Why such a net will exist? G- locally compact group & 
$\lambda(x)f(y) = f(x^{-1}y)\  \forall \ y \in G$. The following condition is called Reiter's finite condition.
$P(G) := \{f \in L^1(G): f \geq 0, \|{f}\|_1 = 1\}.$ 
For every finite subset $K$ of $G$ and every $\epsilon > 0$, there exists $f \in P(G)$ such that $
\sup_{x \in K}\| {\lambda(x)f-f}\|_1 < \epsilon .$
Does this condition ensure the existence of $\{f_i\}_i \in P(G)$ such that $\lim_i\| {\lambda(x)f_i - f_i}\|_1 = 0 $ for all $x \in G$. ?
 A: As often with nets, the key is choosing the "correct" index-set. In this case, let 
$$
I:=\left\{ F\subset G\,\mid\, F\text{ finite}\right\} \times\left(0,\infty\right)
$$
and define a partial order on this set by
$$
\left(F,\varepsilon\right)\leq\left(F',\varepsilon'\right)\qquad:\Longleftrightarrow\qquad F\subset F'\text{ and }\varepsilon'\leq\varepsilon.
$$
Now, by your assumption, for each $i=\left(F,\varepsilon\right)\in I$,
there is a function $f_{i}\in P\left(G\right)$ satisfying
$$
\max_{x\in F}\left\Vert \lambda\left(x\right)f_{i}-f_{i}\right\Vert _{L^{1}}<\varepsilon.
$$
I claim that for each $x\in G$, we have $\left\Vert \lambda\left(x\right)f_{i}-f_{i}\right\Vert _{L^{1}}\to0$.
Indeed, let $\varepsilon>0$ be arbitrary and let $i_{0}:=\left(\left\{ x\right\} ,\varepsilon\right)$.
For $i=\left(F,\delta\right)\in I$ with $i\geq i_{0}$, we have $\delta\leq\varepsilon$
and $\left\{ x\right\} \subset F$ and hence
$$
\left\Vert \lambda\left(x\right)f_{i}-f_{i}\right\Vert _{L^{1}}\le\max_{y\in F}\left\Vert \lambda\left(y\right)f_{i}-f_{i}\right\Vert _{L^{1}}<\delta\leq\varepsilon.
$$
