I'm trying to finish an exercise which I asked about earlier here:
Mapping $G$ into its group algebra as left multiplication. Continuous?
$\bf{\text{The setting:}}$
Let $G$ be a locally compact group, $\mu$ a left Haar measure on $G$, $x\in G,\zeta\in L^p(G)$.
I want to show that the map $x\mapsto L_{x}(\zeta)$ is continuous, where $L_{x}(\zeta)$ is the map $y\mapsto \zeta(xy)$.
To simplify the problem, I am reducing to the case where $\zeta = \chi_{E}$ for some open set $E\subset G$, based on the suggestion of Mariano Suárez-Alvarez.
$\bf{\text{What I have so far:}}$
Taking a net $x_{\alpha}\to 1\in G$, I want to show that $\|L_{x}(\zeta) - \zeta \|_{p} = \int_{G}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu\to 0$.
For every $y\in E$, $x_{\alpha}y$ is eventually in $E$ by continuity of left multiplication since $E$ is open.
Since \begin{eqnarray*} \int_{G}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu &=& \underbrace{\int_{E}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu}_{\to 0} + \int_{G\backslash E}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu\\ \end{eqnarray*}
If $y$ is not a boundary point of $E$, then $x_{\alpha}y$ is eventually in $E$ then similarly we have
\begin{eqnarray*} \int_{G\backslash E}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu &=& \underbrace{\int_{A}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu}_{\to 0} + \int_{B}|\zeta(x_{\alpha}y) - \zeta(y)|^{p}d\mu\\ \end{eqnarray*}
where
$A = \{x\notin E : A\text{ is not a boundary point of }E\},B = \{x\notin E : A\text{ is a boundary point of }E\}$
$\bf{\text{Where I am stuck:}}$ I need to either show that $B$ is countable, or somehow deal with the integral at points $y$ on the boundary of $E$. Can someone give me a suggestion on how to continue or find a better approach?
Thanks!