Usually, in order to define the space of distributions $\mathcal{D}'(\Omega)$ on an open subset $\Omega \subseteq \mathbb{R}^n$, one considers the space $\mathcal{D}(\Omega)$ of $C^\infty$-test functions with compact support in $\Omega \subseteq \mathbb{R}^n$. If we consider Radon measures, then we can also take $\Omega$ to be a locally compact subset, in particular a closed subset $F \subseteq \mathbb{R}^n$. Now, the space of distributions of order 0 on an open set $\Omega$ is precisely the space of Radon measures on $\Omega$. Is there some "natural" way to define distributions on a closed set $F$ such that the distributions of order 0 are precisely the Radon measures on $F$? As an example, we could have some Diracs on the boundary of $F$. If we consider a closed set $F$ then we can retrieve by open sets either from its inner as a closure or from its ambient space by intersections of open sets that contain $F$. The problem with the first "inner" approach seems to be that we miss the boundary of $F$ and thus do not get Diracs on the boundary. In other words, the restriction of a compactly supported map on $F$ is not necessarily compactly supported in the interior of $F$. Thus, it seems that we need to consider the ambient space of $F$ in order to not loose the distributions on the boundary of $F$.
Clearly, we can consider some open set $\Omega \supseteq F$ and the space of distributions on $\Omega$ that are supported in $F$. This should be one possibility. But, the choice of $\Omega$ is not unique. However, any other open $\Omega' \supseteq F$ should lead to a space of distributions with support in $F$ that is isomorphic. What is a "canonical" predual of the space of distributions in some $\Omega \supseteq F$ with support in $F$? Can we choose $C^\infty_c(F)$, where $\varphi \in C^\infty_c(F)$ should be continuous at the boundary of $F$? Note that, if $F$ is compact then $C^\infty_c(F) = C^\infty(F)$!
I hope to have found the answer. Let $A \subseteq \mathbb{R}^n$ be an arbitrary set and choose an arbitrary open set $\Omega \subseteq \mathbb{R}^n$ with $A \subseteq \Omega$ (e.g. $\Omega = \mathbb{R}^n$). Set $\mathcal{D} := \mathcal{D}(\Omega)$ and $\mathcal{D}' := \mathcal{D}'(\Omega)$. We want to find a pre-dual for the space of distributions $T \in \mathcal{D}'$ with $supp(T) \subseteq A$. In fact, since $T\varphi = 0$ for all $\varphi \in \mathcal{D}$ with $supp(\varphi) \subseteq \Omega \setminus A$ it follows that $T$ acts (possibly non-trivially) on functions $\varphi$ for which their support $supp(\varphi)$ intersects $A$. Roughly speaking, we have to forget about functions $\varphi$ with "support" outside $A$. In other words, we identify two functions whenever they are equal on $A$.
Formally, for $x \in \Omega$ consider the evaluation $\delta_x \in \mathcal{D}'$, $\delta_x(\varphi) = \varphi(x)$. Set $N := \bigcap_{x \in A} \delta_x^{-1}(\{ 0 \}) = \{ \varphi \in \mathcal{D} \ | \ \varphi(x) = 0 \text{ for all } x \in A \}$. Since $\delta_x : \mathcal{D} \to \mathbb{R}$ is continuous it follows that $N \subseteq \mathcal{D}$ is a closed linear subspace. Now consider the canonical projection $p : \mathcal{D} \to \mathcal{D} / N$, $\varphi \mapsto \varphi + N$ and equip the range with the quotient topology. Then $p$ is linear, continuous and surjective and thus its transpose $p^t : (\mathcal{D} / N)' \to \mathcal{D}'$ is injective. Check that the image of $p^t$ is precisely the space of distributions with support in $A$.
Alternatively, denote by $V \subseteq \mathcal{D}'$ the subspace of distributions with support in $A$ and let $i : V \to \mathcal{D}'$ be the inclusion map. Give $\mathcal{D}'$ the weak-* topology such that $V$ the induced weak-* topology such that the duals of these spaces are $\mathcal{D}$ respectively $V'$ (under the canonical isomorpism of a space with a subspace of its bidual). Then the transpose $i^t : \mathcal{D} \to V'$ is surjective and its kernel is precisely the subspace $N \subseteq \mathcal{D}$ of functions with support in $\Omega \setminus A$. Thus, we have as above the pre-dual $\mathcal{D} / N$.