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In Wikipedia page of Dirac delta distribution (here), there is the generalization for a property of Dirac delta distribution given by enter image description here

I know this is a generalization of the case when $g(\mathbf{x})$ has finite number of roots (in which case the right hand side becomes a summation over the roots). However, I have difficulty interpreting the integration measure. What is wrong with taking the right hand side to be simply the standard Riemannian integral over $f(\mathbf{x})/|\nabla g|$ over the level set $g^{-1}(0)$, i.e. over $d^{n-1}\mathbf{x}$, instead of what is called Minkowski content measure in the Wikipedia page?

I am looking for intuition and less so from rigorous interpretation in measure theory because I would like to use this for a very practical computation, even numerically perhaps. I would nonetheless appreciate some rigorous explanation on how to bridge the intuition and measure theory.

Evangeline A. K. McDowell
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  • You can take it as a definition, then show it makes sense (that it is compatible with the changes of variables formulas). The alternative approach that I prefer is to define $\int_{R^n} f(x) \delta(g(x))dx = \lim_{n \to \infty} \int_{R^n} f(x) \frac{1_{|g(x)| < 1/n}}{2n}dx$ then your formula is a theorem which follows from $=\lim_{n \to \infty} \frac{1}{2n} \lim_{n \to \infty} \int_{ { x, \exists a, g(a) =0, |x-a| < 1/n}} f(x) \frac{1}{|\nabla g(x)|}dx$. What is your proposal for a measure on $g^{-1}(0)$ alternative to $\sigma$ ? – reuns Aug 02 '19 at 04:57
  • @reuns thanks. I think I am just not seeing what the Minkowski measure is. Naively I would have thought a simple Riemannian integral divided by the gradient will do, restricted to $g^{-1}(0)$. – Evangeline A. K. McDowell Aug 04 '19 at 18:13
  • "A simple Riemannian integral restricted to $g^{-1}(0)$" ?? – reuns Aug 04 '19 at 18:18

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