Let's assume that $G$ is locally compact and Hausdorff topological group, hence it carries a Haar measure, $\mu$. We can than consider space of integrable functions $L^1(G)$ (class of functions to be more precise), and we can define convolution via $f\ast g(x)=\int_{G}f(y)g(y^{-1}x)d\mu(y)$. With usual addition and multiplication by scalars from $\mathbb C$ this makes $L^1(G)$ into algebra.
Is it true that $L^1(G)$ is always Banach algebra? When I tried to prove for example that $\|f\ast g\|\le \|f\| \|g\|$ i had to use Fubini-Tonelli theorem which assumes $\sigma$-finitness of $\mu$. Same goes for associativity of $\ast$. I though that using fact that $C_c(G)$ is dense in $L^1(G)$ might be helpful but after all that didn't seem to work (i was able to prove desired (in)equalities for functions from $C_c(G)$ but without $\|f\ast g\|\le \|f\|\|g\|$ i wasn't able to use this argument for generic $f$)
I'll be greatfull for any help, it's been bugging me for a while.
I've came up with solution, i'll write it in case someone is interested:
(I will identify function with it's class modulo equality almost everywhere, i hope that won't be confusing, also i'll use notation that $A\sqcup B$ stands for regular union $A\cup B$ with condition $A\cap B=\emptyset$)
Lemma: $f\in L^1(G) \; \Rightarrow \exists B\in\mathscr{B}(G)$, $\sigma$-compact such that $f\overset{\cdot}{=}f\cdot \chi_{B}$ where $\overset{\cdot}{=}$ stands for equality almost everywhere.
Proof: Let's take any $\epsilon>0$ and define for $n\in\mathbb N$
$A^{\epsilon}_{n}:=|f|^{-1}([\frac{1}{n+1}, \frac{1}{n}[)$
$A^{\epsilon}_{0}:=|f|^{-1}([1,\infty[)$
These sets are obviously disjoint, $\bigsqcup_{n\in\mathbb Z_+}A^{\epsilon}_{n}=|f|^{-1}(]0,\infty[)$ and with finite measure since $f\in L^1(G)$. Using regularity of haar measure $\mu$ we can find compact sets $K^{\epsilon}_{n}\subset A^{\epsilon}_{n}$ for $n\in\mathbb Z_+$ such that $\mu(A^{\epsilon}_{n} \setminus K^{\epsilon}_{n})\le \frac{\epsilon}{2\cdot 2^n}$ ($n\in\mathbb N$) and $\mu(A^{\epsilon}_{0} \setminus K^{\epsilon}_{0})\le \frac{\epsilon}{2}$.
Let's define $A_{\epsilon}:=\bigsqcup_{n\in\mathbb Z_+}A^{\epsilon}_{n}\setminus K^{\epsilon}_{n}$, $B_{\epsilon}:=\bigsqcup_{n\in \mathbb Z_+}K^{\epsilon}_{n}$. $B_\epsilon$ is $\sigma$-compact set, and $\mu(A_\epsilon)=\sum_{n\in\mathbb Z_+}\mu(A^{\epsilon}_{n}\setminus K^{\epsilon}_{n})\le\epsilon$, moreover $|f|^{-1}(]0,\infty[)=A_\epsilon\sqcup B_\epsilon$.
Let's define $A:=\bigcap_{m\in\mathbb N}A_{\frac{1}{m}}$, $B:=\bigcup_{m\in\mathbb N}B_{\frac{1}{m}}$. B is $\sigma$-compact, A is set of zero measure, $A\cap B=\emptyset$, moreover $|f|^{-1}(]0,\infty[)=A\sqcup B$. Which gives us desired $f\overset{\cdot}{=}f\cdot\chi_{B}$.
If we take $f,g\in L^1(G)$ and $B_f, \; B_g$ sets from lemma we have the following:
$\|f\ast g\|\le\int_{G}\left( \int_{G}\left|f(y)g(y^{-1}x)\right|d\mu(y)\right)d\mu(x)=\int_{B_f B_g}\left( \int_{B_f}\left|f(y)g(y^{-1}x)\right|d\mu(y)\right)d\mu(x)$ where $B_f B_g=\{xy\; |\; x\in B_f, y\in B_g\}$. $B_f$ is $\sigma$-compact, $B_f B_g=m(B_f\times B_g)$ which is also $\sigma$-compact since multiplication is continuous, image of a union is a union of images, and continuous image of compact set is compact. Measure of compact set is finite so both $B_f$ and $B_f B_g$ are $\sigma$-finite, we can use Fubini's theorem