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Reading this I see that the statement:

$ \delta \left( f(x) \right) = \sum_i \dfrac{\delta(x - a_i)}{|f'(a_i)|} $

is equivalent to showing that:

$ \int_{-\infty}^{\infty} g(x)\delta \left( f(x) \right) = \sum_i \dfrac{g(a_i)}{|f'(a_i)|} $

Where $ f(a_i) = 0 \:\: \forall i $

Can somebody explain to me why these two statements are equivalent? Thanks in advance

(I understand some of the proofs in that post, I just don't get why the statements are the same)

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  • $\begingroup$ You misquoted the statement, but if you understood even a single one of those proofs, you understood why it's true (that's the meaning of "proof"). If you didn't understand them, why don't you ask their authors, in that thread? $\endgroup$ – Professor Vector Aug 19 '17 at 14:57
  • $\begingroup$ math.stackexchange.com/questions/2389100/… $\endgroup$ – md2perpe Aug 19 '17 at 17:02
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This is the definition of the Dirac delta : $$\int_{-\infty}^\infty g(x) \delta(x)dx = \lim_{\epsilon \to 0^+} \int_{-\infty}^\infty g(x) \frac{1_{|x| < \epsilon}}{2 \epsilon}dx=g(0)$$ whenever $g$ is continuous, which means $\delta(x)$ is the limit in the sense of distributions of $\frac{1_{|x| < \epsilon}}{2 \epsilon}$ as $\epsilon \to 0^+$.

Thus it is natural to define $\delta(f(x))$ as the limit in the sense of distributions of $\frac{1_{|f(x)| < \epsilon}}{2 \epsilon}$ as $\epsilon \to 0^+$, which means $$\int_{-\infty}^{\infty} g(x)\delta \left( f(x) \right)dx =\lim_{\epsilon \to 0^+}\int_{-\infty}^{\infty} g(x)\frac{1_{|f(x)| < \epsilon}}{2 \epsilon}dx= \sum_{f(\alpha)=0} \dfrac{g(\alpha)}{|f'(\alpha)|}\\=\int_{-\infty}^\infty g(x)\sum_{f(\alpha)=0} \dfrac{\delta(x-\alpha)}{|f'(\alpha)|} dx$$ whenever $g$ is continuous and $f$ is $C^1$ and has finitely many zeros.

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