= Solution
Define two <open sets> by
$$
Z(u)=\bigcup\{U\subseteq\Omega:U\text{ open},\ u|_U=0\},\qquad
G(u)=\bigcup\{U\subseteq\Omega:U\text{ open},\ u|_U\in C^\infty(U)\}.
$$
The <support of a distribution> and its <singular support> are
$$
\boxed{\operatorname{supp}u=\Omega\setminus Z(u),\qquad
\operatorname{sing\,supp}u=\Omega\setminus G(u).}
$$
Both are relatively closed because the removed sets are open. The <locality of distributions> gives $u|_{Z(u)}=0$. On an overlap of two open smoothness regions, the two <smooth functions> represent the same <distribution>, so their difference is zero <almost everywhere> and hence everywhere by <continuity>. They therefore join to one <smooth function> on $G(u)$. The same <partition of unity> argument shows that it represents $u$ there.
For $u=\mathbf1_\omega$, openness of $\omega$ implies
$$
\boxed{\operatorname{supp}\mathbf1_\omega=\overline\omega^{\,\Omega}.}
$$
Indeed every neighborhood of a point in this relative closure meets $\omega$ in an <open set> containing a nonnegative <test function> with positive integral; outside the closure the <indicator function> is identically zero near the point.
The <singular support of an indicator function> needs a measure-sensitive boundary. For a <Lebesgue measurable set> $A\subseteq\Omega$, define its <essential closure> by
$$
\operatorname{esscl}_\Omega A
=\{x\in\Omega:|A\cap B_r(x)|>0\text{ whenever }r>0\text{ and }B_r(x)\subseteq\Omega\}.
$$
Then the general answer is
$$
\boxed{\operatorname{sing\,supp}\mathbf1_\omega
=\overline\omega^{\,\Omega}\cap\operatorname{esscl}_\Omega(\Omega\setminus\omega).}
$$
Outside this intersection, some ball has either no points of $\omega$ or a complement of <Lebesgue measure> zero. The <regular distribution> is respectively the constant zero or one on that ball and is smooth. Conversely, at a point of the intersection, every sufficiently small ball has positive <Lebesgue measure> in both sets. If a <smooth function> represented the <indicator function> there, its values would be in $\{0,1\}$ everywhere: any other value, by <continuity>, would give a ball of positive <Lebesgue measure> on which equality <almost everywhere> fails. On a connected ball such a continuous two-valued <function> is constant, contradicting the two positive measures.
In particular, the answer need not be the full topological <boundary>. For $\Omega=\mathbb R^n$ and $\omega=\mathbb R^n\setminus\{0\}$, the <indicator function> is the constant one <almost everywhere>, so its <singular support> is empty although the <boundary> is $\{0\}$. The full relative <boundary> is the answer when every neighborhood of each boundary point contains a positive <Lebesgue measure> portion of the complement, as for an <open set> with smooth boundary.
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