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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 $\mathrm{supp}(u) = \psi^{-1}(0)$.

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    $\begingroup$ If $A = \{ a_k \}$ is countable, couldn't we then take $\langle u, \varphi \rangle = \sum_k 2^{-k} \varphi(a_k)$ to have $u$ well-defined? $\endgroup$
    – md2perpe
    Aug 3, 2019 at 21:22
  • $\begingroup$ @md2perpe That would indeed work as far as I can see, thanks for the comment $\endgroup$
    – MSDG
    Aug 3, 2019 at 21:26
  • $\begingroup$ Can the intersection of a closed set with a compact set have a countable infinity of points? Will it not then be dense, so that its closure (i.e. it self, since it's closed) contains some interval of points and thus be uncountable? $\endgroup$
    – md2perpe
    Aug 3, 2019 at 21:28
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    $\begingroup$ Yes, there would be no countably infinite compact sets.But I think that I was wrong; take $A = \{ 1/k \mid k = 1, 2, 3, \ldots \} \cup \{0\}.$ That's countably infinite and compact. $\endgroup$
    – md2perpe
    Aug 3, 2019 at 21:50
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    $\begingroup$ You could take a countable dense set inside $A$, which exists by the separability of $\mathbb{R}^d$, and use the idea above $\endgroup$
    – Del
    Aug 6, 2019 at 3:31

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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 $A$.

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    $\begingroup$ I don't understand why you don't just say "let $\{a_k\}$ be a countable subset of $A$ whose closure is $A$ and $T = \sum_{k\ge 1} 2^{-k} \delta_{a_k}$ then its support is $A$" $\endgroup$
    – reuns
    Aug 11, 2019 at 8:46
  • $\begingroup$ I wanted to prove the separability of $A$, as it is not necessarily obvious that a subset of a separable space is separable (actually this is true for any metric separable space but not in general) $\endgroup$
    – FXV
    Aug 11, 2019 at 9:03
  • $\begingroup$ Let $a_0 \in A$, find a sequence of balls such that $\Bbb{R}^d = \bigcup_{k\ge 1} B(x_k,r_k)$ and $r_k \to 0$, for each $k$ pick $a_k \in A \cap B(x_k,r_k)$ if it is non-empty otherwise set $a_k = a_0$. For each $b \in A$ there is some $a_k$ arbitrary close to it. $\endgroup$
    – reuns
    Aug 11, 2019 at 9:11

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