Distribution with support equal to a prescribed set 
Question: Let $A \subset \mathbb{R}^d$ be a closed set. Is there a distribution $u \in D'(\mathbb{R}^d)$ such that $\mathrm{supp}(u) = A$?

Thoughts: If the set $A$ is countable, then one can just take $u$ to be integration over $A$ with respect to the counting measure.
EDIT: Actually this will only be true if the intersection of $A$ with any compact set contains only finitely many points, since $u$ is not well-defined otherwise.
EDIT 2: As suggested in the comments, one could actually take $\sum_{k=1}^\infty 2^{-k}\delta_{a_k}$, if $A = \{a_k\}_{k=1}^\infty$ is (at most) countable.
If $A$ is connected and has interior, then one could take the $u$ to be the indicator function of $A$ (I was not sure what the best way to state the condition on $A$ is in this case). 
Context: I was studying a theorem which says that if $\psi$ is a smooth function and $u$ is a distribution such that $\psi u = 0$, then $\mathrm{supp}(u) \subset \psi^{-1}(0)$.
I was wondering if this result was sharp, in the sense that for each $\psi$, there is a distribution $u$ with $\psi u = 0$ and $\mathrm{supp}(u) = \psi^{-1}(0)$.
 A: Based on the comments, the answer to the question is yes.
Take any closed set $A$, for all $N\in\mathbb N$:
$$A\cap[-N,N]\subset\bigcup_{x\in A} B(x,1/N)$$
so by compacity there exists a finite subset of $A$, say $A_N$, such that:
$$A\cap[-N,N]\subset\bigcup_{x\in A_N} B(x,1/N)$$
$A_\infty=\bigcup A_N$ is a countable subset of $A$ which is dense in $A$, indeed for all $x\in A$ and $\varepsilon > 0$ by taking $N>\max(|x|,1/\varepsilon)$ there exists $x_0 \in A_N\subset A$ such that $x\in B(x_0, 1/N) \subset B(x_0, \varepsilon)$.
Now write $A_\infty = \{ a_k, k\in \mathbb N\}$ and let:
$$T=\sum_{k\in \mathbb N} 2^{-k}\delta_{a_k}$$
By definition, the support of $T$ is the complement of the largest open set where $T$ vanishes.
If $f$ is a test function which support is included in $A^c$, clearly $f(a_k)=0$ for all $k$ and this $T(f)=0$. Hence $T$ vanishes on $A^c$
Conversely $V$ is an open set containing a point $x \in A$, then it contains a point $a_k \in A_\infty$ and one can build a positive test function $f$ such that $f(a_k)>0$ and $\text{supp}(f)\subset V$. Thus $T(f)\geq 2^{-k}f(a_k) >0$ and $T$ does not vanish on $V$.
So we can deduce that the support of a distribution can be any closed set $A$.
