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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:

  1. $\forall V\in\mathcal{I}, \operatorname{supp}(\varphi)\subset V$;
  2. $\forall V\in\mathcal{I}, \int_G\varphi\operatorname{d}\mu=1.$

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?

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  • $\begingroup$ For any $f \in C_c(G)$, for any sequence $V_n \supset V_{n+1}$ of neighborhoods of the identity such that $\bigcap U_n = \{e\}$, for any sequence $\phi_n \in C_c^+(U_n), \int_G\phi_nd\mu = 1$ then $\lim_{n \to \infty} \|f-\phi_n \ast f\|_\infty =0$ (convolution). So the problem reduces to show $f \ast \phi$ can be approximated by finite sums of translates of $\phi_n$, since then that we can approximate uniformly implies we can approximate by above uniformly. $\endgroup$ – reuns Mar 16 at 18:39
  • $\begingroup$ @Bob: "a net converging to $\delta$" - in what topology? $\endgroup$ – Alex M. Mar 23 at 13:39
  • $\begingroup$ It is a definition $\endgroup$ – Bob Mar 23 at 13:42
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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:

  1. 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 $f(x)>0.$
  2. 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 $\|F-F*\phi\|_\infty\leq \delta/2.$
  3. 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 $$\|F*\phi-\sum_ic_i\delta_{g_i}*\phi\|_\infty\leq \delta/2.$$

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  • $\begingroup$ Thanks for the (great) answer, I'm sorry for the delay in reading it. The reason why I was asking myself about this, is to find a motivation in introducing covering numbers in the construction of Haar measure that it's not "try and you'll find that it works". $\endgroup$ – Bob Mar 23 at 18:04

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