# 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:

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?

• 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. – reuns Mar 16 at 18:39
• @Bob: "a net converging to $\delta$" - in what topology? – Alex M. Mar 23 at 13:39
• It is a definition – Bob Mar 23 at 13:42

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

• 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". – Bob Mar 23 at 18:04