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 Incoming links: Singular support of an indicator function

Z (u) = ⋃ {U \subseteq Ω : U open, u ∣_{U} = 0}, G (u) = ⋃ {U \subseteq Ω : U open, u ∣_{U} \in C^{\infty} (U)} .

supp u = Ω ∖ Z (u), sing supp u = Ω ∖ 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 = 1_{ω}

, openness of

ω

implies

supp 1_{ω} = \overline{ω}^{Ω} .

Indeed every neighborhood of a point in this relative closure meets

ω

in an open set containing a nonnegative test function with positive integral; outside the closure the indicator function is identically zero near the point.

A \subseteq Ω

esscl_{Ω} A = {x \in Ω : ∣ A \cap B_{r} (x) ∣ > 0 whenever r > 0 and B_{r} (x) \subseteq Ω} .

Then the general answer is

sing supp 1_{ω} = \overline{ω}^{Ω} \cap esscl_{Ω} (Ω ∖ ω) .

Outside this intersection, some ball has either no points of

ω

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

Ω = R^{n}

and

ω = R^{n} ∖ {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.