Convolution on $L^1({\mathbb{R}}^n)$ is independent of the chosen Borel representatives

Let $N_f:=\{{\tilde{f}}_1\mathrm{\ne }{\tilde{f}}_2\}$ and $N_g:=\{{\tilde{g}}_1\mathrm{\ne }{\tilde{g}}_2\}$. These are Borel null sets. For every fixed $x$, the set where the first factors differ is $x-N_f$, null by reflection [L4] followed by translation [L3]. The second factors differ only on $N_g$, so the two section integrands agree for almost every $y$, outside $(x-N_f)\cup N_g$.

By [L1], both section integrands are measurable, and step 1.1 says they agree almost everywhere in $y$. Thus [L2] gives equality of their integrals at every $x$ where either section product is absolutely integrable.

For either choice of representatives, the nonnegative function $H(x,y):=\mathrm{\mid }{\tilde{f}}_i(x-y){\tilde{g}}_i(y)\mathrm{\mid }$ is product-measurable by [L1]. Tonelli [L5] and translation invariance [L3] give ${\int }_{{\mathbb{R}}^n}\text{\hspace{0.17em}⁣}{\int }_{{\mathbb{R}}^n}H(x,y)dydx={\int }_{{\mathbb{R}}^n}\mathrm{\mid }{\tilde{g}}_i(y)\mathrm{\mid }({\int }_{{\mathbb{R}}^n}\mathrm{\mid }{\tilde{f}}_i(x-y)\mathrm{\mid }dx)dy=\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1<\mathrm{\infty }\mathrm{.}$ The second equality uses [L5] for the measure-preserving translation $x\mapsto x-y$, and the representatives have the same $L^1$ norms as their classes. Hence each section product is absolutely integrable for almost every $x$; intersect the two full-measure sets. Step 2.1 gives equality there, proving the statement.