Let $d\in\mathbb N$ and $D\in\mathcal B(\mathbb R^d)$ be bounded. Moreover, let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$ and $$\hat f(\omega):=\int e^{-{\rm i}2\pi\langle\omega,\;\cdot\;\rangle}f\:{\rm d}\lambda^{\otimes d}\;\;\;\text{for }\omega\in\mathbb R^d$$ for $f\in\mathcal L^1(\lambda^{\otimes d};\mathbb C)$.
In equation (10) on p. 6 of this paper, I've read that $$\int_D|f|^2\:{\rm d}\lambda^{\otimes d}=\frac1{2\pi}\int_{[-\pi,\:\pi]^d}|\hat f|^2\:{\rm d}\lambda^{\otimes d}\tag1.$$ I guess it is assumed hat $f\in\mathcal L^2(\lambda^{\otimes d})$. The author claims that $(1)$ follows from Parseval's identity.
How do we show $(1)$?
Okay, since $D$ is bounded, there is a $r>0$ with $$D\subseteq\left[-\frac r2,\frac r2\right]\tag2.$$ Now, we know that $$e_k:=\frac1{r^{\frac d2}}e^{{\rm i}\frac{2\pi}r\langle k,\;\cdot\;\rangle}\;\;\;\text{for }k\in\mathbb Z^d$$ is an orthonormal basis of $L^2\left(\left[-\frac r2,\frac r2\right]\right)$. Let $$\tilde f(x):=\left.\begin{cases}f(x)&\text{, if }x\in D;\\0&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in D.$$ By Parseval's identity, $$\left\|f\right\|_{L^2(D)}^2=\sum_{k\in\mathbb Z^d}\left|\left\langle\tilde f,e_k\right\rangle_{L^2\left(\left[-\frac r2,\frac r2\right]\right)}\right|^2\tag3.$$ We clearly have $$\hat f(k)=\frac1{r^d}\int e^{-{\rm i}\frac{2\pi}r\langle k,\:x\rangle}f\left(\frac xr\right)\:\lambda^{\otimes d}({\rm d}x)\tag4$$ for all $k\in\mathbb R^d$.
However, I still can't conclude $(1)$ ...
Remark: If I remember correctly, there is a result of the form $(1)$ for functions $f$ belonging to the Schwarz space ...
(1) is obviously not true for all Schwartz functions and all bounded Borel sets $D\subset\Bbb{R}^d$, let alone all $L^2$ functions. Suppose $f$ is a Gaussian, then $\hat{f}$ is again a Gaussian (if I remember correctly, with this convention, $f(x)=e^{-\pi|x|^2}$ implies $\hat{f}=f$), so you cannot have different integration regions on both sides. Even if we don’t know about Gaussians (weird, but ok), we can still see (1) is wrong in general, because the LHS of the equation depends on the domain $D$, while the RHS doesn’t. Hence, we have by the monotone convergence theorem, and Plancherel’s theorem that \begin{align} \|\hat{f}\|_{L^2(\Bbb{R}^d)}^2&=\|f\|_{L^2(\Bbb{R}^d)}^2=\sup\limits_{\text{bounded $D\subset\Bbb{R}^d$}}\int_{D}|f|^2\,dx=\frac{1}{2\pi}\int_{[-\pi,\pi]^d}|\hat{f}|^2\,d\omega. \end{align} With this convention, I don’t think the $\frac{1}{2\pi}$ should be there, but even if we disregard that, then this equation implies $\int_{\Bbb{R}^d\setminus[-\pi,\pi]^d}|\hat{f}|^2\,d\omega=0$, i.e for every Schwartz function $f$, the Fourier transform is compactly supported in $[-\pi,\pi]^d$. This is obviously wrong, since you know the Fourier transform is an isomorphism on Schwartz space (and on $L^2$). If we don’t want to invoke such “abstract” results, we know that Fourier transform converts dilations to modulations and vice-versa, so doing this appropriately, we can easily construct Fourier transforms with arbitrarily large supports (provided of course we know that the Fourier-transform is not identically zero).
So, if you want $(1)$ to be true, you need to make assumptions on the supports of the functions involved.