# Measure of $R_d \times \{0\}$

Let $X=R_d \times R$, where $R_d$ denotes the set of real numbers with the discrete topology and $R$ the set of real numbers with the natural topology. For every $f \in C_c(X)$, one has $f(\{x\} \times R ) \neq 0$ for at most a finite number of $x_1, ...,x_m \in R$. We put $I(f)=\sum_{i=1}^m \int_{R} f(x_i, y)\text{d}y$. Then $I$ is a positive linear functional on $X$. Let $\mu$ be the measure corresponding to $I$ by the Riesz representation theorem.

How to show that $\mu(R_d \times \{0\})=\infty$ (or more generally, $\mu(A \times \{0\})=\infty$ if $A$ is not countable)?

Thanks.

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I removed the tags (fourier-analysis) and (topological-groups) as these are at best marginally related. –  t.b. Aug 8 '11 at 10:34
Note: this gives the Haar measure for the topological group $\mathbb{R}_d \times \mathbb{R}$. –  Mark Aug 8 '11 at 10:35

First define $I$ for all nonnegative continuous functions $g$ like this: $$I(g) = \sup\{I(f):f \le g, f \in C_c(X)\} .$$ Then define $\mu^*$ on open sets $G$ like this: $$\mu^*(G) = \inf\{I(g): g \ge 1_G, g \text{ continuous}\} .$$ Then define $\mu^*$ for all sets $A$ like this: $$\mu^*(A) = \inf\{\mu^*(G): G\supseteq A, G\text{ open}\} .$$ Finally restrict $\mu^*$ to its measurable sets.
Is it true that $\mu^*(G)=sup\{I(g): g\in C_c, 0\leq g \leq 1, supp g \subset G \}$ for open $G$ (measure in the Riesz theorem is defined in such a way for open sets) ? –  Richard Aug 8 '11 at 13:55
@Richard: Even with this, any open set containing $R_d \times \{0\}$ has infinite measure, right? –  GEdgar Aug 8 '11 at 14:31
The measure $\mu$ in Riesz representation theorem is outer regular, i.e. for every measurable set $E$ (in particular for every Borel set) $\mu(E)=\inf\{\mu(G): E \subset G, \ G \ is \ open\}$. My problem is just how to show that for every open set $G$ in $X$ containing $R_d\times 0$ is $\mu(G)=\infty$. –  Richard Aug 8 '11 at 15:14
Thank yor for your help. Now I probably know how to prove it. Since there exists a function $f \in C_c(R)$, $0\leq f \leq 1$, such that $f(0)=1$ and $supp f \subset (-\varepsilon, \varepsilon)$, it is clear that $\mu(J \times (-\varepsilon, \varepsilon))=\infty$, where $J$ is infinite subset of $R_d$, $\varepsilon >0$. Each open set $G$ containing $R_d \times 0$ contains a set of the form $J \times (-\varepsilon, \varepsilon)$. Hence $\mu(R_d \times 0)=inf \{\mu(G): R_d \times 0 \subset G, \ G \ is \ open \}=\infty$. –  Richard Aug 9 '11 at 10:40