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Dirac-delta counting formula for regular zeros
Statement
Assume the Axiom of Countable Choice. Let , let be open, and let be . Suppose every zero is regular, meaning that the real derivative is invertible. For every real continuous compactly supported function , put
where is the open Euclidean ball and is its Lebesgue volume. Then the limit exists and
Only finitely many terms on the right are nonzero. This gives the precise distributional meaning, against compactly supported test functions, of as the counting measure on regular zeros.
Facts & Assumptions
Given: Countable choice, , and the regularity of every zero as in the Statement.
The Euclidean inverse function theorem: a map between open subsets of with invertible derivative at a point is a diffeomorphism between suitable neighborhoods of that point and its image.
A diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: assuming countable choice, a diffeomorphism between open subsets of satisfies for every nonnegative measurable .
The Axiom of Countable Choice (): countable choice is the set-theoretic assumption used by [F2].
Proof
Let , a compact subset of . Every point of is isolated by [F1]. This zero set is finite: if it were infinite, compactness of would give a convergent subsequence of distinct zeros with limit ; continuity gives , while the inverse-function neighborhood of from [F1] contains no other zero, a contradiction.
For each of these finitely many zeros , choose a neighborhood on which is a diffeomorphism onto an open neighborhood . Shrink the neighborhoods to be pairwise disjoint. The compact set contains no zero, so continuity of gives a positive lower bound for there. Choose small enough that for every and that no point of maps into . If there are no such zeros, compactness gives the same positive lower bound on all of , and the integral is zero for sufficiently small .
It is enough first to take ; a real is the difference of its nonnegative continuous compactly supported positive and negative parts. For and as in step 2.1, the integrand vanishes on . Apply [F2], under the countable-choice hypothesis [F3], to with the nonnegative measurable integrand . Change of variables makes the integral over equal to . Summing these finitely many terms gives the original integral.
The function is continuous at and has value . Its integral against is the average over ; continuity implies that this average tends to . The sum is finite by step 1.1, so taking the limit in step 3.1 gives the claimed sum for nonnegative , and subtraction gives it for every real .
The limiting distribution assigns each regular zero the unit mass ; therefore its density notation is exactly , with no additional multiplicity or orientation sign.