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Define two open sets byThe support of a distribution and its singular support areBoth 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 $u=1_{\omega }$, openness of $\omega$ impliesIndeed 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 byThen the general answer isOutside 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.

For any Lebesgue measurable set $A\subseteq \Omega$, the singular support of its indicator function, viewed as a regular distribution, is the intersection of the two essential closures in the displayed formula. Outside this intersection the indicator function is constant almost everywhere on a ball and hence represents a smooth function. At a point in the intersection, each sufficiently small ball has positive Lebesgue measure in both sets. Any continuous representative on that ball must take only the values zero and one everywhere: every other value would persist in a smaller ball and violate equality almost everywhere. By connectedness of the ball it would then be constant, a contradiction.

If $A$ is open, its essential closure is its relative closure. Its singular support can still be smaller than its topological boundary: $A={\mathbb{R}}^n\setminus \{0\}$ represents the constant one and has empty singular support.

Target: Support and singular support rules for convolution. Parent: Convolution of distributions with a compactly supported factor. Intended file: fourier-analysis.bigb.