Problem $2.17$, Rudin's RCA (Dictionary Order Topology) 
Problem $2.17$: Define the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane to be $$|y_1-y_2| \quad \text{if }x_1 = x_2, \quad\quad 1+|y_1 - y_2|\quad \text{if } x_1\ne x_2$$
Show that this is indeed a metric, and that the resulting metric space $X$ is locally compact.
If $f\in C_c(X)$, let $x_1,\ldots,x_n$ be those values of $x$ for which $f(x,y)\ne 0$ for at least one $y$ (there are only finitely many such $x$!), and define $$\Lambda f = \sum_{j=1}^n \int_{-\infty}^\infty f(x_j,y)\ dy$$
Let $\mu$ be the measure associated with this $\Lambda$ by Theorem $2.14$. If $E$ is the $x$-axis, show that $\mu(E) = \infty$ although $\mu(K) = 0$ for every compact $K\subset E$.

Theorem $2.14$ above refers to the Riesz representation theorem, which relates $\mu$ with the linear functional $\Lambda$. I found a solution to the above problem here, which I need some help with understanding. I've reproduced it to the extent necessary below.

Hereafter, the distance defined above (between points in $\mathbb R^2$) is represented by $d$, and it is indeed a metric. It was also proved that $(X,\tau)$ is a LCHS (locally compact Hausdorff space), by identifying it as $(X,\tau) = (\mathbb R, \tau_1) \times (\mathbb R,\tau_2)$ where $\tau_1$ is the discrete topology on $\mathbb R$, and $\tau_2$ is the usual one. If $d_1$ and $d_2$ are the metrics corresponding to these topologies, it is easy to see that the product metric $d = d_1 + d_2$. Now, the real problem begins.

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*If $f\in C_c(X)$, why is it that there are only finitely many $x$ for which $f(x,y)\ne 0$ for at least one $y$? The link I've attached says:


If $K$ is compact in $X$, the first projection $\text{pr}_1(K)$ is compact in $(\mathbb R, τ_1)$. Hence
it is a finite set. Therefore $K$ is a finite union $$\{x_1\} × K_1 ∪ · · · ∪ \{x_n\} × K_n$$ where each $K_i$, $i = 1, 2, . . . , n$, is a compact set in $(\mathbb R, τ_2)$.

Here's my reasoning (please confirm if it is correct): Take $K = \text{supp}(f) = \overline{\{(x,y): f(x,y)\ne 0\}}$. Since $f\in C_c(X)$, $K$ is compact, and its projection $\text{pr}_1(K)$ is also compact. In the discrete topology, sets are compact iff they are finite, so $\text{pr}_1(K)$ is finite. Since projection maps preserve inclusion, we have $\text{pr}_1(\{(x,y): f(x,y)\ne 0\}) \subset \text{pr}_1(K)$ is finite. Now, $\text{pr}_1(\{(x,y): f(x,y)\ne 0\}) = \{x: \exists y, f(x,y)\ne 0\}$ which is finite (as the claim requires).


*In the very next paragraph, they mention that the support of $f\in C_c(X)$ is contained in $\{x_1,x_2,\ldots,x_n\}\times\mathbb R$. How do we use this to deduce that $\Lambda$ is a positive linear functional on $C_c(X)$?


*How do we get $\mu(\{x\} \times K) = m(K)$?


*


Let $V$ be an open set containing $\mathbb R × \{0\}$. Then for $x ∈ \mathbb R$, $(x, 0) ∈ V$ , so
that there exists an $ε_x > 0$ with $\{x\} × [−ε_x, ε_x] ⊂ V$. This implies that there must be an $n$ with uncountably many $ε_x \ge 1/n$. (Otherwise, $ε_x \ge 1/n$ for at most countably many $x$, contradicting the fact that $\mathbb R$ is uncountable.) Let $K_x = \{x\} \times [-ε_x/2,ε_x/2]$ for $ε_x \ge 1/n$. For $K = \bigcup_{j=1}^m K_{x_j}$, we have $\mu(K) \ge m/n$. Hence, if $V ⊃ \mathbb R × \{0\}$ is open, then $\mu(V ) ≥ \sup_{m∈\mathbb N} m/n= ∞$. This implies $\mu(\mathbb R × \{0\}) = ∞$.

How do we get $\{x\} × [−ε_x, ε_x] ⊂ V$? I know the idea is that there exists some $\epsilon_x$ such that the open ball of this radius centered at $x$ is in $V$, but why do open balls in this topology look like this? Also, I didn't really understand what was done after this to show $\mu(\mathbb R × \{0\}) = ∞$, and I'd appreciate if someone could explain in detail.
Thank you!
 A: It is better if you solve this problem on your own. The solution you quote as too confusing and inconsistent with notation.
As you already found out:

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*a. In the topological space  $X=(\mathbb{R}\times\mathbb{R},\tau_d\otimes\tau_e)$, where $\tau_d$ is the discrete topology on $\mathbb{R}$ and $\tau_e$ is the usual Euclidean topology, a compact set $K$ is of the form $\bigcup^n_{j=1}\{x_j\}\times K_j$, where $\{x_1,\ldots,x_n\}\subset\mathbb{R}$, and the $K_1,\ldots,K_n$ are compact sets in $(\mathbb{R},\tau_e)$. It is also clear that $X$ is Hausdorff and locally compact.

