Folland Lemma 8.25 Folland claims in Lemma 8.25 

If $f, g \in L^1(\mathbf{R}^d)$, then $\int f \hat g = \int \hat f g$. 

Note that by the $\hat\cdot$ notation, Folland means the Fourier transform. 
My question is why is this true? The proof (in Folland) appears to be just an application of Fubini Theorem. But why is it obvious that the integrand: 
$$
f(x) g(\theta) e^{-2\pi i \langle x, \theta \rangle} 
$$
is in $L^1(\mathbf{R}^d \times \mathbf{R}^d)$ if $f, g \in L^1(\mathbf{R}^d)$?
 A: First, note that $|f \widehat{g}| \leq |f| {\cdot} |\widehat{g}|_\infty \leq |f| {\cdot} \|g\|_{L^1}$ and similiar $|g\widehat{f}| \leq |g| \|f\|_{L^1}$. Thus, both integrals exist (in the sense of Lebesgue-integrals as usual). Now, observe that $|f(x) g(\theta) e^{- 2\pi \mathrm{i}\langle x, \theta \rangle}| \leq |f(x) g(\theta)|$. And because of the theorem of Fubini (resp. Tonelli), we know that $\int \int |f(x)| {\cdot} |g(\theta)| \, \rm{d}x \, \rm{d} \theta =\|f\|_{L^1} \|g\|_{L^1} < \infty$, i.e. your function is integrable.  
A: Well, you could do this as follows: 
First, you observe that if $E_1$ and $E_2$ are measurable subsets of $\mathbb{R}^d$, then $E_1 \times E_2$ is a measurable subset of $\mathbb{R}^d \times \mathbb{R}^d$. From which you conclude that if $f$ and $g$ are measurable on $\mathbb{R}^d$ then $H(x,y)=f(x)g(y)$ is measurable on $\mathbb{R}^d \times \mathbb{R}^d$, then you're allowed to use Fubini's theorem because $H$ is measurable and you could estimate the integral using the $L^1$ norms of $f$ and $g$ resepectively to conclude that $H \in L^1( \mathbb{R}^d \times \mathbb{R}^d) $.
Apply this reasoning to your problem. 
