y)$ on the $\sigma$-compact products below; the zero convention uses the measurable set where the sequence converges. No equality of the topological and product Borel sigma-algebras is assumed.

(Convolution is a well-defined contraction.) Assume $f,g$ are supported in the $\sigma$-compact sets $S,T$. For these representatives, step 1.1 shows that $\mathrm{\Phi }(y,x):=\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(x-y)\mathrm{\mid }$ is product measurable on $S\times (S+T)$ and is supported in $S\times (S+T)$, a $\sigma$-finite product by [F3], and $g(x-y)=0$ unless $x-y\in T$, that is $x\in y+T\subseteq S+T$. Tonelli's theorem on that $\sigma$-finite product gives ${\int }_{S\times (S+T)}\mathrm{\Phi }d(m_G\otimes m_G)={\int }_S\mathrm{\mid }f(y)\mathrm{\mid }({\int }_{S+T}\mathrm{\mid }g(x-y)\mathrm{\mid }d{m}_G(x))d{m}_G(y)={\int }_S\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_{1}d{m}_G(y)=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1,$ the inner identity being translation invariance of $m_G$ and the fact that $g(x-y)$ vanishes for $x\mathrm{\notin }y+T$. Hence $\mathrm{\Phi }$ is integrable, so by Fubini's theorem the section $x\mapsto {\int }_G\mathrm{\mid }f(y)\mathrm{\mid }\mathrm{\mid }g(x-y)\mathrm{\mid }d{m}_G(y)$ is finite for $m_G$-a.e. $x\in S+T$ (and is $0$ for $x\mathrm{\notin }S+T$, since then no $y$ has simultaneously $f(y)\mathrm{\ne }0$ and $g(x-y)\mathrm{\ne }0$), and its integral is at most $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$. Therefore $(f\ast g)(x)={\int }_{G}f(y)g(x-y)d{m}_G(y)$ converges absolutely for $m_G$-a.e. $x$, the resulting function $f\ast g$ lies in $A$ with $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$, and the class of $f\ast g$ does not depend on the representatives: if $f_1=f_2$ and $g_1=g_2$ a.e., then, for each fixed $x$, the integrands differ only on the union of the null set where the $f_i$ differ and its reflected translate $x-N$, where $N$ is the null set where the $g_i$ differ. Translation and inversion invariance make this union null. Thus the absolute integrals and values agree whenever defined, including for arbitrary representatives before restriction to essential supports.

(Bilinear and commutative.) For $f_1,f_2,g\in A$ supported in $\sigma$-compact sets and $a_1,a_2\in \mathbb{C}$ the identity $(a_1{f}_1+a_2{f}_2)\ast g=a_1(f_1\ast g)+a_2(f_2\ast g)$ holds pointwise for every $x$ at which all three integrals converge, hence a.e. by step 2.1; the same argument on the second variable gives bilinearity. For commutativity let $h\in C_c(G)$ and apply step 2.1 and Tonelli on $S\times T$ ([F3]) to write ${\int }_G(f\ast g)(x)h(x)d{m}_G(x)={\int }_S{\int }_{T}f(y)g(z)h(y+z)d{m}_G(z)d{m}_G(y),$ substituting $x=y+z$ in the inner integral, which is a translation and preserves $m_G$; the right-hand side is symmetric in $f$ and $g$ together with the labels $y,z$, so ${\int }_G(f\ast g)hd{m}_G={\int }_G(g\ast f)hd{m}_G$ for every $h\in C_c(G)$. By [F4] applied to the $L^1$ function $f\ast g-g\ast f$, this gives $f\ast g=g\ast f$ a.e. on $G$.

(Associative.) Let $f,g,u\in A$, all supported in $\sigma$-compact sets, and let $h\in C_c(G)$. Applying step 2.1 twice and Tonelli on the $\sigma$-finite product of the three essential supports (each a countable union of finite-measure sets) gives ${\int }_G((f\ast g)\ast u)(x)h(x)d{m}_G(x)=\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int \text{\hspace{0.17em}⁣}\text{\hspace{0.17em}⁣}\int f(y)g(z)u(w)h(y+z+w)d{m}_G(w)d{m}_G(z)d{m}_G(y),$ where the substitutions $x=y+z+w$ are translations at each stage; the same expression is obtained for ${\int }_G(f\ast (g\ast u))hd{m}_G$. Since $h\in C_c(G)$ was arbitrary, [F4] gives $(f\ast g)\ast u=f\ast (g\ast u)$ a.e.

