Citing Stein and Weiss' "Introduction to Fourier Analysis", if $f,g\in L^1$ then we have the so-called Multiplication formula: $$ \int \hat f g\,dx = \int f\,\hat g\, dx $$ The proof is just a couple of lines:
$$ \int \hat f(x) g(x)\,dx = \int\left( \int f(t)e^{-2\pi i\, tx}\,dt\right) g(x)\, dx = \\ \int\left( \int g(x)e^{-2\pi i\, tx}\,dx\right) f(t)\, dt = \int f(t)\,\hat g(t)\, dt$$
My question is: in order to apply Fubini's theorem, shouldn't we require also that $~fg\in L^1$? I don't see why the hypotheses of Fubini's theorem are met otherwise.
$\int \int |f(t)g(x)|\, dt\, dx=(\int|f(t)|\, dt )(\int |g(x)|\, dx) <\infty$ so Fubini's Theorem is applicable.