Proof, [Theorem 2.45, Fol16]. ($\Rightarrow$): Suppose that $J$ is a left ideal. By Proposition 31.4.10, $L^{1}(G; \complex)$ admits an approximate identity $\angles{\phi_\alpha}_{\alpha \in A}$. In which case, for any $f \in J$ and $x \in G$,

($\Leftarrow$): Suppose that $L_{x}J \subset J$ for all $x \in G$. Let $f \in J$, $A \subset G$ be compact, and $\eps > 0$. By Proposition 31.4.8, for each $y \in A$, there exists $U_{y} \in \cn_{G}(y)$ such that $\norm{L_yf - L_zf}_{L^1(G; \complex)}\le \eps$ for all $z \in U_{y}$. By compactness of $A$, there exists $F \subset A$ finite such that $A \subset \bigcup_{y \in F}U_{y}$. In which case,

where for any $B \in \cb_{G}$, $|B|$ denotes its measure. As the above holds for all $\eps > 0$, $\one_{A} * f \in \ol J = J$.

For any Borel set $B \in \cb_{G}$ with finite measure and $\eps > 0$, by inner regularity of the Haar measure, there exists $A \subset G$ compact such that $\norm{\one_B - \one_A}_{L^1(G; \complex)}< \eps$. By continuity of the convolution product, $\one_{B} * f \in J$ as well.

By linearity, $\phi * f \in J$ for any $\phi \in \Sigma(G; \complex) \cap L^{1}(G; \complex)$. Finally, by density of simple functions, $g * f \in J$ for all $g \in L^{1}(G; \complex)$. Therefore $J$ is a left ideal.$\square$