(The involution.) First let $f\in A$ be supported in a $\sigma$-compact set $S$. The function $f^{\ast }$ is Borel, and [F2] gives $\mathrm{\parallel }{f}^{\ast }{\mathrm{\parallel }}_1={\int }_G\mathrm{\mid }f(-x)\mathrm{\mid }d{m}_G(x)={\int }_G\mathrm{\mid }f(x)\mathrm{\mid }d{m}_G(x)=\mathrm{\parallel }f{\mathrm{\parallel }}_1$, so $f^{\ast }\in A$ and $\ast$ is isometric on $A$; it is conjugate-linear and $f^{\ast \ast }=f$ hold pointwise on representatives. To prove the reversal identity, let first $f,g\in C_c(G)$ and $x\in G$. Substituting $y=-x-w$ in the defining integral and using [F2] (the substitution is inversion followed by a translation), $(f\ast g{)}^{\ast }(x)=\overline{{\int }_{G}f(y)g(-x-y)d{m}_G(y)}={\int }_G\overline{f(-x-w)}\overline{g(w)}d{m}_G(w),$ while substituting $y=-w$ in the defining integral of $g^{\ast }\ast f^{\ast }$ gives $(g^{\ast }\ast f^{\ast })(x)={\int }_G\overline{g(-y)}\overline{f(y-x)}d{m}_G(y)={\int }_G\overline{g(w)}\overline{f(-w-x)}d{m}_G(w)\mathrm{.}$ Since $-x-w=-w-x$ and complex conjugation is additive, the two integrands agree, so $(f\ast g{)}^{\ast }=(g^{\ast }\ast f^{\ast })$ for $f,g\in C_c(G)$. Now choose $f_n,g_n\in C_c(G)$ with $f_n\to f$ and $g_n\to g$ in $A$, possible by [F4], and note $f_n^{\ast }\to f^{\ast }$, $g_n^{\ast }\to g^{\ast }$ by the isometry just proved. By the norm bound of step 2.1, $f_n\ast g_n\to f\ast g$ and $g_n^{\ast }\ast f_n^{\ast }\to g^{\ast }\ast f^{\ast }$, and the isometry gives $(f_n\ast g_n{)}^{\ast }\to (f\ast g{)}^{\ast }$; since $(f_n\ast g_n{)}^{\ast }=g_n^{\ast }\ast f_n^{\ast }$ for every $n$, uniqueness of limits in the normed space $A$ ([F5]) gives $(f\ast g{)}^{\ast }=g^{\ast }\ast f^{\ast }$.

(An identity forces discreteness.) A positive singleton mass $c$ makes every point have mass $c$. A compact neighbourhood $K$ then contains at most $m_G(K)\mathrm{/}c$ distinct points, since every finite subset has that many atoms. Thus $K$ is finite, and an open identity neighbourhood inside $K$ can be intersected with the complements of its finitely many nonidentity points to show $\{0\}$ is open. Hence nondiscreteness implies $m_G(\{0\})=0$. Suppose now that $G$ is nondiscrete and $u\in A$ is an identity. By outer regularity at $\{0\}$ and [F6], there is a symmetric open identity neighbourhood $V$ with ${\int }_V\mathrm{\mid }u\mathrm{\mid }d{m}_G<1$. By continuity of addition and local compactness choose a symmetric open $W$ with compact closure and $W+W\subseteq V$. Then $0<m_G(W)<\mathrm{\infty }$, so ${\mathbf{1}}_W\in A$. For every $x\in W$ the convolution formula gives $\mathrm{\mid }(u\ast {\mathbf{1}}_W)(x)\mathrm{\mid }=\mathrm{\mid }{\int }_{x-W}u(y)d{m}_G(y)\mathrm{\mid }\le {\int }_V\mathrm{\mid }u\mathrm{\mid }d{m}_G<1$, since $x-W\subseteq W+W\subseteq V$. This contradicts $u\ast {\mathbf{1}}_W={\mathbf{1}}_W$ almost everywhere on the positive-measure set $W$. Therefore an identity can exist only when $G$ is discrete.

(Discrete groups have an identity.) If $G$ is discrete, its singleton $\{0\}$ is open and has mass $c=m_G(\{0\})>0$ by [F1]. Then $u=c^{-1}{\mathbf{1}}_{\{0\}}$ is in $A$, and translation invariance gives $m_G(\{x\})=c$. Thus $(f\ast u)(x)=c^{-1}f(x)m_G(\{x\})=f(x)$ wherever defined, for every $f\in A$. Commutativity makes $u$ a two-sided identity.

(Conclusion.) By steps 2.1, 3.1, 3.2 and 3.3, convolution is a well-defined bilinear, associative, commutative product on $A$ with $\mathrm{\parallel }f\ast g{\mathrm{\parallel }}_1\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$, and $\ast$ is an isometric conjugate-linear involution satisfying $(f\ast g{)}^{\ast }=g^{\ast }\ast f^{\ast }$ and $f^{\ast \ast }=f$; no unit is required. Since $A$ is complete in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$ ([F5]), it is a commutative Banach $\ast$-algebra; it has a unit exactly when $G$ is discrete, as proved above.