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Suppose I have a function of the form for $y\in \mathbb{R}$ $$H(y) = \int_{-\infty}^\infty \delta(y-f_1(x)) f_2(x) dx$$ assuming that $f_1, f_2$ are functions which are sufficiently nice that this expression makes sense. I want to argue that $$\int_{-\infty}^\infty H(y) dy = \int_{-\infty}^\infty \left( \int_{-\infty}^\infty \delta(y-f_1(x)) dy \right) f_2(x) dx = \int_{-\infty}^\infty f_2(x) dx$$ I would like to invoke the Fubini-Tonelli theorem but the problem is $\delta$ is a distribution and not a function. Is there a generalisation of Fubini-Tonelli which I can use here?

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    $\begingroup$ It is obvious you need to define $\delta(y-f_1(x))$ and propose some "sufficiently nice" conditions $\endgroup$ – reuns Aug 14 at 22:25
  • $\begingroup$ Let $f_1$ and $f_2$ be test functions on $\mathbb{R}$. Letting $\delta_y$ be the Dirac-delta distribution centred at $y$, we have $H(y) = \delta_y \circ f_1 [f_2] $. I am writing this out because you asked for it but I really think it is unnecessary and these technicalities totally miss the point. If you need such things explained, then you certainly won't be able to answer my question. $\endgroup$ – UtilityMaximiser Aug 15 at 23:08
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    $\begingroup$ You are completely wrong. JG just proposed a definition of $\delta(y-f_1(x))$ and obtained the result trivially from the conditions for exchanging $\lim$ and $\int$. Your definition of $H$ doesn't make sense, I'm asking for a definition of $\delta(y-f_1(x))$. $\endgroup$ – reuns Aug 15 at 23:10
  • $\begingroup$ I would link to the Wikipedia page for the delta function, but I'm a bit puzzled why I should have to do that. I'm asking for help with analysis. My time is not best served explaining the basics of distribution theory to people who are unlikely to answer my question. $\endgroup$ – UtilityMaximiser Aug 15 at 23:25
  • $\begingroup$ Then do you understand that $n 1_{x \in [0,1/n]} \to \delta$ and that $n 1_{y-f_1(x) \in [0,1/n]} \to \delta(y-f_1(x))$ whenever $f_1$ is $C^1$ with finitely many simple zeros ? This is the definition of compositions of distributions with a function and it is clearly compatible with the laws of change of variables in integrals. When everything is continuous we can exchange $\lim$ and $\int$ obtaining your result. $\endgroup$ – reuns Aug 15 at 23:35
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@AbdelmalekAbdesselam notes such a generalization is found in Théorie des distributions 1966 Ch 4, Sec. 3, Thm. IV. Alternatively, you could use nascent deltas $\eta_\epsilon(x):=\frac{1}{\epsilon}\eta\left(\frac{x}{\epsilon}\right)$ so$$\int_{\Bbb R}H(y)dy=\int_{\Bbb R}\left(\lim_{\epsilon\to0^+}\int_{\Bbb R}\eta_\epsilon\left(y-f_1(x)\right)f_2(x)dx\right)dy\\=\int_{\Bbb R}\left(\lim_{\epsilon\to0^+}\int_{\Bbb R}\eta_\epsilon\left(y-f_1(x)\right)dy\right)f_2(x)dx,$$where the second $=$ uses ordinary Fubini-Tonelli together with some limit-integral commutations you'd need to justify. The expression then becomes$$\int_{\Bbb R}\left(\int_{\Bbb R}\delta\left(y-f_1(x)\right)dy\right)f_2(x)dx=\int_{\Bbb R}f_2(x)dx.$$

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  • $\begingroup$ Why "Absent such generalization"? There is a Fubini Theorem for distributions, see Schwartz's book Ch 4, Sec. 3, Thm. IV $\endgroup$ – Abdelmalek Abdesselam Aug 20 at 17:47
  • $\begingroup$ @AbdelmalekAbdesselam It's been a week, but thanks; I've edited that into my answer. (All I ever meant was "should no-one volunteer a generalization"; I for one didn't know of one.) Which book of Schwartz's is it? If I can find the theorem, I'll update my answer again. $\endgroup$ – J.G. Aug 20 at 17:51
  • $\begingroup$ I think your answer is perfect for the question which does not need this Fubini Thm for distributions but 1) a proper definition of $\delta(y-f_1(x))$ and then 2) the usual Fubini of measure theory. The book by Schwartz I was referring to is the very classic "Theorie des distributions" 1966 edition. $\endgroup$ – Abdelmalek Abdesselam Aug 20 at 17:55

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