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?
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:
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 $$\|F*\phi-\sum_ic_i\delta_{g_i}*\phi\|_\infty\leq \delta/2.$$