Fubini-Tonelli theorem for distributions

Question

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?

Answer 1

@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$$\int_{\Bbb R}\left(\int_{\Bbb R}\delta\left(y-f_1(x)\right)dy\right)f_2(x)dx=\int_{\Bbb R}f_2(x)dx.$$