I've been working on this problem for a while but cannot seem to reach a solution. Let $\iota(A)= \bar{\bar{I}}(\xi_A)$ where $\xi_A$ is the characteristic function of the set $A=\{(x,0) | x \in \mathbb{R}\}$ seen as a subset of the set $\mathbb{R}_d \times \mathbb{R}$ where $\mathbb{R}_d$ is the set of real numbers with the discrete topology and $\mathbb{R}$ the set of real numbers with the usual topology. Now $\bar{\bar{I}}$ is given by,
$$\bar{\bar{I}}(g) = inf\{\bar{f} | f \text{ is lower semicontinuous}, f \geq g\}.$$
Now, for $g$ a positive lower semicontinuous function,
$$\bar{I}(g) = sup\{I(f)| f \in C_{00}^+(\mathbb{R}_d \times \mathbb{R}) , f \leq g\},$$
taking the set $C_{00}^+(\mathbb{R}_d \times \mathbb{R})$ as the set of functions with compact support from $\mathbb{R}_d \times \mathbb{R}$ to $\mathbb{C}$ and $I$ a nonnegative linear functional defined on $C_{00}^+(\mathbb{R}_d \times \mathbb{R})$, namely the Radon Measure.
I need to prove that $\iota(A)\neq 0$, however for every compact set $K \subseteq \mathbb{R}_d \times \mathbb{R}$ we have that $\iota(A \cap K)=0$.
My attempt to the solution is noticing that a compact set $K$ in $\mathbb{R}_d \times \mathbb{R}$ can be seen as $K = \{x_1,...,x_n\} \times K'$ where $K' \subseteq \mathbb{R}$ is a compact set. Therefore, if one of the $x_i$ for $i=1,...,n$ is not 0, then $\iota(A \cap K)=\iota(\emptyset) = 0$. However, if one of the $x_i$ is 0 then $A \cap K = \{0\} \times K '$ and I don't know how to prove that $\iota(\{0\} \times K ')=0$, much less argue that $\iota(A) \neq 0$. Any thoughts on this would be deeply appreciated.
Your general form of a compact set of $\mathbb R_d \times \mathbb R$ is not quite right: The correct form would be $\{x_1\} \times K_1 \cup \cdots \cup \{x_n\} \times K_n$ where $K_1, \ldots, K_n$ are compact in $\mathbb R$. Indeed, any such set is clearly compact since finite products and unions of compact sets are compact, and if $K \subset \mathbb R_d \times \mathbb R$ is a compact set, then since the projections are continuous and continuous images of compact sets are compact, $\pi_{\mathbb R_d}(K) = \{x_1, \ldots, x_n\}$ for certain elements $x_1, \ldots, x_n \in \mathbb R$, and for each $j \in \{1, \ldots, n\}$, the set $$ K_j := \{y \in \mathbb R| (x_j,y) \in K\} $$ is compact as a closed subset of a compact set; indeed, it is a closed subset of $\pi_{\mathbb R}(K)$, because vertical open sets are open, since the horizontal space carries the discrete topology.
Now note the following: Suppose that $g$ is already lower semi-continuous. Then $$ \overline I(g) = g, $$ because one obviously can't do any better than $g$ itself. Thus, we obtain $$ \overline{\overline I}(g) = \inf \{f| f \text{ lower semi-continuous}, f \ge g\}. $$ From this we see that whenever $K \cap A \neq \emptyset$ (be $K$ compact or not), then in fact $\iota(A \cap K) \neq 0$ because if $(x,0) \in A \cap K$, then $$ \overline{\overline I}(g)(x,0) \ge 1. $$ Thus, the proposition as stated is false.