In the proof of the Fourier inversion theorem on Wikipedia it says
For $f,g \in L^1(\mathbb{R}^n)$, Fubini's theorem implies that $$ \int g(x, \xi) \hat{f}(\xi) d \xi = \int \hat{g}(x,y) d(y) dy. $$
Denote by $c := (2 \pi)^{-\frac{n}{2}}$, then my calculation with Fubini (F) indeed yields \begin{align} \int_{\mathbb{R}^n} g(x, \xi) \hat{f}(\xi) d \xi & = \int_{\mathbb{R}^n} g(x, \xi) \cdot c \int_{\mathbb{R}^n} f(y) e^{-i \langle y, \xi \rangle} d y d \xi \overset{\text{F}}{=} \int_{\mathbb{R}^n} f(y) \cdot c \int_{\mathbb{R}^n} g(x, \xi) e^{-i \langle y, \xi \rangle} d \xi d y \\ & = \int_{\mathbb{R}^n} f(y) \cdot c \int_{\mathbb{R}^n} g(x, \xi) e^{-i \langle \xi, y \rangle} d \xi d y = \int_{\mathbb{R}^n} f(y) \hat{g}(x,y) dy, \end{align} but how is $g$ to be understood? As $x$ and $\xi$ are $\mathbb R^n$-values, and $g \in L^1(\mathbb{R}^n)$, what is $g(x, \xi)$?
I feel like it can be written less confusingly like this. \begin{align*} \int_\mathbb{R}g(\xi)\hat{f}(\xi)d\xi &= \int_\mathbb{R}g(\xi)\int_\mathbb{R}f(x)e^{-2\pi i\langle x, \xi\rangle}dxd\xi\\ &= \int_\mathbb{R}f(x)\int_\mathbb{R}g(\xi)e^{-2\pi i\langle x, \xi\rangle}d\xi dx\\ &= \int_\mathbb{R}f(x)\hat{g}(x)dx. \end{align*} Fubini was used moving from the first line to the second.