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.
- 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).
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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 $C_c(X)$?
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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!
Best Answer
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:
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 $\mu^*(E)=\infty$.