Is it true that the Haar covering numbers $(f:\varphi)$ converges to $\int_Gf\operatorname{d}\mu$ for $\varphi\to\delta$? Let $(G,\cdot,\tau)$ be a topological group whose topology is Hausdorff and locally compact and whose identity is $e.$
Denote by $\mathcal{B}_\tau$ the family of Borel subsets of $(G,\cdot,\tau)$, i.e. the $\sigma$-algebra of subsets of $G$ generated by $\tau$.
Let $\mu:\mathcal{B}_\tau\to[0,+\infty]$ be a left-Haar measure of $(G,\cdot,\tau)$, i.e. a non-null left-translation-invariant Radon-measure (outer regular on the elements of $\mathcal{B_\tau}$ and inner regular on the elements of $\tau$) defined on $\mathcal{B}_\tau$.
Denote by $\mathcal{I}$ the family of neighborhoods of $e$ in $(G,\cdot,\tau)$ with its natural partial order (i.e. $V\le U$ iff $U\subset V$), that makes it a directed family. 
Denote by $C^+_c(G)$ the family of non-null, non-negative, continuous functions of compact support of $(G,\cdot,\tau)$.
We will say that a net $\varphi: \mathcal{I}\to C^+_c(G)$ converges to $\delta$ in $(G,\cdot,\tau,\mu)$ if:


*

*$\forall V\in\mathcal{I}, \operatorname{supp}(\varphi)\subset V$;

*$\forall V\in\mathcal{I}, \int_G\varphi\operatorname{d}\mu=1.$
If $f,\varphi\in C^+_c(G)$ define:
$$A_{f,\varphi}:=\left\{\alpha>0 : \exists n\in\mathbb{N}, \exists c_1,...,c_n>0, \exists g_1,...,g_n\in G, \left(f\le\sum_{k=1}^nc_k\tau_{g_k}\varphi\right)\land\left(\alpha=\sum_{k=1}^nc_k\right)\right\}$$
where $\tau_h(\varphi)(g):=\varphi(h^{-1}g)$
and define the Haar covering number by:
$$(f:\varphi):=\inf(A_{f,\varphi}).$$

If $\varphi :\mathcal{I}\to C^+_c(G)$ is a net converging to $\delta$ in $(G,\cdot,\tau,\mu)$, is it true that the net $\left((f:\varphi_V)\right)_{V\in\mathcal{I}}$ converges to $\int_G f\operatorname{d}\mu?$

Since $$\forall f,\varphi\in C^+_c(G), \int_G f\operatorname{d}\mu\le(f:\varphi)\int_G \varphi\operatorname{d}\mu,$$ it is clear that, if the net converges at all, then the limit is greater or equal than $\int_G f\operatorname{d}\mu$.
However, I strongly believe that (since the support of $\varphi_V$ gets smaller and smaller as the net index $V$ shrinks to $\{e\}$) the quantity $(f:\varphi_V)$ should be a good approximation of $\int_G f\operatorname{d}\mu$ if $V$ is small enough, so that equality should hold, but I can't prove rigorously this claim.
Any help?
 A: Those covering numbers are used to construct Haar measure, and have no other application. So it's strange to assume the existence of Haar measure when studying them. Anyway, here is a proof.
Consider an arbitrary $\epsilon>0.$ Proof outline:


*

*Find $\delta>0$ and $F\in C_c^+(G)$ with $\int F\;d\mu\leq \int f\;d\mu+\epsilon$ and $F(x)\geq f(x)+\delta$ whenever $f(x)>0.$

*Find $U\in\mathcal I$ such that whenever $\phi\in C_c^+(G)$ is zero outside $U$ and has $\int \phi\;d\mu=1$ then $\|F-F*\phi\|_\infty\leq \delta/2.$

*Given any such $\phi,$ find a finitely-supported measure $\nu=\sum c_i\delta_{g_i}$ with $\nu(G)=\int F\;d\mu$ and such that $\|F*\phi-\nu*\phi\|_\infty\leq\delta/2.$
These ensure $f\leq \nu*\phi$ and $\int \nu*\phi\;d\mu\leq\int f\;d\mu+\epsilon,$ which is enough to show that $(f:\phi_V)_{V\in\mathcal I}$ converges to $\int f\;d\mu.$
For 1. Construct $G\in C_c^+(G)$ that is strictly positive on $\operatorname{supp}(f),$ the closure of $\{x\mid f(x)>0\}.$ Scale $G$ if necessary to get $\int G\;d\mu\leq\epsilon.$ Take $F=f+G$ and $\delta=\min_{x\in\operatorname{supp}(f)}G(x).$
For 2. By uniform continuity of $F$ there is $U\in\mathcal I$ such that $|F(y)-F(x)|\leq\delta/2$ whenever $y^{-1}x\in U.$ This gives $$\left|\int (F(y)-F(x))\phi(y^{-1}x)\;d\mu(y)\right|\leq\delta/2$$
because $\phi(y^{-1}x)=0$ unless $y^{-1}x\in U,$ in which case $|F(y)-F(x)|\leq\delta/2.$ So $\|F*\phi-F\|_\infty\leq\delta/2.$
For 3. Let $A=\int F\;d\mu.$ By uniform continuity of $\phi$ there is $V\in\mathcal I$ such that $|\phi(y^{-1}x)-\phi(x)|\leq\delta/2C$ whenever $y\in V.$
Using a partition of unity write $F=F_1+\dots+F_n$ where each $F_i$ is zero outside some right translate $Vg_i.$ Let $c_i=\int F_i\;d\mu.$
For each $i$ and $x\in G,$
$$\left|\int F_i(y)(\phi(y^{-1}x)-\phi(g_i^{-1}x))\;d\mu\right|\leq c_i\delta/2C$$
because $F_i(y)=0$ unless $yg_i^{-1}\in V,$ in which case $|\phi(y^{-1}x)-\phi(g_i^{-1}x)|\leq \delta/2C.$ This gives
$$\|F_i*\phi-c_i\delta_{g_i}*\phi\|_\infty\leq c_i\delta/2C.$$
Therefore
$$\|F*\phi-\sum_ic_i\delta_{g_i}*\phi\|_\infty\leq \delta/2.$$
