Let $f,g$ and $f\cdot g$ be a functions with $\int_{-\infty}^\infty|h(x)|dx<\infty$ for $h=f,g,fg$.
I have now given in a proof $$\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(u)g(w)\exp(-iwt)du\right)dw=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty f(u)g(w)\exp(-iwt)dw\right)du$$
without saying why this equation is right.
So when is this right?
In other words: When can you change the order of integration? What are the conditions? And why?
Add: I know for Lebesgue integrable functions the theorem of Fubini but above I don't have given functions in $\mathscr L^1$.
Best Answer
This is Fubini's Theorem. Both $f$ and $g$ are integrable functions and $e^{-iwt}$ has absolute value one (assuming $t$ is real). So
$$ \int_{-\infty}^\infty\left(\int_{-\infty}^\infty |f(u)g(w)\exp(-iwt)|du\right)dw = ||f||_1 ||g||_1 $$ where $||\cdot||_1$ is the $L^1$-norm.
First we can take out $g(w)$ since it is independent of $u$ and then take out $||f||_1$ since it is a constant:
$$ \begin{split} &\int_{-\infty}^\infty\left(\int_{-\infty}^\infty |f(u)g(w)|du\right)dw = \int_{-\infty}^\infty |g(w)| \left(\int_{-\infty}^\infty |f(u)|du\right)dw \\ &= \int_{-\infty}^\infty |g(w)| ||f||_1 dw = ||f||_1 \int_{-\infty}^\infty |g(w)| dw = ||f||_1 ||g||_1 \end{split} $$ So $f(u)g(w)$ is thus integrable in $\mathbb{R} \times \mathbb{R}$ (but remember that this is different from $fg$ being integrable in $\mathbb{R}$ which was one your assumptions). I have by the way used the version of Fubini where one checks the iterated integrals (it is in the Wikipedia page).