*b.  If $f\in\mathcal{C}_c(X)$, then there are points $\{x_j:1\leq j\leq m\}$ and a sequence  of functions $\phi_j\in\mathcal{C}_c((\mathbb{R},\tau_e))$, $1\leq j\leq m$ such that
$$ f(x_j,y)=\phi_j(y)\qquad 1\leq j\leq m, \quad y\in\mathbb{R}$$
The map $\Lambda(f):=\sum^m_{j=1}\int_\mathbb{R} \phi_j(y)\,dy$ clearly defines a linear and positive functional on $\mathcal{C}_c(X)$ (why?)

*c.  Form the Riesz representation theorem there is a measure $\mu$ on a $\sigma$-algebra $\mathfrak{M}_\Lambda$ that contains $\mathscr{B}(X)$ such that
$$\Lambda f=\int_X f\,d\mu,\qquad f\in\mathcal{C}_c(X)$$
To determine the marginals of $\mu$ over $(\mathbb{R},\tau_d)$ and $(\mathbb{R},\tau_e)$ first notice that for any fixed $\phi\in\mathcal{C}_c((\mathbb{R},\tau_e))$, and any $x_1,x_2\in\mathbb{R}$,
if $f_i(x,y)=\mathbb{1}_{\{x_i\}}(x)\phi(y)$, $i=1,2$, then $f_i\in\mathcal{C}_c(X)$ and
$$\Lambda f_1=\int_{\mathbb{R}}\phi(y)\,dy=\Lambda f_2$$
The Riesz representation theorem applied to $(\mathbb{R},\tau_e)$ and $\tilde{\Lambda}:\phi\mapsto \int_{\mathbb{R}}\phi(y)\,dy$, $\phi\in\mathcal{C}_c((\mathbb{R},\tau_e))$ implies that for any compact set $B\subset\mathbb{R}$, there is a sequence $\phi_n\in \mathcal{C}_c((\mathbb{R},\tau_e))$ such that $\phi_n\searrow\mathbb{1}_B$ point wise, and
$$\int_\mathbb{R}|\mathbb{1}_B(y)-\phi(y)|\,dy\xrightarrow{n\rightarrow\infty}0$$
Define $g^i_n(x,y)=\mathbb{1}_{\{x_i\}}(x)\phi_n(y)$, $ i = 1,2$. Then each $g^i_n\in\mathcal{C}_c(X)$ $ i = 1,2$, and $n\in\mathbb{N}$. Furthermore
$$
\int_X|\mathbb{1}_{\{x_i\}}(x)\mathbb{1}_B(y)-g^i_n(x,y)|\,\mu(dx,dy)=\int_{\mathbb{R}}|\mathbb{1}_B(y)-\phi_n(y)|\,dy\xrightarrow{n\rightarrow\infty}0$$
and so,
$$\int_X\mathbb{1}_{\{x_1\}}\mathbb{1}_{B}(y)\,\mu(dx,dy)=\int_X\mathbb{1}_{\{x_2\}}\mathbb{1}_{B}(y)\,\mu(dx,dy)=\lim_n\int_{\mathbb{R}}\phi_n(y)\,dy=\lambda(B)$$
That is,
$$\begin{align}\mu(\{x\}\times B)=\lambda(B)\tag{1}\label{one}\end{align}$$
for all $x\in\mathbb{R}$ and any compact set $B$.
(Solution to 3.) If $K\subset\mathbb{R}\times\{0\}=E$ is compact, then $K=\{x_1,\ldots,x_m\}\times\{0\}$ for some points $x_j\in\mathbb{R}$ and so $\mu(K)=\sum^m_{j=1}\mu(\{x_j\}\times\{0\})=0$ by \eqref{one}.
(Solution to 4.) The Riesz representation also implies that for any measurable $A\subset X$
$$ \mu(A)=\inf\{\mu(G):G\,\text{open,}\, A\subset G\}$$
Any open set $G$ that contains $E$, also contains sets of the form $U=\bigcup_{x\in\mathbb{R}}\{x\}\times(-a_x,a_x)$, $a_x>0$. Since $\mathbb{R}$ is uncountable, there is $n_0\in\mathbb{N}$ such that $X_{n_0}=\{x:a_x>\frac{1}{n_0}\}$ is uncountable. Thus, for such $G$
$$\mu(G)\geq\mu^*(U)\geq\sup_{J\subset X_{n_0}}\sum_{x\in J}2a_x=\infty,$$ where the sup is over all finite subsets of $X_{n_0}$. That is $\mu^*(E)=\infty$.
