 Singular support of an indicator function

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.