\in \operatorname{Co}(Y).
 \label{eq:66}\end{aligned}$$

The proof is essentially an application of Theorem 3.5 in [@dadedelastte17]. We first note that convergence with respect to $\|\cdot\|_Y = \|\cdot\|_{L_{r,m}}$ implies convergence in measure. Furthermore, since $m$ is $w$-moderate, it is not hard to see that $Y = L_{r,m}(G)$ is $\lambda$-invariant, and that the restriction of $\lambda$ to $Y$ is a continuous representation of $G$. Therefore, we only need to prove that Assumption 5 and Assumption 6 in [@dadedelastte17] are satisfied.

We first show that $Y\subset \mathcal{U}_w$. By  and  we get for any $x \in G$ that $$\begin{aligned}
\label{eq:weight_e_inequality}
  m(e) = m(x x^{-1}) \leq m(x) \cdot w(x^{-1}) = m(x) \cdot w(x) \, ,\end{aligned}$$ and hence $[w(x)]^{-1} \cdot m(e) \leq m(x)$, whence $Y = L_{r,m}(G) \hookrightarrow L_{r,w^{-1}}(G) \subset \mathcal{U}_w$ since $r > 1$. Since $\mathcal{U}_w = \mathcal{T}_w'$ under the pairing , for all $F \in Y$ and $f \in \mathcal{T}_w$ it holds that $F f \in L_1(G)$. In particular, by assumption , $FK \in L_1(G)$ for all $F \in Y$ and, by construction, $F \,Vv \in L_1(G)$ for all $v \in \mathcal{S}_w$ and $F \in \mathcal M^Y$, so that Assumption 5 and Assumption 6 in [@dadedelastte17] hold true.

Discretization {#sec:discretization}
==============

The aim of this section 