Let $f,g \in L^1 ( \mathbb{R}^d)$ then consider the function $F(\xi, y) = g(\xi)f(y) e^{-2\pi i \xi y } $ defined for $(\xi, y) \in \mathbb{R}^d \times \mathbb{R}^d$. This is a measurable function on $\mathbb{R}^{2d}$ in view of Corollary 3.7.
I am unsure why this is a measurable function.
What I think:
By corollary 3.7, $h_1( \xi ,y ) = g( \xi )$, $h_2 (\xi,y)= f(y)$ and $h_3 (\xi,y) = e^{-2 \pi i \xi y }$ are all measurable functions $\mathbb{R}^{2d}$ where $h_1, h_2, h_3: \mathbb{R}^d \times \mathbb{R}^d \rightarrow \bar{ \mathbb{R} } $, As $F = h_1 \cdot h_2 \cdot h_3$, it is also measurable.
What you think are correct. $g(\xi), f(y)$ and $e^{-2\pi i \xi y}$ are all measurable on $\mathbb{R}^{2d}$, so it their product.
In general, if $f,g$ are measurable, then so is $fg$, even at some point they're infinite.