Borel representatives make the convolution integrand Borel measurable

If $n=0$, every space and function in the statement has a one-point domain, so the conclusion is immediate. For $n\ge 1$, the map $T:{\mathbb{R}}^{2n}\to {\mathbb{R}}^{2n}$, $T(x,y)=(x-y,y)$, is continuous and hence Borel measurable. The coordinate projections are continuous, so [L3] makes $h_1(x,y)=\tilde{f}(x-y)$ and $h_2(x,y)=\tilde{g}(y)$ Borel measurable maps to $\mathbb{C}$.

The pair $(h_1,h_2):{\mathbb{R}}^{2n}\to {\mathbb{C}}^2$ is Borel measurable: the preimage of every open rectangle $U\times V$ is $h_1^{-1}(U)\cap h_2^{-1}(V)$, and such rectangles generate the product Borel sigma-algebra by [L2]. Complex multiplication ${\mathbb{C}}^2\to \mathbb{C}$ is continuous, so [L3] makes $H=h_1{h}_2$ Borel measurable. For fixed $x$, the map $y\mapsto (x,y)$ is continuous, and its composite with $H$ is the stated measurable section.