An application of Fubini’s theorem on Fourier transform

Question

Given $f,g\in L^1(\mathbb{R}^n)$ and we denote the Fourier transform of $f$ by $\widehat{f}$.

I want to prove that $$\int_{\mathbb{R}^n}\widehat{f}(x)g(x)~dx= \int_{\mathbb{R}^n}f(x)\widehat{g}(x)~dx.$$

Here’s my attempt: \begin{align} \int_{\mathbb{R}^n}\widehat{f}(x)g(x)~dx=&\int _{\mathbb{R}^n}\left\{\int _{\mathbb{R}^n} f(t)e^{-2\pi it\cdot x}~dt\right\}g(x)dx\\ =&\int _{\mathbb{R}^n}\left\{\int _{\mathbb{R}^n} g(x)f(t)e^{-2\pi it\cdot x}~dt\right\}dx\\ {\color{red}{=}}&\int _{\mathbb{R}^n}\left\{\int_{\mathbb{R}^n} g(x)f(t)e^{-2\pi it\cdot x}dx\right\}~dt\\ =&\int_{\mathbb{R}^n}\left\{\int_{\mathbb{R}^n} g(x)e^{-2\pi it\cdot x}dx\right\}f(t)~dt\\ =&\int_{\mathbb{R}^n}f(t)\widehat{g}(t)~dt\\ =&\int_{\mathbb{R}^n}f(x)\widehat{g}(x)dx. \end{align}

In the third equation I used Fubini’s theorem. But here’s something that I’m not sure: if we want to apply Fubini’s theorem, then $F(x,t):= g(x)f(t)e^{-2\pi it\cdot x}$ must be $\mathbb{R}^n\times\mathbb{R}^n$ integrable. But I was stuck when trying to prove that $F(x,t)$ is integrable. Could you give me some help? Thanks!

Answer 1

$$\int_{\mathbb{R}^n \times \mathbb{R}^n}|F(x,t)| \lambda(dx,dt)$$ $$=\int_{\mathbb{R}^n \times \mathbb{R}^n}|g(x) f(t)| \lambda(dx,dt)$$ $$= \int_{\mathbb{R}^n}\int_{\mathbb{R}^n} |g(x)||f(t)| \lambda(dx) \lambda(dt)$$ $$= \Vert g \Vert_1 \Vert f \Vert_1 < \infty$$

by Fubini for positive functions.